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Quantum Mechanics 215C Homework Solutions #2

Sam Pinansky

April 25, 2004

1. (Problem 7.3 10 Points) We can again make use of the integral representation for fl derivedon the previous homework to solve for fl’s quickly:

fl = −2m

∫ ∞

jl(kr′)Al(k; r′)V (r′)r′2 dr′ (1)

V (r > R) = 0, and in our approximation V0 ¿ E k ≈ k′. Also, Al(k; r′) = jl(k′r′), as solvedfor in the text for the wavefunction inside a spherical square well. So the integral simplifies to:

fl = −2mV0

(jl(kr′))2r′2 dr′ (2)

Now we use the expansion of jl(ρ) about ρ ¿ 1 from the text (valid here since we are assumingkR ¿ 1)

fl = −2mV0

(kr′)2l

((2l + 1)!!)2r′2 dr′ (3)

= −2mV0R3

(kR)2l

(3 + 2l)((2l + 1)!!)2(4)

Now we can clearly see that in this limit only the small l terms are significant. The lowestorder terms are:

f0 = −2mV0R3

3~2(5)

f1 = −2mV0k2R5

45~2(6)

The differential cross section is |f(θ)|2 where

f(θ) =∑

(2l + 1)flPl(cos θ) (7)

To lowest order, it is simply |f0|2:dσ

4m2R6V 20

9~4(8)

σtot = 4πdσ

)m2V 2

as is stated in the book.

To next lowest order, the cross section is:

dΩ= |f0|2 + 6Re(f∗0 f1) cos θ (10)

Thus, with A = |f0|2 and B = 6Re(f∗0 f1):

25(kR)2 (11)

2. (Problem 7.3too 5 Points) For this, we can no longer use the approximation kR ¿ 1 or V0 ¿ E(That is what is meant by general parameters). In this case, it is simpler to work with theexpression given in the text for tan δ0:

tan δ0 =kRj′0(kR)− β0j0(kR)kRn′0(kR)− β0n0(kR)

We can solve for β0 since we know the solution for A0 inside of R:

β0 =(

= k′R cot(k′R)− 1 (13)

after a little algebra. Now we plug this into the general expression above, and use the definitionsof the bessel functions to get an expression defining δ0:

tan δ0 =1− k′

k cot(k′R) tan(kR)k′k cot(k′R) + tan(kR)

This is valid for general parameters.

The final part asks you to calculate the scattering length a, defined by limk→0 k cot δ0 = −1/a:

k cot δ0 =k′ cot(k′R) + k tan(kR)1− k′

k cot(k′R) tan(kR)(15)

so in the limit as k → 0, we get:

limk→0

=k′ cot(k′R)

1− k′R cot(k′R)(16)

which makesa = R− 1

k′tan(k′R) (17)

or, in terms of V0, since ~2k′2

2m = E − V0:

a = R−√

−2mV0tan

(√−2mV0

-40 -30 -20 -10 10

Figure 1: a vesus V0

We see there are two behaviors. Letting V ≡ |V0|, we get that

R−√

~22mV tan

(√2mV~2 R

)V0 < 0

R +√

~22mV tanh

(√2mV~2 R

)V0 > 0

The plot of this is shown in figure 1. Notice that for a repulsive potential V0 > 0, a > 0, andfor an attractive potential, a < 0 until V0 is negative enough, at which point it has a pole.The locations of the zeros of a represent the energies of the bound states for a particle trappedinside the well.

3. (Problem 7.5 8 Points) We want to show that 〈(∆x)2〉〈(∆px)2〉 ≥ ~24 . We use that 〈(∆x)2〉 =

〈x2〉 − 〈x〉2, and the fact that the state is spherically symmetric gives that 〈x〉 = 0, so wesimply need to show that

〈x2〉〈p2x〉 ≥

We can also use spherical symmetry to rewrite 〈x2〉 = 13 〈r2〉, and likewise for px:

19〈r2〉〈p2

r〉 (21)

in the n = 1, l = 0 state. This state is in an eigenstate of pr, and thus 〈p2r〉 = (~k)2 = ~2π2/a2,

since in the ground state k = π/a where a is the radius of the well. For the other term, weneed to express it in terms of spherical coordinates and do the integral:

〈r2〉 = A2

sin2(πr/a)π2r2/a2

r2r2 dr (22)

= A2 a5(2π2 − 3)12π4

where we have already integrated out the separately normalized angular part. To determinethe normalization factor A:

1 = A2

sin2(πr/a)π2r2/a2

r2 dr (24)

A2 =2π2

a3(25)

So then

〈r2〉 =a2(2π2 − 3)

6π2(26)

So combining with our previous results, we get that

〈(∆x)2〉〈(∆px)2〉 =2π2 − 3

54~2 (27)

And indeed, checking numerically:

2π2 − 354

= 0.31 > 0.25 (28)

thus:2π2 − 3

54~2 >

4. (Problem 7.10 7 Points) The short answer to the first part of this problem, is that since thepotential has only fourier modes at frequencies ±~ω, the change in energy must equal one ofthese freqencies. We can see this in this way: If our potential is V(r, t) = V (r) cos ωt, we cando the calculation leading to 7.11.13. Start with equation 7.11.12, we calculate the transitionprobability (to first order):

〈k′|U (1)I (t,−∞)|k〉 = − i

−∞dt′〈k′|eiH0t′/~(V (r) cos ωt′eηt)e−iH0t′/~|k〉 (30)

= − i

2~〈k′|V (r)|k〉

−∞dt′

(ei(Ek′−Ek+~ω)t′/~ + ei(Ek′−Ek−~ω)t′/~

where we have used the interaction picture for the potential. This integral with t = ∞ is a deltafunction. Now to find the transition rate, we let V have a small real part (for convergence),take the integral, and square:

|〈k′|U (1)I (t,−∞)|k〉|2 =

|〈k′|V (r)|k〉|2∣∣∣∣

e(η+i(Ek′−Ek+~ω)/~)t

η + i(Ek′ − Ek + ~ω)/~+

e(η+i(Ek′−Ek−~ω)/~)t

η + i(Ek′ − Ek − ~ω)/~

∣∣∣∣2

4~2|〈k′|V (r)|k〉|2

(e2ηt

η2 − ((Ek′ − Ek + ~ω)/~)2+

η2 − ((Ek′ − Ek − ~ω)/~)2(33)

+e2ηte2iωt

η2 + 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2+

e2ηte−2iωt

η2 − 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2

Now we differentiate and take the limit as η → 0:

dt|〈k′|U (1)

I (t,−∞)|k〉|2 =1

4~2|〈k′|V (r)|k〉|2

(2ηe2ηt

η2 − ((Ek′ − Ek + ~ω)/~)2+

2ηe2ηt

η2 − ((Ek′ − Ek − ~ω)/~)2(35)

2(η + iω)eηte2iωt

η2 + 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2+

2(η − iω)eηte−2iωt

η2 − 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2

Now to take the limit, we use a definition of the delta function

limη→0

η2 + x2≡ πδ(x) (37)

Note that the limit of the cross terms is perfectly regular:

limη→0

dt|〈k′|U (1)

I (t,−∞)|k〉|2 = (38)

|〈k′|V (r)|k〉|2(

2π~δ(Ek′ − Ek + ~ω) + 2π~δ(Ek′ − Ek − ~ω)− 4ω sin(2ωt)((Ek′ − Ek)2 − ~2ω2)/~2

The time dependent term IS present, but we recall that the first order approximation is onlyapplicable for ωt small, and thus the rate is very small away from the resonance frequencies,and we can simply lump the pole of the final term into the delta functions. Therefore, thetransition rate is simply:

dt|〈k′|U (1)

I (t,−∞)|k〉|2 =π

2~|〈k′|V (r)|k〉|2(δ(Ek′ − Ek + ~ω) + δ(Ek′ − Ek − ~ω)) (40)

So the only energies with appreciable transition rate is when the electron gains or loses ~ω ofenergy.

Using fermi’s golden rule, we see that

2~|〈k′|V |k〉|2

~2(k′+ + k′−)dΩ (41)

k′± =

√2m(Ek ± ~ω)

~2(42)

Then, proceding as in the text, we multiply times the incoming flux to get the scattering crosssection:

(k′+ + k′−

16π2~4

∣∣∣∣∫

d3xV (x)ei(k−k′)·x∣∣∣∣2

Note that if ω → 0, then we recover the result of equation 7.11.27. If we were to go toa higher order approximation, we would find transitions between farther away states, withEk′ − Ek = ±2~ω etc...

Average 25/30.

Sam Pinansky

April 14, 2004

1. (5 Points)

(a) The problem is completely done in the text, on pages 406-408.(b)

2. (a) Given that:

∆(b) = − m

∫ ∞

−∞V

√b2 + z2 dz (1)

we substitute the gaussian potential V = V0 exp(−r2/a2) and integrate:

∆(b) = − m

∫ ∞

−∞V0 exp

(b2 + z2

)dz (2)

= −mV0a

√π exp

(− b2

δl = ∆(b)|b= lk

√π exp

(− l2

This clearly decays exponentially as l increases.(b) Now we substitute V = V0 exp(−µr)/µr:

∆(b) = − m

∫ ∞

−∞V0

e−µ√

b2 + z2(5)

= − mV0

2k~2µ2

∫ ∞

e−µbu

√u2 − 1

du u =

√1 +

= −mV0

k~2µK0(µb) (7)

δl = ∆(b)|b= lk

= −mV0

k~2µK0

where K0 is the zeroth modified Bessel function of the second kind. The asymptoticbehavior of the Bessel functions are known:

K0(z) ≈√

2ze−z as z →∞ (9)

so when l is large δl is suppressed exponentially.

3. (10 Points)

(a) Many people did this problem by following the green’s function argument given in Jackson.But there is a much more direct way making use of the results in the text that uses asimple contour integration argument.First we insert a complete set of states:

2m〈x| 1

E −H0 + iε|x′′〉 =

∫dE′

∫dE′′∑

l′,m′〈x|E′lm〉〈E

′lm|E′′l′m′〉E − E′ + iε

〈E′l′m′|x′〉

∫dE′∑

πk′jl(k′r)Y m

l (r)1

E − E′ + iεjl(k′r′)Y m

l∗(r′)

Y ml (r)Y m

l∗(r′)

∫ ∞

−∞dk′ k′2

jl(k′r)jl(k′r′)k2 − k′2 + iε

We note that the integrand has poles located at k′ = ±(k + iε), thus a contour along thereal axis that closed in the +i direction encloses the positive k′ pole, and if closed alongthe negative imaginary direction, it encloses the negative k′ pole. We need to figure outthe asymptotic behavior of the numerator to figure out which contour to choose:

4jl(k′r)jl(k′r′) = (h(1)l (k′r) + h

(2)l (k′r))(h(1)

l (k′r′) + h(2)l (k′r′)) (13)

∝r large1r2

(eik′r + e−ik′r)(eik′r′ + e−ik′r′) (14)

Therefore, if r > r′ we see that we should close the contour above for the h(1)(k′r)h(1)(k′r′)and h(1)(k′r)h(2)(k′r′) and below for the h(2)(k′r)h(1)(k′r′) and h(2)(k′r)h(2)(k′r′) terms.Since h(1)(z) = −h(2)(−z), each term ends up with the same sign:

= − 12π

∫ ∞

−∞dk′ k′2

h(1)l (k′r)jl(k′r′) + h

(2)l (k′r)jl(k′r′)

(k′ + k + iε)(k′ − k − iε)(15)

= −2πi

(1)l (kr)jl(kr′)

2k− h

(2)l (−kr)jl(−kr′)

)r > r′ (16)

= −ikh(1)l (kr)jl(kr′) r > r′ (17)

For r < r′, simply flip r and r′. Thus we get:

2m〈x| 1

E −H0 + iε|x′′〉 = −ik

Y ml (r)Y m

l∗(r′)jl(kr<)h(1)

l (kr>) (18)

(b) Inner product the equation into the |x〉 basis and insert a complete set of states.

〈x|Elm(+)〉 = 〈x|Elm〉+∫

d3x′〈x| 1E −H0 + iε

|x′〉V (r′)〈x′|Elm(+)〉 (19)

We know that for free states, and general scattered states:

〈x|Elm〉 = cljl(kr)Y ml (r) (20)

〈x|Elm(+)〉 = clAl(k, r)Y ml (r) (21)

Plugging this into the equation, and using our result from part a):

clAl(k, r)Y ml (r) = cljl(kr)Y m

l (r)+ (22)∫

d3x′

−2mik

l′,m′Y m′

l′ (r)Y m′l′

∗(r′)jl′(kr<)h(1)

l′ (kr>)

V (r′)clAl(k, r′)Y m

l (r′)

Now we do the solid angle integral over r′, which because of the completeness of the Y ml ’s

just gives a δll′δmm′ , we get:

Al(k, r)Y ml (r) = jl(kr)Y m

l (r)− 2mik

l (r)∫ ∞

r′2dr′jl(kr<)h(1)l (kr>)V (r′)Al(k, r′)

(24)dividing out by the common spherical harmonic, we get our result:

Al(k; r) = jl(kr)− 2mik

∫ ∞

jl(kr<)h(1)l (kr>)V (r′)Al(k; r′)r′2dr′ (25)

For the second part, if you take r →∞, then r> = r, r< = r′, and writing out:

∫ ∞

jl(kr′)(jl(kr) + inl(kr))V (r′)Al(k; r′)r′2dr′ (26)

But we know thatAl(k; r) = eiδl [cos δljl(kr)− sin δlnl(kr)] (27)

soeiδl

sin δl

k= −1

k(coefficient of nl) (28)

i.e., from above

fl = −2m

∫ ∞

jl(kr′)Al(k; r′)V (r′)r′2dr′ (29)

4. (10 Point)

(a) This problem can be solved using the answer to the previous problem, and it’s quite a biteasier then solving the 1-d shroedinger equation. No one actually did the problem thisway, though.We have a potential

V (r) =~2

2mγδ(r −R) (30)

so we can simply plug this into the integral equation derived above for A0(k; r), and sincethe potential is a delta function, you can easily solve this.

∫ ∞

A0(k; r) = j0(kr)− ikγj0(kR)h(1)0 (kr)A0(k; R)R2 (32)

where we are restricting to the r > R solution. First, we solve for A0(k;R):

A0(k; R) =j0(kR)

1 + ikγj0(kR)h(1)0 (kR)R2

So now we solve for A0(k; r) in general:

A(k; r) = j0(kr)− ikγh(1)0 (kr)j0(kR)j0(kR)R2

Now, factor this into the form:

A0(k; r) = eiδ0 [cos δ0j0(kr)− sin δ0n0(kr)] (35)

A0(k; r) = j0(kr)

(1− kγj0(kR)j0(kR)R2

)+ (36)

n0(kr)

(kγj0(kR)j0(kR)R2

− cot δ0 =1− kγj0(kR)n0(kR)R2

kγj0(kR)j0(kR)R2(38)

cot δ0 = −k

γcsc2(kR)− cot(kR) (39)

using the definition of j0 and n0 from the appendix. Wasn’t this easier then then matchingboundary conditions?

(b) If γ À k, and tan kR is not small, then immediately we see:

cot δ0 ≈ − cot(kR) (40)

so δ0 = −kR, exactly as in the hard shell case.Rewriting:

cot δ0 = csc2(kR)[−k

γ− cos(kR) sin(kR)

since csc2(kR) is never zero, cos δ0 is zero when:

cos(kR) sin(kR) = −k

γ(42)

This is solved (for γ À k), by kR = nπ/2, but only the even n zeros approach zero fromthe positive side. Thus, we expand about these zeros:

γ= kR− nπ (43)

k =nπ

11 + 1

≈ nπ

(1− 1

Corresponding to the energies:

(1− 2

Compare with the bound-state energies of a spherical well:

R2(46)

For the final part, we simply need to calculate this resonance width:

Γ =−2

[d(cot δ0)/dE]|E=Er

This is some messy algebra, but you find after plugging in the previous results that (toorder 1/g2):

Γ = −14~n3π3

mR4γ2+ O(1/γ3) (48)

(All the lower order terms are exactly zero since they depend on sin(nπ)). Thus as γ →∞,Γ → 0, i.e. the resonance peak is very narrow.

Quantum Mechanics 215A Homework Solutions #1

Sam Pinansky

October 8, 2003

People did fine on this assignment. The only mistakes were on the second problem. The averagewas 18.1/20.

1. (5 Points) This can be proved simply by writing out both sides:

−ACD, B+ AC, BD − CD, AB + C, ADB (1)= −ACDB −ACBD + ACBD + ABCD − CDAB − CADB + CADB + ACDB (2)= ABCD − CDAB (3)= [AB, CD] (4)

2. (10 Points) The problem asks you to use bra-ket algebra to prove these, but some of youconverted them into matrix problems. This was not incorrect, but was not exactly in the spiritof the problem.

(a) By the definition of the trace:

tr(XY ) ≡∑

〈a|XY |a〉 (5)

where |a〉 is some orthonormal basis. Now we insert a complete set of states:∑

〈a|XY |a〉 =∑

a′〈a|X|a′〉〈a′|Y |a〉 (6)

〈a′|Y |a〉〈a|X|a′〉 (7)

a′〈a′|Y X|a′〉 (8)

= tr(Y X) (9)

where we have used the fact that 〈a′|Y |a〉 and 〈a|X|a′〉 are just numbers so commute, andthe completeness of the |a〉. Challenge: Generalize this problem to prove the cyclicity ofthe trace, namely: tr(X1X2 . . . Xn) = tr(XnX1 . . . Xn−1). (Well, not much of a challengereally).

(b) There are two ways to prove this: 1. Using the dual correspondence. 2. Use the identitythat 〈a′′|X|a′〉∗ = 〈a′|X†|a′′〉. Actually, these methods are completely equivalent, onejust uses a specific basis representation while the other is basis independent. Here is thedual correspondance proof:

|β〉 ≡ X†|α〉 ↔ 〈α|X = 〈β| (10)

Y †X†|α〉 = Y †|β〉 ↔ 〈β|Y = 〈α|XY ↔ (XY )†|α〉 (11)

⇒ Y †X† = (XY )† (12)

where ↔ means dual correspondence. Essentially you are applying the dual correspon-dence twice.The other method goes like this: Take a basis of states, |a〉, and look at the general matrixelement of (XY )†:

〈a|(XY )†|b〉 = 〈b|XY |a〉∗ (13)

〈b|X|c〉∗〈c|Y |a〉∗ (14)

〈c|Y |a〉∗〈b|X|c〉∗ (15)

〈a|Y †|c〉〈c|X†|b〉 (16)

= 〈a|Y †X†|b〉 (17)

Thus (XY )† = Y †X†. This proof is a little more direct then the above, but it chooses aparticular basis to work in so is a bit less elegant.

(trivial iteration proof). Now we note that we expand f(A) in power series for any nicefunction f :

f(A) =∞∑

fnAn (18)

So taking a general ket expanded out in the eigenbasis of A: |α〉 =∑

j αj |aj〉 (whereαj = 〈aj |α〉), we get:

f(A)|α〉 =∑

fnαjAn|aj〉 (19)

fnanj αj |aj〉 (20)

αjf(aj)|aj〉 (21)

So now we apply the operator exp(if(A)) to a general ket |α〉:

exp(if(A))|α〉 =∞∑

n!f(A)n|α〉 (22)

=∞∑

n!f(aj)nαj |aj〉 (23)

exp(if(aj))αj |aj〉 (24)

exp(if(aj))|aj〉〈aj |α〉 (25)

So the operator is exp(if(A)) =∑

j exp(if(aj))|aj〉〈aj |, i.e. it projects onto the eigen-states of A and multiplies by some phase.On a side note, if A = A†, i.e. A is hermition, then the operator exp(if(A)) is alwaysunitary.

(d) This is simply substitution:∑

a′ψ∗a′(x

′)ψa′(x′′) =∑

a′〈x′|a′〉∗〈x′′|a′〉 (26)

a′〈x′′|a′〉〈a′|x′〉 (27)

= 〈x′′|x′〉 (28)= δ(x′′ − x′) (29)

where in the last step we used the orthogonality of the continuous x basis.

3. (5 points) Its easiest to do these calculation (at least, for me) by using a matrix representation,with

|+〉 =(

|−〉 =(

Then we see that:

(0 11 0

(0 −ii 0

(1 00 −1

Now the verifications are just matrix multiplications:

[Sx, Sy] =h

(0 11 0

(0 −ii 0

)− h

(0 −ii 0

(0 11 0

(1 00 −1

)= ihSz

(35)You can easily use these representation to prove all the components of [Si, Sj ] = iheijkSk andSi, Sj = h2/2δij . (No one got this wrong, so I don’t feel the urge to show you trivial matrixalgebra).

Sam Pinansky

October 14, 2003

Average score: 23/25.

10. The hamilonian (in the basis |1〉, |2〉) is represented by the matrix:

(1 11 −1

From the characteristic equation, λ2 − 2a2 = 0, so the eigenvalues are λ± = ±√2a. Theeigenvectors are similarly found by elementary linear algebra to be:

|λ+〉 =1√

4 +√

|λ−〉 =1√

4−√2

(1−√2

or, in stadard ket notation:

|λ+〉 =1√

4 +√

√2)|1〉+ |2〉

|λ−〉 =1√

4−√2

((1−

√2)|1〉+ |2〉

12. First, we need to identify what state the system starts in. You can just look immediately abovethe problem statement in problem 11 to see the answer, but here’s the proof for completeness:We are told that the state is in an eigenstate of the operator S · n with eigenvalue +h/2, wheren = sin γx + cos γz. So we must find eigenstates of:

S · n =h

(cos γ sin γsin γ − cos γ

(Hmmm, seem familiar?). The eigenvalues and eigenvectors are readily solved for, giving thestate

|n; +〉 = cosγ

2|+〉+ sin

2|−〉 (7)

(where |±〉 are the eigenvectors in the Sz basis) with eigenvalue +h/2. Note that you mightnaively expect the state to be cos γ|+〉+sin γ|x; +〉, which is incorrect. Note first that this stateisn’t properly normalized. Second, it is only true that you can linearly superpose eigenvectorslike this if the operators commute.

(a) With that starting state, we want to calculate:

|〈x; +|n; +〉|2 =∣∣∣∣

1√2(〈+|+ 〈−|)

2|+〉+ sin

2|−〉

)∣∣∣∣2

2+ sin

=12(1 + sin γ) (10)

(b) Using 〈(Sx − 〈Sx〉)2〉 = 〈S2x〉 − 〈Sx〉2, we first see that since S2

x = h2

4 I, 〈S2x〉 = h2

4 (in anystate). For the other term:

〈Sx〉2 =h2

4〈n; +|Sx|n; +〉2 (11)

2〈+| sin γ

2〈−|

)(|+〉〈−|+ |−〉〈+|)

2|+〉 sin γ

2|−〉

= h2 sin2 γ

4sin2 γ (14)

So therefore:

〈(Sx − 〈Sx〉)2〉 = 〈S2x〉 − 〈Sx〉2 =

4(1− sin2 γ) =

4cos2 γ (15)

This has the correct limits as γ → 0, π/2π, namely h2/4, 0, h2/4, respectively.

15. (Dropping primes for clarity) Expand a general ket into the complete basis |a, b〉:

|ψ〉 =∑

cij |a, b〉 (16)

Now apply [A,B] to this general ket:

[A,B]|ψ〉 = AB∑

cij |a, b〉 −BA∑

cij |a, b〉 (17)

cij(bjai − aibj)|a, b〉 (18)

= 0 (19)

So since for any |ψ〉, [A, B]|ψ〉 = 0, [A,B] must be the zero operator. (Many of you onlyshowed this for an eigenket |a, b〉 and not for a general superposition. This doesn’t prove that[A,B] = 0 directly without arguing that any ket can be made of linear superpositions of theeigenkets).

19. (a) First, we want 〈Sx〉2 (taken in the state |+〉)

〈Sx〉 =h

2(〈+|(|+〉〈−|+ |−〉〈+|)|+〉) = 0 (20)

and from the reasoning in the last part 〈S2x〉 = h2

4 , so 〈(∆Sx)2〉 = h2/4, From symmetrywe know that 〈(∆Sy)2〉 must be equal to that, when taken in an eigenvector of Sz. Theright side of the uncertainty relation is calculated as:

14|〈[Sx, Sy]〉|2 =

14|〈ihSz〉|2 (21)

=14| ih

2|2 (22)

16(23)

So the uncertainty relation is saturated.(b) We still have 〈S2

x〉 = h2/4, but now since we are taking the expectation value in the state|x; +〉, we also have 〈Sx〉 = h/2, so that 〈(∆Sx)2〉 = 0. Checking the other side of therelation,we need to find:

〈Sz〉 = 〈x; +|Sz|x; +〉 (24)

∝ (1 1

)(1 00 −1

)(1−1

)= 0 (25)

So again the uncertainty relation is true, and is saturated. Note that the fact that bothsides are 0 is obvious when you think about the definition. If we have a state in aneigenstate of Sx, then obviously it has 0 uncertainty in a measurement of Sx.

23. (a) The characteristic equation for B is (b− λ)(λ2 − b2) = 0, so the eigenvalues are b,±b. Sothe b eigenvalue is degenerate.

(b) By straight matrix calculation, [A,B] = 0.(c) By straightforward means you can find a set of eigenvectors for B:

λ = b

λ = b1√2

0−i1

λ = −b1√2

Applying A to these vectors, we see that they are also eigenstates of A, with eigenvaluesa,−a,−a respectively. Therefore these spectrum is no longer degenerate since specifyingthe eigenvalues completely specify’s the eigenvector.

Prob 13:

This is more or less a continuation of Ex-9Using the eigenkets for the z,x,y referance frame.

a. The prepared beam can be represented by

)b. The filter|a′ >< a′| which will selecta′and rejecta′′ . Using Ex 9 the repre-

sentation of this in z eigenket space is:[cos(β

2 )2 cos(β2 )sin(β

2 )e−i∗α

cos(β2 )sin(β

2 )ei∗α sin(β2 )2

]multiplying[

cos(β2 )2 cos(β

2 )sin(β2 )e−i∗α

cos(β2 )sin(β

)evaluating with respect to the other eigenket(

0 1) (

cos(β2 )2

cos(β2 )sin(β

2 )ei∗α

)= cos(β

2 )sin(β2 )ei∗α

and using trigonometrycos(β

2 )sin(β2 )ei∗α = sin(β)

2 ei∗α

Give the survivors:(sin(β)

2 ei∗α) (

sin(β)2 ei∗α

(sin(β)

which is maximum forβ = ±π2

And we can recapture14 of the atoms.

Sakurai problem solutions (Chapter 1)

Joseph Geddes

[13] Three Stern–Gerlach filters are placed in series. The first apparatus, ori-ented in the z direction, passes atoms having sz = +h/2 and stops others. Thesecond apparatus is oriented at an angle β to the z axis; it passes atoms hav-ing positive spin (+h/2) with respect to that orientation. The third apparatusis also oriented in the z direction, but passes atoms having sz = −h/2 andstops others. Atoms that pass through the first apparatus will pass throughthe second with probability amplitude cos(β/2); atoms that pass through thesecond apparatus will pass through the third with a probability amplitude of− sin(β/2). Therefore, the fraction sin2(β/2) cos2(β/2) = 1

4 sin2 β of atoms thatpass through the first apparatus will pass through the third. When β = ±π

2 ,a maximum fraction of 1

4 the atoms that pass through the first apparatus willpass through the second.

[15] Suppose the simultaneous eigenkets of A and B, denoted by |γi〉 , form anorthonormal set. The equations

A |γi〉 = ai |γi〉 , (1)B |γi〉 = bi |γi〉 , (2)

define the eigenvalues ai and bi, and orthonormality dictates that the complete-ness relation ∑

|γi〉〈γi| = 1 (3)

holds. Therefore,

A = A(1) = A∑

|γi〉〈γi| =∑

ai |γi〉〈γi| (4)

B = B(1) = B∑

|γi〉〈γi| =∑

bi |γi〉〈γi| (5)

Now, we compute the commutator

[A,B] =

ai |γi〉〈γi|

] ∑j

bj |γj〉〈γj |

bi |γi〉〈γi|

] ∑j

aj |γj〉〈γj |

which, when we consider that

〈γi | γj〉 =

0 i 6= j1 i = j

simplifies to[A,B] =

aibi (|γi〉〈γi| − |γi〉〈γi|) = 0 . (9)

Hence, under the foregoing conditions, the commutator [A,B] must be nullvalued.

[16] We are told thatA,B = AB + BA = 0 , (1)

where A and B are Hermitian operators. Suppose that

A |γ〉 = a |γ〉 , (2)B |γ〉 = b |γ〉 . (3)

A,B |γ〉 = bA |γ〉 + aB |γ〉 (4)= 2ab , (5)

and this result implies that either a = 0 or b = 0. So the operators A and B canshare a simultaneous eigenket if either or both the eigenvalues corresponding tothat eigenket are 0.

Problem h

The sequence of measurements produces a sequence of eigenstates for the corresponding observables Sz S n and Sz

ji jn ihn ji jihjn ihn jiie the amplitude of going from ji to ji is hjn ihn ji where problem

jn i cos

ji sin

Thus the probability is sin cos

sin which has a maximum for

This is reasonable since this angle corresponds to the intermediate state jn ibeing symmetric with respect to ji in fact an eigenstate for Sx correspondingto eigenvalue

Problem h

This problem is of course trivially solved by using the representation of rotationmatrices for spin particles given in chapter However it is instructive and theintention since the problem belongs to chapter to solve it without this knowledge

a ibia b

In order that this operation can be viewed as the rotation of the twocomponentspinor we have to associate a direction to the normalized spinor with componentsa b This is done via the formula in problem

aji bji eijn i eicos

ji ei sin

which denes and thus the direction vector

n cos sin sin sin cos

The interpretation is as usual that our spinor with components a b is an eigen

vector for the operator S n If the operator ixp represents a rotation of

around the x axis then the vector jn i should be rotated to eijn i where is a phase factor and n is obtained from n by the mentioned rotation xcomponentof n unchanged while y zcomponents are rotated in the y zplane

n cos sin sin sin cos cos sin cos sin sin

orcos sin sin ei

sin i cos cos sin

From eq we have

ei sin

iei sin

ei sin

If two complex twocomponent vectors only dier by a phase

i jaj jcj ii ab cd

If we apply i and ii in we get

or precisely eq thus showing that the linear operator in question rotates jn ito jn i up to a phase factor where n is obtained from n by a rotation of around the x axis

An alternative and simpler way to solve the problem is as follows Denote

ip by R Ry i

p Since Rjn i is an eigenvector for S n

correspoding to eigenvalue and similarly for jni j n i correspoding toeigenvalue we have

Rjni R

such thatS n R

The rhs of is calculated by using Si i and the relation

ij ij iijkk

After some algebra reads

S n nS ikniSk

orn n n

which precisely means that n is obtained from n by a rotation by around thexaxis

Problem h

Let jai be the spinor with components and jy i be the spinor correspondingto the eigenvalue above ie with components i

p It is an eigenstate of the

spin operator Sy with eigenvalue h Thus the probability of measuring sy h in state jai is

jhy jaij

jj jj i

Problem

U a ia a ia U y

a ia a ia U

det a ia deta ia ia

ia a a ia

a det a ia

Thus detU

We can assume aa in the expression for U since we can scale numerator and

denominator with the same factor In this case a ia are themselves unitaryand unimodular

a ia Thus

U a ia

From we know that a ia represents a rotation around the axis aa a

pa with angle given by

Thus U represents a rotation with angle around the unit vector a

"! # $

0 11 0

] % σ2 =

0 −i

] % σ3 =

1 00 −1

&')(+* ,-/.101"-2 *3 &4*tr(X) = 2a05 ** 6!#78 *3 9:; * 68;66< &=* 6-/! *>& .;6

I σ1 σ2 σ3

σ1 I iσ3 −iσ2

σ2 −iσ3 I iσ1

σ3 iσ2 −iσ1 I

? ! @A 6!#BCD< * "-2! & 'FE 6G;,@tr (σkX) = 2a. 'H *C* "!#

X11 X12X21 X22

] & ! @IJC6!# &'a0 = tr(X)

2 = X11+X222

%a1 = tr(σ0·X)

X22 X21X12 X11

2 = X21+X212

a2 = tr(σ2·X)2 =

−iX22 −iX21iX12 iX11

2 = i · X12−X212

a3 = tr(σ3·X)2 =

X11 X12−X21 −X22

2 = X11−X222

KLM6!#

det(AB) N 3J& 0/ det(eXSe−X) = det(eX)det(S)det(e−X)& ;6;O-4P *3 Q@R * 6! &=* & S<T-2 8;"GUV!W :.XG> *3 &=* <T-2QS * '9Y3 J N : &ZE & & !#2 *3 det() '. *

eXe−X = IC-det(σ · a′) = det(σ · a)( ! * G8 *>&=* 6-/![$ Y3 B@ * 6! &=* -4P & !\-28J &4* -2,]"!10 & & ! * ! @RG^LM! *& E* & !J_P`-27a.JD< & C

UU† = 1b E -/;6 * "-2! * - *3 ^<T-2! @c8 & * -4P *3 98-/. ;"d6! - * 0/ E ;"# & ! *D' "!# *3 SeCfhgI<G-/DC8X-/! @ * -a3 = 1 & !J@ *3 I- *3 :<T-2 8J-2!G! * & Bi %*3 9TUR8J-2!G! * .JD<T-/ iθσ3

C6! <T *3 6j6j@ & #/-/! & ;k$e

iθσ3

0 e−iθ

] l e− iθσ3

e−iθ

0 eiθ

mj- NnN Q< & !o6!10/D * "# &=* & 88; E "! # *3 6 * & !J_P`-2 * -A & < 3 -4P *3 axσx

C8JCp&=* G; E "! <G *3 9Dq2 &=* 6-/!A,r;"6! & "! *3

ax'(a& d6! <T;6 @R6!#

a0.X< & ] * G & C-2! & . ;" & ! @c & E2'

a0 → a′

0 sut * : ; * "8; EB*3 9 &=* ,<TD * -/#2 *3 Ghv

a3 → a′

3 s & Q & a0v:; * 68; E

σ3* - *3 x;"GP *a3 & ! @ * G % < 3 & !#/6!# *3 ]C6#/!B-/P*3 9; & * <T-2;" Q! % & ! @ *3 G!y:; * 68; EB*3 9DC ; *D'

σ2, σ3& x -29<T-2 8;",< &4* D@ 'O( N 6;";zGUR8 & ! @ *3 * & ! CP`-/7 * -

eiθσ3

cos( θ2 ) + i · sin( θ

2 ) 00 cos( θ

2 ) − i · sin( θ2 )

e−iθσ3

cos( θ2 ) − i · sin( θ

2 ) 00 cos( θ

2 ) + i · sin( θ2 )

TUR8 & ! @R6!# & # & "![$e

iθσ3

2 = cos( θ2 )I + iσ3sin( θ

e−iθσ3

2 = cos( θ2 )I − iσ3sin( θ

2 )0 & ;6 &=* "! # *3 9; * -4Pa2 C6!# *3 *& . ;"96!cUy|

cos( θ2 )I + iσ3sin( θ

cos( θ2 )I − iσ3sin( θ

cos( θ2 )σ2 + σ1sin( θ

2 )) (

a2(cos2( θ

2 ) − sin2( θ2 ))σ2 + a22(sin( θ

2 )cos( θ2 ))σ1

a2cos(θ)σ2 + a2sin(θ)σ15 6Q6;6 & ; E P`-/σ1

cos( θ2 )σ1 − σ2sin( θ

2 )) (

a1(cos2( θ

2 ) − sin2( θ2 ))σ1 − a22(sin( θ

2 )cos( θ2 ))σ2

a1cos(θ)σ1 − a1sin(θ)σ2 j<T-2:.6!6!# $a′

1 = a1cos(θ) + a2sin(θ)a′

2 = a2cos(θ) − a1sin(θ)

~& vbV; * 68; E 6!# *3 -/# 3 . E]& ! E 0/< * -2[0 & ! @] 6!#j;6"! & *_E* -@R-j * <T-2 8J-2!G! * . E<T-2 8J-2!G! * % "! *3 ^. & C,P`-/ @A. EB*3 ^G6#/!/ * % E ";6@ jiQP`-/j0/ E <G-/ 8X-/!! *P`-/7-/ * G-4P *3 y8-R@R < *D'Y3 ,SP`-/;6;"- N S. E <T-/!J * < * 6-/! 'Y3 c6! @R6016@ & ;<T-2 8J-2!G! * & M"#2G!10/D< * -2C- N 6;6;X.Xx8DC02@ & & < 37* -/P *3 x8-1@ < * 68-R<T@ % N *3o& < & ; & r: ; * "8;66G % ! * 6; *3 E & < 3A*3 * *3J&=* 0 & ;" &=* D * -fG- '.Jv Y3 6 N ";6; E 6G;,@7i]P`-/ & ;";X<T-2Q8X-/! G! * )GU<TG8 * P`-/0XW<T-28J-2! @R6!# * - & l *3 6P`-/;6;6- N jP`-/ *3 & -/!6!#I-/P & v ' 0<T-28X-/! @R6!# * - & N ";6;).XS8G0/D@ & !J@3 & 02 & "#2G!10 & ;6j-4P w C6! <G *3 x<G-/ 8 * @B &4* 6< 3 & *3 ]@RG! -/ "! &=* -2 * G * -!-2 & ;66fG *3 9; *' -2 & ;"; E

Sx, Sy, Sz

< & !o.JQG8G! * D@y. E h2 σx, h

2 σy , h2 σz

8X< * "02G; E/']Y3 G! *3 & !J N GaP`-/;6;6- N P`-/ *3 ]S; * "8 ;",< &4* 6-/! *& .;"9-/P8-/. ;"| '

Sam Pinansky

October 23, 2003

Average score: 33.4/40. Quite a bit lower then the last two.

