Tor KjellssonStockholm University

Chapter 6

6.1

Q. Suppose we put a delta-function bump in the center of the infinite squarewell:

H ′ = αδ(x− a/2) (1)

where α is a constant.

a)

Find the first-order correction to the allowed energies. Explain why the energiesare not perturbed for even n.

Sol:For the infinite square well we have that the energies are given by:

En =n2π2~2

2ma2n = 1, 2, 3 . . . (2)

since each energy level is unique the spectrum is non-degenerate. So when weseek the first order corrections to the energy levels we can use non-degenerateperturbation theory:

E1n = 〈ψ0

n|H ′|ψ0n〉 (3)

where ψ0n are the unperturbed eigenfunctions of H. For the infinite square

well these are:

ψ0n(x) =

√2

asin(nπax). (4)

Using eq.(3) we now get:

E1n = 〈ψ0

n|H ′|ψ0n〉 = 〈ψ0

n|αδ(x− a/2)|ψ0n〉

= α

∫ ∞0

(ψ0n(x)

)∗δ(x− a/2) ψ0

n(x) dx

=2

a· α∫ ∞0

sin(nπax)δ(x− a/2) sin

(nπax)dx

=2

a· α sin2

(nπa

a

2

)=

{0 if n = even2a · α else

For even n the correction is 0 at x = a/2. This is reasonable since the wavefunctions for even n are 0 (check it for at least one case!) where the delta-function is located - they are thus not affected by the perturbation.

1

b)

Find the first three nonzero terms in the expansion (eq.(6.13) in the text book)of the correction to the ground state ψ1

1.

Sol:Eq.(6.13) in the book is:

ψ1n =

∑n 6=m

〈ψ0m|H ′|ψ0

n〉E0n − E0

m

ψ0m =⇒ ψ1

1 =∑m 6=1

〈ψ0m|H ′|ψ0

1〉E0

1 − E0m

ψ0m. (5)

First we evaluate 〈ψ0m|H ′|ψ0

1〉:

〈ψ0m|H ′|ψ0

1〉 =

∫ a

0

(ψ0m(x)

)∗αδ(x− a

2 ) ψ01(x) dx

= α

∫ a

0

√2

asin(mπax)δ(x− a

2 )

√2

asin(πax)dx

= α2

asin(mπ

2

)sin(π

2

)= α

2

asin(mπ

2

)

Now we insert this together with ψ0m =

√2a sin

(mπa x)

into eq.(5):

ψ11(x) =

∑m 6=1

α 2a sin

(mπ2

)E0

1 − E0m

√2

asin(mπax).

we can now write out the first terms1 of the sum:

ψ11(x) = α

(2

a

)3/2 [-1

E01−E0

3sin(3πa x)

+ 1E0

1−E05

sin(5πa x)

+ -1E0

1−E07

sin(7πa x)

+ . . .].

Using eq.(2) for the energies and performing some basic algebra we obtain:

ψ11(x) =

√a2αmπ2~2

[sin(3πa x)− 1

3 sin(5πa x)

+ 16 sin

(7πa x)

+ . . .]

(6)

1Try to understand why they are m = 3, 5, 7!

2

6.2

Q. For the harmonic oscillator [V (x) = (1/2)kx2], the allowed energies are:

En = (n+ 1/2)~ω (n = 0, 1, 2, . . .) (7)

where ω =√k/m is the classical frequency. Now suppose the spring constant

increases slightly: k → (1 + ε)k.

a)

Q. Find the exact new energies. Expand your formula as a power series in ε,up to second order.

Sol:The change in the spring constant also gives a change in the frequency:

k = (1 + ε)k =⇒ ω =

√(1+ε)km =

√1 + ε ω (8)

The new energies are obtained by using the new frequency ω:

En = (n+ 12 )~ω = (n+ 1

2 )~ω√

1 + ε. (9)

Now make a power series expansion of this to second order in ε:

√1 + ε ≈ 1 +

1

2ε− 1

8ε2 + . . .

to see that the new energies are approximately:

En ≈ (n+ 1/2)~ω(

1 +1

2ε− 1

8ε2 + . . .

). (10)

b)

Q. Now calculate the first-order perturbation in energy using eq.(6.9) in the textbook. What is H ′ here? Compare your result with part (a). Hint: do not cal-culate any integral but use information from p.48-49 when suitable.

Sol:First-order energy perturbation is given by:

E1n = 〈ψ0

n|H ′|ψ0n〉

and to calculate it we must know the perturbation H ′. For the Harmonicoscillator (in one dimension) we have the Hamiltonian:

H = − ~2

2m

d2

dx2+

1

2kx2 (11)

3

so with the new spring constant k = (1 + ε)k we get:

H +H ′ = − ~2

2m

d2

dx2+

1

2kx2 +

1

2εkx2︸ ︷︷ ︸H′

(12)

which gives us the first-order energy perturbation as:

E1n = 〈ψ0

n|H ′|ψ0n〉 =

1

2εk 〈ψ0

n|x2|ψ0n〉.