22. (5 Points) This problem was hard for some people to figure out what Sakurai was asking. Iclaim that there is only one interpretation that makes sense, and I shall “prove” this statementas such: Assume that the problem has a well defined answer (i.e. there is a certain order ofmagnitude that is the one Sakurai expected you to get). The issue hinges on what you thinkSakurai means when he says the icepick is SHARP. By assuming a finite point, the problembecomes trivial, i.e. it simply reduces to how long it would take the center of mass to movehalf the width of the point. But the crucial point is that the order of magnitude of your answerdepends directly on the order of magnitude of the size of the tip. Since you can reasonablyinterpret the tip to be anywhere from 1 atom to a few micrometers in width as SHARP, youcan reasonably get answers which differ by 7 orders of magnitude! Since by assumption, thereis one “correct” answer, this interpretation cannot be the one Sakurai intended (and thus, ifyou did the problem this way, you didn’t get full credit). Other, less formal arguments arethat this interpretation trivializes the calculations in the problem, and using the democraticprinciple, since most people did the problem my way, it must be correct :-).

OK, My interpretation was that the point of the ice pick was infinitesimal in size, and did notslip along the surface. Also I took the rod to be thin, and uniform, with mass m and lengthl. Then, the langrangian is:

L =16ml2θ2 −mg

2cos θ (1)

(recall that the moment of inertia of a rod about one end is 1/3ml2). The equation of motionis then:

θ = λ2 sin θ (2)

where λ2 = 3g2l . We will take the approximation where sin θ ≈ θ. This is valid up until about

π/4 or so. We will assume that after the rod has fallen by π/4 it will continue to fall prettyfast till it hits the ground. Physical intuition tells us that it will only take less then 1 sec or soto fall the rest of the way, so if we find a max time greater then 1 sec, we know that ignoringthis part of the motion won’t change the order of magnitude of the answer.

Then, the solution of this differential equation is

θ(t) = θ0 cosh(λt) +1λ

θ0 sinh(λt) (3)

where θ0 and θ0 are the initial conditions. Now purely classically if you choose θ0 = θ0 = 0,then θ(t) = 0 for all t, but the uncertainty principle tells us that:

∆x∆p ≥ ~2

∆θ∆θ ≥ ~2ml2

Clearly, to maximize the time it takes the pick to fall, we want to saturate the inequality, sowe can eliminate θ0 from the equation of motion:

θ0 =~

2ml2θ0(6)

Now we can use this, and invert the solution of the equation of motion to solve for t:

tf =1λ

cosh−1

(µθfθ30 − ~

√µθ2

0(θ2f − θ2

0) + ~2)

(µθ40 − ~2)

where µ = 4l4m2λ2. (Note you can simplify this a bit by noting that θf − θ0 ≈ θf ). Nowall we need to do is to maximize this with respect to θ0. This is not nice symbolically, so weplug in numbers and maximize this numerically. Taking l = .5m, m = .5kg, g = 10m/s2, and~ = 10−34J · s and θf = π/4, we can plot tf : See attached mathematica printout, where it isseen that tf has a maximum of about 7 seconds. We should also note that the maximum is atθ0 ≈ 10−17, which is when θ0 ≈ θ0.

One thing to notice is that this fall time is basically independent of the size of ~. This issomewhat nontrivial to show, with some simplification you can get that:

tmax ∝ cosh−1(~−1/2) (8)

This gives a range of 60 to about .1 seconds over a range of ~ from 10−300 to 1 (300 ordersof magnitude versus 2!). Thus, the scale of quantum mechanics is completely irrelevant to theproblem. The only thing that matters is that the uncertainty principle holds at all, not thespecific value of ~.

26. (5 Points) We want to find an operator U such that Sz = USxU†. Expressing the operatorsin the Sz basis, this becomes:

(1 00 −1

(0 11 0

)U† (9)

Multiply by Sz on the left, and U on the right:

1 00 −1

(0 11 0

Letting:

a bc d

we calculate: (a bc d

(b a−d −c

so a = b and d = −c. We also have the unitarity requirement, which forces a2+b2 = c2+d2 = 1and ac+bd = 0. Choosing a = 1/

√2 then b = 1/

√2, so c+d = 0, which along with c2 +d2 = 1

fixes c = 1/√

2, and d = −1/√

2. So we get that:

U =1√2

(1 11 −1

Note: A lot of people didn’t really solve for U this way, and in some way just used the identitythat you were supposed to show instead. The problem was kind of vague. Now to compare tothe formula U =

∑r |b(r)〉〈a(r)|, so:

U = |+〉〈Sx; +|+ |−〉〈Sx;−| (14)

=1√2

(|+〉〈+|+ |+〉〈−|+ |−〉〈+| − |−〉〈−|) (15)

=1√2

(1 11 −1

as we found above.

30. (5 Points)

(a) There were tons of ways to prove this, and it seemed everyone used a different one. I willgive two ways. The first uses the fact that the translation operators are generated by themomentum operators, i.e. that T (~l) = e−i~p·~l/~, as well as this relation:Given that [[A,B], B] = 0, then [A,Bn] = n[A, B]Bn−1. Proof:

[A,Bn] =n∑

Bn−j [A, B]Bj (17)

[A, B]Bn−1 (18)

= n[A,B]Bn−1 (19)

where the third step is valid since you can commute B’s with [A,B]’s. Then we see that:

[A, f(B)] = [A,

∞∑n=0

f (n)(B)|0n!

Bn] (20)

= [A, f(0) +∞∑

f (n)(B)|0n!

Bn] (21)

Applying the lemma (and the fact that constants commute):

∞∑n=1

f (n)(B)|0n!

Bn] (22)

= [A,B]n∞∑

f (n)(B)|0n!

Bn−1] (23)

= [A,B]f ′(B) (24)

Q.E.D.So therefore:

[X, T (~l)] = [X, P ]−i~l

~e−i~p·~l/~ = ~lT (~l) (25)

which is what we were trying to prove. You could also prove the relation assuming onlythat the translation operator was the operator which translates the states (i.e. thatT (~l)|~x〉 = |~x + ~l〉). Then the proof goes as follows: Let [Xi, T (~l)] act on a positioneigenstate |xi〉. Then:

= liT (~l)|xi〉 (28)

and the theorem is proven because everything is linear and the |xi〉’s are a completeeigenbasis (unlike the last assignment, I didn’t take off points for not mentioning thisfact, but its still important). You could also proven things by taylor expanding theexponential or similar things, but you end up proving special cases of the relation provedabove. The second proof is “better” then the first since it does not rely on any specialform of the translation operator other then its definition.

(b) From part a, we have that T (~l)−1 ~XT (~l) = ~X +~l Then the proof is simply:

〈T (~l)−1XT (~l)〉 = 〈α| ~X +~l|α〉 = 〈 ~X〉+~l (29)

where |α〉 is the state of the system.

33. (5 Points)

(a) i. This is a simple proof by insertion of a complete set of position eigenkets:

〈p′|X|α〉 =∫

dx′〈p′|X|x′〉〈x′|α〉 (30)

dx′ x′〈p′|x′〉〈x′|α〉 (31)

= i~1√2π~

∫dx′

∂p′e−ip′·x′/~〈x′|α〉 (32)

= i~∂

∂p′

∫dx′〈p′|x′〉〈x′|α〉 (33)

= i~∂

∂p′〈p′|α〉 (34)

which is what we were trying to prove. There were other ways of proving this whichwere less straightforward involving representations of the delta function, but this wasthe quickest route.

ii. This is obvious in light of part i)

〈β|X|α〉 =∫

dp′〈β|p′〉i~ ∂

∂p′〈p′|α〉 (35)

dp′φ∗β(p′)i~∂

∂p′φα(p′) (36)

(b) This is the translation operator in momentum space (also called a boost operator inrelativistic terminology). Let’s show that this interpretation is a good one: Take aneigenstate of p, |p〉:

)|p〉 =

∫dx exp

(iXΞ~

)|x〉〈x|p〉 (37)

=1√2π~

∫dx exp

(iX(Ξ + P )

)|x〉 (38)

dx 〈x|p + Ξ〉|x〉 (39)

= |p + Ξ〉 (40)

So indeed it translates a momentum state by Ξ.

Problem 3-5 (10 Points)

(a) (Note, in this problem I will be working in the standard cartesian representation of theX and P operators). You have three canonical commutation relations to check:

[Xj , Pk] = i~δjk (41)[Xj , Xk] = 0 (42)[Pj , Pk] = 0 (43)

The second is clearly still true, the first gives (with the new Pk):

[Xj , Pk] = i~δjk + [Xj , fk(x)] = i~δjk (44)

since [Xj , fk(x)] = 0 (fk’s are just numbers, not operators, so any operator commuteswith them). So this commutation relation is also always satisfied. The third relation,however, gives:

[Pj , Pk] = [Pj , Pk] + [fj(x), fk(x)] + [fj(x), Pk] + [Pj , fk(x)] (45)

(∂fk

∂xj− ∂fj

∂xk=

∂xj(47)

This can be stated in differential vector calculus as ∇× ~f = 0, or the curl of f vanishesover all space.

(b) We know that any unitary operator can be expressed as eiA where A is hermitian. Nowwe can apply this general unitary operation and assume the unitary equivalence to solvefor A:

〈x|Pj |x′〉 = 〈x|Pj |x′〉 (49)

= 〈x|e−iA ~i

∂xjeiA|x′〉 (50)

= 〈x|e−iAeiA

∂xjA +

)|x′〉 (51)

So to have equality requires that

∂xjA = fj(x) (53)

From part a), we have that ∇× ~f = 0, which implies that f is the gradient of some scalarfunction φ: ∇φ = ~f . Therefore if we set A = φ/~, the required equation is satisfied, andthe unitary equivalence is proven. So under the unitary transformation U = eiφ/~, whereφ is defined by ∇φ = ~f , the new momentum operator is unitarily transformed into theold. φ always exists because of the requirement on f from part a.

(c) If you remove an infinite cylinder around the z-axis (or any axis actually), the reasoningin part b) does not apply, because the implication ∇ × ~f ⇒ ∃φ : ∇φ = ~f (also calledthe Helmholtz theorem) is no longer true. It is no longer true because its proof relies onStoke’s Theorem, which if you recall says that the line integral around a closed loop ofsome vector field is equal to the surface integral on the surface bounded by that closedloop of the curl of that vector field. In math notation:

(∇× F) · da =∫

F · ds (54)

But with an infinite cylinder removed, if you choose a loop which winds around thatcylinder there do not exist any surfaces with the loop as the boundary (note, the notation∂S means “the boundary” of the surface S). Thus Stoke’s theorem does not apply tothose loops and you can’t prove the implication above.Note, another way of putting this is that the Helmholtz theorem is only true if space issimply connected, e.g. has the topology of Rn or Sn. By removing an infinite cylinderyou change the topology of space, thus making it not simply connected globally (notehowever, that it is still simply connected locally). So if the cylinder was finite, part b)would in fact still apply because you could topologically shrink the missing space downto a point at the origin. Note how none of this really says anything about rotationalsymmetry.

NRQM Class

R. Rogers

23 ,26————————————————————-

Prob 23:

The characteristic polynomials are:

A:(λ−a)(λ+a)2 with rootsa,−a and eigenvectors

B: (λ−b)2(λ+b) with rootsb,-b and eigenvectorsm

0−i1

a. Yes

The product is:

a · b 0 00 0 i · a · b0 −i · a · b 0

from either side.

Obviously

works for one eigenket and has eigenvlaues a,b respectively. 0−i1

works with eigenvalues -a,b 0i1

works with eigenvalues -a,-b

Any vector v

can satisfy Av=-av .

but it is impossible for it to satisfy Bv= -bv=bvSimiliary we can satisfy Bv=bv but then Av =-av=avThus, because of the combinatorial properties for the (or any) common eigenvalues

there is only one set of solutions.

Prob 26:

SinceSz is already diagonal we can use the eigenkets ofSx

(1−1

)to form

the Unitary matrixU = 1√2

[1 1−1 1

]andU † SxU = Sz

I will take a moment to digress into a matrix representation interpretation of 1.5.4

(for those of us who have problems interpreting the symbolism)Let Az, Bxbe a column collection the normalized eigenkets ofAz, Bx for instance[

1 00 1

]= Sz for Sz

Similiarly for Sx Sz = 1√2

[1 1−1 1

]The use ofz, x in Az, Bx is just for identification, the following is general.(Az)†Az = I and the same holds forBx because they are composed of normalized

eigenkets.Then considerU = Bz · (Az)† (or Sz(Sx)† in our case), note thatUU† = I.Now we prove that|bl >= U |al > in matrix representationMultiplying U on the right by eigenkets ofA ,v will move a one around the output

of(Az)†v depending upon the ordering of B. Now the wandering one will extract

exactly one of the eigenkets ofB from Bz

Now evaluatingSz(Sx)†[

1 00 1

] [1 1−1 1

[1 1−1 1

]I guess I should apologize for using a crutch, matrix representation, but I haveproblems following the flying summations and superscripts. Hopefully I will beable to dispense with the crutch after a while; although it took me a long time todo Linear Algebra in the abstract (and proper) form.

Problem h

SxSy i

jihj jihj

SySx i

jihj jihj

Sx Sy SxSy SySx i

jihj jihj

fSx Syg SxSSy SySx

jihj jihj

Similarly for all the other combinations of Si Sj

Note that in the basis ji ji the operator

Ajihj Ajihj Ajihj Ajihjis given by the matrix

Problem h

S n Sx cos sin Sy sin sin Sz cos

sin eijihj sin eijihj cos jihj jihj

Expressed as a matrix in the ji ji basis this operator reads

cos sin ei

sin ei cos

and the eigenvalue equation becomes

cos sin ei

sin ei cos

which has the solutionxy

sinei cos

or normalized

ei sin

Note that by changing n n ie we get the eigenvector

for S n corresponding to eigenvalue xy

ei cos

Problem h

In the basis ji ji the operator H is given by the matrix

with eigenvalues

qH H H

Eigenvectors when H

H HqH H H

with the normalization constant C is given by

H HH H

qH H H

One can check that the limit H makes sense For instance if H H theeigenvector for goes to for H

Alternatively one can use the results in problem to nd eigenvalues and eigenvectors Decompose the matrix for H as follows

qH H H

cos sinsin cos

cos H Hq

This corresponds to the situation in problem with and we can now reado the eigenvalues and eigenvectors

Problem h

In the notation of problem we have and thus the eigenvectorcorresponding to eigenvalue h is

j n i cos

ji sin

The operator Sx corresponds to S n with ie the eigenvector of Sx witheigenvalue is

j x i p

a well known result of course

The probability of nding the system in the state j x i after a measurement of Sxgiven that it initially was in the state j n i is

jh x j n ij

The expectation value of Sx in the state j n i can be calculated using matrixnotation or the ketbra notation of problem Using the ketbra notation oneobtains

hSxi h n j

jihj jihj

hSx hSxii hS

xi hSxi

Prob 9:Using the identication of Sxfrom problem 8 and applying the diagram:

sin(β)cos(α) h2

[0 11 0

]+ sin(β)sin(α) h

[0 −ii 0

]+ cos(β) h

[1 00 −1

[cos(β) sin(β)cos(α) − isin(β)cos(β)

sin(β)cos(α) + isin(β)cos(β) −cos(β)

[cos(β) sin(β)e−iα

sin(β)eiα −cos(β)

]Now we could use the last part of problem 7 to identify the eigenvector, this

leads to some inobvious manipulation; division by sqrt's to normalize to 1. I willdo a simple extraction (one that would lead to headaches for a larger numberof states).

Let the eigenket be

]and do the simple evaluation of

[cos(β) − λ sin(β)e−iα

sin(β)eiα −cos(β) − λ

])to determine the eigenvalues h

2 ,−h2

Choosing h2

a = a · cos(β) + b · sin(β)e−iα

b = a · sin(β)eiα − b · cos(β)Where the h

2 term cancels out.

]is normalized and that we can introduce a normalizing phase

for the rst term; we can assume the form

[cos(γ)

sin(γ)eiτ

]but more dramatically

a= tan(γ)eiτ (1)

Solving the above conditions on a,bba = (1−cos(β))eiα

sin(β) or ba = sin(β)eiα

1+cos(β)

Which fortunatly are consistant and yield ba = tan(β

2 )eiα

Resolving this with 1 yields the answer.For a greater number of states we just have to walk through using 7.b; not

to obvious but mechanical.

Sam Pinansky

October 26, 2003

Average score: 25.5/30.

1. (5 Points) First, a useful lemma: From the definition of the Heisenberg representation, AH(t) =U†(t)AU(t). Therefore:

[AH , BH ] = [U†(t)AU(t), U†(t)BU(t)] (1)= U†(t)CU(t) (2)= CH (3)

So the commutation relation between the Heisenberg operators is the same at equal times.Note that the relation is NOT true for the commutation relation between operators at differenttimes.Then with H = ωSz, the heisenberg equations of motion are:

dSx(t)dt

i~[Sx(t), Sz(t)] (4)

dSy(t)dt

i~[Sy(t), Sz(t)] (5)

dSz(t)dt

= 0 (6)

Clearly, Sz(t) = Sz(0). Now using the previous lemma, we know that [Sx(t), Sz(t)] = −i~Sy(t),etc.. so we get:

dSx(t)dt

= −ωSy(t) (7)

dSy(t)dt

= ωSx(t) (8)

We can solve these coupled differential equations by elementary means, giving:

Sx(t) = A cos ωt + B sin ωt (9)Sy(t) = C cos ωt + D sin ωt (10)

Now we apply the initial conditions Sx(0) = Sx(0) and Sx(0) = −ωSy(0), etc... giving:

Sx(t) = Sx(0) cos ωt− Sy(0) sin ωt (11)Sy(t) = Sy(0) cos ωt + Sx(0) sin ωt (12)Sz(t) = Sz(0) (13)

4. (5 Points) Just solve the Heisenberg equation of motion for X(t):

[X, H] =1i~

[X,P 2

2P (14)

So X(t) = P (0)tm + X(0), and therefore:

[X(t), X(0)] = − i~tm

5. (5 Points) Using the canonical commutation relations, you see that

[H,X] = − i~Pm

[[H, X], X] = −~2

Now we can work backwards to the theorem:

2m= −1

2〈a′′|[[H, X], X]|a′′〉 (19)

= −12

(〈a′′|HX2|a′′〉 − 2〈a′′|XHX|a′′〉+ 〈a′′|X2H|a′′〉) (20)

= −12

(2Ea′′〈a′′|X2|a′′〉 − 2

a′|〈a′′|X|a′〉|2

a′|〈a′′|X|a′〉|2(Ea′ − Ea′′) (22)

where we have inserted a complete set of states.

11. (5 Points) Recall X(t) = X(0) cos ωt + P (0)mω sin ωt. Clearly 〈P 〉 = 〈0|eiPa/~Pe−iPa/~|0〉 = 0.

Also, using the Baker Hausdorff Lemma (or your result from a previous homework):

〈X〉 = 〈0|eiPa/~Xe−iPa/~|0〉 (23)= 〈0|X|0〉+ a (24)

So then

〈X(t)〉 = 〈X〉 cos ωt +〈P 〉mω

sin ωt + a cosωt (25)

= a cosωt (26)

since 〈0|X|0〉 = 〈0|P |0〉 = 0 (expand out P and X in terms of a and a† to see this).

12. (10 Points)

(a) e−ipa/~|0〉 is just the translation operator T (a), so we know that this operator translatesthe system by a in the positive direction. So then the wave function in position space is(by definition):

〈x|e−ipa/~|0〉 = 〈x− a|0〉 (27)

=1√√πx0

e−(x−a)2/2x20 (28)

(b) First, we note that in general, the time dependence drops out, as can be seen:

|ψ(t)〉 = e−iHt/~e−iPa/~|0〉 (29)

|〈0|ψ(t)〉|2 =∣∣∣e−iE0t/~〈0|e−iPa/~|0〉

∣∣∣2

=∣∣∣〈0|e−iPa/~|0〉

∣∣∣2

So all the time dependence drops out.To calculate the probability:

〈0|ψ〉 =1√πx0

∫dxe−x2/2x2

0−(x−a)2/2x20 (32)

= e−a2/4x20 (33)

|〈0|ψ〉|2 = e−a2/2x20 (34)

Sam Pinansky

November 4, 2003

15. (5 Points) We know that

x(t) = x(0) cos ωt +p(0)mω

sin ωt (1)

So then, the correlation function is:

〈x(t)x(0)〉 = 〈0|x(0)2 cosωt +p(0)x(0)

mωsin ωt |0〉 (2)

Now we expand x(0) and p(0) in terms of a and a†:

2mω〈0| (a + a†)2 cosωt + i(a− a†)2 sin ωt |0〉 (3)

2mω〈0| aa† cos ωt− iaa† sin ωt |0〉 (4)

2mωe−iωt (5)

where the aa a†a† and a†a terms vanish in the expectation.

16. (10 Points)

(a) Let’s say that we have a state α |0〉+ β |1〉, with the constraint |α|2 + |β|2 = 1, then

〈x〉 = (α∗ 〈0|+ β∗ 〈1|)(√

(a + a†)

)(α |0〉+ β |1〉) (6)

2mω(αβ∗ + βα∗) (7)

Going to polar coordinates, we can choose α is real (since there is an arbitrary phaseambiguity), then applying the normalization constraint, we get that α =

√1− r2, β =

reiφ. Plugging in:

2mω(2r

√1− r2 cos φ) (8)

We can maximize each part individually, giving φ = 0, and r = 1/√

2. So the maximumstate is:

1√2(|0〉+ |1〉) (9)

(b) The time evolved state is:

(e−iωt/2 |0〉+ e−3iωt/2 |1〉

i. Schroedinger picture:

〈x〉 (t) =

(eiωt/2 〈0|+ e3iωt/2 〈1|

)(a + a†)

(e−iωt/2 |0〉+ e3iωt/2 |1〉

(eiωt + e−iωt

2mωcos ωt (13)

ii. Heisenberg:

〈x(t)〉 =

12(〈0|+ 〈1|)((a + a†) cos ωt + i(a− a†) sin ωt)(|0〉+ |1〉) (14)

2mωcosωt (15)

(c) We already know from the last part that

〈x〉2 =~

2mωcos2 ωt (16)

so all we need now is⟨x2

⟩. Using the schroedinger picture:

(eiωt/2 〈0|+ e3iωt/2 〈1|

)(a + a†)2

(e−iωt/2 |0〉+ e−3iωt/2 |1〉

(eiωt/2 〈0|+ e3iωt/2 〈1|

)(e−iωt/2 |0〉+ 3e−3iωt/2 |1〉

mω(19)

So therefore ⟨(∆x)2

(2− cos2 ωt) =~

2mω(1 + sin2 ωt) (20)

21. (5 Points) You should recall from previous quantum mechanics that the eigenfunctions of theinfinite square well are:

ψn(x) =

sinnπx

with energy En = ~2n2π2

2mL2 . You could also look these up in appendix A.2 in the textbook. Ifthe particle is known to be exactly at x = L/2, then it’s wave function (not properly normal-ized) would be δ(x − L/2). So measuring the energy of the particle would have probabilities

| 〈n|x = L/2〉 |2, or inserting an x basis:

〈n|x = L/2〉 =∫ ∞

−∞

sinnπx

Lδ(x− L/2) dx (22)

sinnπ

√2L n odd

0 n even(24)

so the relative probability of being in a state n versus a state n′ is:

Pn′=

0 n even n odd1 n and n′ even or odd∞ n odd n′ even

i.e. all states of even n can’t occur, and all states of odd n have equal probability. The timeevolution is simply:

ψ(t) =∑

(−1)ne−iEnt/~ sinnπx

Note that this state is badly unrenormalized.

22. (5 Points) (Let λ = v0, so I can easily reuse my solutions from last year). The Schroedingerequation in this case is just

− ~2

dx2− λδ(x)ψ = Eψ (27)

Looking at this differential equation, we can see that there is an infinite jump in the potentialat x = 0. So any solution to this equation will need to have a similar jump in d2ψ/dx2, i.e.dψ/dx will be discontinuous. To obtain the discontinuity in ψ′, we integrate the Schroedingerequation around an epsilon neighborhood about the origin:

− ~2

∫ ε

dx2dx− λ

∫ ε

δ(x)ψ(x) dx = E

∫ ε

ψ(x) dx (28)

In the ε → 0 limit, the first term is the change in the slope of ψ at 0 by the fundamentaltheorem of calculus, the second term is λψ(0) from the definition of the delta function, andthe right term is zero since we restrict ψ(x) to be finite at x = 0. This gives:

ψ′(0+)− ψ′(0−) = −2mλ

~2ψ(0) (29)

The matching condition of ψ(0) is clearly that ψ need to be continuous at zero. AssumingE < 0 (which restricts our view to bound states), we have that in the region x < 0:

ψ(x) = Aekx (30)

and in the region x > 0ψ(x) = Be−kx (31)

where k =√−2mE/~2. (The other exponential terms with opposite exponents are zero

from the requirement that the wavefunction not blow up at infinity). The continuity of ψ(x)is already set by setting B = A. Then the last matching condition at 0 (the one on thederivatives) gives:

ψ′(0+)− ψ′(0−) = −2mλ

~2ψ(0) (32)

−2Ak = −2mλ

~2A (33)

k =mλ

~2(34)

√−2mE

~2(35)

E = −mλ2

2~2(36)

Which is the only bound state energy. Normalizing ψ:

ψ(x) =

√mλ

~2e−

mλ~2 |x| (37)

Extra Problem (15 Points)

(a) We will make the reasonable guess that a simple rotation will decouple the coordinates:

Xx =X1 + X2√

X ′y =

X1 −X2√2

(Reason for the prime will be clear shortly). And the same for the momentum states:

Px =P1 + P2√

Py =P1 − P2√

Inserting these into the hamiltonian, and simplifying CAREFULLY (many people mademistakes here), we arrive at:

H =P 2

x + P 2y

12mω2

x + (1 + 4λ)X ′2y − 2

√2hX ′

y + 2h2)

So we have decoupled the hamiltonian. But considering the next parts, let’s changevariables again by completing the square:

H =P 2

x + P 2y

12mω2

x + (1 + 4λ)

(Xy −

1 + 4λ

+8λh2

1 + 4λ

So now we let

Xy = X ′y −

1 + 4λ(44)

and the hamiltonian is simply:

H =P 2

x + P 2y

12mω2X2

x +12mω′2X2

y +4mω2λh

1 + 4λ(45)

with ω′ = ω√

1 + 4λ. This is obviously decoupled.

(b) Clearly the X’s and P ’s commute with themselves since the original X’s and P ’s allcommuted with themselves, the only ones we need to check are [Xx, Px] = i~, [Xx, Py] = 0(and the two that are almost identical).

[Xx, Px] =12[X1 + X2, P1 + P2] (46)

=12([X1, P1] + [X2, P2]) (47)

=12(2i~) (48)

= i~ (49)

[Xx, Py] =12[X1 + X2, P1 − P2] (50)

=12([X1, P1]− [X2, P2]) (51)

= 0 (52)

where we’ve dropped all zero terms. The additive constant in the definition of Xy obvi-ously won’t change anything since constants commute with everything. So they satisfythe standard commutation relations. The hamiltonian looks like two separate decoupledharmonic oscillators, one with frequency ω, one with frequency ω′, and an extra energyoffset of 4mω2λh/(1 + 4λ).

(c) Using the one dimensional oscillator as an example, we take:

ax =√

(Xx + i

a†x =√

(Xx − i

√mω′

(Xy + i

mω′

a†y =

√mω′

(Xy − i

mω′

These must satisfy the standard commutation relations [ax, a†x] = 1 etc.. since the X’sand P ’s satisfy the standard ones. We can rewrite the hamiltonian in terms of theseoperators, (specifically Nx = a†xax and Ny = a†yay) just as in the 1-d case:

H = ~ω(

Nx +12

)+ ~ω′

4mω2λh

1 + 4λ(57)

Since H is linear in both Nx and Ny, it can be diagonalized in both simultaneously. Let|nx, ny〉 denote a simultaneously eigenket of both Nx and Ny, such that Nx |nx, ny〉 =nx |nx, ny〉 and Ny |nx, ny〉 = ny |nx, ny〉, (or equivalently by taking two different 1-dharmonic oscialltors and taking their tensor product |nx〉 ⊗ |xy〉 ≡ |nxny〉). Then thesestates are also eigenstates of the full hamiltonian with eigenvalues:

Enx,ny = ~ω(

nx +12

)+ ~ω′

4mω2λh

1 + 4λ(58)

We can construct the full space of eigenstates by multiple applications of the raisingoperators:

|nx, ny〉 =(a†x)nx(a†y)ny

√nx!

√ny!

|0, 0〉 (59)

properly normalized. The ground state wavefunction is simply the multiplication of theground state wave functions of the two oscillators:

〈xx, xy| 0, 0〉 =

√m√

ωω′√π~

exp(−1

ymω′

(d) You can have degeneracy in the energy only if Enx,ny = En′x,n′y for some pairs (nx, ny)and (n′x, n′y). Therefore we want:

nx +12

)+ ~ω′

)= ~ω

(n′x +

)+ ~ω′

(n′y +

nx +√

1 + 4λny = n′x +√

1 + 4λn′y (62)√

1 + 4λ =n′x − nx

ny − n′y(63)

Thus the only way this can occur is if√

1 + 4λ is a rational number. In the specialcase λ = 0 you simply have an isotropic 2-d harmonic osscillator, where every state isdegenerate except the ground state.

"!$#%'&(*),+.-/ 0,1234*576*89:6;)=<

|λ >= c∑

λ′n

√n!|n > > 1@?

AB85 λ′),C6 85),+EDF9GIHJ9+KMLJ),*NOJ+ NP9:G,)IQ@9R6 )I+.<S9LT6*

a|λ >= c∑

λ′λ′n

√n!|n >

a|λ >= λ′|λ >

(*)I+U-/ 0V-/1

|λ >= c∑

λ′n

(a†)n

= c∑

λ′n(a†)n

n!|0 >

= ceλ′a†|0 >

W XY8+*NP9:G,),Q9R6 )I+UZ /LT@*C)[;9L\8),5DKU]_^

< λ|λ > ` 9:+K1

< λ|λ >= c2∑

|λ′|2n

n!< n|n > > -?

> W ),+L5J6*8 |n > a ;9*7*6*8+9:Gb?c +Kd*T6*6*),+O6*8)[Y6 e+f

c2e|λ′|2

c = e−1|λ′|2/2

X$ONOg9:+:6*85;h_Hi6*),+M)[5fkjl9OLT85 5+_6;i6 9R6 l6\9:]G,@m(*)I+O6*8gK/)[ LTH i),+M9: H+nKM-I1 on16 8),Y)[Y6 *HJ)I<

λ)[;9:+d5),+p56C<

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a † a +1

6**Nq),C6 *),DE),9GrXY8

a † a|λ >6**Nq89C6 e]sgLT+*),K/*@K.6 5 Nt]E^P6**Ndf

a † aλ′n

√n!|n >= a †

λ′n

√n|n − 1 >=

AB8)[L\8MNO@9:+C6 8gLT85 5+_6;l6\9R6 7)[Y+6l6\9:]G,uv &w)I\i6Y3CJ+5@Kx6*ey+K

< λ|x|λ >, < λ|p|λ >(*)I+U-/ 0V-Ro

2mω(a + a†), p =

2(−a + a†)

w< x >

9:ZZGI^x9 6 O6*8 ),8_6 ` 9+K a†6 e6*8G,T<z6 ` ^E)IG,K/),+

2mω< λ|(a + a†)|λ >

< x >= (λ′ + λ′∗)

2mω< λ|λ >

< x >2= 4 · (Re(λ′))2h

2mωW ),Ne),G[9: GI^

= (−λ′ + λ′∗)

2= 4 · (Im(λ′))2mωh

2w< x2 >, < p2 >

< λ|x2|λ >=h

2mω< λ|(a2 + a †2 +a † a + aa†)|λ >

;R3|Hi),+O6 8K/Ty+)=6 )I+d9+KMGI),+@9: )=6l^R*HZs5 Zn_i)I6*),+$~1@?Y9ZZG,^P6*8Jyn i6Y6*5 Nt6 e6*8 )I8_6

λ′2 < λ|λ >-?Y9ZZG,^P6*8g*L5+KU6**Nt6*O6*8GI5<z6λ′∗2 < λ|λ >0?Y9ZZG,^P6*8J6 8), Kx6**Nt6*O6*8*),8_69:+KMG,T<z6

= |λ′|2 < λ|λ >= λ′∗λ′

o_?46 8J<H*6*8x6**Nq),9eG,)=6*6*G,gNO 7L5NOZG,),L9R6 K'f(*)I+U-/ 0 0

aa† = 1 + a † a

XY85+< λ|aa † |λ >= 1 + λ′∗λ′

9:+Kx6 8_Hn

< λ|x2|λ >=h

2mω(1 + (λ′ + λ′∗)2) =

2mω(1 + 4 · (Re(λ′))2)

< x2 > − < x >2=h

w< λ|p2|λ > ` p = i

mhω2 (−a + a†)

< λ|p2|λ >= −mhω

2< λ|(a2 + a †2 −a † a − aa†)|λ >

),5G[K1@?

λ′2-?λ′∗20?−λ′∗λ′o_?1 − λ′∗λ′;*

< λ|p2|λ >=mhω

2(1 + (λ′ − λ′∗)2) =

2(1 + 4 · (Im(λ′))2)

< p2 > − 2=mhω

2W << ∆x2 >< ∆p2 >>=

&XY8), *+M< N),K/+)I+Z9:*69x-XY8gNO_l67Z*]9:]GI^MDR9:G,H > ;DR9:G,H@ ?9: LTNOZH/6*@K]E^U6\9:pE),+.6 8 \9R6 )I+:<*HL5L5 i),DJ6 5 NO Ti

= n+1|λ|2

38)[L\8),D@<NO/K/FRNe_l6Z ]n9:]G,DR9GIHf

n+1|λ|2

> 1 ),+PK/R3+8),GIG ` i6 9:6*),+ n > |λ|2 − 1nλ

< 1 ),+OHZ8),G,G ` i6 9:6*),+ |λ|2 < n

|λ|2 − 1 < n < λ

AB8)[L\8M),+iNOJL9*C89C6l34P*G,H/6*),+ &c ZZGI^U6*8gKTy+)I6*),+d<k485 5+_6fAB89:6;),C6 8DR9:G,H:<

a · e−ipl/h|0 >mHGI6*),ZG,^_),+P]E^eipl/h

eipl/ha · e−ipl/h = a + ilh[p, a] + (il)2

2! [p, [p, a]] . . .

[p, a] = i

mωh2 ((a † −a)a − a(a † −a)) = i

mωh2 (a † a − aa†) = −i

mωh2 IXY8EH8),8Y\K/5Y6 5 NO9*7Q5*P9+K

eipl/ha · e−ipl/h = a + l√

eipl/ha · e−ipl/h|0 >=(

a + l√

|0 >= l√

mωh|0 >XY8EH5f

a · e−ipl/h|0 >= l

he−ipl/h|0 >

jlK/5+_6 )=<^E),+|λ >= e−ipl/h|0 >

),D@a|λ >= l

mωh|λ

! #"%$%&('*),+.-%/10,$ 2"%&,354%67 98

:;=<?>A@BDCFE

J± = a†

±a∓ G Jz = h

2 (N+ − N−) G N = N+ + N−

[Jz, J±] =h

2([N+, J±] − [N−, J±])

[N+, J±] = [N+, a†

±a∓] = [N+, a

±]a∓ = a

±a∓

[N−, a †± a∓] = a†

±[N−, a∓] = −a

±a∓HI$KJ)

[Jz, J±] =h

2([N+, J±] − [N−, J±]) = ha

±a∓ = ±hJ±

:;=<?>A@BDCPO

J2 = J2z + 1

2 (J+J− + J−J+)H #&,8Q4KRS0, #&,8UTV 4KWX+73*J=)Y67RZT

[JzJz, Jz] = 0[),+.-/T[BC, A] = [B, A]C + B[C, A]

8\3K]^3`_Ia*b c%bed`f* [J+J−, Jz] = [J+, Jz]J− + J+[J−, Jz] = −hJ+J− − J+(−)hJ− = 0g +78h+767+7'*&,67RS_i3*&I0j$% 26k'*)Y0!0, 9&j8Ub

:;=<?>A@BDCl

+ + N2−− N+N− − N+N−)

J+J− = h2a†

+a−a†

−a+ = N+a−a

−= N+(I + N−) = N+ + N+N−

J−J+ = h2a†

−a+a

+a− = N−a+a†

+ = N−(I + N+) = N− + N−N+

J2 = h2

4 (N2+ + N2

−− N+N− − N+N− + 2N+ + 2N+N− + 2N− + 2N−N+)

J2 = h2

N2 + 2N)

Sam Pinansky

November 10, 2003

23. (7 Points) We know from the previous homework that the wavefunction at t < 0 is:

ψ(x) =√

κe−κ|x| (1)

where κ ≡ mλ/~2. To find the time evolution of this wave function, we note that:

U |t < 0〉 =∫

dk U |k〉〈k| t < 0〉 (2)

∫dx U |k〉〈k|x〉〈x| t < 0〉 (3)

where U is the time evolution operator and |k〉 are a complete set of states for a free particle,(i.e. eigenstates of H). Therefore U |k〉 = e−i~k2t/2m |k〉, and we just need to evaluate:

∫dx 〈k|x〉〈x| t < 0〉 =

∫ ∞

−∞dx

1√2π

e−ikx√

κe−κ|x| (4)

κ + ik+

1κ− ik

k2 + κ2

Plugging back into the full state:

|t > 0〉 =∫

dk e−i~k2t

k2 + κ2

)|k〉 (7)

So the wave function at times t > 0 is then:

〈x| t > 0〉 =κ3/2

e−i~k22m t+ikx

k2 + κ2(8)

Try as I might, I can’t seem to be able to do this integral. Even numerically it seems toresist evaluation. Anyway, there was also a different way to do this problem using the freeparticle propagator, which gives a seemingly different undoable intergral, but the two answersare simply related by a fourier transform, I believe.

-10 -5 5 10

Figure 1: Wave function

24. (8 Points)

(a) In this problem, you were expected to use the WKB approximation to find approximateexpressions for the eigenfunctions. It is possible to solve this exactly in terms of specialfunctions called Airy functions, but it wasn’t required. The energy spectrum is continuous,since there is only one boundary condition at infinity (or in other words, because thereare no bound states). Using the WKB approximation, in the region where E > V , youget (after plugging in for V and doing the trivial integral):

ψ(x, t) =A

(E − λx)1/4exp

(± i√

3λ~(E − λx)3/2 − i

for x < E/λ. This is simply a wave with a slightly decreasing amplitude and increasingfrequency as x → −∞. In the other domain of validity:

(E − λx)1/4exp

(−√

3λ~(λx− E)3/2 − i

which is simply a decreasing exponential as x →∞.

(b) With the extra boundary condition at −∞, the states become bound, and the energiesdiscrete. You now have an extra matching condition as well. Again this can be solvedexactly in terms of special airy functions. Also, you can use the same method as thebouncing ball system in the text to derive the energy spectrum, namely that:

3π~mλ(n + 12 )

2m(11)

but this wasn’t required (the derivation is exactly as in the text for the bouncing ball).

Bonus Challenge: Given a generic potential well of the form f(|x|) where f is polynomialof degree d, how does the energy scale with the energy levels at large n, i.e. find a whereE ∝ na. We already know that if d = 1, a = 2/3, and if d = 2, a = 1. What aboutfractional d? Answer this question correctly and I’ll be very impressed.

Sam Pinansky

November 17, 2003

Average score: 24.8/25. No this isn’t a typo. People did extremely well on this assignment sinceall the answers were given (all you had to do was figure out the calculation). There’s almost nopoint in me writing solutions.

28. Let’s do both parts of the problem at once by finding the propagator in D dimensions. Weknow from Eq. 1.7.50 (and a little extrapolation) that

〈x|p〉 =1

(2π~)D/2exp

(ip · x~

Combining this with equation 2.5.8 and the energy p2/2m you generalize equation 2.5.15:

K(x′′, t;x′, t0) =(

)D ∫ ∞

−∞dp′ exp

(ip′ · (x′′ − x′)

~− ip′2(t− t0)

where dp′ ≡ dp1dp2 . . . dpD. Now all we need to do is to do the integral. First we completethe square:

)D ∫ ∞

−∞dp′ exp

(− i(t− t0)

(p′ − m(x′′ − x′)

(t− t0)

im(x′′ − x′)2

2~(t− t0)

Shifting our variables, let p = p′ −m(x′′ − x′)/(t− t0), then:

im(x′′ − x′)2

2~(t− t0)

) ∫ ∞

−∞dp exp

(− i(t− t0)

The integral is simply D copies of a basic gausian integral∫

dx exp(−ax2) =√

K(x′′, t;x′, t0) =(

2πi~(t− t0)

im(x′′ − x′)2

2~(t− t0)

Which is eq. 2.5.16 if D = 1, and is also true for a free particle in any dimension.

29. Calculating:

− 1Z

∂β=

∑a Eae−βEa

∑a e−βEa

Now by assumption there is a lowest energy, E0, such that E0 ≤ Ea : ∀a. So we factor thatout:

− 1Z

∂β=

E0 +∑

a>0 Eae−β(Ea−E0)

1 +∑

a>0 e−β(Ea−E0)(7)

Now we take the limit β →∞, noticing that Ea − E0 > 0 so all the exponentials go to zero:

limβ→∞

E0 +∑

1 +∑

a>0 e−β(Ea−E0)= E0 (8)

Illustrating it for the particle in the box is really pointless, and I won’t bother to do it.

36. (a) First, note that the P ’s and A′s commute with themselves. Then

[Πi, Πj ] = [Pi − eAi

c, Pj − eAj

c] (9)

= −e

c[Pi, Aj ]− e

c[Ai, Pj ] (10)

= −e

c(PiAj − PjAi) (11)

= − i~ec

εijkBk (12)

from B = i~p×A. In this specific case, B = Bz so:

[Πx, Πy] =i~ec

B (13)

with all the other commutators zero.(b) Defining ω = |eB|

mc . Then H = H1 + H2 where H1 = Π2z

2m and H2 = 12m (Π2

x + Π2y). Now

define P ′ = Πy and X ′ = Πx/(mω) then [X ′, P ′] = i~, i.e. they have the standardcommutation relations of the harmonic oscillator, and H2 = P ′2

2m + 12mω2X ′2. Therefore

the energy of H2 is just E2 = ~ω(n + 1/2). The contribution from H1 is just the freeparticle energy E1 = ~2k2

2m . Since [H1,H2] = 0, then E = E1 + E2.

37. One particle precedes to the detector unhindered, whereas the other particle passes througha region of constant magnetic field. While the neutron is in the field, it’s phase processes atfrequency ω = gn|e|B/(2mnc) (from equations 2.1.52 and 2.1.56 in the text). From classicalconsiderations, the neutron is in the magnetic field for l/v seconds, or t = mλl/h. The thetotal phase difference is

mλlgn|e|B2mnc

Now whenever this phase difference changes by 2π, the count rate will be the same, so thechange in B needed to give successive maxima is:

2π =mnλlgn|e|

2mnhc∆B (15)

∆B =4πhc

|e|gnλl(16)

Sam Pinansky

November 24, 2003

Average score: 27.8/30. Common errors were forgetting that α and β are complex, and missinga subtle sign issue in 2b.

1. We’ve all done the first part a ton of times now, I’ll just quote the answers: Eigenvaluesλ = ±1, with eigenvectors:

|1〉 =1√2

|−1〉 =1√2

(There is an arbitrary choice of phase here...). Now if you are in a state(

), then the

probability of being spin up in the y direction is:

∣∣∣∣(

1 −i) (

)∣∣∣∣2

=12− Im(αβ∗) (4)

assuming that the initial state is normalized.

2. (a) First, note that U = A(A†)−1 where A = a0 + iσ · a, since σ†i = σi, and also note that(A†)−1 = (A−1)†, provable direcetly by taking that dagger of AA−1 = I. Since thenU = A(A−1)†, then U† = A−1A†. What is A−1? Well note that:

AA† = (a20 + |~a|2)I (5)

A−1 =A†

a20 + |~a|2 (6)

So then

UU† = A(A−1)†A−1A† (7)

=AAA†A†

(a20 + |~a|2)2 (8)

=AA†

a20 + |~a|2 (9)

= I (10)

So U is unitary. To show that it is unimodular, note that det(U) = det(A) det((A†)−1) =det Adet A† , now all we need is:

det A = a20 + |~a|2 = det A† (11)

after a simple calculation. So therefore det(U) = 1.

(b) For this part we need an explicit expression for U . Cranking through the algebra, youarrive at:

a20 + |~a|2

(a20 − |~a|2 + 2ia0a3 2a0a2 + 2ia0a1

−2a0a2 + 2ia0a1 a20 − |~a|2 − 2ia0a3

Now comparing with Eq. 3.2.45, we immediately see from the real part of the first entrythat:

φ = 2 cos−1

(a20 − |~a|2

a20 + |~a|2

Now equating the imaginary parts of the first component, we see that:

−nz sin(

a20 + |~a|2 (14)

Now we calculate what the sin is (note the ± from the square root... many people missedthis)

√1− cos2

= ± 2a0|~a|a20 + |~a|2 (16)

Inserting this into the above, and solving for nz:

nz = ∓ a3

|~a| (17)

Similarly, you can show that:

ny = ∓ a2

|~a| (18)

nx = ∓ a1

|~a| (19)

So simply n = ∓a, where the sign is determined by the sign of sin(φ/2).

3. (a) In the limit A → 0 (we will always list e− and e+ in that order when operators or ketsare next to each other):

Hχ+χ− =eB

mc(SzI − ISz)χ+χ− (20)

=eB~mc

χ+χ− (21)

so it IS an eigenstate, with eigenvalue E = eB~mc .

Hχ+χ− = A(SxSx + SySy + SzSz)χ+χ− (22)

4(χ−χ+ + χ−χ+ − χ+χ−) (23)

(This is most easily calculated by expression Sx = (S−+ S+)/2 and Sy = i(S−−S+)/2).Thus it is not an eigenstate, and has an expectation value −A~2

4. (a) Note that since all three elements Sz, Sz−~ and Sz +~ are diagonal, so they all commute.Now notice that

Sz |0〉 = 0 (25)(Sz − ~) |1〉 = 0 (26)

(Sz + ~) |−1〉 = 0 (27)

So then Sz(Sz + ~)(Sz − ~) |i〉 = 0 where |i〉 is any eigenstate of z. Since any ket isexpressible in terms of this basis, then Sz(Sz + ~)(Sz − ~) |ψ〉 = 0 for any state, thus ismust be the zero matrix.

(b) Rotate your axis such that x goes to z. Then Sx → Sz, and Sx(Sx + ~)(Sx − ~) =RSz(Sz + ~)(Sz − ~)R−1 where R is the rotation matrix. But of course R0R−1 = 0, so itis still simply the zero matrix.Note that in general, given any matrix A:

(A− ai) = 0 (28)

where ai are the eigenvalues of A. This theorem has some name I forget.

Problem

a Yes l since x y and z can all be expanded in Y ml s with l More

explicitly

where crp

b From we get

cfr i p

i pY Y

and the relative probabilities for measuring m are

c H E where

is an eigenstate of L corresponding to eigenvalue ll l Thus usinghr rfr

V rh Eh or V r E

Problem

Thus hlmjLxyjlmi since L change jlmi to jlm iSince L

x Ly L L

z we get

hlmjLxjlmi hlmjL

yjlmi ll m

On the other hand we have hlmjLxjlmi hlmjL

yjlmi and in fact

hlmjfLxjlmi hlmjfLyjlmi

for any function f since we can write

Ly eiLzLxeiLz ie fLy eiLzfLxe

and eiLzjlmi eimjlmiSemiclassically we have a vector of length

pll rotating around the zaxis with

the zcomponent being of length m The average values of the x and ycomponentsare zero but the projection of the vector on the xyplane a constant length equalpll m ie precisely eq

Problem

In principle this problem can be solved by using the rotation matrices relatedto the irreducible j representation of the rotation group in much the same wayas done for j in the texteq and in problem We have that a ketjlmi is rotated to

ji DRjlmisuch that

hlmji hlmjDRjlmi Dlmm

where are the Euler angles associated with the rotation So the probabilityamplitude for nding ji in state jlmi is simply Dl

mm In general thismatrix is complicated but in the special case where m and we knowSakurai that the matrix elements are simply

Dlm Y m

In our case we have in addition plain rotation about the yaxis ie thematrix elements become for l m r

sin cos

However this method will only work for m and it quite instructive to use thedenition of rotation of the wave function This method will work for any m andin fact for any wave function x

x x x Rx or x R x

In our case we know that x only depends on the direction x n of vector x Infact

The rotation R is a rotation with about the yaxis It rotates

nx ny nz nx ny nz cos nx sinnz ny cos nz sinnx

Thus from we get

cosnz sin nx

where nz cos and nx cos sin in spherical coordinates We know that thisfunction can be expressed in terms of Y m

s and after a little calculation one obtains

cos sin sin cos cos sin cos cos

sin cos

In a more systematic but in this case more tedious approach one can of coursedetermine the coecients of Y m

in the expansion of by projecting Y m on

cmjmi cm hmji Z

sin dd Y m

Problem

This problem is just intended to ll out the details of The onlysubtlety is Since J

y Jy expanding the exponential we have evenpowers of Jy which all give J

y expect power just giving I the identity matrixand odd powers of Jy which just give the matrix Jy The coeents in front of thepowers of Jn

y will be the standard coecients of the exponential function We thusget I from the rst term in the power expansion iJy sin from all the odd powersand J

y cos from all the even powers except power zero

expiJy I iJy sin Jy cos

Sam Pinansky

December 3, 2003

15. (a) Converting into spherical coordintes:

ψ(x) = rf(r)(sin θ(cosφ + sin φ) + 3 cos θ) (1)

Now using the expression for the spherical harmonic functions

Y 01 =

√34π

cos θ (2)

Y ±11 = ∓

√38π

sin θ(cos φ± i sin φ) (3)

we find by inspection that the angular part of the wave function is:√

3((i− 1)Y 1

1 + (i + 1)Y −11 ) +

√12πY 0

Since all three spherical harmonic functions are l = 1 functions (we didn’t need any fromany other set, for instance a constant term would have left us with a Y 0

0 part) the wholefunction is clearly also an eigenfunction of L2 with eigenvalue 2~2.Note that there were many other ways of doing this problem, for instance operating onthe wavefuction directly by L2 (either in polar OR square coordinates) and see what youget. I chose this method since the results were necessary for the next part anyway.

(b) From the previous result, we see that m = −1, 0, 1 with different weights. We need tonormalize things to get the probabilities:

∣∣∣∣∣

√2π

3(1− i)

∣∣∣∣∣

12π|2 = 12π (6)

So divide bother results by 44π/3 and you see that P (m = ±1) = 1/11 and P (m = 0) =9/11.

(c) The point here is to notice that you have the wave function now in terms of its angularand radial variables:

ψ(x) = rf(r)Y ml (θ, φ) (7)

Now we can take the Schroedinger equation in spherical coordinates:

− ~2

(r2 ∂ψ

]+ V (r)ψ = Eψ (8)

Now we can operate the L2 on the ψ (just gives a l(l + 1)~2 = 2~2 factor) and divideout by the angular part, leaving a one-variable differential equation for ψ(r). But in theproblem we are given ψ(r), so we can just solve for V (r) algebraically:

V (r) = − ~2

1rf(r)

[rEf(r) +

dr2(rf(r))

18. There are many, many ways of doing this problem. I will choose to utilize the formula alreadyderived in the book for the rotation of the spherical harmonics, Eq. 3.10.20. So therefore sincethe initial state is Y 0

2 , the rotated state would be

Y 02′=

m′Y m′

2 d(2)mm′(β) (10)

Thus, simply by the nature of this formula, we know that the probability of being in the |2,m〉state is (

d(2)0m(β)

We have a nice (ok, hideous) formula for these, Eq. 3.8.33. Using this formula, you calculatethat (after a bunch of trig identities):

d(2)00 (β) =

14(1 + 3 cos(2β)) (12)

d(2)0±1(β) = ∓

sin β cosβ (13)

d(2)0±2(β) =

sin2 β (14)

Squaring, we find the probabilities:

m = 0 :116

(1 + 3 cos(2β))2 (15)

m = ±1 :32

sin2 β cos2 β (16)

m = ±2 :38

sin4 β (17)

You can show that adding all these together gives 1, plus notice the limits: β = 0 gives thatm = 0 of course, β = π is 1/4 in m = 0, and 3/4 in |m| = 2 (with no m = ±1). And whenβ = π you’re back to where you started.

20. Nothing complicated here, just simple computation using S± |j,m〉 =√

(j ∓m)(j ±m + 1)~ |j, m± 1〉.Start with the highest state, where there’s only one possible choice:

|2, 2〉 = |+,+〉 (18)

Now apply the S− operator to get |2, 1〉 (don’t forget to normalize!)

|2, 1〉 =1√2(|0,+〉+ |+, 0〉) (19)

and we continue in this manner:

|2, 0〉 =1√6(|−, +〉+ 2 |0, 0〉+ |+,−〉) (20)

|2,−1〉 =1√2(|−, 0〉+ |0,−〉) (21)

|2,−2〉 = |−−〉 (22)

Now to find |1, 1〉, we know it must be a linear combination of |+, 0〉 and |0, +〉, and orthogonalto all the other states so far. The orthogonality with |2, 1〉 therefore forces it to have this form:

|1, 1〉 =1√2(|+, 0〉 − |0, +〉) (23)

Now we go back to applying S− again:

|1, 0〉 =1√2(|+,−〉 − |−, +〉) (24)

|1,−1〉 =1√2(|0,−〉 − |−, 0〉) (25)

And of course you can’t forget the |0, 0〉 state. In general it can have the form a |+,−〉 +b |−,+〉+ c |0, 0〉, but orthogonality with the |1, 0〉 and |2, 0〉 states restrict a = b and c = −a,respectively. So therefore:

|0, 0〉 =1√3

(|−,+〉+ |+,−〉 − |0, 0〉) (26)

And there you have it.

21. (a) The trick here is to use equation 3.5.51 and to rewrite the quantity in terms of matrixelements of operators. Trying to use 3.8.33 is not tenable. Therefore:

m=−j

|d(j)mm′(β)|2m =

〈j,m′| e− iJyβ

~ |j, m〉〈j,m| eiJyβ

β |j, m′〉m (27)

= 〈j, m′| e− iJyβ

(m |j,m〉〈j, m|) eiJyβ

~ |j, m′〉 (28)

Note that the sum is simply the operator Jz:

~ JzeiJyβ

~ |j, m′〉 (29)

Now we just need to rotate the operator Jz. You can do this with the Baker Hausdorfflemma, like in the chapter. I’ll simply note that J is a vector, and so transforms like one:

= 〈j,m′| (sin βJx + cos βJz) |j, m′〉 (30)= 〈j,m′| cos βJz |j, m′〉 (31)

m=−j

|d(j)mm′(β)|2m = m′ cosβ (32)

Checking this against Eq. 3.5.52 shows it holds.

(b) The same manipulations apply as in the last part, except now you get J2z instead. So:

m=−j

m2|d(j)mm′(β)|2 = 〈j, m′| e− iJyβ

~ J2z e

~ |j, m′〉 (33)

= 〈j, m′| (sinβJx + cos βJz)2 |j,m′〉 (34)

= 〈j, m′| (sin2 βJ2x + 2 sin β cosβ(JxJz + JzJx) + cos2 βJ2

z ) |j,m′〉(35)

= sin2 βj(j + 1)−m′2

2+ m′2 cos2 β (36)

=j(j + 1)

2sin2 β + m′2 1

2(3 cos2 β − 1) (37)

(Note that both cross terms just give 0).

24. (a) First, some notation: Denote the eigenstates of Sz with eigenvalue −~/2, ~/2 as |0〉, |1〉respectively, and the eigenstates of Sx with eigenvalue −~/2, ~/2 as |−〉, |+〉 repectively.Then the initial state of the particle is

|10〉 − |01〉√2

Clearly, the probability of measuring the first particle in the state |1〉 is clearly 1/2. Usingthe fact that

|1〉 =|+〉+ |−〉√

|0〉 =|+〉 − |−〉√

Rewriting the initial state of the system

− |+−〉+ |−+〉√2

So clearly the probability of measuring the first particle in the state |+〉 is clearly 1/2 aswell.

(b) After the measurement by B, the state of the particle is simply:

− |01〉 (42)

So clearly observer A will find the particle is always spin down when measuring Sz, andwill find it in the state |+〉 with probably 1/2. Notice that B measuring in the z directiondoesn’t change the measurement in the x direction by A. That’s because the direction isorthogonal.

Problem

a In principle the construction of Tq from Uq and Vq is given by where

Uq and Vq are the spherical tensor with components of vector U is given by

U pUx iUy U Uz

and similarly for vector V According to we have

U V x iU V y

pUVz UzV

ipU V z

ipUxVy UyVx

pUV UV

b The construction in terms the irreducible components of U and V is again givenby We have

UxVx UyVy iUxVy UyVx

UxVz UzVx iUyVz UzVy

UzVz UxVx UyVy

Problem

a x is an vector operator with standard spherical components

V px iy V z

We are asked to relate the matrix elements of this vector operator According toWE we have

l l and m m m m for V

Strictly speaking WE also allows l l It is ruled out by the fact that the vector operator x has odd parity Let us assume that the reduced matrix elementhnljjV jjlni is dierent from zero Then WE gives

hn l m jVjn lmihn l m jVjn lmi

hlm jl m ihlm jl mi

hlmjl m ihlm jl mi

where we assume m is in an appropriate range not leading to the division with zeroThese relations imply that for given n n l l all matrix elements hn l mjVqjn lmican be expressed in terms of known CG coecients and one of the matrix elementsfor instance hn l l jVjn l li assuming it is dierent from zero since WEalso gives

hl m q jVqjl m i hlm q qjl m q ihlm q qjl m qi hl m qjVqjl mi

In order to make it explicit one has of course to calculate or nd in other waysthe CG coecients

b The only additional information we obtain by using the explicit wave functionsfor jnlmi is that we can calculate the matrix elements if we can do the integralsinvolved and in this way obtain the absolute value In particular we can check ifsome of the reduced matrix elements are zero Since Vq can be expressed in termsof Y m

by we can use to write explicitly

hn l mjVqjn lmi F n l n l hl jl i hlm qjl mi

F n l n l

Zrdr Rnlr r Rnl

Note as already remarked in the book that this explicit calculation reproduces asit should the WE theorem in the case of a vector operator Note also that some ofthe indices of CG coecients are interchanged in order to agree with the notationof WE One has

hjjm mjjmi jjjhjjm mjjmiwhich showns that the matrix element hn l mjVqjn lmi is zero unless l l in agreement with the parity convention for Y m

l and the parity of x

Problem

a Let us introduce as usual the notation Vq for the spherical components of vectorx We have using

iV VV V

xz pV VV

xy V Vp

b We have again according to and the discussion afterwards

z r p T

WE gives

h j j jT j j ji

h j jjT j j ji hj jjj ji h jjjT jj ji

h j jjT j j ji hj j jj ji h jjjT jj ji

Thus we have that only for m j is the matrix element of x y dierentfrom zero and

eh j jjxyjj ji Qp

hj jjj jihj j jj ji

Problem

a A implies H eBmc

is an eigenfunction for H corre

sponding to eigenvalue eBmc

Acting on

with ASe Se results in the

which shows that e

is not an eigenvector of H for B hHi is obtained asthe scalar product of vectors and

Problem

For any N N hermitian matrix A with eigenvalues i

because every eigenvector jvii to A satis es M jvii Since any vector can beexpanded on the eigenvectors of A the matrix M must be zero Sz and Sx both haveeigenvalues and

Problem a

Let the state jvi be given by the spinor

We assume as usual that the state is

normalized jaj jbj Further the expectation value of the spin matrices areclearly unchanged if we multiplied jvi by a phase factor Thus our vector can onlybe determined up to a phase and we can assume that Re a

ab ba hSyi i

ba ab hSzi

jaj jbj

or if a

hSxi a Re b hSyi a Im b hSzi a

If a we have already that jvi ji up to a phase Thus

hSzi Re b

Im b pjbj Re b

p jaj Re b

or expressed entirely in terms of expectation values of Sx and Sz

hSzi Re b

Im b s

hSzi hSxi

where we should choose the plus sign since we know that hSyi and a are positive

hSyi is known up to a sign since the three components hSxi hSyi hSzi transform likean ordinary vector under rotations Thus the norm is preserved under rotations andif we rotate jvi to ji the vector hSxi hSyi hSzi rotates to with norm Thus

hSyi r

hSxi hSzi

Problem

We want to prove that Gi Gj iijk Gk Let us consider the l nth entry of thismatrix eequation

Gi Gjln iijk

or in terms of the ijk symbols

iilmijmn ijlmiimn iijkiklnThe relation follows from the identity

ijkilm jlkm jmkl

The matrices Gi are the generators for rotations around the coordinate axis xi asshould be clear from the lectures expressed in the standard Cartesian coordinatesystem where

A Vjxi Vjxi Vjxi

Indeed we can with write

V V n V V j

I i ni Gi

The eigenvalues for G are and with corresponding eigenvectors

pjxi ip

jxi ji jxi

A coordinate change from basis vectors jxii to basis vectors ji willchange the matrix G into the diagonal matrix J with diagonal elements and More explicitly we can write in braket notation for the abstract operator Gcorresponding to G and J

hxij Gjxji hxijihj Gj ih jxji

where we have summation over The matrix J has entries hj Gj iand the transformation matrix Ti i which connects G and Jhas entries hxiji Thus reads in ordinary matrix notation

Gij TiJTyj

The above change of vectors

jxi jxi ji ji

is similar to the superposition of photons travelling in the x direction and linearpolarized in x and x directions into right and left circularly polarized photons seeSakurai formula It is well know that even classically properly normalizedright and left circularly polarized plane waves of light carry an angular momentumof one unit in the direction of propagation We see that this classicalinterpretation of plane electromagnetic waves has many similarities to the quantum

interpretation of the photon as a spin particle Read Sakurai sec in particular

GRADUATE QUANTUM MECHANICS: 502 Spring 2002

Solutions to Assignment 1.

1. (a) To construct an eigenket of τ~a, we take the combination

|~k〉 =∑

e−i~k·~r|~r〉, (1)

where ~k = (kx, ky, kz). Now

τ |~k〉 =∑

e−i~k·~rτ~a|~r〉

e−i~k·~r|~r + ~a〉

=∑~r′e−i~k·(~r′−~a)|~r〉

= ei~k·a|~k〉. (2)

(b) The action of H on the state |r〉 is

H |r〉 = Eo|~r〉 −∆∑

~a=(x,y,z)

[|~r − ~a〉+ |~r + ~a〉] (3)

so that the action of H on |~k〉 is

H |~k〉 =∑

e−i~k·~rH |~r〉

e−i~k·~r

Eo|~r〉 −∆

∑~a=(x,y,z)

[|~r − ~a〉

e−i~k·~r

Eo −∆

∑~a=(x,y,z)

(ei~k·~a + e−i~k·~a)

|~r〉

= E(~k)|~k〉. (4)

E(~k) = Eo − 2∆∑

~a=(x,y,z)

cos(~k · a)

= Eo − 2∆(cos kx + cos ky + cos kz) (5)

is the corresponding energy eigenstate.

2. (a) Since momentum operators always commute, any function of these operators also commutes, so that

[τ~d, τ~d′ ] = [e−i~P ·~d/h, e−i~P ·~d′/h] = 0 (6)

Translation operators commute.

(b) Rotations about different axes do not commute, so that

[D(n, φ), D(n′, φ′)] 6= 0 (7)

(c) The inverstion operator reverses the direction of all translation, so that

πτ~dπ−1 = τ−~d (8)

Consequently, the inversion operator does not commute with the translation operator.

[π, τ~d] 6= 0. (9)

(d) Under the inversion operation, angular momentum operators are invariant, π ~Jπ−1 = ~J so that [π, ~J ] = 0.Consequently, the inversion operation commutes with functions of the angular momentum operator, andthus commutes with the rotation operator.

[π,D(R)] = 0. (10)

3. Sakurai problem 9. When we time reverse a momentum eigenstate, we reverse the sign of the momentum, inaddition to complex conjugating the state. We therefore expect that the time reversal of φ(p) is φ(−p)∗. Toshow this explicitly,

〈p|Θ|α〉 = 〈p|Θ(∫

dDp′|p′〉φ(p′))

= 〈p|∫dDp′Θ|p′〉φ∗(p′)

= 〈p|∫dDp′| − p′〉φ∗(p′)

=∫dDp′

δ(D)(p+p′)︷︸︸︷〈p| − p′〉 φ∗(p′) = φ∗(−p) (11)

4. Sakurai problem 12. We can rewrite the matrix as

H = AS2z +

+ + S2−] (12)

where S± = Sx ± iSy. Written out explicitly for S = 1 we have

H ≡A 0 B

0 0 0B 0 A

where I have taken h = 1. Taking det[E1−H ] = E((E −A)2 −B2) we see that the energy eigenvalues are

E = A±B, 0 (14)

The corresponding eigenkets are

|±〉 =|+ 1〉 ± | − 1〉√

2, (E = A±B)

and for E = 0, |0〉 = |ms = 0〉.The Hamiltonian is invariant under time-reversal, since Θ~SΘ−1 = ~S is unchanged by time-reversal. SinceΘ|mJ〉 = (i)2mJ | −mJ 〉, we have

Θ|±〉 = ∓|±〉, Θ|0〉 = |0〉, (16)

i.e the lower and upper eigenstates are odd-parity under time reversal, whereas the central state is even-parityunder time-reversal.

5. Sakurai, chapter 4, Q 6. This is a tricky problem. There are two ways you could do it: (i) solving the completeproblem but to exponential accuracy or (ii) by directly calculating the matrix elements between the states onthe left, and right hand side. I shall illustrate method (ii). To begin, let us consider the problem when thelength a is infinitely large. In this case, the wavefunction for the left, and right hand ground-states are

ψR(x) = 〈x|ψR〉 =

0 (x > a+ b)A sin[k(a+ b− x)] (a < x < b)Beκx (x < a)

ψL(x) = 〈x|ψL〉 =

0 (x < −a− b)A sin[k(a+ b+ x)] (−b < x < −a)Be−κx (x > −a)

where κ =√

2mh2 (Vo + E) ≈

√2mh2 Vo.

Now the tricky bit is that we need to construct orthogonalized wavefunctions. To do this, we construct

|ψR〉 =1

[1− |〈ψL|ψR〉|2] 12[|ψR〉 − |ψL〉〈ψL|ψR〉]

|ψL〉 = |ψL〉 (18)

These states are now orthogonal and normalized.

We shall now approximate the complete wavefunction in the form

|ψ〉 = αR|ψR〉+ αL|ψL〉 (19)

Applying the Hamiltonian to this expression, and demanding that H |ψ〉 = E|ψ〉, we obtain the eigenvalueequation Habαb = Eαb, (a, b ∈ R,L), where

Hab ≡[ 〈ψR|H |ψR〉〈ψR|H |ψL〉〈ψL|H |ψR〉〈ψL|H |ψL〉

]. (20)

To evaluate this matrix, it is helpful to realize that the complete Hamiltonian can be written

H = HR + VL = HL + VR (21)

where HL is the Hamiltonian for the left-hand well and HR is the Hamiltonian for the right-hand well and

VR = −Vo[θ(x− a)− θ(x − a− b)],VL = −Vo[θ(x+ a+ b)− θ(x + a)], (22)

ψ( )x

-(a+b) -a-

(a+b) x

Fig. 1.: Showing ψR(x) and the potential VL(x).