From pages 48-49 in the text book we now find2

x2 =~

2mω

[a2+ + a+a− + a−a+ + a2−

](13)

a+ψn =√n+ 1ψn+1, a−ψn =

√nψn−1. (14)

Using eq.(13) the energy perturbation becomes:

E1n =

1

2εk

~2mω

〈ψ0n|a2+ + a+a− + a−a+a

2−|ψ0

n〉. (15)

E1n =

1

2εk

~2mω

[〈ψ0n|a2+|ψ0

n〉+ 〈ψ0n|a+a−|ψ0

n〉+ 〈ψ0n|a−a+|ψ0

n〉+ 〈ψ0n|a2−|ψ0

n〉]

(16)

Now we will focus on the four scalar products:

〈ψ0n|a2+|ψ0

n〉+ 〈ψ0n|a+a−|ψ0

n〉+ 〈ψ0n|a−a+|ψ0

n〉+ 〈ψ0n|a2−|ψ0

n〉

let the first operator act on the ket (use eq.(14)) to give:

=√n+ 1〈ψ0

n|a+|ψ0n+1〉+

√n〈ψ0

n|a+|ψ0n−1〉+

√n+ 1〈ψ0

n|a−|ψ0n+1〉+

√n〈ψ0

n|a−|ψ0n−1〉

now let the remaining operator act on the new ket:

=√n+ 1

√n+ 2 〈ψ0

n|ψ0n+2〉︸ ︷︷ ︸

0

+√n√n 〈ψ0

n|ψ0n〉︸ ︷︷ ︸

1

+√n+ 1

√n+ 1 〈ψ0

n|ψ0n〉︸ ︷︷ ︸

1

+√n√n− 1 〈ψ0

n|ψ0n−2〉︸ ︷︷ ︸

0

= 0 + n+ (n+ 1) + 0 = 2n+ 1

Inserting this into eq.(17) gives:

E1n =

1

2εk

~2mω

(2n+ 1) =1

2εmω2 ~

mω

(n+

1

2

)=ε

2

(n+

1

2

)~ω (17)

Note: for n = 0 the second and fourth scalar product should become 0 right afterthe first step, since |ψn−1〉 does not exist. This is indeed also the case since√n =√

0.

2There is a famous quote: ”physics is just learning how to solve the harmonic oscillatorproblem at ever increasing level”.

4

6.4

a)

Q. Find the second-order correction to the energies (E2n) for the potential in

problem 6.1. Comment: You can sum the series explicitly, obtaining −2m(α/π~n)2

for odd n.

Sol:The second-order energy correction (eq.(6.15) in the text book) is:

E2n =

∑n 6=m

|〈ψ0m|H ′|ψ0

n〉|2

E0n − E0

m

(18)

On p.2 in this chapter of solutions we found3

〈ψ0m|H ′|ψ0

n〉 = 〈ψ0m|αδ(x− a/2)|ψ0

n〉 =2

a· α sin

(mπ2

)sin(nπ

2

)so:

|〈ψ0m|H ′|ψ0

n〉|2 =4α2

a2sin2

(mπ2

)sin2

(nπ2

)=

{0 for even n or m4α2

a2 else.

Since E0n − E0

m = π2~2

2ma2 (n2 −m2) we obtain:

E2n =

4α2

a22ma2

π2~2∑n 6=mn=odd

1

n2 −m2=

8mα2

π2~2· −1

4n2= −2m

( α

π~n

)2(19)

where we have looked up the sum in a table. Note that m outside the sum isthe mass, not the summation index in the sum.

3Revisit problem 1 where we computed 〈ψ0n|H′|ψ0

n〉.

5

b)

Q. Calculate the second-order correction to the ground state energy (E20) for

the potential in problem 6.2. Check that your result is consistent with the exactresult obtained in problem 6.2 a) (eq.(9) in these solutions!).

Sol:The second order correction to the ground state is:

E20 =

∑m 6=0

|〈ψ0m|H ′|ψ0

0〉|2

E00 − E0

m

and we are asked to consider the potential in problem 6.2. There we saw thatthe perturbation was:

H ′ =1

2εkx2

and we exploited the ladder operators4

x2 =~

2mω

[a2+ + a+a− + a−a+ + a2−

](20)

a+ψn =√n+ 1ψn+1, a−ψn =

√nψn−1. (21)

Now we have all the tools for solving the problem. Start by finding 〈ψ0m|H ′|ψ0

n〉:

〈ψ0m|H ′|ψ0

n〉 =1

2εk

~2mω

⟨ψ0m|a2+|ψ0

n〉+ 〈ψ0m|a+a−|ψ0

n〉+ 〈ψ0m|a−a+|ψ0

n〉+ 〈ψ0m|a2−|ψ0

n〉]

which gives:

〈ψ0m|H ′|ψ0

n〉 =1

2εk

~2mω

[√n+ 1

√n+ 2 〈ψ0

m|ψ0n+2〉︸ ︷︷ ︸

δm,n+2

+√n√n 〈ψ0

m|ψ0n〉︸ ︷︷ ︸

δm,n

+

+√n+ 1

√n+ 1 〈ψ0

m|ψ0n〉︸ ︷︷ ︸

δm,n

+√n√n− 1 〈ψ0

m|ψ0n−2〉︸ ︷︷ ︸

δm,n−2

].

(If you couldn’t follow - revisit page 4 in this chapter of solutions for details.)

Now, before we start computing |〈ψ0m|H ′|ψ0

n〉|2 we pause a moment and think.If we just square the expression above we will obtain 16 terms - lets first findout if we can skip some of them by thinking about their structure.

The 16 terms mentioned above will all contain the product of two (Kronecker)delta functions. If you now consider the product of two such that are not equal,say δm,n · δm,n+2 this has to be 0. For fun, try to prove this - it is a one lineargument that is provided in the footnote-section on the next page.

4Note that the mass is denoted m to avoid confusion with the index m.

6

Since we don’t need to consider terms containing products of two distinct deltafunctions we obtain:

|〈ψ0m|H ′|ψ0

n〉|2 =

=ε2k2~2

16m2ω2

[(n+1)(n+2) (δm,n+2)

2+n2 (δm,n)

2+(n+1) (δm,n)

2+n(n−1) (δm,n−2)

2

].

Now we can calculate E20 :

E20 =

ε2k2~2

16m2ω2

∑m 6=0

1

E00 − E0

m

[(0 + 1)(0 + 2) (δm,0+2)

2+ 0(0− 1) (δm,0+2)

2

]

E20 =

ε2k2~2

16m2ω2

∑m6=0

1

E00 − E0

m

2 (δm,2)2

(22)

one of the delta functions collapses the sum:

E20 =

ε2k2~2

16m2ω2

2δm,2E0

0 − E02

=ε2k2~2

8m2ω2

1

E00 − E0

2

while the other one can be thrown away since it does nothing - there are nomore indices m to force to the value 2.

Now we use:

E00 − E0

2 =(0 + 1

2

)~ω −

(2 + 1

2

)~ω = −2~ω and k2 = m2ω4 (23)

to obtain:

E20 =

ε2m2ω4~2

8m2ω2· 1

−(2~ω)= − ε

2

16~ω (24)

Note that the both the first order correction (eq.(17)) and second order correc-tion (eq.(24)) are in agreement with the exact result obtained in eq.(9)!

Finally, here is the footnote for why the product of two distinct delta functionsmust be zero.5

5Consider δm,n+2 · δm,n. The first delta function sets m = n + 2 while the second setsm = n. Combining these would mean n+ 2 = n which has no solutions.

7

6.7

Q. Consider a particle of mass m that is free to move in a one-deminsionalregion of length L that closes on itself. An example of this would be problem2.46 in the text book, where a bead slides frictionlessly on a circular wire.

a)Q. Show that the stationary states can be written in the form:

ψn(x) =1√Le2πin/L, (−L/2, < x < L/2) (25)

where n = 0,±1,±2, . . . and the energies are given by:

En =2

m

(nπ~L

)2

. (26)

Notice that the spectrum is doubly degenerate for all n but n=0.

Sol:The stationary states are the eigenstates of H and to find these we solve theeigenvalue equation:

H|ψ〉 = E|ψ〉. (27)

What is our H? Recall that it normally has the form H = T + V , where T isthe kinetic energy and V the potential.

In this particular case, the particle is moving freely so the potential is zero.Therefore, the Hamiltonian only contains the usual kinetic part:

H = − ~2

2m

d2

dx2. (28)

The eigenvalue equation to solve is thus6

Hψ(x) = Eψ(x)

− ~2

2m

d2

dx2ψ(x) = Eψ(x)

which has solutions of the form:

ψ(x) = Ce±ikx (29)

with k =√

2mE/~2. These eigenfunctions are not yet fully determined - to dothis we must know the values of k and C. Here k is connected to the eigenvalues Eand C is the normalization constant. We start by finding C from the condition:

〈ψ(x)|ψ(x)〉 = 1 (30)

6You may have noticed that we went from H|ψ〉 = E|ψ〉 in eq.(27) to H(x)ψ(x) = Eψ(x)and you might wonder why. It is not because |ψ〉 = ψ(x), this is never the case! The

reason is that in eq.(27), we did not specify in what basis H was expressed. As soon as weused the position basis (x) the whole equation (27) was transformed into that basis.