With this set-up, we note that HL,R|ψL,R〉 = Eo|ψL,R〉 , where Eo is the energy of an isolated well. If you nowcompute the matrix element 〈ψR|H |ψL〉, you obtain

〈ψR|H |ψL〉 = E〈ψR|ψL〉+〈ψR|VR|ψL〉√1− |〈ψL|ψR〉|2

=〈ψR|VR|ψL〉√1− |〈ψL|ψR〉|2

≈ 〈ψR|VR|ψL〉. (23)

In the last step, we have noted that |〈ψL|ψR〉| is exponentially smaller than unity, so that terms containingthis quantity have been dropped. The splitting between the two states is then going to be simply

±∆ = ±|〈ψR|VR|ψL〉| (24)

Now to calculate this, we need to compute the exponential tail in ψL. Applying continuity of the wavefunctionand continuity of the logarithmic derivative, we obtain

A sin kb = Beκa, k tan(kb) = −κ (25)

To leading exponential accuracy, this gives

√2b,

κbe−κa (26)

Carrying out the integral, we then obtain

〈ψR|VR|ψL〉 = −Vo

∫ a+b

sin[k(a+ b− x)]Be−κx

= −Vo

κe−κ(2a+b)

) ∫ b

dx sin[kx]eκx

= −Vo

κe−κ(2a+b)

) ∫ b

dxIme(κ+ik)x

≈ −Vo

κe−κ(2a)

≈2ko/κ︷︸︸︷[eikb

κ+ ik

≈ −Vo

)e−κ2a

(h2π2

)e−2κa (27)

The splitting between the two levels is then

∆E = 2∆ =h2

mκb3e−2κa (28)

where κ =√

2mh2 Vo for large Vo.

(a+b) x

-(a+b) -a a

(a+b) x

-(a+b) -a

ψ + ψ / 2R L

ψ − ψ / 2R L

Fig. 2.: Showing the even and odd wavefunctions for the symmetric potential well.

Physics 512 Winter 2003

Homework Set #5 – Solutions

1. This problem is essentially Merzbacher, Chapter 18, Problem 4 [or Sakurai, Chapter 5,Problem 1]. Consider a one-dimensional harmonic oscillator perturbed by a constantforce

12mω2x2 − Fx

a) Show that the first order perturbation in the energy levels vanishes

Since we will need to find the second order perturbation in part b), let us considerthe general matrix element

Vmn = −F 〈m(0)|x|n(0)〉

The operator x is given in terms of creation and annihilation operators by x =√h/(2mω)(a† + a). Thus

Vmn = −F√

2mω〈m(0)|(a† + a)|n(0)〉

= −F√

2mω[√n+ 1 δm,n+1 +

√n δj,n−1]

This indicates that the x operator changes the oscillator number by ±1. In par-ticular, Vnn = 0, which demonstrates that the first order energy perturbationvanishes.

b) Now calculate the eigenenergies En up to second order in the perturbation.

Using the matrix element calculated above, we find

E(2)n =

∑k =n

|Vkn|2E

(0)n −E

=|Vn+1,n|2

E(0)n −E

(0)n+1

+|Vn−1,n|2

E(0)n − E

(0)n−1

= F 2 h

[n+ 1−hω +

]= − F 2

Combined with the zeroth order energy, this gives

En = (n+ 12 )hω − F 2

c) Show that the second-order perturbation result gives the exact eigenenergies(which may be obtained by completing the square in H). Explain why thishappens.

By completing the square, we find

12mω2

(x− F

− F 2

12mω2x2 − F 2

where x = x− F/(mω2) is a shifted coordinate. In terms of this new coordinate,we have an ordinary harmonic oscillator, however with a constant offset in thepotential. We easily read off the eigenenergies as

En = (n+ 12 )hω − F 2

2mω2(1)

which is identical to the perturbation result of b). Of course, in general we wouldapply perturbation theory only when we do not know the exact answer. How-ever, in this toy example, after completing the square, we see that F enters theHamiltonian only quadratically as an overall offset. In particular, we know thatthe correct energies, (1), only has F 2, and no other powers of F in it. Now re-call that each order in the perturbation theory corresponds to precisely that orderin a series expansion of the small perturbation parameter (which would be F inthis case); first order perturbation gives O(F ), second order gives O(F 2), and soon. As a result, only the second order perturbation to the energy can contributetowards the correct answer, and this is in fact what happens.

2. Sakurai, Chapter 5, Problem 2. In nondegenerate time-independent perturbationtheory, what is the probability of finding in a perturbed energy eigenstate |ψn〉 thecorresponding unperturbed eigenstate |ψ(0)

n 〉? Solve this up to terms of order g2.We recall that the perturbed eigenstate is not necessarily normalized, and may bewritten as

|ψn〉 = |ψ(0)n 〉 + g|ψ(1)

n 〉 + g2|ψ(2)n 〉 + · · ·

Since the probability is the amplitude squared, we want to calculate (to second orderin g)

∣∣∣∣∣ 〈ψ(0)n |ψn〉√〈ψn|ψn〉

∣∣∣∣∣ = |〈ψ(0)n |ψn〉|2

〈ψn|ψn〉

Since all the higher order |ψ(i>0)n 〉 states are orthogonal to |ψ(0)

n 〉, it is easy to see that

〈ψ(0)n |ψn〉 = 〈ψ(0)

n |ψ(0)n 〉 = 1

and〈ψn|ψn〉 = 〈ψ(0)

n |ψ(0)n 〉 + g

(〈ψ(0)

n |ψ(1)n 〉 + 〈ψ(1)

n |ψ(0)n 〉)

+ g2(〈ψ(0)

n |ψ(2)n 〉 + 〈ψ(1)

n |ψ(1)n 〉 + 〈ψ(2)

n |ψ(0)n 〉)

+ · · ·

= 1 + g2〈ψ(1)n |ψ(1)

n 〉 + · · · = 1 + g2∑k =n

|Vkn|2|E(0)

n − E(0)k |2

+ · · ·

Using 1/(1 + x) = 1 − x+ · · ·, we find

P = 1 − g2∑k =n

|Vkn|2|E(0)

n − E(0)k |2

+ · · ·

which we verify is less than or equal to one. Clearly, the only possibility of finding100% probability is if |ψn〉 is identical to |ψ(0)

n 〉, indicating that the initial eigenstateremains an eigenstate of the perturbation.[Yes, this is the same as Wave-function Renormalization on p. 293 of Sakurai!]

3. Consider the 2p levels of hydrogen (m = 0,±1) subject to a perturbation

V = Ax2 + By2 − (A+B)z2

a) Write V in terms of components of a rank-2 spherical tensor.Using the spherical harmonic ‘trick’, we can write out the appropriate rank-2tensor as

T(2)±2 = 1

2(x± iy)2

T(2)±1 = ∓(x± iy)z

T(2)0 = − 1√

6(x2 + y2 − 2z2)

Of course, the actual normalization is arbitrary. I have removed a factor of√15π/8 when compared to the Y m

l=2 spherical harmonics. In this case, we mayreexpress

x2 − y2 = T(2)2 + T

(2)−2 , x2 + y2 − 2z2 = −

√6T (2)

This allows us to rewrite the perturbation

V = 12 (A+ B)(x2 + y2) + 1

2(A− B)(x2 − y2) − (A+ B)z2

= 12(A+ B)(x2 + y2 − 2z2) + 1

2(A− B)(x2 − y2)

= −√

32 (A+ B)T (2)

0 + 12 (A− B)(T (2)

2 + T(2)−2 )

Note that the coefficients of x2, y2 and z2 had to work out just right for this tobe possible. If instead of (A + B)z2, we had something like Cz2, then we wouldhave an added term left over which would be proportional to r2 = x2 + y2 + z2.This would correspond to a scalar perturbation in addition to a pure rank-2 tensorperturbation.

b) Neglecting electron spin, find the “correct” zeroth-order eigenstates and theircorresponding energies. It is sufficient to give the energies in terms of a reducedmatrix element 〈2p||T (2)||2p〉.We denote the 2p states as |nlm〉 = |211〉, |210〉 and |21 −1〉. We first observethat the perturbation will not mix |210〉 with the other two states [because of the

∆m = 0,±2 selection rule evident from (1)]. So, although all three states aredegenerate, the perturbation is block diagonal in the |210〉 and |211〉, |21 −1〉subspaces. This simplification allows us to immediately write down, for the |210〉state

∆E210 = 〈210|V |210〉 = −√

32 (A+B)〈210|T (2)

0 |210〉

= −√

32 (A+B)〈1200|1210〉〈2p||T (2)||2p〉

35 (A+ B)〈2p||T (2)||2p〉

For the remaining two states, we write out the matrix

V =( 〈211|V |211〉〈211|V |21 −1〉〈21 −1|V |211〉〈21 −1|V |21 −1〉

= −√

32 (A+B)

( 〈211|T (2)0 |211〉 00 〈21 −1|T (2)

0 |21 −1〉

+ 12(A− B)

(0 〈211|T (2)

2 |21 −1〉〈21 −1|T (2)

−2 |211〉 0

= −√

32(A+B)〈2p||T (2)||2p〉

( 〈1210|1211〉 00 〈12 −10|121 −1〉

+ 12(A− B)〈2p||T (2)||2p〉

(0 〈12 −12|1211〉

〈121 −2|121 −1〉 0

√35〈2p||T (2)||2p〉

(−(A+B) (A− B)(A− B) −(A+B)

Working out the eigenvalues and eigenstates, we find a simple result

∆E21x = −√

35A〈2p||T (2)||2p〉 |21x〉 ≡ 1√

(|211〉 − |21 −1〉)∆E21y = −

√35B〈2p||T (2)||2p〉 |21y〉 ≡ 1√

(|211〉 + |21 −1〉)Here, the labels x and y were chosen to be suggestive of px and py orbitals. How-ever, of course, this is just an arbitrary labeling of the eigenstates. To summarize,the perturbation splits all three 2p states (ignoring spin) to

∆E210 =√

35(A+ B)〈2p||T (2)||2p〉, |210〉

∆E21x = −√

35A〈2p||T (2)||2p〉, |21x〉 = 1√

(|211〉 − |21 −1〉)∆E21y = −

√35B〈2p||T (2)||2p〉, |21y〉 = 1√

(|211〉 + |21 −1〉)Note that the energy shifts are proportional to A, B and −(A + B), the factorsmultiplying x2, y2 and z2, respectively. The corresponding eigenstates, |21x〉,

|21y〉 and |210〉 are simply the px, py and pz orbitals (which are aligned along thex, y and z axis). So this makes perfect physical sense.

c) Show that the expectation value of Lz vanishes in each eigenstate found above.Of course the expectation values of Lz take on the values 0,±h in the originalbasis states |210〉 and |21 ±1〉. Here, however, we consider

〈210|Lz|210〉 = 0

〈21x|Lz|21x〉 = 12

(〈211| − 〈21 −1|)Lz

(|211〉 − |21 −1〉)= 1

(〈211|Lz|211〉 + 〈21 −1|Lz|21 −1〉) = 0

〈21y|Lz|21y〉 = 12

(〈211| + 〈21 −1|)Lz

(|211〉 + |21 −1〉)= 1

(〈211|Lz|211〉 + 〈21 −1|Lz|21 −1〉) = 0

so that the expectation values indeed vanish. It is important to keep in mind thatthe expectation values are not the same thing as eigenvalues! The eigenvalues ofLz are always 0,±h for the 2p states.

4. Merzbacher, Chapter 18, Problem 14.For the Hamiltonian H = AL 2 + Bh2 cos 2ϕ with A B, we may take the secondterm as a perturbation. Hence

H0 = AL 2 = Ah2(+ 1)

for states of angular momentum . For the perturbation, we find it convenient to write

V = Bh2 cos 2ϕ = 12Bh

2(e2iϕ + e−2iϕ)

Although the perturbation breaks rotational invariance (after all, it is a hindered rota-tion), we may nevertheless consider its effect on angular momentum states |m〉. Inthis case, since Y m

(θ, ϕ) ∼ eimϕPm (cos θ), we see that

〈′m′|V |m〉 =∫

(Y m′′ )∗ V Y m

dΩ ∼∫ 2π

e−im′ϕ(e2iϕ + e−2iϕ)eimϕ dϕ× (rest)

Hence we may deduce that m′ and m obey a selection rule ∆m = ±2. This providesan important simplification, as it indicates that most of the matrix elements of theperturbation actually vanish. We are now ready to examine the S, P and D levels oneat a time.

For the S level ( = 0)This state is non-degenerate, and has vanishing zeroth order energy, E (0) = 0.In addition, the first order shift in the energy also vanishes since 〈00|V |00〉 = 0by the ∆m selection rule. We thus have, for the S level

E = 0, |00〉

Note that, while the energy vanishes up to first order, |00〉 is not an eigenstate ofH.

For the P levels ( = 1)

At zeroth order, the P levels are three-fold degenerate with E (0) = 2Ah2. For thefirst order perturbation, use of the ∆m selection rule indicates that |10〉 cannotmix with the other states. Hence it is already the “correct” eigenstate, and doesnot get shifted in energy

E = 2Ah2, |10〉For the remaining |11〉 and |1 −1〉 states, we have to diagonalize the perturbation

0 〈11|V |1 −1〉〈1 −1|V |11〉 0

The two non-trivial matrix elements are complex conjugates of each other. So weonly need to compute

〈11|V |1 −1〉 = 12Bh

∫(Y 1

1 )∗ e2iϕ Y −11 dϕ d cos θ

= 12Bh

(− 3

)∫sin2 θ dϕ d cos θ = −1

V = −12Bh2

(0 11 0

By now, it should be obvious what the eigenvalues and eigenvectors of this matrixare. Including the zeroth order energy, we have

E = 2Ah2 − 12Bh

2, 1√2(|11〉 + |1 −1〉)

E = 2Ah2 + 12Bh2, 1√

2(|11〉 − |1 −1〉)

We see that the perturbation completely lifts the degeneracy of the P states.

For the D levels ( = 2)

The D levels are five-fold degenerate at the zeroth order, with E (0) = 6Ah2. Ingeneral, we may have to work out a 5 × 5 matrix for the perturbation. However,the ∆m selection rule again helps, and indicates that the five states split up intotwo subspaces, one with |21〉, |2 −1〉 and the other with |22〉, |20〉, |2 −2〉. Forthe former, we have

0 〈21|V |2 −1〉〈2 −1|V |21〉 0

〈21|V |2 −1〉 = 12Bh

∫(Y 1

= 12Bh

(− 15

)∫(sin θ cos θ)2dϕ d cos θ = −1

The eigenenergies and eigenstates in this subspace are hence

E = 6Ah2 − 12Bh

2, 1√2(|21〉 + |2 −1〉)

E = 6Ah2 + 12Bh

2, 1√2(|21〉 − |2 −1〉)

Finally, turning to the three-dimensional subspace, we have

0 〈22|V |20〉 0

〈20|V |22〉 0 〈20|V |2 −2〉0 〈2 −2|V |20〉 0

We evaluate

〈22|V |20〉 = 12Bh2

∫(Y 2

2 )∗ e2iϕ Y 02 dϕ d cos θ

= 12Bh2

)∫sin2 θ(3 cos2 θ − 1)dϕ d cos θ = − 1

and note that 〈20|V |2 −2〉 gives the same result. Thus

V = − 12√

1 0 10 1 0

It is easy to check that the matrix of ones and zeros has eigenvalues√

2, 0and −√

2, with eigenvectors ( 1√

2 1 )T , ( 1 0 −1 )T and ( 1 −√2 1 )T .

Hence the correct eigenenergies and eigenstates are

E = 6Ah2 − 12√

3Bh2, 1

2 (|22〉 +√

2|20〉 + |2 −2〉)E = 6Ah2, 1√

2(|21〉 − |2 −2〉)

E = 6Ah2 + 12√

3Bh2, 1

2 (|22〉 −√

2|20〉 + |2 −2〉)

Finally, to summarize, we find, for the S, P and D levels

S : E = 0, |00〉P : E = 2Ah2 − 1

2Bh2, 1√

2(|11〉 + |1 −1〉)

E = 2Ah2, |10〉E = 2Ah2 + 1

2Bh2, 1√

2(|11〉 − |1 −1〉)

D : E = 6Ah2 − 12Bh2, 1√

2(|21〉 + |2 −1〉)

E = 6Ah2 − 12√

3Bh2, 1

2(|22〉 +√

2|20〉 + |2 −2〉)E = 6Ah2, 1√

2(|21〉 − |2 −2〉)

E = 6Ah2 + 12√

3Bh2, 1

2(|22〉 −√

2|20〉 + |2 −2〉)E = 6Ah2 + 1

2Bh2, 1√

2(|21〉 − |2 −1〉)

Note that the P states are split just as they were in problem 3).

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Problem

jti Xn

eiEntcntjni Pnt jcntj

where Pnt is the probability to nd the system in state jni at time t

Assume system is in ground state ji at t For n we have to lowest order

Pnt jcn j

cn t i

dt eint

Aet hnjxji i

hnjxjiA eint

Thus we obtain for n

Pnt Ajhnjxjij

hnjxji Xk

hnjxjkihkjxji hnjxjihjxji

It is seen that the matrix element hjxji Thus c t However a formula

like is not correct in this case since c by assumption

dtAet hnjxji i

hnjxjiA

Note that c is purely imaginary Thus there is no rst order contribution to

Pt jc c tj jc tj

and the calculation is clearly inconsistent It requires that we also include the secondorder contribution c

t such that to second order

Pt jc c t c

tj jc tj Re c

By consistency this has to agree to second order with

Pnt for potential x Pt

Check it

The reason that we do not have the same kind of problem for Pnt n is of

course that cn so is the lowest order contribution

Problem

Notation EE jsi

First order perturbation theory

Pskt jhkjUItjsij hkjUItjsi i

dt eithkjV tjsi

hkjV tjsi

hkjUItjsi

eit cost ieit sint

h cost cost sint sint

sint cost cost sint

The term hkjUItjsi has to be small in order that we trust rst order perturbationtheory Let us consider as xed and assume we can vary an external eldwith an adjustable frequency Let us further assume that rst order perturbationtheory is reliable when is not close to ie However when the term e

in hkjUItjsi can be of order

j jand the condition might not be satised However usingsinxx

we obtain that it is satised for t

Problem

dt eiskthkjV tjsi

where Ek Es ks In our case jsi ji jki ji and

hjV tji F

thjxji

dt eit

Let us consider the classical system a particle oscillating forth and back with frequency and acted upon by an external force F

p mxt F t p

Zdt mxt F t

Zdt F t

The energy transfer by F to the harmonic oscillator is

Zdt xtF t

If F t is contant clearly no average energy is transferred to the oscillator since aconstant F just means a translation of the center of oscillation from x to x Fmand the E calculated over one period is zero Now consider a very slowly varyingF t xt will consist of a slow motion xt F tm with the fast oscillationssuperimposed and the fast oscillations will only add a small average energy If F tstarts out being zero and ends up being zero in the end only a small amount ofenergy has been transferred to the classical system provided the variation of F tis suciently slow In the quantum system which for large positive and negativetimes is just a harmonic oscillator a possible jump in energy cannot have been lessthan and the probability of this becomes smaller if the energy transfer in theclassical system becomes smaller

The above classical observations can be made precise in quantum mechanics as welland is called the adiabatic theorem Let Ht be the Hamiltonian Although it isnot stationary we can of course solve

Htjnit Entjnitfor all times t as if we had a stationy Hamiltonian The eigenvalues Ent willbe continuous functions of t and also the eigenvectors jnit can be chosen to be

continuous functions of t if the time evolution of H is sucently nice Assume thatthe system is in the eigenstate jni corresponding to En at t The adiabatictheorem states that in the limit of an innitely slow evolution of Ht the systemwill be in an eigenstate of Ht corresponding to Ent at time t ie

jti eitjnit

where t is some phase which one can try to calculate in the adiabatic approxi

mation In our case the system remains in the ground state since at time t Ht isjust a harmonic oscillation displaced from x to xF tm As jnit we canthus take the displaced eigenfunctions of the harmonic oscillators and

jnit jnit jni

where jni is the standard eigenvector of the harmonic oscillator simply becauseF t for jtj

Solutions to assignment 3.

1. (Solution to Sakurai, problem 4, ch 5) The isotropic Harmonic oscillator in two dimensions

Ho =p2

2m+mω2

2(x2 + y2). (1)

can be re-written as

Ho = hω(a†a+ b†b+ 1) (2)

a =√mω

√mω

(y + i

are the annihilation operators for modes in the x and y directions, respectively. The eigenvalues of the numberoperators na = a†a and nb = b†b are non-negative integers. The corresponding eigenkets |na, nb〉 are energyeigenkets, with energy

Enanb= hω(1 + na + nb) (4)

(a) The ground-state is then

|0, 0〉 Eo = hω (5)

and the first excited state is doubly degenerate, corresponding to |1, 0〉 and |0, 1〉, with energy E10 =E01 = 2hω.

(b) The perturbation to the Hamiltonian can be written

V = δmω2xy = δhω

2(a+ a†)(b + b†) (6)

We apply non-degenerate perturbation theory to the ground-state, to obtain

|0, 0〉′ = |0, 0〉+1

Eo −HoV |0, 0〉+O(δ2)

= |0, 0〉 − δ|1, 1〉 (7)

with energy

E′0,0 = Eoo +

0︷︸︸︷〈0, 0|V |0, 0〉+ |〈1, 1|V |0, 0〉|2

E00 − E11+O(δ3)

= E00 − δ2hω

8+O(δ3) (8)

Next we apply degenerate perturbation theory to the degenerate excited states |1, 0〉 and |0, 1〉. In thismanifold of states, the perturbation has matrix elements

Vab =δhω

[0 11 0

To zeroth order, the new energy eigenkets are the eigenkets of Vab, given by

|±〉 =1√2(|10〉 ± |01〉) (10)

with energy eigenvalues

E± = 2hω ± δhω

(c) To solve this problem exactly, we merely have to rotate our spatial axes through 45o, writing

p± =px ± py√

2, X± =

x± y√2. (12)

2m(P 2

+ + P 2−) +

+ +mω2

X2− (13)

where ω± = ω(1± δ)1/2, so that if we write

α =√mω+

(X+ + i

√mω−2h

(X− + i

p−mω−

we may cast the Hamiltonian in the form

H = hω+(n+ +12) + hω−(n− +

12) (15)

where n+ = α†α and n− = β†β. The ground-state energy is now

E00 =12(hω+ + hω−) = hω(1− δ2

8) +O(δ3) (16)

confirming (8). The excited states have one α or β quanta, and have energies

E± = Eo + hω± =12(hω+ + hω−) + hω± (17)

To leading order in δ, this gives

E± = (2 ± δ

2)hω (18)

confirming (11).Let us now construct the new ground-state in terms of the unperturbed ground-state. We first note that

α|0, 0〉′ = β|0, 0〉 = 0 (19)

With a bit of work, by rewriting (14) in terms of the original creation and annihilation operators, youmay confirm that

α = u+

(a+ b√

(a† + b†√

β = u−

(a− b√

)+ v−

(a† − b†√

u± =12

[√ω±ω

], v± =

[√ω±ω

−√

]. (21)

Note that u2± − v2± = 1. Now to construct the new ground-state, consider the state |ψ〉 = eAa†a† |0〉. Nowsince [a, eAa†a† ] = 2Aa†eAa†a† , it follows that

(ua+ va†)eAa†a† |0〉 = (2Au + v)a†eAa†a† |0〉 (22)

so that

(ua+ va†)e−v2u a†a† |0〉 = 0 (23)

Using this result, we can satisfy conditions (19) by writing

|0, 0〉 = exp

[− v+

(a† + b†√

− v−2u−

(a† − b†√

)2]|0, 0〉 (24)

Expanding this expression to leading order, gives

|0, 0〉 = |0, 0〉 −[v+2u+

(a† + b†√

+v−2u−

(a† − b†√

)2]|0, 0〉 (25)

Now to leading order, u± = 1 + O(δ2), v± = ±δ/2 + O(δ3). Substituting into the above expression, weobtain

|0, 0〉 = |0, 0〉 − δa†b†|0, 0〉 = |0, 0〉 − δ|1, 1〉+O(δ2) (26)

which confirms the perturbative result (7). Finally, the eigenkets of the excited states are obtained bycreating quanta:

α†|0, 0〉′ =1√2(a† + b†)|0, 0〉+O(δ2)

β†|0, 0〉′ =1√2(a† − b†)|0, 0〉+O(δ2) (27)

confirming (10).

2. (a) In the degenerate manifold of p-states,

V = λ(x2 − y2) (28)

has the matrix elements

〈m|λ(x2 − y2)|m′〉 = c〈m|J2x − J2

y |m′〉 =c

2〈m|J2

+ − J2−|m′〉 (29)

where |m〉 ≡ |l = 1,m〉, and we have used the Wigner Eckart theorem to relate the matrix elements tothose of the angular momentum operator. The quantity “c” is a constant. From the above, we see thatthe two only non-vanishing matrix elements are

〈+1|V | − 1〉 = 〈−1|V |1〉 = λVo (30)

where Vo is real, due to time-reversal invariance. The zeroth order energy eigenstates are the eigenstates|ξ〉 of

〈a|V |b〉 =

0 0 λVo

0 0 0λVo 0 0

with energy E = Eo + 〈ξ|V |ξ〉+O(λ3) which gives

|+〉 =1√2(|+ 1〉+ | − 1〉), E = E0 + λV0 +O(λ2)

|−〉 =1√2(|+ 1〉 − | − 1〉), E = E0 − λV0 +O(λ2)

|0〉 E = E0. (32)

showing that the degeneracy of the p-states is completely removed.

(b) Under the time-reversal transformation,

Θ|l,ml〉 = (i)2ml |l,−ml〉 = (−1)ml |l,−ml〉 (33)

so that

Θ| ± 1〉 = −| ∓ 1〉Θ|0〉 = |0〉 (34)

which means that

Θ( |+ 1〉 ± | − 1〉√

)= ∓

( |+ 1〉 ± | − 1〉√2

Θ|0〉 = |0〉 (35)

proving that the energy eigenstates are eigenkets of the time-reversal operator.

3. Sakurai 5.17

(a) Let us write the Hamiltonian as H = Ho + CV , where

Ho = AL2 +BLz

V = Ly =(L+ − L−

The eigenkets of Ho are simultaneous eigenkets of L2 and Lz, |lm〉, with energy Elm = Ah2l(l+1)+Bhm.We can apply non-degenerate perturbation theory. Now since V is off-diagonal in the |lm〉 basis, the firstorder energy shift vanishes, and we are left with

E′lm = E

(o)lm + C2

∑m′

〈〈lm′|(

L+−L−2i

)|lm〉|2

E(o)lm − E

(o)lm′

+O(C3)

= E(o)lm + (Ch)2

(l −m)(l +m+ 1)−4Bh

+ (Ch)2(l +m)(l −m+ 1)

4Bh+O(C3)

= E(o)lm + hm

2B.+O(C3) (37)

Problem

Notation x y z

H px py pz

mx y z

z mz O

V Lx ypz zpy eB

nx ny ny

n is the ground state n the rst excited state It is three times degenerate

We denote the energy eigenstates of H as jnxnynzi jnxijnyijnzi We use thisbasis in the rst order degenerate perturbation theory

Recall that either problem or Libo

hnzjzjnzi pm

z nz pnz n

Similarly one has Libo

hnzjpzjnzi m

these formulas can readily be proven by using the a ay representation of z pz andjnzi

We want to calculate the matrix of V V

in the basis ji ji ji

and nd its eigenvalues The superscript is introduced to remind us that we justdiscuss the matrices and not the complete operators on LR

Because z has even parity and jnzi has parity nz the V has only diagonal

elements Thus we have

hnzjzjnzi Xk

hnzjzjkihkjzjnzi nz

In d dimensions the nth level of a harmonic oscillator is

times degenerate

number of distinguishable ways to put n identical balls in d boxes

y pz z py has only odiagonal elements dierent from zero since y and pz etchave odd parity and since it does not involve x px in fact only matrix elementsbetween ji and ji can be dierent from zero From and we obtain eBmc

with eigenvalues

Let us now assume that Then we obtain

E O E O

If we assume we obtain

The leading terms in these expressions are expected if we can solve exactlyfor H V and we obtain

Enxnynz nx

If we can also solve HV exactly by using thatH is rotational symmetricThe eigenfunctions can thus be chosen to be simultaneous eigenfunctions of H L

and Lx The energy Enl of such an eigenfunction jn l lxi will change as by lx ifLx is added to H In our case we have the three eigenfunctions ji ji andji They all corresponds to l why look at their x y z dependence Thus

the diagonalization of V is precisely the diagonalization of Lx in the subspace

corresponding to l and the eigenvalues are

Problem

V eE z z is the zeroth component of a vector operator with odd parity ThusWignerEckart and parity tell us that of matrix elements hnlmjV jnlmi we have tohave m m and l l The energy levels depend only on the radial quantumnumber n for the hydrogen atom in the approximation mentioned in the Problemand the degeneracy is n ignoring spin In rst order degenerate perturbationtheory we thus have to diagonalize the nn matrix hnlmjV jnlmi However dueto the conservation of the quantum number m mentioned above which is of coursea consequence of rotational symmetry around the zaxis dierent mvalues will notmix and it will be blockdiagonal afterm We can thus treat eachm value separatelyn implies l spd The matrix splits in blocks

m only one state ji No linear Stark eect because of parity

m same remarks

m Double degenerate ji and ji Again diagonal elements are zerobecause of parity We thus obtain a hjzji

with eigenvalues and eigenvectors

E eE jaj jvi p

m Same as with obvious changes Note that the matrix elements arethe same as in except a a since Y m

l mY m

m Triple degenerate ji ji and ji but only the mixing mentionin the start b hjzji and c hjzji We thus obtain matrix

b b c c

with eigenvaluesE E

pjbj jcj

and the corresponding eigenvectors

jvi pjbj jcjc b

ppjbj jcj

bpjbj jcj c

In principle the matrix elements a b and c can be calculated from the explicit wavefunctions given in Appendix A

Problem uge

Let jjmi and jjmi denote the angular momentum states for particle and respectively We have

jJMi X

jjmijjmihjjmmjJMi

jJMi X

jjmijjmihjjmmjJMi

or since jaijbi has the same meaning as jbijai after a renaming m andm

jJMi X

jjmijjmihjjmmjJMi

ie according to the symmetries of CGcoe cients

jJMi jJ jJMiFor two spin particles it reects the well known fact that the singlet state J isantisymmtric while the triplet state J is symmetric with respect to interchangeof particles

Thus jLMLijSMSi contructed from two electrons both with angular momenta l isonly allowed according to the Pauli principle if jLj jSj is even

This leave us with the S P and D states ie a total of states This isexactly what we would expect from having two electrons in a pshell we have dierent oneelectron states in the pshell three from lz for the orbitalangular momentum and sz for the spin We have to put the electrons intwo dierent oneelectron states according to the Pauli principle It can be done in

ways ie ways

Sakurai We use the eigenfunctions A corresponding to En n

ox s x s p

bpx s x s xx s s

where and denote a symmetric triplet state and the antisymmetric singletstate

c order perturbation theory leads to

Zdxdxjox xjV x x

Zdxjox xj

because of the antisymmetry of the spatial part of ox x

Zdxdxjpx xjV x x

Zdxjpx xj

For an attrative potential V x x peaked around xx the energy of the spatialsymmetric wave function be lowered more than the energy of the spatial antisymmetric wave function because the wave function is allowed to be dierent from zeroat coinciding points Had the potential been repulsive the energy of thespatial antisymmetric wave function increases less than the energy of the spatialsymmetric wave function

If we consider the repulsion of the two electrons in the pshell of the carbon atom assu ciently peaked around coinciding points to applay the above reasoning one wouldexpect that the P states to have lower energy that the S and the D states sincethe P states are antisymmetric with respect to interchange of the spatial coordinatesof the particles

If we include spinorbit coupling only the P states have S and are aectedSince only the total angular momentum J L S is now conserved L and S arestill conserved though the P states are split according to the possible values ofJ J according to

hJM jL SjJMi

JJ LL SS

see in the case of S The coe cient coming from the radial integralsinvolved is positive see Thus the states corresponding to L andS split in one state corresponding to J three states corrending to J and ve states corresponding to J increasing in energy with J The split issmall compared to the P SD split

Problem uge

V r Vera V q

j V qj

If we assume that mVa

is of order one it is seen that the Born approximation is

reliable for ka ie the wavelength of the incoming particle is much larger thanthe range of the potential If we want the approximation to valid for all k we haveto assume mVa

Problem uge

rdrsin qr

aq cos aq sin aq

j V qj a

aq cos aq sin aq

We have aq cos aq sin aq

for aq Note that it falls o slower with large q than cross section in problem the reason being the potential in problem is smoother For aq small we have

aq cos aq sin aq

The total cross section is

aq cos aq sin aq

sin ka

ka sinka

We see again that the approximation is reliable of ka which means that thewavelength of the particle is much less that a the range of the potential

In the limit ka we have

sin ka

ka sinka

ie we have

tot mVa

In order to assure the validity of the Born approximation for all values of k we haveto assume

Sam Pinansky

April 25, 2004

fl = −2m

∫ ∞

jl(kr′)Al(k; r′)V (r′)r′2 dr′ (1)

fl = −2mV0

(jl(kr′))2r′2 dr′ (2)

fl = −2mV0

(kr′)2l

((2l + 1)!!)2r′2 dr′ (3)

= −2mV0R3

(kR)2l

(3 + 2l)((2l + 1)!!)2(4)

f0 = −2mV0R3

3~2(5)

f1 = −2mV0k2R5

45~2(6)

f(θ) =∑

(2l + 1)flPl(cos θ) (7)

4m2R6V 20

9~4(8)

σtot = 4πdσ

)m2V 2

dΩ= |f0|2 + 6Re(f∗0 f1) cos θ (10)

25(kR)2 (11)

β0 =(

= k′R cot(k′R)− 1 (13)

tan δ0 =1− k′

limk→0

=k′ cot(k′R)

1− k′R cot(k′R)(16)

k′tan(k′R) (17)

2m = E − V0:

a = R−√

−2mV0tan

(√−2mV0

-40 -30 -20 -10 10

R−√

~22mV tan

(√2mV~2 R

)V0 < 0

R +√

~22mV tanh

(√2mV~2 R

)V0 > 0

〈x2〉〈p2x〉 ≥

19〈r2〉〈p2

r〉 (21)

〈r2〉 = A2

sin2(πr/a)π2r2/a2

r2r2 dr (22)

= A2 a5(2π2 − 3)12π4

1 = A2

sin2(πr/a)π2r2/a2

r2 dr (24)

A2 =2π2

a3(25)

So then

〈r2〉 =a2(2π2 − 3)

6π2(26)

〈(∆x)2〉〈(∆px)2〉 =2π2 − 3

54~2 (27)

2π2 − 354

= 0.31 > 0.25 (28)

thus:2π2 − 3

54~2 >

〈k′|U (1)I (t,−∞)|k〉 = − i

= − i

2~〈k′|V (r)|k〉

−∞dt′

|〈k′|U (1)I (t,−∞)|k〉|2 =

|〈k′|V (r)|k〉|2∣∣∣∣

η + i(Ek′ − Ek + ~ω)/~+

η + i(Ek′ − Ek − ~ω)/~

∣∣∣∣2

4~2|〈k′|V (r)|k〉|2

(e2ηt

η2 − ((Ek′ − Ek + ~ω)/~)2+

η2 − ((Ek′ − Ek − ~ω)/~)2(33)

+e2ηte2iωt

η2 + 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2+

e2ηte−2iωt

η2 − 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2

dt|〈k′|U (1)

I (t,−∞)|k〉|2 =1

4~2|〈k′|V (r)|k〉|2

(2ηe2ηt

η2 − ((Ek′ − Ek + ~ω)/~)2+

2ηe2ηt

η2 − ((Ek′ − Ek − ~ω)/~)2(35)

η2 + 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2+

η2 − 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2

limη→0

η2 + x2≡ πδ(x) (37)

limη→0

dt|〈k′|U (1)

I (t,−∞)|k〉|2 = (38)

|〈k′|V (r)|k〉|2(

dt|〈k′|U (1)

I (t,−∞)|k〉|2 =π

2~|〈k′|V |k〉|2

~2(k′+ + k′−)dΩ (41)

k′± =

√2m(Ek ± ~ω)

~2(42)

(k′+ + k′−

16π2~4

∣∣∣∣∫

d3xV (x)ei(k−k′)·x∣∣∣∣2

Average 25/30.

Sam Pinansky

April 14, 2004

1. (5 Points)

2. (a) Given that:

∆(b) = − m

∫ ∞

−∞V

√b2 + z2 dz (1)

∆(b) = − m

∫ ∞

−∞V0 exp

(b2 + z2

)dz (2)

= −mV0a

√π exp

(− b2

δl = ∆(b)|b= lk

√π exp

(− l2

∆(b) = − m

∫ ∞

−∞V0

e−µ√

b2 + z2(5)

= − mV0

2k~2µ2

∫ ∞

e−µbu

√u2 − 1

du u =

√1 +

= −mV0

k~2µK0(µb) (7)

δl = ∆(b)|b= lk

= −mV0

k~2µK0

K0(z) ≈√

2ze−z as z →∞ (9)

3. (10 Points)

2m〈x| 1

E −H0 + iε|x′′〉 =

∫dE′

∫dE′′∑

′lm|E′′l′m′〉E − E′ + iε

〈E′l′m′|x′〉

∫dE′∑

πk′jl(k′r)Y m

l (r)1

l∗(r′)

Y ml (r)Y m

l∗(r′)

∫ ∞

−∞dk′ k′2

4jl(k′r)jl(k′r′) = (h(1)l (k′r) + h

(2)l (k′r))(h(1)

l (k′r′) + h(2)l (k′r′)) (13)

∝r large1r2

= − 12π

∫ ∞

−∞dk′ k′2

h(1)l (k′r)jl(k′r′) + h

(2)l (k′r)jl(k′r′)

(k′ + k + iε)(k′ − k − iε)(15)

= −2πi

(1)l (kr)jl(kr′)

2k− h

(2)l (−kr)jl(−kr′)

)r > r′ (16)

= −ikh(1)l (kr)jl(kr′) r > r′ (17)

2m〈x| 1

E −H0 + iε|x′′〉 = −ik

Y ml (r)Y m

l (kr>) (18)

〈x|Elm(+)〉 = 〈x|Elm〉+∫

d3x′〈x| 1E −H0 + iε

|x′〉V (r′)〈x′|Elm(+)〉 (19)

〈x|Elm(+)〉 = clAl(k, r)Y ml (r) (21)

l (r)+ (22)∫

d3x′

−2mik

l′,m′Y m′

l′ (r)Y m′l′

∗(r′)jl′(kr<)h(1)

l′ (kr>)

l (r′)

l (r)− 2mik

l (r)∫ ∞

∫ ∞

soeiδl

sin δl

k= −1

i.e., from above

fl = −2m

∫ ∞

jl(kr′)Al(k; r′)V (r′)r′2dr′ (29)

4. (10 Point)

V (r) =~2

2mγδ(r −R) (30)

∫ ∞

A0(k; R) =j0(kR)

A0(k; r) = j0(kr)

)+ (36)

n0(kr)

(kγj0(kR)j0(kR)R2

cot δ0 = −k

cot δ0 ≈ − cot(kR) (40)

γ(42)

γ= kR− nπ (43)

k =nπ

11 + 1

≈ nπ

(1− 1

(1− 2

R2(46)

Γ =−2

Γ = −14~n3π3

mR4γ2+ O(1/γ3) (48)

Sam Pinansky

October 8, 2003

People did fine on this assignment. The only mistakes were on the second problem. The averagewas 18.1/20.

1. (5 Points) This can be proved simply by writing out both sides:

−ACD, B+ AC, BD − CD, AB + C, ADB (1)= −ACDB −ACBD + ACBD + ABCD − CDAB − CADB + CADB + ACDB (2)= ABCD − CDAB (3)= [AB, CD] (4)

2. (10 Points) The problem asks you to use bra-ket algebra to prove these, but some of youconverted them into matrix problems. This was not incorrect, but was not exactly in the spiritof the problem.

(a) By the definition of the trace:

tr(XY ) ≡∑

〈a|XY |a〉 (5)

where |a〉 is some orthonormal basis. Now we insert a complete set of states:∑

〈a|XY |a〉 =∑

a′〈a|X|a′〉〈a′|Y |a〉 (6)

〈a′|Y |a〉〈a|X|a′〉 (7)

a′〈a′|Y X|a′〉 (8)

= tr(Y X) (9)

where we have used the fact that 〈a′|Y |a〉 and 〈a|X|a′〉 are just numbers so commute, andthe completeness of the |a〉. Challenge: Generalize this problem to prove the cyclicity ofthe trace, namely: tr(X1X2 . . . Xn) = tr(XnX1 . . . Xn−1). (Well, not much of a challengereally).

(b) There are two ways to prove this: 1. Using the dual correspondence. 2. Use the identitythat 〈a′′|X|a′〉∗ = 〈a′|X†|a′′〉. Actually, these methods are completely equivalent, onejust uses a specific basis representation while the other is basis independent. Here is thedual correspondance proof:

|β〉 ≡ X†|α〉 ↔ 〈α|X = 〈β| (10)

Y †X†|α〉 = Y †|β〉 ↔ 〈β|Y = 〈α|XY ↔ (XY )†|α〉 (11)

⇒ Y †X† = (XY )† (12)

where ↔ means dual correspondence. Essentially you are applying the dual correspon-dence twice.The other method goes like this: Take a basis of states, |a〉, and look at the general matrixelement of (XY )†:

〈a|(XY )†|b〉 = 〈b|XY |a〉∗ (13)

〈b|X|c〉∗〈c|Y |a〉∗ (14)

〈c|Y |a〉∗〈b|X|c〉∗ (15)

〈a|Y †|c〉〈c|X†|b〉 (16)

= 〈a|Y †X†|b〉 (17)

Thus (XY )† = Y †X†. This proof is a little more direct then the above, but it chooses aparticular basis to work in so is a bit less elegant.

(trivial iteration proof). Now we note that we expand f(A) in power series for any nicefunction f :

f(A) =∞∑

fnAn (18)

So taking a general ket expanded out in the eigenbasis of A: |α〉 =∑

j αj |aj〉 (whereαj = 〈aj |α〉), we get:

f(A)|α〉 =∑

fnαjAn|aj〉 (19)

fnanj αj |aj〉 (20)

αjf(aj)|aj〉 (21)

So now we apply the operator exp(if(A)) to a general ket |α〉:

exp(if(A))|α〉 =∞∑

n!f(A)n|α〉 (22)

=∞∑

n!f(aj)nαj |aj〉 (23)

exp(if(aj))αj |aj〉 (24)

exp(if(aj))|aj〉〈aj |α〉 (25)

So the operator is exp(if(A)) =∑

j exp(if(aj))|aj〉〈aj |, i.e. it projects onto the eigen-states of A and multiplies by some phase.On a side note, if A = A†, i.e. A is hermition, then the operator exp(if(A)) is alwaysunitary.

(d) This is simply substitution:∑

a′ψ∗a′(x

′)ψa′(x′′) =∑

a′〈x′|a′〉∗〈x′′|a′〉 (26)

a′〈x′′|a′〉〈a′|x′〉 (27)

= 〈x′′|x′〉 (28)= δ(x′′ − x′) (29)

where in the last step we used the orthogonality of the continuous x basis.

3. (5 points) Its easiest to do these calculation (at least, for me) by using a matrix representation,with

|+〉 =(

|−〉 =(

Then we see that:

(0 11 0

(0 −ii 0

(1 00 −1

Now the verifications are just matrix multiplications:

[Sx, Sy] =h

(0 11 0

(0 −ii 0

)− h

(0 −ii 0

(0 11 0

(1 00 −1

)= ihSz

(35)You can easily use these representation to prove all the components of [Si, Sj ] = iheijkSk andSi, Sj = h2/2δij . (No one got this wrong, so I don’t feel the urge to show you trivial matrixalgebra).

Sam Pinansky

October 14, 2003

Average score: 23/25.

10. The hamilonian (in the basis |1〉, |2〉) is represented by the matrix:

(1 11 −1

From the characteristic equation, λ2 − 2a2 = 0, so the eigenvalues are λ± = ±√2a. Theeigenvectors are similarly found by elementary linear algebra to be:

|λ+〉 =1√

4 +√

|λ−〉 =1√

4−√2

(1−√2

or, in stadard ket notation:

|λ+〉 =1√

4 +√

√2)|1〉+ |2〉

|λ−〉 =1√

4−√2

((1−

√2)|1〉+ |2〉

12. First, we need to identify what state the system starts in. You can just look immediately abovethe problem statement in problem 11 to see the answer, but here’s the proof for completeness:We are told that the state is in an eigenstate of the operator S · n with eigenvalue +h/2, wheren = sin γx + cos γz. So we must find eigenstates of:

S · n =h

(cos γ sin γsin γ − cos γ

(Hmmm, seem familiar?). The eigenvalues and eigenvectors are readily solved for, giving thestate

|n; +〉 = cosγ

2|+〉+ sin

2|−〉 (7)

(where |±〉 are the eigenvectors in the Sz basis) with eigenvalue +h/2. Note that you mightnaively expect the state to be cos γ|+〉+sin γ|x; +〉, which is incorrect. Note first that this stateisn’t properly normalized. Second, it is only true that you can linearly superpose eigenvectorslike this if the operators commute.

(a) With that starting state, we want to calculate:

|〈x; +|n; +〉|2 =∣∣∣∣

1√2(〈+|+ 〈−|)

2|+〉+ sin

2|−〉

)∣∣∣∣2

2+ sin

=12(1 + sin γ) (10)

(b) Using 〈(Sx − 〈Sx〉)2〉 = 〈S2x〉 − 〈Sx〉2, we first see that since S2

x = h2

4 I, 〈S2x〉 = h2

4 (in anystate). For the other term:

〈Sx〉2 =h2

4〈n; +|Sx|n; +〉2 (11)

2〈+| sin γ

2〈−|

)(|+〉〈−|+ |−〉〈+|)

2|+〉 sin γ

2|−〉

= h2 sin2 γ

4sin2 γ (14)

So therefore:

〈(Sx − 〈Sx〉)2〉 = 〈S2x〉 − 〈Sx〉2 =

4(1− sin2 γ) =

4cos2 γ (15)

This has the correct limits as γ → 0, π/2π, namely h2/4, 0, h2/4, respectively.

15. (Dropping primes for clarity) Expand a general ket into the complete basis |a, b〉:

|ψ〉 =∑

cij |a, b〉 (16)

Now apply [A,B] to this general ket:

[A,B]|ψ〉 = AB∑

cij |a, b〉 −BA∑

cij |a, b〉 (17)

cij(bjai − aibj)|a, b〉 (18)

= 0 (19)

So since for any |ψ〉, [A, B]|ψ〉 = 0, [A,B] must be the zero operator. (Many of you onlyshowed this for an eigenket |a, b〉 and not for a general superposition. This doesn’t prove that[A,B] = 0 directly without arguing that any ket can be made of linear superpositions of theeigenkets).

19. (a) First, we want 〈Sx〉2 (taken in the state |+〉)

〈Sx〉 =h

2(〈+|(|+〉〈−|+ |−〉〈+|)|+〉) = 0 (20)

and from the reasoning in the last part 〈S2x〉 = h2

4 , so 〈(∆Sx)2〉 = h2/4, From symmetrywe know that 〈(∆Sy)2〉 must be equal to that, when taken in an eigenvector of Sz. Theright side of the uncertainty relation is calculated as:

14|〈[Sx, Sy]〉|2 =

14|〈ihSz〉|2 (21)

=14| ih

2|2 (22)

16(23)

So the uncertainty relation is saturated.(b) We still have 〈S2

x〉 = h2/4, but now since we are taking the expectation value in the state|x; +〉, we also have 〈Sx〉 = h/2, so that 〈(∆Sx)2〉 = 0. Checking the other side of therelation,we need to find:

〈Sz〉 = 〈x; +|Sz|x; +〉 (24)

∝ (1 1

)(1 00 −1

)(1−1

)= 0 (25)

So again the uncertainty relation is true, and is saturated. Note that the fact that bothsides are 0 is obvious when you think about the definition. If we have a state in aneigenstate of Sx, then obviously it has 0 uncertainty in a measurement of Sx.

23. (a) The characteristic equation for B is (b− λ)(λ2 − b2) = 0, so the eigenvalues are b,±b. Sothe b eigenvalue is degenerate.

(b) By straight matrix calculation, [A,B] = 0.(c) By straightforward means you can find a set of eigenvectors for B:

λ = b

λ = b1√2

0−i1

λ = −b1√2

Applying A to these vectors, we see that they are also eigenstates of A, with eigenvaluesa,−a,−a respectively. Therefore these spectrum is no longer degenerate since specifyingthe eigenvalues completely specify’s the eigenvector.

Prob 13:

This is more or less a continuation of Ex-9Using the eigenkets for the z,x,y referance frame.

a. The prepared beam can be represented by

)b. The filter|a′ >< a′| which will selecta′and rejecta′′ . Using Ex 9 the repre-

sentation of this in z eigenket space is:[cos(β

2 )2 cos(β2 )sin(β

2 )e−i∗α

cos(β2 )sin(β

]multiplying[

cos(β2 )2 cos(β

2 )sin(β2 )e−i∗α

cos(β2 )sin(β

)evaluating with respect to the other eigenket(

0 1) (

cos(β2 )2

cos(β2 )sin(β

2 )ei∗α

)= cos(β

2 )sin(β2 )ei∗α

and using trigonometrycos(β

2 )sin(β2 )ei∗α = sin(β)

2 ei∗α

Give the survivors:(sin(β)

2 ei∗α) (

sin(β)2 ei∗α

(sin(β)

which is maximum forβ = ±π2

And we can recapture14 of the atoms.

Sakurai problem solutions (Chapter 1)

Joseph Geddes

[13] Three Stern–Gerlach filters are placed in series. The first apparatus, ori-ented in the z direction, passes atoms having sz = +h/2 and stops others. Thesecond apparatus is oriented at an angle β to the z axis; it passes atoms hav-ing positive spin (+h/2) with respect to that orientation. The third apparatusis also oriented in the z direction, but passes atoms having sz = −h/2 andstops others. Atoms that pass through the first apparatus will pass throughthe second with probability amplitude cos(β/2); atoms that pass through thesecond apparatus will pass through the third with a probability amplitude of− sin(β/2). Therefore, the fraction sin2(β/2) cos2(β/2) = 1

4 sin2 β of atoms thatpass through the first apparatus will pass through the third. When β = ±π

2 ,a maximum fraction of 1

4 the atoms that pass through the first apparatus willpass through the second.

[15] Suppose the simultaneous eigenkets of A and B, denoted by |γi〉 , form anorthonormal set. The equations

A |γi〉 = ai |γi〉 , (1)B |γi〉 = bi |γi〉 , (2)

define the eigenvalues ai and bi, and orthonormality dictates that the complete-ness relation ∑

|γi〉〈γi| = 1 (3)

holds. Therefore,

A = A(1) = A∑

|γi〉〈γi| =∑

ai |γi〉〈γi| (4)

B = B(1) = B∑

|γi〉〈γi| =∑

bi |γi〉〈γi| (5)

Now, we compute the commutator

[A,B] =

ai |γi〉〈γi|

] ∑j

bj |γj〉〈γj |

bi |γi〉〈γi|

] ∑j

aj |γj〉〈γj |

which, when we consider that

〈γi | γj〉 =

0 i 6= j1 i = j

simplifies to[A,B] =

aibi (|γi〉〈γi| − |γi〉〈γi|) = 0 . (9)

Hence, under the foregoing conditions, the commutator [A,B] must be nullvalued.

[16] We are told thatA,B = AB + BA = 0 , (1)

where A and B are Hermitian operators. Suppose that

A |γ〉 = a |γ〉 , (2)B |γ〉 = b |γ〉 . (3)

A,B |γ〉 = bA |γ〉 + aB |γ〉 (4)= 2ab , (5)

and this result implies that either a = 0 or b = 0. So the operators A and B canshare a simultaneous eigenket if either or both the eigenvalues corresponding tothat eigenket are 0.

Problem h

The sequence of measurements produces a sequence of eigenstates for the corresponding observables Sz S n and Sz

ji jn ihn ji jihjn ihn jiie the amplitude of going from ji to ji is hjn ihn ji where problem

jn i cos

ji sin

Thus the probability is sin cos

sin which has a maximum for

This is reasonable since this angle corresponds to the intermediate state jn ibeing symmetric with respect to ji in fact an eigenstate for Sx correspondingto eigenvalue

Problem h

This problem is of course trivially solved by using the representation of rotationmatrices for spin particles given in chapter However it is instructive and theintention since the problem belongs to chapter to solve it without this knowledge

a ibia b

In order that this operation can be viewed as the rotation of the twocomponentspinor we have to associate a direction to the normalized spinor with componentsa b This is done via the formula in problem

aji bji eijn i eicos

ji ei sin

which denes and thus the direction vector

The interpretation is as usual that our spinor with components a b is an eigen

vector for the operator S n If the operator ixp represents a rotation of

around the x axis then the vector jn i should be rotated to eijn i where is a phase factor and n is obtained from n by the mentioned rotation xcomponentof n unchanged while y zcomponents are rotated in the y zplane

n cos sin sin sin cos cos sin cos sin sin

orcos sin sin ei

sin i cos cos sin

From eq we have

ei sin

iei sin

ei sin

If two complex twocomponent vectors only dier by a phase

i jaj jcj ii ab cd

If we apply i and ii in we get

or precisely eq thus showing that the linear operator in question rotates jn ito jn i up to a phase factor where n is obtained from n by a rotation of around the x axis

An alternative and simpler way to solve the problem is as follows Denote

ip by R Ry i

p Since Rjn i is an eigenvector for S n

correspoding to eigenvalue and similarly for jni j n i correspoding toeigenvalue we have

Rjni R

such thatS n R

The rhs of is calculated by using Si i and the relation

ij ij iijkk

After some algebra reads

S n nS ikniSk

orn n n

which precisely means that n is obtained from n by a rotation by around thexaxis

Problem h

Let jai be the spinor with components and jy i be the spinor correspondingto the eigenvalue above ie with components i

p It is an eigenstate of the

spin operator Sy with eigenvalue h Thus the probability of measuring sy h in state jai is

jhy jaij

jj jj i

Problem

U a ia a ia U y

a ia a ia U

det a ia deta ia ia

ia a a ia

a det a ia

Thus detU

We can assume aa in the expression for U since we can scale numerator and

denominator with the same factor In this case a ia are themselves unitaryand unimodular

a ia Thus

U a ia

From we know that a ia represents a rotation around the axis aa a

pa with angle given by

Thus U represents a rotation with angle around the unit vector a

"! # $

0 11 0

] % σ2 =

0 −i

] % σ3 =

1 00 −1

&')(+* ,-/.101"-2 *3 &4*tr(X) = 2a05 ** 6!#78 *3 9:; * 68;66< &=* 6-/! *>& .;6

I σ1 σ2 σ3

σ1 I iσ3 −iσ2

σ2 −iσ3 I iσ1

σ3 iσ2 −iσ1 I

? ! @A 6!#BCD< * "-2! & 'FE 6G;,@tr (σkX) = 2a. 'H *C* "!#

X11 X12X21 X22

] & ! @IJC6!# &'a0 = tr(X)

2 = X11+X222

%a1 = tr(σ0·X)

X22 X21X12 X11

2 = X21+X212

a2 = tr(σ2·X)2 =

−iX22 −iX21iX12 iX11

2 = i · X12−X212

a3 = tr(σ3·X)2 =

X11 X12−X21 −X22

2 = X11−X222

KLM6!#

det(AB) N 3J& 0/ det(eXSe−X) = det(eX)det(S)det(e−X)& ;6;O-4P *3 Q@R * 6! &=* & S<T-2 8;"GUV!W :.XG> *3 &=* <T-2QS * '9Y3 J N : &ZE & & !#2 *3 det() '. *

eXe−X = IC-det(σ · a′) = det(σ · a)( ! * G8 *>&=* 6-/![$ Y3 B@ * 6! &=* -4P & !\-28J &4* -2,]"!10 & & ! * ! @RG^LM! *& E* & !J_P`-27a.JD< & C

UU† = 1b E -/;6 * "-2! * - *3 ^<T-2! @c8 & * -4P *3 98-/. ;"d6! - * 0/ E ;"# & ! *D' "!# *3 SeCfhgI<G-/DC8X-/! @ * -a3 = 1 & !J@ *3 I- *3 :<T-2 8J-2!G! * & Bi %*3 9TUR8J-2!G! * .JD<T-/ iθσ3

C6! <T *3 6j6j@ & #/-/! & ;k$e

iθσ3

0 e−iθ

] l e− iθσ3

e−iθ

0 eiθ

mj- NnN Q< & !o6!10/D * "# &=* & 88; E "! # *3 6 * & !J_P`-2 * -A & < 3 -4P *3 axσx

C8JCp&=* G; E "! <G *3 9Dq2 &=* 6-/!A,r;"6! & "! *3

ax'(a& d6! <T;6 @R6!#

a0.X< & ] * G & C-2! & . ;" & ! @c & E2'

a0 → a′

0 sut * : ; * "8; EB*3 9 &=* ,<TD * -/#2 *3 Ghv

a3 → a′

3 s & Q & a0v:; * 68; E

σ3* - *3 x;"GP *a3 & ! @ * G % < 3 & !#/6!# *3 ]C6#/!B-/P*3 9; & * <T-2;" Q! % & ! @ *3 G!y:; * 68; EB*3 9DC ; *D'

σ2, σ3& x -29<T-2 8;",< &4* D@ 'O( N 6;";zGUR8 & ! @ *3 * & ! CP`-/7 * -

eiθσ3

cos( θ2 ) + i · sin( θ

2 ) 00 cos( θ

2 ) − i · sin( θ2 )

e−iθσ3

cos( θ2 ) − i · sin( θ

2 ) 00 cos( θ

2 ) + i · sin( θ2 )

TUR8 & ! @R6!# & # & "![$e

iθσ3

2 = cos( θ2 )I + iσ3sin( θ

e−iθσ3

2 = cos( θ2 )I − iσ3sin( θ

2 )0 & ;6 &=* "! # *3 9; * -4Pa2 C6!# *3 *& . ;"96!cUy|

cos( θ2 )σ2 + σ1sin( θ

2 )) (

a2(cos2( θ

2 ) − sin2( θ2 ))σ2 + a22(sin( θ

2 )cos( θ2 ))σ1

a2cos(θ)σ2 + a2sin(θ)σ15 6Q6;6 & ; E P`-/σ1

cos( θ2 )σ1 − σ2sin( θ

2 )) (

a1(cos2( θ

2 ) − sin2( θ2 ))σ1 − a22(sin( θ

2 )cos( θ2 ))σ2

a1cos(θ)σ1 − a1sin(θ)σ2 j<T-2:.6!6!# $a′

1 = a1cos(θ) + a2sin(θ)a′

2 = a2cos(θ) − a1sin(θ)

~& vbV; * 68; E 6!# *3 -/# 3 . E]& ! E 0/< * -2[0 & ! @] 6!#j;6"! & *_E* -@R-j * <T-2 8J-2!G! * . E<T-2 8J-2!G! * % "! *3 ^. & C,P`-/ @A. EB*3 ^G6#/!/ * % E ";6@ jiQP`-/j0/ E <G-/ 8X-/!! *P`-/7-/ * G-4P *3 y8-R@R < *D'Y3 ,SP`-/;6;"- N S. E <T-/!J * < * 6-/! 'Y3 c6! @R6016@ & ;<T-2 8J-2!G! * & M"#2G!10/D< * -2C- N 6;6;X.Xx8DC02@ & & < 37* -/P *3 x8-1@ < * 68-R<T@ % N *3o& < & ; & r: ; * "8;66G % ! * 6; *3 E & < 3A*3 * *3J&=* 0 & ;" &=* D * -fG- '.Jv Y3 6 N ";6; E 6G;,@7i]P`-/ & ;";X<T-2Q8X-/! G! * )GU<TG8 * P`-/0XW<T-28J-2! @R6!# * - & l *3 6P`-/;6;6- N jP`-/ *3 & -/!6!#I-/P & v ' 0<T-28X-/! @R6!# * - & N ";6;).XS8G0/D@ & !J@3 & 02 & "#2G!10 & ;6j-4P w C6! <G *3 x<G-/ 8 * @B &4* 6< 3 & *3 ]@RG! -/ "! &=* -2 * G * -!-2 & ;66fG *3 9; *' -2 & ;"; E

Sx, Sy, Sz

< & !o.JQG8G! * D@y. E h2 σx, h

2 σy , h2 σz

8X< * "02G; E/']Y3 G! *3 & !J N GaP`-/;6;6- N P`-/ *3 ]S; * "8 ;",< &4* 6-/! *& .;"9-/P8-/. ;"| '

Sam Pinansky

October 23, 2003

Average score: 33.4/40. Quite a bit lower then the last two.

22. (5 Points) This problem was hard for some people to figure out what Sakurai was asking. Iclaim that there is only one interpretation that makes sense, and I shall “prove” this statementas such: Assume that the problem has a well defined answer (i.e. there is a certain order ofmagnitude that is the one Sakurai expected you to get). The issue hinges on what you thinkSakurai means when he says the icepick is SHARP. By assuming a finite point, the problembecomes trivial, i.e. it simply reduces to how long it would take the center of mass to movehalf the width of the point. But the crucial point is that the order of magnitude of your answerdepends directly on the order of magnitude of the size of the tip. Since you can reasonablyinterpret the tip to be anywhere from 1 atom to a few micrometers in width as SHARP, youcan reasonably get answers which differ by 7 orders of magnitude! Since by assumption, thereis one “correct” answer, this interpretation cannot be the one Sakurai intended (and thus, ifyou did the problem this way, you didn’t get full credit). Other, less formal arguments arethat this interpretation trivializes the calculations in the problem, and using the democraticprinciple, since most people did the problem my way, it must be correct :-).

OK, My interpretation was that the point of the ice pick was infinitesimal in size, and did notslip along the surface. Also I took the rod to be thin, and uniform, with mass m and lengthl. Then, the langrangian is:

L =16ml2θ2 −mg

2cos θ (1)

(recall that the moment of inertia of a rod about one end is 1/3ml2). The equation of motionis then:

θ = λ2 sin θ (2)

where λ2 = 3g2l . We will take the approximation where sin θ ≈ θ. This is valid up until about

π/4 or so. We will assume that after the rod has fallen by π/4 it will continue to fall prettyfast till it hits the ground. Physical intuition tells us that it will only take less then 1 sec or soto fall the rest of the way, so if we find a max time greater then 1 sec, we know that ignoringthis part of the motion won’t change the order of magnitude of the answer.

Then, the solution of this differential equation is

θ(t) = θ0 cosh(λt) +1λ

θ0 sinh(λt) (3)

where θ0 and θ0 are the initial conditions. Now purely classically if you choose θ0 = θ0 = 0,then θ(t) = 0 for all t, but the uncertainty principle tells us that:

∆x∆p ≥ ~2

∆θ∆θ ≥ ~2ml2

Clearly, to maximize the time it takes the pick to fall, we want to saturate the inequality, sowe can eliminate θ0 from the equation of motion:

θ0 =~

2ml2θ0(6)

Now we can use this, and invert the solution of the equation of motion to solve for t:

tf =1λ

cosh−1

(µθfθ30 − ~

√µθ2

0(θ2f − θ2

0) + ~2)

(µθ40 − ~2)

where µ = 4l4m2λ2. (Note you can simplify this a bit by noting that θf − θ0 ≈ θf ). Nowall we need to do is to maximize this with respect to θ0. This is not nice symbolically, so weplug in numbers and maximize this numerically. Taking l = .5m, m = .5kg, g = 10m/s2, and~ = 10−34J · s and θf = π/4, we can plot tf : See attached mathematica printout, where it isseen that tf has a maximum of about 7 seconds. We should also note that the maximum is atθ0 ≈ 10−17, which is when θ0 ≈ θ0.

One thing to notice is that this fall time is basically independent of the size of ~. This issomewhat nontrivial to show, with some simplification you can get that:

tmax ∝ cosh−1(~−1/2) (8)

This gives a range of 60 to about .1 seconds over a range of ~ from 10−300 to 1 (300 ordersof magnitude versus 2!). Thus, the scale of quantum mechanics is completely irrelevant to theproblem. The only thing that matters is that the uncertainty principle holds at all, not thespecific value of ~.

26. (5 Points) We want to find an operator U such that Sz = USxU†. Expressing the operatorsin the Sz basis, this becomes:

(1 00 −1

(0 11 0

)U† (9)

Multiply by Sz on the left, and U on the right:

1 00 −1

(0 11 0

Letting:

a bc d

we calculate: (a bc d

(b a−d −c

so a = b and d = −c. We also have the unitarity requirement, which forces a2+b2 = c2+d2 = 1and ac+bd = 0. Choosing a = 1/

√2 then b = 1/

√2, so c+d = 0, which along with c2 +d2 = 1

fixes c = 1/√

2, and d = −1/√

2. So we get that:

U =1√2

(1 11 −1

Note: A lot of people didn’t really solve for U this way, and in some way just used the identitythat you were supposed to show instead. The problem was kind of vague. Now to compare tothe formula U =

∑r |b(r)〉〈a(r)|, so:

U = |+〉〈Sx; +|+ |−〉〈Sx;−| (14)

=1√2

(|+〉〈+|+ |+〉〈−|+ |−〉〈+| − |−〉〈−|) (15)

=1√2

(1 11 −1

as we found above.

30. (5 Points)

(a) There were tons of ways to prove this, and it seemed everyone used a different one. I willgive two ways. The first uses the fact that the translation operators are generated by themomentum operators, i.e. that T (~l) = e−i~p·~l/~, as well as this relation:Given that [[A,B], B] = 0, then [A,Bn] = n[A, B]Bn−1. Proof:

[A,Bn] =n∑

Bn−j [A, B]Bj (17)

[A, B]Bn−1 (18)

= n[A,B]Bn−1 (19)

where the third step is valid since you can commute B’s with [A,B]’s. Then we see that:

[A, f(B)] = [A,

∞∑n=0

f (n)(B)|0n!

Bn] (20)

= [A, f(0) +∞∑

f (n)(B)|0n!

Bn] (21)

Applying the lemma (and the fact that constants commute):

∞∑n=1

f (n)(B)|0n!

Bn] (22)

= [A,B]n∞∑

f (n)(B)|0n!

Bn−1] (23)

= [A,B]f ′(B) (24)

Q.E.D.So therefore:

[X, T (~l)] = [X, P ]−i~l

~e−i~p·~l/~ = ~lT (~l) (25)

which is what we were trying to prove. You could also prove the relation assuming onlythat the translation operator was the operator which translates the states (i.e. thatT (~l)|~x〉 = |~x + ~l〉). Then the proof goes as follows: Let [Xi, T (~l)] act on a positioneigenstate |xi〉. Then:

= liT (~l)|xi〉 (28)

and the theorem is proven because everything is linear and the |xi〉’s are a completeeigenbasis (unlike the last assignment, I didn’t take off points for not mentioning thisfact, but its still important). You could also proven things by taylor expanding theexponential or similar things, but you end up proving special cases of the relation provedabove. The second proof is “better” then the first since it does not rely on any specialform of the translation operator other then its definition.

(b) From part a, we have that T (~l)−1 ~XT (~l) = ~X +~l Then the proof is simply:

〈T (~l)−1XT (~l)〉 = 〈α| ~X +~l|α〉 = 〈 ~X〉+~l (29)

where |α〉 is the state of the system.

33. (5 Points)

(a) i. This is a simple proof by insertion of a complete set of position eigenkets:

〈p′|X|α〉 =∫

dx′〈p′|X|x′〉〈x′|α〉 (30)

dx′ x′〈p′|x′〉〈x′|α〉 (31)

= i~1√2π~

∫dx′

∂p′e−ip′·x′/~〈x′|α〉 (32)

= i~∂

∂p′

∫dx′〈p′|x′〉〈x′|α〉 (33)

= i~∂

∂p′〈p′|α〉 (34)

which is what we were trying to prove. There were other ways of proving this whichwere less straightforward involving representations of the delta function, but this wasthe quickest route.

ii. This is obvious in light of part i)

〈β|X|α〉 =∫

dp′〈β|p′〉i~ ∂

∂p′〈p′|α〉 (35)

dp′φ∗β(p′)i~∂

∂p′φα(p′) (36)

(b) This is the translation operator in momentum space (also called a boost operator inrelativistic terminology). Let’s show that this interpretation is a good one: Take aneigenstate of p, |p〉:

)|p〉 =

∫dx exp

(iXΞ~

)|x〉〈x|p〉 (37)

=1√2π~

∫dx exp

(iX(Ξ + P )

)|x〉 (38)

dx 〈x|p + Ξ〉|x〉 (39)

= |p + Ξ〉 (40)

So indeed it translates a momentum state by Ξ.

Problem 3-5 (10 Points)

(a) (Note, in this problem I will be working in the standard cartesian representation of theX and P operators). You have three canonical commutation relations to check:

[Xj , Pk] = i~δjk (41)[Xj , Xk] = 0 (42)[Pj , Pk] = 0 (43)

The second is clearly still true, the first gives (with the new Pk):

[Xj , Pk] = i~δjk + [Xj , fk(x)] = i~δjk (44)

since [Xj , fk(x)] = 0 (fk’s are just numbers, not operators, so any operator commuteswith them). So this commutation relation is also always satisfied. The third relation,however, gives:

[Pj , Pk] = [Pj , Pk] + [fj(x), fk(x)] + [fj(x), Pk] + [Pj , fk(x)] (45)

(∂fk

∂xj− ∂fj

∂xk=

∂xj(47)

This can be stated in differential vector calculus as ∇× ~f = 0, or the curl of f vanishesover all space.

(b) We know that any unitary operator can be expressed as eiA where A is hermitian. Nowwe can apply this general unitary operation and assume the unitary equivalence to solvefor A:

〈x|Pj |x′〉 = 〈x|Pj |x′〉 (49)

= 〈x|e−iA ~i

∂xjeiA|x′〉 (50)

= 〈x|e−iAeiA

∂xjA +

)|x′〉 (51)

So to have equality requires that

∂xjA = fj(x) (53)

From part a), we have that ∇× ~f = 0, which implies that f is the gradient of some scalarfunction φ: ∇φ = ~f . Therefore if we set A = φ/~, the required equation is satisfied, andthe unitary equivalence is proven. So under the unitary transformation U = eiφ/~, whereφ is defined by ∇φ = ~f , the new momentum operator is unitarily transformed into theold. φ always exists because of the requirement on f from part a.

(c) If you remove an infinite cylinder around the z-axis (or any axis actually), the reasoningin part b) does not apply, because the implication ∇ × ~f ⇒ ∃φ : ∇φ = ~f (also calledthe Helmholtz theorem) is no longer true. It is no longer true because its proof relies onStoke’s Theorem, which if you recall says that the line integral around a closed loop ofsome vector field is equal to the surface integral on the surface bounded by that closedloop of the curl of that vector field. In math notation:

(∇× F) · da =∫

F · ds (54)

But with an infinite cylinder removed, if you choose a loop which winds around thatcylinder there do not exist any surfaces with the loop as the boundary (note, the notation∂S means “the boundary” of the surface S). Thus Stoke’s theorem does not apply tothose loops and you can’t prove the implication above.Note, another way of putting this is that the Helmholtz theorem is only true if space issimply connected, e.g. has the topology of Rn or Sn. By removing an infinite cylinderyou change the topology of space, thus making it not simply connected globally (notehowever, that it is still simply connected locally). So if the cylinder was finite, part b)would in fact still apply because you could topologically shrink the missing space downto a point at the origin. Note how none of this really says anything about rotationalsymmetry.

NRQM Class

R. Rogers

23 ,26————————————————————-

Prob 23:

The characteristic polynomials are:

A:(λ−a)(λ+a)2 with rootsa,−a and eigenvectors

B: (λ−b)2(λ+b) with rootsb,-b and eigenvectorsm

0−i1

a. Yes

The product is:

a · b 0 00 0 i · a · b0 −i · a · b 0

from either side.

Obviously

works for one eigenket and has eigenvlaues a,b respectively. 0−i1

works with eigenvalues -a,b 0i1

works with eigenvalues -a,-b

Any vector v

can satisfy Av=-av .

but it is impossible for it to satisfy Bv= -bv=bvSimiliary we can satisfy Bv=bv but then Av =-av=avThus, because of the combinatorial properties for the (or any) common eigenvalues

there is only one set of solutions.

Prob 26:

SinceSz is already diagonal we can use the eigenkets ofSx

(1−1

)to form

the Unitary matrixU = 1√2

[1 1−1 1

]andU † SxU = Sz

I will take a moment to digress into a matrix representation interpretation of 1.5.4

(for those of us who have problems interpreting the symbolism)Let Az, Bxbe a column collection the normalized eigenkets ofAz, Bx for instance[

1 00 1

]= Sz for Sz

Similiarly for Sx Sz = 1√2

[1 1−1 1

]The use ofz, x in Az, Bx is just for identification, the following is general.(Az)†Az = I and the same holds forBx because they are composed of normalized

eigenkets.Then considerU = Bz · (Az)† (or Sz(Sx)† in our case), note thatUU† = I.Now we prove that|bl >= U |al > in matrix representationMultiplying U on the right by eigenkets ofA ,v will move a one around the output

of(Az)†v depending upon the ordering of B. Now the wandering one will extract

exactly one of the eigenkets ofB from Bz

Now evaluatingSz(Sx)†[

1 00 1

] [1 1−1 1

[1 1−1 1

]I guess I should apologize for using a crutch, matrix representation, but I haveproblems following the flying summations and superscripts. Hopefully I will beable to dispense with the crutch after a while; although it took me a long time todo Linear Algebra in the abstract (and proper) form.

Problem h

SxSy i

jihj jihj

SySx i

jihj jihj

Sx Sy SxSy SySx i

jihj jihj

fSx Syg SxSSy SySx

jihj jihj

Similarly for all the other combinations of Si Sj

Note that in the basis ji ji the operator

Ajihj Ajihj Ajihj Ajihjis given by the matrix

Problem h

S n Sx cos sin Sy sin sin Sz cos

sin eijihj sin eijihj cos jihj jihj

Expressed as a matrix in the ji ji basis this operator reads

cos sin ei

sin ei cos

and the eigenvalue equation becomes

cos sin ei

sin ei cos

which has the solutionxy

sinei cos

or normalized

ei sin

Note that by changing n n ie we get the eigenvector

for S n corresponding to eigenvalue xy

ei cos

Problem h

In the basis ji ji the operator H is given by the matrix

with eigenvalues

qH H H

Eigenvectors when H

H HqH H H

with the normalization constant C is given by

H HH H

qH H H

One can check that the limit H makes sense For instance if H H theeigenvector for goes to for H

Alternatively one can use the results in problem to nd eigenvalues and eigenvectors Decompose the matrix for H as follows

qH H H

cos sinsin cos

cos H Hq

This corresponds to the situation in problem with and we can now reado the eigenvalues and eigenvectors

Problem h

In the notation of problem we have and thus the eigenvectorcorresponding to eigenvalue h is

j n i cos

ji sin

The operator Sx corresponds to S n with ie the eigenvector of Sx witheigenvalue is

j x i p

a well known result of course

The probability of nding the system in the state j x i after a measurement of Sxgiven that it initially was in the state j n i is

jh x j n ij

The expectation value of Sx in the state j n i can be calculated using matrixnotation or the ketbra notation of problem Using the ketbra notation oneobtains

hSxi h n j

jihj jihj

hSx hSxii hS

xi hSxi

Prob 9:Using the identication of Sxfrom problem 8 and applying the diagram:

sin(β)cos(α) h2

[0 11 0

]+ sin(β)sin(α) h

[0 −ii 0

]+ cos(β) h

[1 00 −1

[cos(β) sin(β)cos(α) − isin(β)cos(β)

sin(β)cos(α) + isin(β)cos(β) −cos(β)

[cos(β) sin(β)e−iα

sin(β)eiα −cos(β)

]Now we could use the last part of problem 7 to identify the eigenvector, this

leads to some inobvious manipulation; division by sqrt's to normalize to 1. I willdo a simple extraction (one that would lead to headaches for a larger numberof states).

Let the eigenket be

]and do the simple evaluation of

[cos(β) − λ sin(β)e−iα

sin(β)eiα −cos(β) − λ

])to determine the eigenvalues h

2 ,−h2

Choosing h2

a = a · cos(β) + b · sin(β)e−iα

b = a · sin(β)eiα − b · cos(β)Where the h

2 term cancels out.

]is normalized and that we can introduce a normalizing phase

for the rst term; we can assume the form

[cos(γ)

sin(γ)eiτ

]but more dramatically

a= tan(γ)eiτ (1)

Solving the above conditions on a,bba = (1−cos(β))eiα

sin(β) or ba = sin(β)eiα

1+cos(β)

Which fortunatly are consistant and yield ba = tan(β

2 )eiα

Resolving this with 1 yields the answer.For a greater number of states we just have to walk through using 7.b; not

to obvious but mechanical.

Sam Pinansky

October 26, 2003

1. (5 Points) First, a useful lemma: From the definition of the Heisenberg representation, AH(t) =U†(t)AU(t). Therefore:

[AH , BH ] = [U†(t)AU(t), U†(t)BU(t)] (1)= U†(t)CU(t) (2)= CH (3)

So the commutation relation between the Heisenberg operators is the same at equal times.Note that the relation is NOT true for the commutation relation between operators at differenttimes.Then with H = ωSz, the heisenberg equations of motion are:

dSx(t)dt

i~[Sx(t), Sz(t)] (4)

dSy(t)dt

i~[Sy(t), Sz(t)] (5)

dSz(t)dt

= 0 (6)

Clearly, Sz(t) = Sz(0). Now using the previous lemma, we know that [Sx(t), Sz(t)] = −i~Sy(t),etc.. so we get:

dSx(t)dt

= −ωSy(t) (7)

dSy(t)dt

= ωSx(t) (8)

We can solve these coupled differential equations by elementary means, giving:

Sx(t) = A cos ωt + B sin ωt (9)Sy(t) = C cos ωt + D sin ωt (10)

Now we apply the initial conditions Sx(0) = Sx(0) and Sx(0) = −ωSy(0), etc... giving:

Sx(t) = Sx(0) cos ωt− Sy(0) sin ωt (11)Sy(t) = Sy(0) cos ωt + Sx(0) sin ωt (12)Sz(t) = Sz(0) (13)

4. (5 Points) Just solve the Heisenberg equation of motion for X(t):

[X, H] =1i~

[X,P 2

2P (14)

So X(t) = P (0)tm + X(0), and therefore:

[X(t), X(0)] = − i~tm

5. (5 Points) Using the canonical commutation relations, you see that

[H,X] = − i~Pm

[[H, X], X] = −~2

Now we can work backwards to the theorem:

2m= −1

2〈a′′|[[H, X], X]|a′′〉 (19)

= −12

(〈a′′|HX2|a′′〉 − 2〈a′′|XHX|a′′〉+ 〈a′′|X2H|a′′〉) (20)

= −12

(2Ea′′〈a′′|X2|a′′〉 − 2

a′|〈a′′|X|a′〉|2

a′|〈a′′|X|a′〉|2(Ea′ − Ea′′) (22)

where we have inserted a complete set of states.

11. (5 Points) Recall X(t) = X(0) cos ωt + P (0)mω sin ωt. Clearly 〈P 〉 = 〈0|eiPa/~Pe−iPa/~|0〉 = 0.

Also, using the Baker Hausdorff Lemma (or your result from a previous homework):

〈X〉 = 〈0|eiPa/~Xe−iPa/~|0〉 (23)= 〈0|X|0〉+ a (24)

So then

〈X(t)〉 = 〈X〉 cos ωt +〈P 〉mω

sin ωt + a cosωt (25)

= a cosωt (26)

since 〈0|X|0〉 = 〈0|P |0〉 = 0 (expand out P and X in terms of a and a† to see this).

12. (10 Points)

(a) e−ipa/~|0〉 is just the translation operator T (a), so we know that this operator translatesthe system by a in the positive direction. So then the wave function in position space is(by definition):

〈x|e−ipa/~|0〉 = 〈x− a|0〉 (27)

=1√√πx0

e−(x−a)2/2x20 (28)

(b) First, we note that in general, the time dependence drops out, as can be seen:

|ψ(t)〉 = e−iHt/~e−iPa/~|0〉 (29)

|〈0|ψ(t)〉|2 =∣∣∣e−iE0t/~〈0|e−iPa/~|0〉

∣∣∣2

=∣∣∣〈0|e−iPa/~|0〉

∣∣∣2

So all the time dependence drops out.To calculate the probability:

〈0|ψ〉 =1√πx0

∫dxe−x2/2x2

0−(x−a)2/2x20 (32)

= e−a2/4x20 (33)

|〈0|ψ〉|2 = e−a2/2x20 (34)

Sam Pinansky

November 4, 2003

15. (5 Points) We know that

x(t) = x(0) cos ωt +p(0)mω

sin ωt (1)

So then, the correlation function is:

〈x(t)x(0)〉 = 〈0|x(0)2 cosωt +p(0)x(0)

mωsin ωt |0〉 (2)

Now we expand x(0) and p(0) in terms of a and a†:

2mω〈0| (a + a†)2 cosωt + i(a− a†)2 sin ωt |0〉 (3)

2mω〈0| aa† cos ωt− iaa† sin ωt |0〉 (4)

2mωe−iωt (5)

where the aa a†a† and a†a terms vanish in the expectation.

16. (10 Points)

(a) Let’s say that we have a state α |0〉+ β |1〉, with the constraint |α|2 + |β|2 = 1, then

〈x〉 = (α∗ 〈0|+ β∗ 〈1|)(√

(a + a†)

)(α |0〉+ β |1〉) (6)

2mω(αβ∗ + βα∗) (7)

Going to polar coordinates, we can choose α is real (since there is an arbitrary phaseambiguity), then applying the normalization constraint, we get that α =

√1− r2, β =

reiφ. Plugging in:

2mω(2r

√1− r2 cos φ) (8)

We can maximize each part individually, giving φ = 0, and r = 1/√

2. So the maximumstate is:

1√2(|0〉+ |1〉) (9)

(b) The time evolved state is:

(e−iωt/2 |0〉+ e−3iωt/2 |1〉

i. Schroedinger picture:

〈x〉 (t) =

(eiωt/2 〈0|+ e3iωt/2 〈1|

)(a + a†)

(e−iωt/2 |0〉+ e3iωt/2 |1〉

(eiωt + e−iωt

2mωcos ωt (13)

ii. Heisenberg:

〈x(t)〉 =

12(〈0|+ 〈1|)((a + a†) cos ωt + i(a− a†) sin ωt)(|0〉+ |1〉) (14)

2mωcosωt (15)

(c) We already know from the last part that

〈x〉2 =~

2mωcos2 ωt (16)

so all we need now is⟨x2

⟩. Using the schroedinger picture:

(eiωt/2 〈0|+ e3iωt/2 〈1|

)(a + a†)2

(e−iωt/2 |0〉+ e−3iωt/2 |1〉

(eiωt/2 〈0|+ e3iωt/2 〈1|

)(e−iωt/2 |0〉+ 3e−3iωt/2 |1〉

mω(19)

So therefore ⟨(∆x)2

(2− cos2 ωt) =~

2mω(1 + sin2 ωt) (20)

21. (5 Points) You should recall from previous quantum mechanics that the eigenfunctions of theinfinite square well are:

ψn(x) =

sinnπx

with energy En = ~2n2π2

2mL2 . You could also look these up in appendix A.2 in the textbook. Ifthe particle is known to be exactly at x = L/2, then it’s wave function (not properly normal-ized) would be δ(x − L/2). So measuring the energy of the particle would have probabilities

| 〈n|x = L/2〉 |2, or inserting an x basis:

〈n|x = L/2〉 =∫ ∞

−∞

sinnπx

Lδ(x− L/2) dx (22)

sinnπ

√2L n odd

0 n even(24)

so the relative probability of being in a state n versus a state n′ is:

Pn′=

0 n even n odd1 n and n′ even or odd∞ n odd n′ even

i.e. all states of even n can’t occur, and all states of odd n have equal probability. The timeevolution is simply:

ψ(t) =∑

(−1)ne−iEnt/~ sinnπx

Note that this state is badly unrenormalized.

22. (5 Points) (Let λ = v0, so I can easily reuse my solutions from last year). The Schroedingerequation in this case is just

− ~2

dx2− λδ(x)ψ = Eψ (27)

Looking at this differential equation, we can see that there is an infinite jump in the potentialat x = 0. So any solution to this equation will need to have a similar jump in d2ψ/dx2, i.e.dψ/dx will be discontinuous. To obtain the discontinuity in ψ′, we integrate the Schroedingerequation around an epsilon neighborhood about the origin:

− ~2

∫ ε

dx2dx− λ

∫ ε

δ(x)ψ(x) dx = E

∫ ε

ψ(x) dx (28)

In the ε → 0 limit, the first term is the change in the slope of ψ at 0 by the fundamentaltheorem of calculus, the second term is λψ(0) from the definition of the delta function, andthe right term is zero since we restrict ψ(x) to be finite at x = 0. This gives:

ψ′(0+)− ψ′(0−) = −2mλ

~2ψ(0) (29)

The matching condition of ψ(0) is clearly that ψ need to be continuous at zero. AssumingE < 0 (which restricts our view to bound states), we have that in the region x < 0:

ψ(x) = Aekx (30)

and in the region x > 0ψ(x) = Be−kx (31)

where k =√−2mE/~2. (The other exponential terms with opposite exponents are zero

from the requirement that the wavefunction not blow up at infinity). The continuity of ψ(x)is already set by setting B = A. Then the last matching condition at 0 (the one on thederivatives) gives:

ψ′(0+)− ψ′(0−) = −2mλ

~2ψ(0) (32)

−2Ak = −2mλ

~2A (33)

k =mλ

~2(34)

√−2mE

~2(35)

E = −mλ2

2~2(36)

Which is the only bound state energy. Normalizing ψ:

ψ(x) =

√mλ

~2e−

mλ~2 |x| (37)

Extra Problem (15 Points)

(a) We will make the reasonable guess that a simple rotation will decouple the coordinates:

Xx =X1 + X2√

X ′y =

X1 −X2√2

(Reason for the prime will be clear shortly). And the same for the momentum states:

Px =P1 + P2√

Py =P1 − P2√

Inserting these into the hamiltonian, and simplifying CAREFULLY (many people mademistakes here), we arrive at:

H =P 2

x + P 2y

12mω2

x + (1 + 4λ)X ′2y − 2

√2hX ′

y + 2h2)

So we have decoupled the hamiltonian. But considering the next parts, let’s changevariables again by completing the square:

H =P 2

x + P 2y

12mω2

x + (1 + 4λ)

(Xy −

1 + 4λ

+8λh2

1 + 4λ

So now we let

Xy = X ′y −

1 + 4λ(44)

and the hamiltonian is simply:

H =P 2

x + P 2y

12mω2X2

x +12mω′2X2

y +4mω2λh

1 + 4λ(45)

with ω′ = ω√

1 + 4λ. This is obviously decoupled.

(b) Clearly the X’s and P ’s commute with themselves since the original X’s and P ’s allcommuted with themselves, the only ones we need to check are [Xx, Px] = i~, [Xx, Py] = 0(and the two that are almost identical).

[Xx, Px] =12[X1 + X2, P1 + P2] (46)

=12([X1, P1] + [X2, P2]) (47)

=12(2i~) (48)

= i~ (49)

[Xx, Py] =12[X1 + X2, P1 − P2] (50)

=12([X1, P1]− [X2, P2]) (51)

= 0 (52)

where we’ve dropped all zero terms. The additive constant in the definition of Xy obvi-ously won’t change anything since constants commute with everything. So they satisfythe standard commutation relations. The hamiltonian looks like two separate decoupledharmonic oscillators, one with frequency ω, one with frequency ω′, and an extra energyoffset of 4mω2λh/(1 + 4λ).

(c) Using the one dimensional oscillator as an example, we take:

ax =√

(Xx + i

a†x =√

(Xx − i

√mω′

(Xy + i

mω′

a†y =

√mω′

(Xy − i

mω′

These must satisfy the standard commutation relations [ax, a†x] = 1 etc.. since the X’sand P ’s satisfy the standard ones. We can rewrite the hamiltonian in terms of theseoperators, (specifically Nx = a†xax and Ny = a†yay) just as in the 1-d case:

H = ~ω(

Nx +12

)+ ~ω′

4mω2λh

1 + 4λ(57)

Since H is linear in both Nx and Ny, it can be diagonalized in both simultaneously. Let|nx, ny〉 denote a simultaneously eigenket of both Nx and Ny, such that Nx |nx, ny〉 =nx |nx, ny〉 and Ny |nx, ny〉 = ny |nx, ny〉, (or equivalently by taking two different 1-dharmonic oscialltors and taking their tensor product |nx〉 ⊗ |xy〉 ≡ |nxny〉). Then thesestates are also eigenstates of the full hamiltonian with eigenvalues:

Enx,ny = ~ω(

nx +12

)+ ~ω′

4mω2λh

1 + 4λ(58)

We can construct the full space of eigenstates by multiple applications of the raisingoperators:

|nx, ny〉 =(a†x)nx(a†y)ny

√nx!

√ny!

|0, 0〉 (59)

properly normalized. The ground state wavefunction is simply the multiplication of theground state wave functions of the two oscillators:

〈xx, xy| 0, 0〉 =

√m√

ωω′√π~

exp(−1

ymω′

(d) You can have degeneracy in the energy only if Enx,ny = En′x,n′y for some pairs (nx, ny)and (n′x, n′y). Therefore we want:

nx +12

)+ ~ω′

)= ~ω

(n′x +

)+ ~ω′

(n′y +

nx +√

1 + 4λny = n′x +√

1 + 4λn′y (62)√

1 + 4λ =n′x − nx

ny − n′y(63)

Thus the only way this can occur is if√

1 + 4λ is a rational number. In the specialcase λ = 0 you simply have an isotropic 2-d harmonic osscillator, where every state isdegenerate except the ground state.

"!$#%'&(*),+.-/ 0,1234*576*89:6;)=<

|λ >= c∑

λ′n

√n!|n > > 1@?

AB85 λ′),C6 85),+EDF9GIHJ9+KMLJ),*NOJ+ NP9:G,)IQ@9R6 )I+.<S9LT6*

a|λ >= c∑

λ′λ′n

√n!|n >

a|λ >= λ′|λ >

(*)I+U-/ 0V-/1

|λ >= c∑

λ′n

(a†)n

= c∑

λ′n(a†)n

n!|0 >

= ceλ′a†|0 >

W XY8+*NP9:G,),Q9R6 )I+UZ /LT@*C)[;9L\8),5DKU]_^

< λ|λ > ` 9:+K1

< λ|λ >= c2∑

|λ′|2n

n!< n|n > > -?

> W ),+L5J6*8 |n > a ;9*7*6*8+9:Gb?c +Kd*T6*6*),+O6*8)[Y6 e+f

c2e|λ′|2

c = e−1|λ′|2/2

X$ONOg9:+:6*85;h_Hi6*),+M)[5fkjl9OLT85 5+_6;i6 9R6 l6\9:]G,@m(*)I+O6*8gK/)[ LTH i),+M9: H+nKM-I1 on16 8),Y)[Y6 *HJ)I<

λ)[;9:+d5),+p56C<

H = hω

a † a +1

6**Nq),C6 *),DE),9GrXY8

a † a|λ >6**Nq89C6 e]sgLT+*),K/*@K.6 5 Nt]E^P6**Ndf

a † aλ′n

√n!|n >= a †

λ′n

√n|n − 1 >=

AB8)[L\8MNO@9:+C6 8gLT85 5+_6;l6\9R6 7)[Y+6l6\9:]G,uv &w)I\i6Y3CJ+5@Kx6*ey+K

< λ|x|λ >, < λ|p|λ >(*)I+U-/ 0V-Ro

2mω(a + a†), p =

2(−a + a†)

w< x >

9:ZZGI^x9 6 O6*8 ),8_6 ` 9+K a†6 e6*8G,T<z6 ` ^E)IG,K/),+

2mω< λ|(a + a†)|λ >

< x >= (λ′ + λ′∗)

2mω< λ|λ >

< x >2= 4 · (Re(λ′))2h

2mωW ),Ne),G[9: GI^

= (−λ′ + λ′∗)

2= 4 · (Im(λ′))2mωh

2w< x2 >, < p2 >

< λ|x2|λ >=h

2mω< λ|(a2 + a †2 +a † a + aa†)|λ >

;R3|Hi),+O6 8K/Ty+)=6 )I+d9+KMGI),+@9: )=6l^R*HZs5 Zn_i)I6*),+$~1@?Y9ZZG,^P6*8Jyn i6Y6*5 Nt6 e6*8 )I8_6

λ′2 < λ|λ >-?Y9ZZG,^P6*8g*L5+KU6**Nt6*O6*8GI5<z6λ′∗2 < λ|λ >0?Y9ZZG,^P6*8J6 8), Kx6**Nt6*O6*8*),8_69:+KMG,T<z6

= |λ′|2 < λ|λ >= λ′∗λ′

o_?46 8J<H*6*8x6**Nq),9eG,)=6*6*G,gNO 7L5NOZG,),L9R6 K'f(*)I+U-/ 0 0

aa† = 1 + a † a

XY85+< λ|aa † |λ >= 1 + λ′∗λ′

9:+Kx6 8_Hn

< λ|x2|λ >=h

2mω(1 + (λ′ + λ′∗)2) =

2mω(1 + 4 · (Re(λ′))2)

< x2 > − < x >2=h

w< λ|p2|λ > ` p = i

mhω2 (−a + a†)

< λ|p2|λ >= −mhω

2< λ|(a2 + a †2 −a † a − aa†)|λ >

),5G[K1@?

λ′2-?λ′∗20?−λ′∗λ′o_?1 − λ′∗λ′;*

< λ|p2|λ >=mhω

2(1 + (λ′ − λ′∗)2) =

2(1 + 4 · (Im(λ′))2)

< p2 > − 2=mhω

2W << ∆x2 >< ∆p2 >>=

&XY8), *+M< N),K/+)I+Z9:*69x-XY8gNO_l67Z*]9:]GI^MDR9:G,H > ;DR9:G,H@ ?9: LTNOZH/6*@K]E^U6\9:pE),+.6 8 \9R6 )I+:<*HL5L5 i),DJ6 5 NO Ti

= n+1|λ|2

38)[L\8),D@<NO/K/FRNe_l6Z ]n9:]G,DR9GIHf

n+1|λ|2

> 1 ),+PK/R3+8),GIG ` i6 9:6*),+ n > |λ|2 − 1nλ

< 1 ),+OHZ8),G,G ` i6 9:6*),+ |λ|2 < n

|λ|2 − 1 < n < λ

AB8)[L\8M),+iNOJL9*C89C6l34P*G,H/6*),+ &c ZZGI^U6*8gKTy+)I6*),+d<k485 5+_6fAB89:6;),C6 8DR9:G,H:<

a · e−ipl/h|0 >mHGI6*),ZG,^_),+P]E^eipl/h

eipl/ha · e−ipl/h = a + ilh[p, a] + (il)2

2! [p, [p, a]] . . .

[p, a] = i

mωh2 ((a † −a)a − a(a † −a)) = i

mωh2 (a † a − aa†) = −i

mωh2 IXY8EH8),8Y\K/5Y6 5 NO9*7Q5*P9+K

eipl/ha · e−ipl/h = a + l√

eipl/ha · e−ipl/h|0 >=(

a + l√

|0 >= l√

mωh|0 >XY8EH5f

a · e−ipl/h|0 >= l

he−ipl/h|0 >

jlK/5+_6 )=<^E),+|λ >= e−ipl/h|0 >

),D@a|λ >= l

mωh|λ

! #"%$%&('*),+.-%/10,$ 2"%&,354%67 98

:;=<?>A@BDCFE

J± = a†

±a∓ G Jz = h

2 (N+ − N−) G N = N+ + N−

[Jz, J±] =h

2([N+, J±] − [N−, J±])

[N+, J±] = [N+, a†

±a∓] = [N+, a

±]a∓ = a

±a∓

[N−, a †± a∓] = a†

±[N−, a∓] = −a

±a∓HI$KJ)

[Jz, J±] =h

2([N+, J±] − [N−, J±]) = ha

±a∓ = ±hJ±

:;=<?>A@BDCPO

J2 = J2z + 1

2 (J+J− + J−J+)H #&,8Q4KRS0, #&,8UTV 4KWX+73*J=)Y67RZT

[JzJz, Jz] = 0[),+.-/T[BC, A] = [B, A]C + B[C, A]

8\3K]^3`_Ia*b c%bed`f* [J+J−, Jz] = [J+, Jz]J− + J+[J−, Jz] = −hJ+J− − J+(−)hJ− = 0g +78h+767+7'*&,67RS_i3*&I0j$% 26k'*)Y0!0, 9&j8Ub

:;=<?>A@BDCl

+ + N2−− N+N− − N+N−)

J+J− = h2a†

+a−a†

−a+ = N+a−a

−= N+(I + N−) = N+ + N+N−

J−J+ = h2a†

−a+a

+a− = N−a+a†

+ = N−(I + N+) = N− + N−N+

J2 = h2

4 (N2+ + N2

−− N+N− − N+N− + 2N+ + 2N+N− + 2N− + 2N−N+)

J2 = h2

N2 + 2N)

Sam Pinansky

November 10, 2003

23. (7 Points) We know from the previous homework that the wavefunction at t < 0 is:

ψ(x) =√

κe−κ|x| (1)

where κ ≡ mλ/~2. To find the time evolution of this wave function, we note that:

U |t < 0〉 =∫

dk U |k〉〈k| t < 0〉 (2)

∫dx U |k〉〈k|x〉〈x| t < 0〉 (3)

where U is the time evolution operator and |k〉 are a complete set of states for a free particle,(i.e. eigenstates of H). Therefore U |k〉 = e−i~k2t/2m |k〉, and we just need to evaluate:

∫dx 〈k|x〉〈x| t < 0〉 =

∫ ∞

−∞dx

1√2π

e−ikx√

κe−κ|x| (4)

κ + ik+

1κ− ik

k2 + κ2

Plugging back into the full state:

|t > 0〉 =∫

dk e−i~k2t

k2 + κ2

)|k〉 (7)

So the wave function at times t > 0 is then:

〈x| t > 0〉 =κ3/2

e−i~k22m t+ikx

k2 + κ2(8)

Try as I might, I can’t seem to be able to do this integral. Even numerically it seems toresist evaluation. Anyway, there was also a different way to do this problem using the freeparticle propagator, which gives a seemingly different undoable intergral, but the two answersare simply related by a fourier transform, I believe.

-10 -5 5 10

Figure 1: Wave function

24. (8 Points)

(a) In this problem, you were expected to use the WKB approximation to find approximateexpressions for the eigenfunctions. It is possible to solve this exactly in terms of specialfunctions called Airy functions, but it wasn’t required. The energy spectrum is continuous,since there is only one boundary condition at infinity (or in other words, because thereare no bound states). Using the WKB approximation, in the region where E > V , youget (after plugging in for V and doing the trivial integral):

ψ(x, t) =A

(E − λx)1/4exp

(± i√

3λ~(E − λx)3/2 − i

for x < E/λ. This is simply a wave with a slightly decreasing amplitude and increasingfrequency as x → −∞. In the other domain of validity:

(E − λx)1/4exp

(−√

3λ~(λx− E)3/2 − i

which is simply a decreasing exponential as x →∞.

(b) With the extra boundary condition at −∞, the states become bound, and the energiesdiscrete. You now have an extra matching condition as well. Again this can be solvedexactly in terms of special airy functions. Also, you can use the same method as thebouncing ball system in the text to derive the energy spectrum, namely that:

3π~mλ(n + 12 )

2m(11)

but this wasn’t required (the derivation is exactly as in the text for the bouncing ball).

Bonus Challenge: Given a generic potential well of the form f(|x|) where f is polynomialof degree d, how does the energy scale with the energy levels at large n, i.e. find a whereE ∝ na. We already know that if d = 1, a = 2/3, and if d = 2, a = 1. What aboutfractional d? Answer this question correctly and I’ll be very impressed.

Sam Pinansky

November 17, 2003

Average score: 24.8/25. No this isn’t a typo. People did extremely well on this assignment sinceall the answers were given (all you had to do was figure out the calculation). There’s almost nopoint in me writing solutions.

28. Let’s do both parts of the problem at once by finding the propagator in D dimensions. Weknow from Eq. 1.7.50 (and a little extrapolation) that

〈x|p〉 =1

(2π~)D/2exp

(ip · x~

Combining this with equation 2.5.8 and the energy p2/2m you generalize equation 2.5.15:

K(x′′, t;x′, t0) =(

)D ∫ ∞

−∞dp′ exp

(ip′ · (x′′ − x′)

~− ip′2(t− t0)

where dp′ ≡ dp1dp2 . . . dpD. Now all we need to do is to do the integral. First we completethe square:

)D ∫ ∞

−∞dp′ exp

(− i(t− t0)

(p′ − m(x′′ − x′)

(t− t0)

im(x′′ − x′)2

2~(t− t0)

Shifting our variables, let p = p′ −m(x′′ − x′)/(t− t0), then:

im(x′′ − x′)2

2~(t− t0)

) ∫ ∞

−∞dp exp

(− i(t− t0)

The integral is simply D copies of a basic gausian integral∫

dx exp(−ax2) =√

K(x′′, t;x′, t0) =(

2πi~(t− t0)

im(x′′ − x′)2

2~(t− t0)

Which is eq. 2.5.16 if D = 1, and is also true for a free particle in any dimension.

29. Calculating:

− 1Z

∂β=

∑a Eae−βEa

∑a e−βEa

Now by assumption there is a lowest energy, E0, such that E0 ≤ Ea : ∀a. So we factor thatout:

− 1Z

∂β=

E0 +∑

1 +∑

a>0 e−β(Ea−E0)(7)

Now we take the limit β →∞, noticing that Ea − E0 > 0 so all the exponentials go to zero:

limβ→∞

E0 +∑

1 +∑

a>0 e−β(Ea−E0)= E0 (8)

Illustrating it for the particle in the box is really pointless, and I won’t bother to do it.

36. (a) First, note that the P ’s and A′s commute with themselves. Then

[Πi, Πj ] = [Pi − eAi

c, Pj − eAj

c] (9)

= −e

c[Pi, Aj ]− e

c[Ai, Pj ] (10)

= −e

c(PiAj − PjAi) (11)

= − i~ec

εijkBk (12)

from B = i~p×A. In this specific case, B = Bz so:

[Πx, Πy] =i~ec

B (13)

with all the other commutators zero.(b) Defining ω = |eB|

mc . Then H = H1 + H2 where H1 = Π2z

2m and H2 = 12m (Π2

x + Π2y). Now

define P ′ = Πy and X ′ = Πx/(mω) then [X ′, P ′] = i~, i.e. they have the standardcommutation relations of the harmonic oscillator, and H2 = P ′2

2m + 12mω2X ′2. Therefore

the energy of H2 is just E2 = ~ω(n + 1/2). The contribution from H1 is just the freeparticle energy E1 = ~2k2

2m . Since [H1,H2] = 0, then E = E1 + E2.

37. One particle precedes to the detector unhindered, whereas the other particle passes througha region of constant magnetic field. While the neutron is in the field, it’s phase processes atfrequency ω = gn|e|B/(2mnc) (from equations 2.1.52 and 2.1.56 in the text). From classicalconsiderations, the neutron is in the magnetic field for l/v seconds, or t = mλl/h. The thetotal phase difference is

mλlgn|e|B2mnc

Now whenever this phase difference changes by 2π, the count rate will be the same, so thechange in B needed to give successive maxima is:

2π =mnλlgn|e|

2mnhc∆B (15)

∆B =4πhc

|e|gnλl(16)

Sam Pinansky

November 24, 2003

Average score: 27.8/30. Common errors were forgetting that α and β are complex, and missinga subtle sign issue in 2b.

1. We’ve all done the first part a ton of times now, I’ll just quote the answers: Eigenvaluesλ = ±1, with eigenvectors:

|1〉 =1√2

|−1〉 =1√2

(There is an arbitrary choice of phase here...). Now if you are in a state(

), then the

probability of being spin up in the y direction is:

∣∣∣∣(

1 −i) (

)∣∣∣∣2

=12− Im(αβ∗) (4)

assuming that the initial state is normalized.

2. (a) First, note that U = A(A†)−1 where A = a0 + iσ · a, since σ†i = σi, and also note that(A†)−1 = (A−1)†, provable direcetly by taking that dagger of AA−1 = I. Since thenU = A(A−1)†, then U† = A−1A†. What is A−1? Well note that:

AA† = (a20 + |~a|2)I (5)

A−1 =A†

a20 + |~a|2 (6)

So then

UU† = A(A−1)†A−1A† (7)

=AAA†A†

(a20 + |~a|2)2 (8)

=AA†

a20 + |~a|2 (9)

= I (10)

So U is unitary. To show that it is unimodular, note that det(U) = det(A) det((A†)−1) =det Adet A† , now all we need is:

det A = a20 + |~a|2 = det A† (11)

after a simple calculation. So therefore det(U) = 1.

(b) For this part we need an explicit expression for U . Cranking through the algebra, youarrive at:

a20 + |~a|2

(a20 − |~a|2 + 2ia0a3 2a0a2 + 2ia0a1

−2a0a2 + 2ia0a1 a20 − |~a|2 − 2ia0a3

Now comparing with Eq. 3.2.45, we immediately see from the real part of the first entrythat:

φ = 2 cos−1

(a20 − |~a|2

a20 + |~a|2

Now equating the imaginary parts of the first component, we see that:

−nz sin(

a20 + |~a|2 (14)

Now we calculate what the sin is (note the ± from the square root... many people missedthis)

√1− cos2

= ± 2a0|~a|a20 + |~a|2 (16)

Inserting this into the above, and solving for nz:

nz = ∓ a3

|~a| (17)

Similarly, you can show that:

ny = ∓ a2

|~a| (18)

nx = ∓ a1

|~a| (19)

So simply n = ∓a, where the sign is determined by the sign of sin(φ/2).

3. (a) In the limit A → 0 (we will always list e− and e+ in that order when operators or ketsare next to each other):

Hχ+χ− =eB

mc(SzI − ISz)χ+χ− (20)

=eB~mc

χ+χ− (21)

so it IS an eigenstate, with eigenvalue E = eB~mc .

Hχ+χ− = A(SxSx + SySy + SzSz)χ+χ− (22)

4(χ−χ+ + χ−χ+ − χ+χ−) (23)

(This is most easily calculated by expression Sx = (S−+ S+)/2 and Sy = i(S−−S+)/2).Thus it is not an eigenstate, and has an expectation value −A~2

4. (a) Note that since all three elements Sz, Sz−~ and Sz +~ are diagonal, so they all commute.Now notice that

Sz |0〉 = 0 (25)(Sz − ~) |1〉 = 0 (26)

(Sz + ~) |−1〉 = 0 (27)

So then Sz(Sz + ~)(Sz − ~) |i〉 = 0 where |i〉 is any eigenstate of z. Since any ket isexpressible in terms of this basis, then Sz(Sz + ~)(Sz − ~) |ψ〉 = 0 for any state, thus ismust be the zero matrix.

(b) Rotate your axis such that x goes to z. Then Sx → Sz, and Sx(Sx + ~)(Sx − ~) =RSz(Sz + ~)(Sz − ~)R−1 where R is the rotation matrix. But of course R0R−1 = 0, so itis still simply the zero matrix.Note that in general, given any matrix A:

(A− ai) = 0 (28)

where ai are the eigenvalues of A. This theorem has some name I forget.

Problem

a Yes l since x y and z can all be expanded in Y ml s with l More

explicitly

where crp

b From we get

cfr i p

i pY Y

and the relative probabilities for measuring m are

c H E where

is an eigenstate of L corresponding to eigenvalue ll l Thus usinghr rfr

V rh Eh or V r E

Problem

Thus hlmjLxyjlmi since L change jlmi to jlm iSince L

x Ly L L

z we get

hlmjLxjlmi hlmjL

yjlmi ll m

On the other hand we have hlmjLxjlmi hlmjL

yjlmi and in fact

hlmjfLxjlmi hlmjfLyjlmi

for any function f since we can write

Ly eiLzLxeiLz ie fLy eiLzfLxe

and eiLzjlmi eimjlmiSemiclassically we have a vector of length

pll rotating around the zaxis with

the zcomponent being of length m The average values of the x and ycomponentsare zero but the projection of the vector on the xyplane a constant length equalpll m ie precisely eq

Problem

In principle this problem can be solved by using the rotation matrices relatedto the irreducible j representation of the rotation group in much the same wayas done for j in the texteq and in problem We have that a ketjlmi is rotated to

ji DRjlmisuch that

hlmji hlmjDRjlmi Dlmm

where are the Euler angles associated with the rotation So the probabilityamplitude for nding ji in state jlmi is simply Dl

mm In general thismatrix is complicated but in the special case where m and we knowSakurai that the matrix elements are simply

Dlm Y m

In our case we have in addition plain rotation about the yaxis ie thematrix elements become for l m r

sin cos

However this method will only work for m and it quite instructive to use thedenition of rotation of the wave function This method will work for any m andin fact for any wave function x

x x x Rx or x R x

In our case we know that x only depends on the direction x n of vector x Infact

The rotation R is a rotation with about the yaxis It rotates

nx ny nz nx ny nz cos nx sinnz ny cos nz sinnx

Thus from we get

cosnz sin nx

where nz cos and nx cos sin in spherical coordinates We know that thisfunction can be expressed in terms of Y m

s and after a little calculation one obtains

cos sin sin cos cos sin cos cos

sin cos

In a more systematic but in this case more tedious approach one can of coursedetermine the coecients of Y m

in the expansion of by projecting Y m on

cmjmi cm hmji Z

sin dd Y m

Problem

This problem is just intended to ll out the details of The onlysubtlety is Since J

y Jy expanding the exponential we have evenpowers of Jy which all give J

y expect power just giving I the identity matrixand odd powers of Jy which just give the matrix Jy The coeents in front of thepowers of Jn

y will be the standard coecients of the exponential function We thusget I from the rst term in the power expansion iJy sin from all the odd powersand J

y cos from all the even powers except power zero

expiJy I iJy sin Jy cos

Sam Pinansky

December 3, 2003

15. (a) Converting into spherical coordintes:

ψ(x) = rf(r)(sin θ(cosφ + sin φ) + 3 cos θ) (1)

Now using the expression for the spherical harmonic functions

Y 01 =

√34π

cos θ (2)

Y ±11 = ∓

√38π

sin θ(cos φ± i sin φ) (3)

we find by inspection that the angular part of the wave function is:√

3((i− 1)Y 1

1 + (i + 1)Y −11 ) +

√12πY 0

Since all three spherical harmonic functions are l = 1 functions (we didn’t need any fromany other set, for instance a constant term would have left us with a Y 0

0 part) the wholefunction is clearly also an eigenfunction of L2 with eigenvalue 2~2.Note that there were many other ways of doing this problem, for instance operating onthe wavefuction directly by L2 (either in polar OR square coordinates) and see what youget. I chose this method since the results were necessary for the next part anyway.

(b) From the previous result, we see that m = −1, 0, 1 with different weights. We need tonormalize things to get the probabilities:

∣∣∣∣∣

√2π

3(1− i)

∣∣∣∣∣

12π|2 = 12π (6)

So divide bother results by 44π/3 and you see that P (m = ±1) = 1/11 and P (m = 0) =9/11.

(c) The point here is to notice that you have the wave function now in terms of its angularand radial variables:

ψ(x) = rf(r)Y ml (θ, φ) (7)

Now we can take the Schroedinger equation in spherical coordinates:

− ~2

(r2 ∂ψ

]+ V (r)ψ = Eψ (8)

Now we can operate the L2 on the ψ (just gives a l(l + 1)~2 = 2~2 factor) and divideout by the angular part, leaving a one-variable differential equation for ψ(r). But in theproblem we are given ψ(r), so we can just solve for V (r) algebraically:

V (r) = − ~2

1rf(r)

[rEf(r) +

dr2(rf(r))

18. There are many, many ways of doing this problem. I will choose to utilize the formula alreadyderived in the book for the rotation of the spherical harmonics, Eq. 3.10.20. So therefore sincethe initial state is Y 0

2 , the rotated state would be

Y 02′=

m′Y m′

2 d(2)mm′(β) (10)

Thus, simply by the nature of this formula, we know that the probability of being in the |2,m〉state is (

d(2)0m(β)

We have a nice (ok, hideous) formula for these, Eq. 3.8.33. Using this formula, you calculatethat (after a bunch of trig identities):

d(2)00 (β) =

14(1 + 3 cos(2β)) (12)

d(2)0±1(β) = ∓

sin β cosβ (13)

d(2)0±2(β) =

sin2 β (14)

Squaring, we find the probabilities:

m = 0 :116

(1 + 3 cos(2β))2 (15)

m = ±1 :32

sin2 β cos2 β (16)

m = ±2 :38

sin4 β (17)

You can show that adding all these together gives 1, plus notice the limits: β = 0 gives thatm = 0 of course, β = π is 1/4 in m = 0, and 3/4 in |m| = 2 (with no m = ±1). And whenβ = π you’re back to where you started.

20. Nothing complicated here, just simple computation using S± |j,m〉 =√

(j ∓m)(j ±m + 1)~ |j, m± 1〉.Start with the highest state, where there’s only one possible choice:

|2, 2〉 = |+,+〉 (18)

Now apply the S− operator to get |2, 1〉 (don’t forget to normalize!)

|2, 1〉 =1√2(|0,+〉+ |+, 0〉) (19)

and we continue in this manner:

|2, 0〉 =1√6(|−, +〉+ 2 |0, 0〉+ |+,−〉) (20)

|2,−1〉 =1√2(|−, 0〉+ |0,−〉) (21)

|2,−2〉 = |−−〉 (22)

Now to find |1, 1〉, we know it must be a linear combination of |+, 0〉 and |0, +〉, and orthogonalto all the other states so far. The orthogonality with |2, 1〉 therefore forces it to have this form:

|1, 1〉 =1√2(|+, 0〉 − |0, +〉) (23)

Now we go back to applying S− again:

|1, 0〉 =1√2(|+,−〉 − |−, +〉) (24)

|1,−1〉 =1√2(|0,−〉 − |−, 0〉) (25)

And of course you can’t forget the |0, 0〉 state. In general it can have the form a |+,−〉 +b |−,+〉+ c |0, 0〉, but orthogonality with the |1, 0〉 and |2, 0〉 states restrict a = b and c = −a,respectively. So therefore:

|0, 0〉 =1√3

(|−,+〉+ |+,−〉 − |0, 0〉) (26)

And there you have it.

21. (a) The trick here is to use equation 3.5.51 and to rewrite the quantity in terms of matrixelements of operators. Trying to use 3.8.33 is not tenable. Therefore:

m=−j

|d(j)mm′(β)|2m =

〈j,m′| e− iJyβ

~ |j, m〉〈j,m| eiJyβ

β |j, m′〉m (27)

(m |j,m〉〈j, m|) eiJyβ

~ |j, m′〉 (28)

Note that the sum is simply the operator Jz:

~ JzeiJyβ

~ |j, m′〉 (29)

Now we just need to rotate the operator Jz. You can do this with the Baker Hausdorfflemma, like in the chapter. I’ll simply note that J is a vector, and so transforms like one:

= 〈j,m′| (sin βJx + cos βJz) |j, m′〉 (30)= 〈j,m′| cos βJz |j, m′〉 (31)

m=−j

|d(j)mm′(β)|2m = m′ cosβ (32)

Checking this against Eq. 3.5.52 shows it holds.

(b) The same manipulations apply as in the last part, except now you get J2z instead. So:

m=−j

m2|d(j)mm′(β)|2 = 〈j, m′| e− iJyβ

~ J2z e

~ |j, m′〉 (33)

= 〈j, m′| (sinβJx + cos βJz)2 |j,m′〉 (34)

= 〈j, m′| (sin2 βJ2x + 2 sin β cosβ(JxJz + JzJx) + cos2 βJ2

z ) |j,m′〉(35)

= sin2 βj(j + 1)−m′2

2+ m′2 cos2 β (36)

=j(j + 1)

2sin2 β + m′2 1

2(3 cos2 β − 1) (37)

(Note that both cross terms just give 0).

24. (a) First, some notation: Denote the eigenstates of Sz with eigenvalue −~/2, ~/2 as |0〉, |1〉respectively, and the eigenstates of Sx with eigenvalue −~/2, ~/2 as |−〉, |+〉 repectively.Then the initial state of the particle is

|10〉 − |01〉√2

Clearly, the probability of measuring the first particle in the state |1〉 is clearly 1/2. Usingthe fact that

|1〉 =|+〉+ |−〉√

|0〉 =|+〉 − |−〉√

Rewriting the initial state of the system

− |+−〉+ |−+〉√2

So clearly the probability of measuring the first particle in the state |+〉 is clearly 1/2 aswell.

(b) After the measurement by B, the state of the particle is simply:

− |01〉 (42)

So clearly observer A will find the particle is always spin down when measuring Sz, andwill find it in the state |+〉 with probably 1/2. Notice that B measuring in the z directiondoesn’t change the measurement in the x direction by A. That’s because the direction isorthogonal.

Problem

a In principle the construction of Tq from Uq and Vq is given by where

Uq and Vq are the spherical tensor with components of vector U is given by

U pUx iUy U Uz

and similarly for vector V According to we have

U V x iU V y

pUVz UzV

ipU V z

ipUxVy UyVx

pUV UV

b The construction in terms the irreducible components of U and V is again givenby We have

UxVx UyVy iUxVy UyVx

UxVz UzVx iUyVz UzVy

UzVz UxVx UyVy

Problem

a x is an vector operator with standard spherical components

V px iy V z

We are asked to relate the matrix elements of this vector operator According toWE we have

l l and m m m m for V

Strictly speaking WE also allows l l It is ruled out by the fact that the vector operator x has odd parity Let us assume that the reduced matrix elementhnljjV jjlni is dierent from zero Then WE gives

hlm jl m ihlm jl mi

hlmjl m ihlm jl mi

where we assume m is in an appropriate range not leading to the division with zeroThese relations imply that for given n n l l all matrix elements hn l mjVqjn lmican be expressed in terms of known CG coecients and one of the matrix elementsfor instance hn l l jVjn l li assuming it is dierent from zero since WEalso gives

hl m q jVqjl m i hlm q qjl m q ihlm q qjl m qi hl m qjVqjl mi

In order to make it explicit one has of course to calculate or nd in other waysthe CG coecients

b The only additional information we obtain by using the explicit wave functionsfor jnlmi is that we can calculate the matrix elements if we can do the integralsinvolved and in this way obtain the absolute value In particular we can check ifsome of the reduced matrix elements are zero Since Vq can be expressed in termsof Y m

by we can use to write explicitly

hn l mjVqjn lmi F n l n l hl jl i hlm qjl mi

F n l n l

Zrdr Rnlr r Rnl

Note as already remarked in the book that this explicit calculation reproduces asit should the WE theorem in the case of a vector operator Note also that some ofthe indices of CG coecients are interchanged in order to agree with the notationof WE One has

hjjm mjjmi jjjhjjm mjjmiwhich showns that the matrix element hn l mjVqjn lmi is zero unless l l in agreement with the parity convention for Y m

l and the parity of x

Problem

a Let us introduce as usual the notation Vq for the spherical components of vectorx We have using

iV VV V

xz pV VV

xy V Vp

b We have again according to and the discussion afterwards

z r p T

WE gives

h j j jT j j ji

h j jjT j j ji hj jjj ji h jjjT jj ji

h j jjT j j ji hj j jj ji h jjjT jj ji

Thus we have that only for m j is the matrix element of x y dierentfrom zero and

eh j jjxyjj ji Qp

hj jjj jihj j jj ji

Problem

a A implies H eBmc

is an eigenfunction for H corre

sponding to eigenvalue eBmc

Acting on

with ASe Se results in the

which shows that e

is not an eigenvector of H for B hHi is obtained asthe scalar product of vectors and

Problem

For any N N hermitian matrix A with eigenvalues i

because every eigenvector jvii to A satis es M jvii Since any vector can beexpanded on the eigenvectors of A the matrix M must be zero Sz and Sx both haveeigenvalues and

Problem a

Let the state jvi be given by the spinor

We assume as usual that the state is

normalized jaj jbj Further the expectation value of the spin matrices areclearly unchanged if we multiplied jvi by a phase factor Thus our vector can onlybe determined up to a phase and we can assume that Re a

ab ba hSyi i

ba ab hSzi

jaj jbj

or if a

hSxi a Re b hSyi a Im b hSzi a

If a we have already that jvi ji up to a phase Thus

hSzi Re b

Im b pjbj Re b

p jaj Re b

or expressed entirely in terms of expectation values of Sx and Sz

hSzi Re b

Im b s

hSzi hSxi

where we should choose the plus sign since we know that hSyi and a are positive

hSyi is known up to a sign since the three components hSxi hSyi hSzi transform likean ordinary vector under rotations Thus the norm is preserved under rotations andif we rotate jvi to ji the vector hSxi hSyi hSzi rotates to with norm Thus

hSyi r

hSxi hSzi

Problem

We want to prove that Gi Gj iijk Gk Let us consider the l nth entry of thismatrix eequation

Gi Gjln iijk

or in terms of the ijk symbols

iilmijmn ijlmiimn iijkiklnThe relation follows from the identity

ijkilm jlkm jmkl

The matrices Gi are the generators for rotations around the coordinate axis xi asshould be clear from the lectures expressed in the standard Cartesian coordinatesystem where

A Vjxi Vjxi Vjxi

Indeed we can with write

V V n V V j

I i ni Gi

The eigenvalues for G are and with corresponding eigenvectors

pjxi ip

jxi ji jxi

A coordinate change from basis vectors jxii to basis vectors ji willchange the matrix G into the diagonal matrix J with diagonal elements and More explicitly we can write in braket notation for the abstract operator Gcorresponding to G and J

hxij Gjxji hxijihj Gj ih jxji

where we have summation over The matrix J has entries hj Gj iand the transformation matrix Ti i which connects G and Jhas entries hxiji Thus reads in ordinary matrix notation

Gij TiJTyj

The above change of vectors

jxi jxi ji ji

is similar to the superposition of photons travelling in the x direction and linearpolarized in x and x directions into right and left circularly polarized photons seeSakurai formula It is well know that even classically properly normalizedright and left circularly polarized plane waves of light carry an angular momentumof one unit in the direction of propagation We see that this classicalinterpretation of plane electromagnetic waves has many similarities to the quantum

interpretation of the photon as a spin particle Read Sakurai sec in particular

Solutions to Assignment 1.

1. (a) To construct an eigenket of τ~a, we take the combination

|~k〉 =∑

e−i~k·~r|~r〉, (1)

where ~k = (kx, ky, kz). Now

τ |~k〉 =∑

e−i~k·~rτ~a|~r〉

e−i~k·~r|~r + ~a〉

=∑~r′e−i~k·(~r′−~a)|~r〉

= ei~k·a|~k〉. (2)

(b) The action of H on the state |r〉 is

H |r〉 = Eo|~r〉 −∆∑

~a=(x,y,z)

[|~r − ~a〉+ |~r + ~a〉] (3)

so that the action of H on |~k〉 is

H |~k〉 =∑

e−i~k·~rH |~r〉

e−i~k·~r

Eo|~r〉 −∆

∑~a=(x,y,z)

[|~r − ~a〉

e−i~k·~r

Eo −∆

∑~a=(x,y,z)

(ei~k·~a + e−i~k·~a)

|~r〉

= E(~k)|~k〉. (4)

E(~k) = Eo − 2∆∑

~a=(x,y,z)

cos(~k · a)

= Eo − 2∆(cos kx + cos ky + cos kz) (5)

is the corresponding energy eigenstate.

2. (a) Since momentum operators always commute, any function of these operators also commutes, so that

[τ~d, τ~d′ ] = [e−i~P ·~d/h, e−i~P ·~d′/h] = 0 (6)

Translation operators commute.

(b) Rotations about different axes do not commute, so that

[D(n, φ), D(n′, φ′)] 6= 0 (7)

(c) The inverstion operator reverses the direction of all translation, so that

πτ~dπ−1 = τ−~d (8)

Consequently, the inversion operator does not commute with the translation operator.

[π, τ~d] 6= 0. (9)

(d) Under the inversion operation, angular momentum operators are invariant, π ~Jπ−1 = ~J so that [π, ~J ] = 0.Consequently, the inversion operation commutes with functions of the angular momentum operator, andthus commutes with the rotation operator.

[π,D(R)] = 0. (10)

3. Sakurai problem 9. When we time reverse a momentum eigenstate, we reverse the sign of the momentum, inaddition to complex conjugating the state. We therefore expect that the time reversal of φ(p) is φ(−p)∗. Toshow this explicitly,

〈p|Θ|α〉 = 〈p|Θ(∫

dDp′|p′〉φ(p′))

= 〈p|∫dDp′Θ|p′〉φ∗(p′)

= 〈p|∫dDp′| − p′〉φ∗(p′)

=∫dDp′

δ(D)(p+p′)︷︸︸︷〈p| − p′〉 φ∗(p′) = φ∗(−p) (11)

4. Sakurai problem 12. We can rewrite the matrix as

H = AS2z +

+ + S2−] (12)

where S± = Sx ± iSy. Written out explicitly for S = 1 we have

H ≡A 0 B

0 0 0B 0 A

where I have taken h = 1. Taking det[E1−H ] = E((E −A)2 −B2) we see that the energy eigenvalues are

E = A±B, 0 (14)

The corresponding eigenkets are

|±〉 =|+ 1〉 ± | − 1〉√

2, (E = A±B)

and for E = 0, |0〉 = |ms = 0〉.The Hamiltonian is invariant under time-reversal, since Θ~SΘ−1 = ~S is unchanged by time-reversal. SinceΘ|mJ〉 = (i)2mJ | −mJ 〉, we have

Θ|±〉 = ∓|±〉, Θ|0〉 = |0〉, (16)

i.e the lower and upper eigenstates are odd-parity under time reversal, whereas the central state is even-parityunder time-reversal.

5. Sakurai, chapter 4, Q 6. This is a tricky problem. There are two ways you could do it: (i) solving the completeproblem but to exponential accuracy or (ii) by directly calculating the matrix elements between the states onthe left, and right hand side. I shall illustrate method (ii). To begin, let us consider the problem when thelength a is infinitely large. In this case, the wavefunction for the left, and right hand ground-states are

ψR(x) = 〈x|ψR〉 =

0 (x > a+ b)A sin[k(a+ b− x)] (a < x < b)Beκx (x < a)

ψL(x) = 〈x|ψL〉 =

0 (x < −a− b)A sin[k(a+ b+ x)] (−b < x < −a)Be−κx (x > −a)

where κ =√

2mh2 (Vo + E) ≈

√2mh2 Vo.

Now the tricky bit is that we need to construct orthogonalized wavefunctions. To do this, we construct

|ψR〉 =1

[1− |〈ψL|ψR〉|2] 12[|ψR〉 − |ψL〉〈ψL|ψR〉]

|ψL〉 = |ψL〉 (18)

These states are now orthogonal and normalized.

We shall now approximate the complete wavefunction in the form

|ψ〉 = αR|ψR〉+ αL|ψL〉 (19)

Applying the Hamiltonian to this expression, and demanding that H |ψ〉 = E|ψ〉, we obtain the eigenvalueequation Habαb = Eαb, (a, b ∈ R,L), where

Hab ≡[ 〈ψR|H |ψR〉〈ψR|H |ψL〉〈ψL|H |ψR〉〈ψL|H |ψL〉

]. (20)

To evaluate this matrix, it is helpful to realize that the complete Hamiltonian can be written

H = HR + VL = HL + VR (21)

where HL is the Hamiltonian for the left-hand well and HR is the Hamiltonian for the right-hand well and

VR = −Vo[θ(x− a)− θ(x − a− b)],VL = −Vo[θ(x+ a+ b)− θ(x + a)], (22)

ψ( )x

-(a+b) -a-

(a+b) x

Fig. 1.: Showing ψR(x) and the potential VL(x).

With this set-up, we note that HL,R|ψL,R〉 = Eo|ψL,R〉 , where Eo is the energy of an isolated well. If you nowcompute the matrix element 〈ψR|H |ψL〉, you obtain

〈ψR|H |ψL〉 = E〈ψR|ψL〉+〈ψR|VR|ψL〉√1− |〈ψL|ψR〉|2

=〈ψR|VR|ψL〉√1− |〈ψL|ψR〉|2

≈ 〈ψR|VR|ψL〉. (23)

In the last step, we have noted that |〈ψL|ψR〉| is exponentially smaller than unity, so that terms containingthis quantity have been dropped. The splitting between the two states is then going to be simply

±∆ = ±|〈ψR|VR|ψL〉| (24)

Now to calculate this, we need to compute the exponential tail in ψL. Applying continuity of the wavefunctionand continuity of the logarithmic derivative, we obtain

A sin kb = Beκa, k tan(kb) = −κ (25)

To leading exponential accuracy, this gives

√2b,

κbe−κa (26)

Carrying out the integral, we then obtain

〈ψR|VR|ψL〉 = −Vo

∫ a+b

sin[k(a+ b− x)]Be−κx

= −Vo

κe−κ(2a+b)

) ∫ b

dx sin[kx]eκx

= −Vo

κe−κ(2a+b)

) ∫ b

dxIme(κ+ik)x

≈ −Vo

κe−κ(2a)

≈2ko/κ︷︸︸︷[eikb

κ+ ik

≈ −Vo

)e−κ2a

(h2π2

)e−2κa (27)

The splitting between the two levels is then

∆E = 2∆ =h2

mκb3e−2κa (28)

where κ =√

2mh2 Vo for large Vo.

(a+b) x

-(a+b) -a a

(a+b) x

-(a+b) -a

ψ + ψ / 2R L

ψ − ψ / 2R L

Fig. 2.: Showing the even and odd wavefunctions for the symmetric potential well.

Physics 512 Winter 2003

Homework Set #5 – Solutions

1. This problem is essentially Merzbacher, Chapter 18, Problem 4 [or Sakurai, Chapter 5,Problem 1]. Consider a one-dimensional harmonic oscillator perturbed by a constantforce

12mω2x2 − Fx

a) Show that the first order perturbation in the energy levels vanishes

Since we will need to find the second order perturbation in part b), let us considerthe general matrix element

Vmn = −F 〈m(0)|x|n(0)〉

The operator x is given in terms of creation and annihilation operators by x =√h/(2mω)(a† + a). Thus

Vmn = −F√

2mω〈m(0)|(a† + a)|n(0)〉

= −F√

2mω[√n+ 1 δm,n+1 +

√n δj,n−1]

This indicates that the x operator changes the oscillator number by ±1. In par-ticular, Vnn = 0, which demonstrates that the first order energy perturbationvanishes.

b) Now calculate the eigenenergies En up to second order in the perturbation.

Using the matrix element calculated above, we find

E(2)n =

∑k =n

|Vkn|2E

(0)n −E

=|Vn+1,n|2

E(0)n −E

(0)n+1

+|Vn−1,n|2

E(0)n − E

(0)n−1

= F 2 h

[n+ 1−hω +

]= − F 2

Combined with the zeroth order energy, this gives

En = (n+ 12 )hω − F 2

c) Show that the second-order perturbation result gives the exact eigenenergies(which may be obtained by completing the square in H). Explain why thishappens.

By completing the square, we find

12mω2

(x− F

− F 2

12mω2x2 − F 2

where x = x− F/(mω2) is a shifted coordinate. In terms of this new coordinate,we have an ordinary harmonic oscillator, however with a constant offset in thepotential. We easily read off the eigenenergies as

En = (n+ 12 )hω − F 2

2mω2(1)

which is identical to the perturbation result of b). Of course, in general we wouldapply perturbation theory only when we do not know the exact answer. How-ever, in this toy example, after completing the square, we see that F enters theHamiltonian only quadratically as an overall offset. In particular, we know thatthe correct energies, (1), only has F 2, and no other powers of F in it. Now re-call that each order in the perturbation theory corresponds to precisely that orderin a series expansion of the small perturbation parameter (which would be F inthis case); first order perturbation gives O(F ), second order gives O(F 2), and soon. As a result, only the second order perturbation to the energy can contributetowards the correct answer, and this is in fact what happens.

2. Sakurai, Chapter 5, Problem 2. In nondegenerate time-independent perturbationtheory, what is the probability of finding in a perturbed energy eigenstate |ψn〉 thecorresponding unperturbed eigenstate |ψ(0)

n 〉? Solve this up to terms of order g2.We recall that the perturbed eigenstate is not necessarily normalized, and may bewritten as

|ψn〉 = |ψ(0)n 〉 + g|ψ(1)

n 〉 + g2|ψ(2)n 〉 + · · ·

Since the probability is the amplitude squared, we want to calculate (to second orderin g)

∣∣∣∣∣ 〈ψ(0)n |ψn〉√〈ψn|ψn〉

∣∣∣∣∣ = |〈ψ(0)n |ψn〉|2

〈ψn|ψn〉

Since all the higher order |ψ(i>0)n 〉 states are orthogonal to |ψ(0)

n 〉, it is easy to see that

〈ψ(0)n |ψn〉 = 〈ψ(0)

n |ψ(0)n 〉 = 1

and〈ψn|ψn〉 = 〈ψ(0)

n |ψ(0)n 〉 + g

(〈ψ(0)

n |ψ(1)n 〉 + 〈ψ(1)

n |ψ(0)n 〉)

+ g2(〈ψ(0)

n |ψ(2)n 〉 + 〈ψ(1)

n |ψ(1)n 〉 + 〈ψ(2)

n |ψ(0)n 〉)

+ · · ·

= 1 + g2〈ψ(1)n |ψ(1)

n 〉 + · · · = 1 + g2∑k =n

|Vkn|2|E(0)

n − E(0)k |2

+ · · ·

Using 1/(1 + x) = 1 − x+ · · ·, we find

P = 1 − g2∑k =n

|Vkn|2|E(0)

n − E(0)k |2

+ · · ·

which we verify is less than or equal to one. Clearly, the only possibility of finding100% probability is if |ψn〉 is identical to |ψ(0)

n 〉, indicating that the initial eigenstateremains an eigenstate of the perturbation.[Yes, this is the same as Wave-function Renormalization on p. 293 of Sakurai!]

3. Consider the 2p levels of hydrogen (m = 0,±1) subject to a perturbation

V = Ax2 + By2 − (A+B)z2

a) Write V in terms of components of a rank-2 spherical tensor.Using the spherical harmonic ‘trick’, we can write out the appropriate rank-2tensor as

T(2)±2 = 1

2(x± iy)2

T(2)±1 = ∓(x± iy)z

T(2)0 = − 1√

6(x2 + y2 − 2z2)

Of course, the actual normalization is arbitrary. I have removed a factor of√15π/8 when compared to the Y m

l=2 spherical harmonics. In this case, we mayreexpress

x2 − y2 = T(2)2 + T

(2)−2 , x2 + y2 − 2z2 = −

√6T (2)

This allows us to rewrite the perturbation

V = 12 (A+ B)(x2 + y2) + 1

2(A− B)(x2 − y2) − (A+ B)z2

= 12(A+ B)(x2 + y2 − 2z2) + 1

2(A− B)(x2 − y2)

= −√

32 (A+ B)T (2)

0 + 12 (A− B)(T (2)

2 + T(2)−2 )

Note that the coefficients of x2, y2 and z2 had to work out just right for this tobe possible. If instead of (A + B)z2, we had something like Cz2, then we wouldhave an added term left over which would be proportional to r2 = x2 + y2 + z2.This would correspond to a scalar perturbation in addition to a pure rank-2 tensorperturbation.

b) Neglecting electron spin, find the “correct” zeroth-order eigenstates and theircorresponding energies. It is sufficient to give the energies in terms of a reducedmatrix element 〈2p||T (2)||2p〉.We denote the 2p states as |nlm〉 = |211〉, |210〉 and |21 −1〉. We first observethat the perturbation will not mix |210〉 with the other two states [because of the

∆m = 0,±2 selection rule evident from (1)]. So, although all three states aredegenerate, the perturbation is block diagonal in the |210〉 and |211〉, |21 −1〉subspaces. This simplification allows us to immediately write down, for the |210〉state

∆E210 = 〈210|V |210〉 = −√

32 (A+B)〈210|T (2)

0 |210〉

= −√

32 (A+B)〈1200|1210〉〈2p||T (2)||2p〉

35 (A+ B)〈2p||T (2)||2p〉

For the remaining two states, we write out the matrix

V =( 〈211|V |211〉〈211|V |21 −1〉〈21 −1|V |211〉〈21 −1|V |21 −1〉

= −√

32 (A+B)

( 〈211|T (2)0 |211〉 00 〈21 −1|T (2)

0 |21 −1〉

+ 12(A− B)

(0 〈211|T (2)

2 |21 −1〉〈21 −1|T (2)

−2 |211〉 0

= −√

32(A+B)〈2p||T (2)||2p〉

( 〈1210|1211〉 00 〈12 −10|121 −1〉

+ 12(A− B)〈2p||T (2)||2p〉

(0 〈12 −12|1211〉

〈121 −2|121 −1〉 0

√35〈2p||T (2)||2p〉

(−(A+B) (A− B)(A− B) −(A+B)

Working out the eigenvalues and eigenstates, we find a simple result

∆E21x = −√

35A〈2p||T (2)||2p〉 |21x〉 ≡ 1√

(|211〉 − |21 −1〉)∆E21y = −

√35B〈2p||T (2)||2p〉 |21y〉 ≡ 1√

(|211〉 + |21 −1〉)Here, the labels x and y were chosen to be suggestive of px and py orbitals. How-ever, of course, this is just an arbitrary labeling of the eigenstates. To summarize,the perturbation splits all three 2p states (ignoring spin) to

∆E210 =√

35(A+ B)〈2p||T (2)||2p〉, |210〉

∆E21x = −√

35A〈2p||T (2)||2p〉, |21x〉 = 1√

(|211〉 − |21 −1〉)∆E21y = −

√35B〈2p||T (2)||2p〉, |21y〉 = 1√

(|211〉 + |21 −1〉)Note that the energy shifts are proportional to A, B and −(A + B), the factorsmultiplying x2, y2 and z2, respectively. The corresponding eigenstates, |21x〉,

|21y〉 and |210〉 are simply the px, py and pz orbitals (which are aligned along thex, y and z axis). So this makes perfect physical sense.

c) Show that the expectation value of Lz vanishes in each eigenstate found above.Of course the expectation values of Lz take on the values 0,±h in the originalbasis states |210〉 and |21 ±1〉. Here, however, we consider

〈210|Lz|210〉 = 0

〈21x|Lz|21x〉 = 12

(〈211| − 〈21 −1|)Lz

(|211〉 − |21 −1〉)= 1

(〈211|Lz|211〉 + 〈21 −1|Lz|21 −1〉) = 0

〈21y|Lz|21y〉 = 12

(〈211| + 〈21 −1|)Lz

(|211〉 + |21 −1〉)= 1

(〈211|Lz|211〉 + 〈21 −1|Lz|21 −1〉) = 0

so that the expectation values indeed vanish. It is important to keep in mind thatthe expectation values are not the same thing as eigenvalues! The eigenvalues ofLz are always 0,±h for the 2p states.

4. Merzbacher, Chapter 18, Problem 14.For the Hamiltonian H = AL 2 + Bh2 cos 2ϕ with A B, we may take the secondterm as a perturbation. Hence

H0 = AL 2 = Ah2(+ 1)

for states of angular momentum . For the perturbation, we find it convenient to write

V = Bh2 cos 2ϕ = 12Bh

2(e2iϕ + e−2iϕ)

Although the perturbation breaks rotational invariance (after all, it is a hindered rota-tion), we may nevertheless consider its effect on angular momentum states |m〉. Inthis case, since Y m

(θ, ϕ) ∼ eimϕPm (cos θ), we see that

〈′m′|V |m〉 =∫

(Y m′′ )∗ V Y m

dΩ ∼∫ 2π

e−im′ϕ(e2iϕ + e−2iϕ)eimϕ dϕ× (rest)

Hence we may deduce that m′ and m obey a selection rule ∆m = ±2. This providesan important simplification, as it indicates that most of the matrix elements of theperturbation actually vanish. We are now ready to examine the S, P and D levels oneat a time.

For the S level ( = 0)This state is non-degenerate, and has vanishing zeroth order energy, E (0) = 0.In addition, the first order shift in the energy also vanishes since 〈00|V |00〉 = 0by the ∆m selection rule. We thus have, for the S level

E = 0, |00〉

Note that, while the energy vanishes up to first order, |00〉 is not an eigenstate ofH.

For the P levels ( = 1)

At zeroth order, the P levels are three-fold degenerate with E (0) = 2Ah2. For thefirst order perturbation, use of the ∆m selection rule indicates that |10〉 cannotmix with the other states. Hence it is already the “correct” eigenstate, and doesnot get shifted in energy

E = 2Ah2, |10〉For the remaining |11〉 and |1 −1〉 states, we have to diagonalize the perturbation

0 〈11|V |1 −1〉〈1 −1|V |11〉 0

The two non-trivial matrix elements are complex conjugates of each other. So weonly need to compute

〈11|V |1 −1〉 = 12Bh

∫(Y 1

= 12Bh

(− 3

)∫sin2 θ dϕ d cos θ = −1

V = −12Bh2

(0 11 0

By now, it should be obvious what the eigenvalues and eigenvectors of this matrixare. Including the zeroth order energy, we have

E = 2Ah2 − 12Bh

2, 1√2(|11〉 + |1 −1〉)

E = 2Ah2 + 12Bh2, 1√

2(|11〉 − |1 −1〉)

We see that the perturbation completely lifts the degeneracy of the P states.

For the D levels ( = 2)

The D levels are five-fold degenerate at the zeroth order, with E (0) = 6Ah2. Ingeneral, we may have to work out a 5 × 5 matrix for the perturbation. However,the ∆m selection rule again helps, and indicates that the five states split up intotwo subspaces, one with |21〉, |2 −1〉 and the other with |22〉, |20〉, |2 −2〉. Forthe former, we have

0 〈21|V |2 −1〉〈2 −1|V |21〉 0

〈21|V |2 −1〉 = 12Bh

∫(Y 1

= 12Bh

(− 15

)∫(sin θ cos θ)2dϕ d cos θ = −1

The eigenenergies and eigenstates in this subspace are hence

E = 6Ah2 − 12Bh

2, 1√2(|21〉 + |2 −1〉)

E = 6Ah2 + 12Bh

2, 1√2(|21〉 − |2 −1〉)

Finally, turning to the three-dimensional subspace, we have

0 〈22|V |20〉 0

〈20|V |22〉 0 〈20|V |2 −2〉0 〈2 −2|V |20〉 0

We evaluate

〈22|V |20〉 = 12Bh2

∫(Y 2

2 )∗ e2iϕ Y 02 dϕ d cos θ

= 12Bh2

)∫sin2 θ(3 cos2 θ − 1)dϕ d cos θ = − 1

and note that 〈20|V |2 −2〉 gives the same result. Thus

V = − 12√

1 0 10 1 0

It is easy to check that the matrix of ones and zeros has eigenvalues√

2, 0and −√

2, with eigenvectors ( 1√

2 1 )T , ( 1 0 −1 )T and ( 1 −√2 1 )T .

Hence the correct eigenenergies and eigenstates are

E = 6Ah2 − 12√

3Bh2, 1

2 (|22〉 +√

2|20〉 + |2 −2〉)E = 6Ah2, 1√

2(|21〉 − |2 −2〉)

E = 6Ah2 + 12√

3Bh2, 1

2 (|22〉 −√

2|20〉 + |2 −2〉)

Finally, to summarize, we find, for the S, P and D levels

S : E = 0, |00〉P : E = 2Ah2 − 1

2Bh2, 1√

2(|11〉 + |1 −1〉)

E = 2Ah2, |10〉E = 2Ah2 + 1

2Bh2, 1√

2(|11〉 − |1 −1〉)

D : E = 6Ah2 − 12Bh2, 1√

2(|21〉 + |2 −1〉)

E = 6Ah2 − 12√

3Bh2, 1

2(|22〉 +√

2|20〉 + |2 −2〉)E = 6Ah2, 1√

2(|21〉 − |2 −2〉)

E = 6Ah2 + 12√

3Bh2, 1

2(|22〉 −√

2|20〉 + |2 −2〉)E = 6Ah2 + 1

2Bh2, 1√

2(|21〉 − |2 −1〉)

Note that the P states are split just as they were in problem 3).

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Problem

jti Xn

eiEntcntjni Pnt jcntj

where Pnt is the probability to nd the system in state jni at time t

Assume system is in ground state ji at t For n we have to lowest order

Pnt jcn j

cn t i

dt eint

Aet hnjxji i

hnjxjiA eint

Thus we obtain for n

Pnt Ajhnjxjij

hnjxji Xk

hnjxjkihkjxji hnjxjihjxji

It is seen that the matrix element hjxji Thus c t However a formula

like is not correct in this case since c by assumption

dtAet hnjxji i

hnjxjiA

Note that c is purely imaginary Thus there is no rst order contribution to

Pt jc c tj jc tj

and the calculation is clearly inconsistent It requires that we also include the secondorder contribution c

t such that to second order

Pt jc c t c

tj jc tj Re c

By consistency this has to agree to second order with

Pnt for potential x Pt

Check it

The reason that we do not have the same kind of problem for Pnt n is of

course that cn so is the lowest order contribution

Problem

Notation EE jsi

First order perturbation theory

dt eithkjV tjsi

hkjV tjsi

hkjUItjsi

eit cost ieit sint

h cost cost sint sint

sint cost cost sint

The term hkjUItjsi has to be small in order that we trust rst order perturbationtheory Let us consider as xed and assume we can vary an external eldwith an adjustable frequency Let us further assume that rst order perturbationtheory is reliable when is not close to ie However when the term e

in hkjUItjsi can be of order

j jand the condition might not be satised However usingsinxx

we obtain that it is satised for t

Problem

dt eiskthkjV tjsi

where Ek Es ks In our case jsi ji jki ji and

hjV tji F

thjxji

dt eit

Let us consider the classical system a particle oscillating forth and back with frequency and acted upon by an external force F

p mxt F t p

Zdt mxt F t

Zdt F t

The energy transfer by F to the harmonic oscillator is

Zdt xtF t

If F t is contant clearly no average energy is transferred to the oscillator since aconstant F just means a translation of the center of oscillation from x to x Fmand the E calculated over one period is zero Now consider a very slowly varyingF t xt will consist of a slow motion xt F tm with the fast oscillationssuperimposed and the fast oscillations will only add a small average energy If F tstarts out being zero and ends up being zero in the end only a small amount ofenergy has been transferred to the classical system provided the variation of F tis suciently slow In the quantum system which for large positive and negativetimes is just a harmonic oscillator a possible jump in energy cannot have been lessthan and the probability of this becomes smaller if the energy transfer in theclassical system becomes smaller

The above classical observations can be made precise in quantum mechanics as welland is called the adiabatic theorem Let Ht be the Hamiltonian Although it isnot stationary we can of course solve

Htjnit Entjnitfor all times t as if we had a stationy Hamiltonian The eigenvalues Ent willbe continuous functions of t and also the eigenvectors jnit can be chosen to be

continuous functions of t if the time evolution of H is sucently nice Assume thatthe system is in the eigenstate jni corresponding to En at t The adiabatictheorem states that in the limit of an innitely slow evolution of Ht the systemwill be in an eigenstate of Ht corresponding to Ent at time t ie

jti eitjnit

where t is some phase which one can try to calculate in the adiabatic approxi

mation In our case the system remains in the ground state since at time t Ht isjust a harmonic oscillation displaced from x to xF tm As jnit we canthus take the displaced eigenfunctions of the harmonic oscillators and

jnit jnit jni

where jni is the standard eigenvector of the harmonic oscillator simply becauseF t for jtj

Solutions to assignment 3.

1. (Solution to Sakurai, problem 4, ch 5) The isotropic Harmonic oscillator in two dimensions

Ho =p2

2m+mω2

2(x2 + y2). (1)

can be re-written as

Ho = hω(a†a+ b†b+ 1) (2)

a =√mω

√mω

(y + i

are the annihilation operators for modes in the x and y directions, respectively. The eigenvalues of the numberoperators na = a†a and nb = b†b are non-negative integers. The corresponding eigenkets |na, nb〉 are energyeigenkets, with energy

Enanb= hω(1 + na + nb) (4)

(a) The ground-state is then

|0, 0〉 Eo = hω (5)

and the first excited state is doubly degenerate, corresponding to |1, 0〉 and |0, 1〉, with energy E10 =E01 = 2hω.

(b) The perturbation to the Hamiltonian can be written

V = δmω2xy = δhω

2(a+ a†)(b + b†) (6)

We apply non-degenerate perturbation theory to the ground-state, to obtain

|0, 0〉′ = |0, 0〉+1

Eo −HoV |0, 0〉+O(δ2)

= |0, 0〉 − δ|1, 1〉 (7)

with energy

E′0,0 = Eoo +

0︷︸︸︷〈0, 0|V |0, 0〉+ |〈1, 1|V |0, 0〉|2

E00 − E11+O(δ3)

= E00 − δ2hω

8+O(δ3) (8)

Next we apply degenerate perturbation theory to the degenerate excited states |1, 0〉 and |0, 1〉. In thismanifold of states, the perturbation has matrix elements

Vab =δhω

[0 11 0

To zeroth order, the new energy eigenkets are the eigenkets of Vab, given by

|±〉 =1√2(|10〉 ± |01〉) (10)

with energy eigenvalues

E± = 2hω ± δhω

(c) To solve this problem exactly, we merely have to rotate our spatial axes through 45o, writing

p± =px ± py√

2, X± =

x± y√2. (12)

2m(P 2

+ + P 2−) +

+ +mω2

X2− (13)

where ω± = ω(1± δ)1/2, so that if we write

α =√mω+

(X+ + i

√mω−2h

(X− + i

p−mω−

we may cast the Hamiltonian in the form

H = hω+(n+ +12) + hω−(n− +

12) (15)

where n+ = α†α and n− = β†β. The ground-state energy is now

E00 =12(hω+ + hω−) = hω(1− δ2

8) +O(δ3) (16)

confirming (8). The excited states have one α or β quanta, and have energies

E± = Eo + hω± =12(hω+ + hω−) + hω± (17)

To leading order in δ, this gives

E± = (2 ± δ

2)hω (18)

confirming (11).Let us now construct the new ground-state in terms of the unperturbed ground-state. We first note that

α|0, 0〉′ = β|0, 0〉 = 0 (19)

With a bit of work, by rewriting (14) in terms of the original creation and annihilation operators, youmay confirm that

α = u+

(a+ b√

(a† + b†√

β = u−

(a− b√

)+ v−

(a† − b†√

u± =12

[√ω±ω

], v± =

[√ω±ω

−√

]. (21)

Note that u2± − v2± = 1. Now to construct the new ground-state, consider the state |ψ〉 = eAa†a† |0〉. Nowsince [a, eAa†a† ] = 2Aa†eAa†a† , it follows that

(ua+ va†)eAa†a† |0〉 = (2Au + v)a†eAa†a† |0〉 (22)

so that

(ua+ va†)e−v2u a†a† |0〉 = 0 (23)

Using this result, we can satisfy conditions (19) by writing

|0, 0〉 = exp

[− v+

(a† + b†√

− v−2u−

(a† − b†√

)2]|0, 0〉 (24)

Expanding this expression to leading order, gives

|0, 0〉 = |0, 0〉 −[v+2u+

(a† + b†√

+v−2u−

(a† − b†√

)2]|0, 0〉 (25)

Now to leading order, u± = 1 + O(δ2), v± = ±δ/2 + O(δ3). Substituting into the above expression, weobtain

|0, 0〉 = |0, 0〉 − δa†b†|0, 0〉 = |0, 0〉 − δ|1, 1〉+O(δ2) (26)

which confirms the perturbative result (7). Finally, the eigenkets of the excited states are obtained bycreating quanta:

α†|0, 0〉′ =1√2(a† + b†)|0, 0〉+O(δ2)

β†|0, 0〉′ =1√2(a† − b†)|0, 0〉+O(δ2) (27)

confirming (10).

2. (a) In the degenerate manifold of p-states,

V = λ(x2 − y2) (28)

has the matrix elements

〈m|λ(x2 − y2)|m′〉 = c〈m|J2x − J2

y |m′〉 =c

2〈m|J2

+ − J2−|m′〉 (29)

where |m〉 ≡ |l = 1,m〉, and we have used the Wigner Eckart theorem to relate the matrix elements tothose of the angular momentum operator. The quantity “c” is a constant. From the above, we see thatthe two only non-vanishing matrix elements are

〈+1|V | − 1〉 = 〈−1|V |1〉 = λVo (30)

where Vo is real, due to time-reversal invariance. The zeroth order energy eigenstates are the eigenstates|ξ〉 of

〈a|V |b〉 =

0 0 λVo

0 0 0λVo 0 0

with energy E = Eo + 〈ξ|V |ξ〉+O(λ3) which gives

|+〉 =1√2(|+ 1〉+ | − 1〉), E = E0 + λV0 +O(λ2)

|−〉 =1√2(|+ 1〉 − | − 1〉), E = E0 − λV0 +O(λ2)

|0〉 E = E0. (32)

showing that the degeneracy of the p-states is completely removed.

(b) Under the time-reversal transformation,

Θ|l,ml〉 = (i)2ml |l,−ml〉 = (−1)ml |l,−ml〉 (33)

so that

Θ| ± 1〉 = −| ∓ 1〉Θ|0〉 = |0〉 (34)

which means that

Θ( |+ 1〉 ± | − 1〉√

)= ∓

( |+ 1〉 ± | − 1〉√2

Θ|0〉 = |0〉 (35)

proving that the energy eigenstates are eigenkets of the time-reversal operator.

3. Sakurai 5.17

(a) Let us write the Hamiltonian as H = Ho + CV , where

Ho = AL2 +BLz

V = Ly =(L+ − L−

The eigenkets of Ho are simultaneous eigenkets of L2 and Lz, |lm〉, with energy Elm = Ah2l(l+1)+Bhm.We can apply non-degenerate perturbation theory. Now since V is off-diagonal in the |lm〉 basis, the firstorder energy shift vanishes, and we are left with

E′lm = E

(o)lm + C2

∑m′

〈〈lm′|(

L+−L−2i

)|lm〉|2

E(o)lm − E

(o)lm′

+O(C3)

= E(o)lm + (Ch)2

(l −m)(l +m+ 1)−4Bh

+ (Ch)2(l +m)(l −m+ 1)

4Bh+O(C3)

= E(o)lm + hm

2B.+O(C3) (37)

Problem

Notation x y z

H px py pz

mx y z

z mz O

V Lx ypz zpy eB

nx ny ny

n is the ground state n the rst excited state It is three times degenerate

We denote the energy eigenstates of H as jnxnynzi jnxijnyijnzi We use thisbasis in the rst order degenerate perturbation theory

Recall that either problem or Libo

hnzjzjnzi pm

z nz pnz n

Similarly one has Libo

hnzjpzjnzi m

these formulas can readily be proven by using the a ay representation of z pz andjnzi

We want to calculate the matrix of V V

in the basis ji ji ji

and nd its eigenvalues The superscript is introduced to remind us that we justdiscuss the matrices and not the complete operators on LR

Because z has even parity and jnzi has parity nz the V has only diagonal

elements Thus we have

hnzjzjnzi Xk

hnzjzjkihkjzjnzi nz

In d dimensions the nth level of a harmonic oscillator is

times degenerate

number of distinguishable ways to put n identical balls in d boxes

y pz z py has only odiagonal elements dierent from zero since y and pz etchave odd parity and since it does not involve x px in fact only matrix elementsbetween ji and ji can be dierent from zero From and we obtain eBmc

with eigenvalues

Let us now assume that Then we obtain

E O E O

If we assume we obtain

The leading terms in these expressions are expected if we can solve exactlyfor H V and we obtain

Enxnynz nx

If we can also solve HV exactly by using thatH is rotational symmetricThe eigenfunctions can thus be chosen to be simultaneous eigenfunctions of H L

and Lx The energy Enl of such an eigenfunction jn l lxi will change as by lx ifLx is added to H In our case we have the three eigenfunctions ji ji andji They all corresponds to l why look at their x y z dependence Thus

the diagonalization of V is precisely the diagonalization of Lx in the subspace

corresponding to l and the eigenvalues are

Problem

V eE z z is the zeroth component of a vector operator with odd parity ThusWignerEckart and parity tell us that of matrix elements hnlmjV jnlmi we have tohave m m and l l The energy levels depend only on the radial quantumnumber n for the hydrogen atom in the approximation mentioned in the Problemand the degeneracy is n ignoring spin In rst order degenerate perturbationtheory we thus have to diagonalize the nn matrix hnlmjV jnlmi However dueto the conservation of the quantum number m mentioned above which is of coursea consequence of rotational symmetry around the zaxis dierent mvalues will notmix and it will be blockdiagonal afterm We can thus treat eachm value separatelyn implies l spd The matrix splits in blocks

m only one state ji No linear Stark eect because of parity

m same remarks

m Double degenerate ji and ji Again diagonal elements are zerobecause of parity We thus obtain a hjzji

with eigenvalues and eigenvectors

E eE jaj jvi p

m Same as with obvious changes Note that the matrix elements arethe same as in except a a since Y m

l mY m

m Triple degenerate ji ji and ji but only the mixing mentionin the start b hjzji and c hjzji We thus obtain matrix

b b c c

with eigenvaluesE E

pjbj jcj

and the corresponding eigenvectors

jvi pjbj jcjc b

ppjbj jcj

bpjbj jcj c

In principle the matrix elements a b and c can be calculated from the explicit wavefunctions given in Appendix A

Problem uge

Let jjmi and jjmi denote the angular momentum states for particle and respectively We have

jJMi X

jjmijjmihjjmmjJMi

jJMi X

jjmijjmihjjmmjJMi

or since jaijbi has the same meaning as jbijai after a renaming m andm

jJMi X

jjmijjmihjjmmjJMi

ie according to the symmetries of CGcoe cients

jJMi jJ jJMiFor two spin particles it reects the well known fact that the singlet state J isantisymmtric while the triplet state J is symmetric with respect to interchangeof particles

Thus jLMLijSMSi contructed from two electrons both with angular momenta l isonly allowed according to the Pauli principle if jLj jSj is even

This leave us with the S P and D states ie a total of states This isexactly what we would expect from having two electrons in a pshell we have dierent oneelectron states in the pshell three from lz for the orbitalangular momentum and sz for the spin We have to put the electrons intwo dierent oneelectron states according to the Pauli principle It can be done in

ways ie ways

Sakurai We use the eigenfunctions A corresponding to En n

ox s x s p

bpx s x s xx s s

where and denote a symmetric triplet state and the antisymmetric singletstate

c order perturbation theory leads to

Zdxdxjox xjV x x

Zdxjox xj

because of the antisymmetry of the spatial part of ox x

Zdxdxjpx xjV x x

Zdxjpx xj

For an attrative potential V x x peaked around xx the energy of the spatialsymmetric wave function be lowered more than the energy of the spatial antisymmetric wave function because the wave function is allowed to be dierent from zeroat coinciding points Had the potential been repulsive the energy of thespatial antisymmetric wave function increases less than the energy of the spatialsymmetric wave function

If we consider the repulsion of the two electrons in the pshell of the carbon atom assu ciently peaked around coinciding points to applay the above reasoning one wouldexpect that the P states to have lower energy that the S and the D states sincethe P states are antisymmetric with respect to interchange of the spatial coordinatesof the particles

If we include spinorbit coupling only the P states have S and are aectedSince only the total angular momentum J L S is now conserved L and S arestill conserved though the P states are split according to the possible values ofJ J according to

hJM jL SjJMi

JJ LL SS

see in the case of S The coe cient coming from the radial integralsinvolved is positive see Thus the states corresponding to L andS split in one state corresponding to J three states corrending to J and ve states corresponding to J increasing in energy with J The split issmall compared to the P SD split

Problem uge

V r Vera V q

j V qj

If we assume that mVa

is of order one it is seen that the Born approximation is

reliable for ka ie the wavelength of the incoming particle is much larger thanthe range of the potential If we want the approximation to valid for all k we haveto assume mVa

Problem uge

rdrsin qr

aq cos aq sin aq

j V qj a

aq cos aq sin aq

We have aq cos aq sin aq

for aq Note that it falls o slower with large q than cross section in problem the reason being the potential in problem is smoother For aq small we have

aq cos aq sin aq

The total cross section is

aq cos aq sin aq

sin ka

ka sinka

We see again that the approximation is reliable of ka which means that thewavelength of the particle is much less that a the range of the potential

In the limit ka we have

sin ka

ka sinka

ie we have

tot mVa

In order to assure the validity of the Born approximation for all values of k we haveto assume

Sam Pinansky

April 25, 2004

fl = −2m

∫ ∞

jl(kr′)Al(k; r′)V (r′)r′2 dr′ (1)

fl = −2mV0

(jl(kr′))2r′2 dr′ (2)

fl = −2mV0

(kr′)2l

((2l + 1)!!)2r′2 dr′ (3)

= −2mV0R3

(kR)2l

(3 + 2l)((2l + 1)!!)2(4)

f0 = −2mV0R3

3~2(5)

f1 = −2mV0k2R5

45~2(6)

f(θ) =∑

(2l + 1)flPl(cos θ) (7)

4m2R6V 20

9~4(8)

σtot = 4πdσ

)m2V 2

dΩ= |f0|2 + 6Re(f∗0 f1) cos θ (10)

25(kR)2 (11)

β0 =(

= k′R cot(k′R)− 1 (13)

tan δ0 =1− k′

limk→0

=k′ cot(k′R)

1− k′R cot(k′R)(16)

k′tan(k′R) (17)

2m = E − V0:

a = R−√

−2mV0tan

(√−2mV0

-40 -30 -20 -10 10

R−√

~22mV tan

(√2mV~2 R

)V0 < 0

R +√

~22mV tanh

(√2mV~2 R

)V0 > 0

〈x2〉〈p2x〉 ≥

19〈r2〉〈p2

r〉 (21)

〈r2〉 = A2

sin2(πr/a)π2r2/a2

r2r2 dr (22)

= A2 a5(2π2 − 3)12π4

1 = A2

sin2(πr/a)π2r2/a2

r2 dr (24)

A2 =2π2

a3(25)

So then

〈r2〉 =a2(2π2 − 3)

6π2(26)

〈(∆x)2〉〈(∆px)2〉 =2π2 − 3

54~2 (27)

2π2 − 354

= 0.31 > 0.25 (28)

thus:2π2 − 3

54~2 >

〈k′|U (1)I (t,−∞)|k〉 = − i

= − i

2~〈k′|V (r)|k〉

−∞dt′

|〈k′|U (1)I (t,−∞)|k〉|2 =

|〈k′|V (r)|k〉|2∣∣∣∣

η + i(Ek′ − Ek + ~ω)/~+

η + i(Ek′ − Ek − ~ω)/~

∣∣∣∣2

4~2|〈k′|V (r)|k〉|2

(e2ηt

η2 − ((Ek′ − Ek + ~ω)/~)2+

η2 − ((Ek′ − Ek − ~ω)/~)2(33)

+e2ηte2iωt

η2 + 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2+

e2ηte−2iωt

η2 − 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2

dt|〈k′|U (1)

I (t,−∞)|k〉|2 =1

4~2|〈k′|V (r)|k〉|2

(2ηe2ηt

η2 − ((Ek′ − Ek + ~ω)/~)2+

2ηe2ηt

η2 − ((Ek′ − Ek − ~ω)/~)2(35)

η2 + 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2+

η2 − 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2

limη→0

η2 + x2≡ πδ(x) (37)

limη→0

dt|〈k′|U (1)

I (t,−∞)|k〉|2 = (38)

|〈k′|V (r)|k〉|2(

dt|〈k′|U (1)

I (t,−∞)|k〉|2 =π

2~|〈k′|V |k〉|2

~2(k′+ + k′−)dΩ (41)

k′± =

√2m(Ek ± ~ω)

~2(42)

(k′+ + k′−

16π2~4

∣∣∣∣∫

d3xV (x)ei(k−k′)·x∣∣∣∣2

Average 25/30.

Sam Pinansky

April 14, 2004

1. (5 Points)

2. (a) Given that:

∆(b) = − m

∫ ∞

−∞V

√b2 + z2 dz (1)

∆(b) = − m

∫ ∞

−∞V0 exp

(b2 + z2

)dz (2)

= −mV0a

√π exp

(− b2

δl = ∆(b)|b= lk

√π exp

(− l2

∆(b) = − m

∫ ∞

−∞V0

e−µ√

b2 + z2(5)

= − mV0

2k~2µ2

∫ ∞

e−µbu

√u2 − 1

du u =

√1 +

= −mV0

k~2µK0(µb) (7)

δl = ∆(b)|b= lk

= −mV0

k~2µK0

K0(z) ≈√

2ze−z as z →∞ (9)

3. (10 Points)

2m〈x| 1

E −H0 + iε|x′′〉 =

∫dE′

∫dE′′∑

′lm|E′′l′m′〉E − E′ + iε

〈E′l′m′|x′〉

∫dE′∑

πk′jl(k′r)Y m

l (r)1

l∗(r′)

Y ml (r)Y m

l∗(r′)

∫ ∞

−∞dk′ k′2

4jl(k′r)jl(k′r′) = (h(1)l (k′r) + h

(2)l (k′r))(h(1)

l (k′r′) + h(2)l (k′r′)) (13)

∝r large1r2

= − 12π

∫ ∞

−∞dk′ k′2

h(1)l (k′r)jl(k′r′) + h

(2)l (k′r)jl(k′r′)

(k′ + k + iε)(k′ − k − iε)(15)

= −2πi

(1)l (kr)jl(kr′)

2k− h

(2)l (−kr)jl(−kr′)

)r > r′ (16)

= −ikh(1)l (kr)jl(kr′) r > r′ (17)

2m〈x| 1

E −H0 + iε|x′′〉 = −ik

Y ml (r)Y m

l (kr>) (18)

〈x|Elm(+)〉 = 〈x|Elm〉+∫

d3x′〈x| 1E −H0 + iε

|x′〉V (r′)〈x′|Elm(+)〉 (19)

〈x|Elm(+)〉 = clAl(k, r)Y ml (r) (21)

l (r)+ (22)∫

d3x′

−2mik

l′,m′Y m′

l′ (r)Y m′l′

∗(r′)jl′(kr<)h(1)

l′ (kr>)

l (r′)

l (r)− 2mik

l (r)∫ ∞

∫ ∞

soeiδl

sin δl

k= −1

i.e., from above

fl = −2m

∫ ∞

jl(kr′)Al(k; r′)V (r′)r′2dr′ (29)

4. (10 Point)

V (r) =~2

2mγδ(r −R) (30)

∫ ∞

A0(k; R) =j0(kR)

A0(k; r) = j0(kr)

)+ (36)

n0(kr)

(kγj0(kR)j0(kR)R2

cot δ0 = −k

cot δ0 ≈ − cot(kR) (40)

γ(42)

γ= kR− nπ (43)

k =nπ

11 + 1

≈ nπ

(1− 1

(1− 2

R2(46)

Γ =−2

Γ = −14~n3π3

mR4γ2+ O(1/γ3) (48)

Sam Pinansky

April 25, 2004

fl = −2m

∫ ∞

jl(kr′)Al(k; r′)V (r′)r′2 dr′ (1)

fl = −2mV0

(jl(kr′))2r′2 dr′ (2)

fl = −2mV0

(kr′)2l

((2l + 1)!!)2r′2 dr′ (3)

= −2mV0R3

(kR)2l

(3 + 2l)((2l + 1)!!)2(4)

f0 = −2mV0R3

3~2(5)

f1 = −2mV0k2R5

45~2(6)

f(θ) =∑

(2l + 1)flPl(cos θ) (7)

4m2R6V 20

9~4(8)

σtot = 4πdσ

)m2V 2

dΩ= |f0|2 + 6Re(f∗0 f1) cos θ (10)

25(kR)2 (11)

β0 =(

= k′R cot(k′R)− 1 (13)

tan δ0 =1− k′

limk→0

=k′ cot(k′R)

1− k′R cot(k′R)(16)

k′tan(k′R) (17)

2m = E − V0:

a = R−√

−2mV0tan

(√−2mV0

-40 -30 -20 -10 10

R−√

~22mV tan

(√2mV~2 R

)V0 < 0

R +√

~22mV tanh

(√2mV~2 R

)V0 > 0

〈x2〉〈p2x〉 ≥

19〈r2〉〈p2

r〉 (21)

〈r2〉 = A2

sin2(πr/a)π2r2/a2

r2r2 dr (22)

= A2 a5(2π2 − 3)12π4

1 = A2

sin2(πr/a)π2r2/a2

r2 dr (24)

A2 =2π2

a3(25)

So then

〈r2〉 =a2(2π2 − 3)

6π2(26)

〈(∆x)2〉〈(∆px)2〉 =2π2 − 3

54~2 (27)

2π2 − 354

= 0.31 > 0.25 (28)

thus:2π2 − 3

54~2 >

〈k′|U (1)I (t,−∞)|k〉 = − i

= − i

2~〈k′|V (r)|k〉

−∞dt′

|〈k′|U (1)I (t,−∞)|k〉|2 =

|〈k′|V (r)|k〉|2∣∣∣∣

η + i(Ek′ − Ek + ~ω)/~+

η + i(Ek′ − Ek − ~ω)/~

∣∣∣∣2

4~2|〈k′|V (r)|k〉|2

(e2ηt

η2 − ((Ek′ − Ek + ~ω)/~)2+

η2 − ((Ek′ − Ek − ~ω)/~)2(33)

+e2ηte2iωt

η2 + 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2+

e2ηte−2iωt

η2 − 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2

dt|〈k′|U (1)

I (t,−∞)|k〉|2 =1

4~2|〈k′|V (r)|k〉|2

(2ηe2ηt

η2 − ((Ek′ − Ek + ~ω)/~)2+

2ηe2ηt

η2 − ((Ek′ − Ek − ~ω)/~)2(35)

η2 + 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2+

η2 − 2iωη + ((Ek′ − Ek)2 − ~2ω2)/~2

limη→0

η2 + x2≡ πδ(x) (37)

limη→0

dt|〈k′|U (1)

I (t,−∞)|k〉|2 = (38)

|〈k′|V (r)|k〉|2(

dt|〈k′|U (1)

I (t,−∞)|k〉|2 =π

2~|〈k′|V |k〉|2

~2(k′+ + k′−)dΩ (41)

k′± =

√2m(Ek ± ~ω)

~2(42)

(k′+ + k′−

16π2~4

∣∣∣∣∫

d3xV (x)ei(k−k′)·x∣∣∣∣2

Average 25/30.

Sam Pinansky

April 14, 2004

1. (5 Points)

2. (a) Given that:

∆(b) = − m

∫ ∞

−∞V

√b2 + z2 dz (1)

∆(b) = − m

∫ ∞

−∞V0 exp

(b2 + z2

)dz (2)

= −mV0a

√π exp

(− b2

δl = ∆(b)|b= lk

√π exp

(− l2

∆(b) = − m

∫ ∞

−∞V0

e−µ√

b2 + z2(5)

= − mV0

2k~2µ2

∫ ∞

e−µbu

√u2 − 1

du u =

√1 +

= −mV0

k~2µK0(µb) (7)

δl = ∆(b)|b= lk

= −mV0

k~2µK0

K0(z) ≈√

2ze−z as z →∞ (9)

3. (10 Points)

2m〈x| 1

E −H0 + iε|x′′〉 =

∫dE′

∫dE′′∑

′lm|E′′l′m′〉E − E′ + iε

〈E′l′m′|x′〉

∫dE′∑

πk′jl(k′r)Y m

l (r)1

l∗(r′)

Y ml (r)Y m

l∗(r′)

∫ ∞

−∞dk′ k′2

4jl(k′r)jl(k′r′) = (h(1)l (k′r) + h

(2)l (k′r))(h(1)

l (k′r′) + h(2)l (k′r′)) (13)

∝r large1r2

= − 12π

∫ ∞

−∞dk′ k′2

h(1)l (k′r)jl(k′r′) + h

(2)l (k′r)jl(k′r′)

(k′ + k + iε)(k′ − k − iε)(15)

= −2πi

(1)l (kr)jl(kr′)

2k− h

(2)l (−kr)jl(−kr′)

)r > r′ (16)

= −ikh(1)l (kr)jl(kr′) r > r′ (17)

2m〈x| 1

E −H0 + iε|x′′〉 = −ik

Y ml (r)Y m

l (kr>) (18)

〈x|Elm(+)〉 = 〈x|Elm〉+∫

d3x′〈x| 1E −H0 + iε

|x′〉V (r′)〈x′|Elm(+)〉 (19)

〈x|Elm(+)〉 = clAl(k, r)Y ml (r) (21)

l (r)+ (22)∫

d3x′

−2mik

l′,m′Y m′

l′ (r)Y m′l′

∗(r′)jl′(kr<)h(1)

l′ (kr>)

l (r′)

l (r)− 2mik

l (r)∫ ∞

∫ ∞

soeiδl

sin δl

k= −1

i.e., from above

fl = −2m

∫ ∞

jl(kr′)Al(k; r′)V (r′)r′2dr′ (29)

4. (10 Point)

V (r) =~2

2mγδ(r −R) (30)

∫ ∞

A0(k; R) =j0(kR)

A0(k; r) = j0(kr)

)+ (36)

n0(kr)

(kγj0(kR)j0(kR)R2

cot δ0 = −k

cot δ0 ≈ − cot(kR) (40)

γ(42)

γ= kR− nπ (43)

k =nπ

11 + 1

≈ nπ

(1− 1

(1− 2

R2(46)

Γ =−2

Γ = −14~n3π3

mR4γ2+ O(1/γ3) (48)

hep.fcfm.buap.mxhep.fcfm.buap.mx/ecursos/MCII/mec/solucionarios/708 Solutions Sa… · Quantum...

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