8

which gives:

∫ L2

−L2

(Ceikx

)∗Ceikxdx = |C|2

∫ L2

−L2

1 dx =⇒ C =1√L

(31)

Now we find the allowed values for k (and thus also for E ). We do this byapplying the boundary condition which in this case is that the functions areperiodic7:

ψ(−L2 ) = ψ(L2 ) =⇒ Ce−ikL2 = Ceik

L2 =⇒ 1 = eikL

Recall that 1 = ei2πn for all integers n. Inserting this into the box gives:

eikL = ei2πn ∀n ∈ Z =⇒ kL = 2nπ ∀n ∈ Z.

Combining this result with k =√

2mE/~2 gives:

k =2nπ

L=⇒

√2mE

~2=

2nπ

L=⇒ En =

2

m

(~nπL

)2

∀n ∈ Z. (32)

Thus:

ψn(x) =1√L· ei 2nπxL ∀n ∈ Z (33)

En =2

m

(~nπL

)2

∀n ∈ Z. (34)

7since the problem statement says that the region closes on itself

9

b)

Q. Now suppose we introduce the perturbation:

H ′ = −V0e−x2/a2 (35)

where a << L.

Find the first-order correction to En using eq.(6.27) in the text book. Hint: toevaluate the integral, exploit the fact that a << L to extend the limitsfrom ±L/2 to ±∞.

Sol:

The first-order correction for degenerate perturbation theory is given by eq.(6.27)in the text book:

E1± =

1

2

[Waa +Wbb ±

√(Waa −Wbb)2 + 4|Wab|2

](36)

where:

Wab ≡ 〈ψ0a|H ′|ψ0

b 〉 (37)

and the indices a, b denote that indices for the degenerate energy levels. In thisexample the energy levels are degenerate for ±n, so a = n, b = −n.

Now we calculate Wnn:

Wnn = 〈ψ0n|H ′|ψ0

n〉 =

∫ L2

−L2

1√Le−i

2nπxL

[−V0e−x

2/a2]

1√Lei

2nπxL = −V0

L

∫ L2

−L2

e−x2/a2dx

If we now use the hint a << L we obtain8 an integral that we can look up:

Wnn = −V0L

∫ L2

−L2

e−x2/a2dx ≈ −V0

L

∫ ∞−∞

e−x2/a2dx = −V0

La√π (38)

Looking at eq.(36) we see that we also have to calculate Wn,-n:

〈ψ0n|H ′|ψ0

−n〉 = −V0L

∫ L/2

−L/2e−i

2nπxL e−x

2/a2e−i2nπxL dx = −V0

L

∫ L/2

−L/2e−

x2

a2−i 4nπxL dx

Once again, we use the approximation given in the hint:

〈ψ0n|H ′|ψ0

−n〉 = −V0L

∫ L2

−L2

e−x2

a2−i 4nπxL dx ≈ −V0

L

∫ ∞−∞

e−x2

a2−i 4nπxL dx

8Why can we do this? If you don’t see it, think of the integral as a sum. What happens

to e−x2/a2for large values of x and small values of a?

10

This integral is a bit complicated. If the QM course you are following does notrequire you to have taken a mathematics course in complex analysis you aremost likely getting the following result on a sheet of formulae:∫ ∞

−∞e−

x2

a2−i 4nπxL dx = e

−(2nπaL

)2

a√π (39)

using this we obtain:

Wn,-n ≈ −V0Le−(2nπaL

)2

a√π (40)

which together with eq.(38) inserted into eq.(36) gives:

E1± = −V0

La√π

(1∓ e−

(2nπaL

)2)(41)

For the interested reader we provide some details about eq.(39). Remember thatthese are most likely overkill if you are taking a QM course using this text bookso skip them if you are not interested.

Completing the square9 gives:∫ ∞−∞

e−x2

a2−i 4nπxL dx =

∫ ∞−∞

e−( xa+i2nπaL )

2−( 2nπaL )

2

= e−( 2nπaL )

2∫ ∞−∞

e−( xa+i2nπaL )

2

dx

Make the variable substitution:

x

a+ i

2nπa

L= y, dx = ady, −∞+ i

2nπa

L< y <∞+ i

2nπa

L

which gives: ∫ ∞−∞

e−( xa+i2nπaL )

2

dx = a

∫ ∞+i2nπaL

−∞+i2nπaL

e−y2

dy. (42)

The right hand side is a very famous integral that is tabulated in most places.However, the limits are now complex which might be a first sight for most ofyou. Using something called the Cauchy residue theorem10 one sees that thisintegral is the same as if it had been computed on the real line. Thus, in thiscase, you can just ignore the imaginary parts in the integration limits to obtain:

∫ ∞−∞

e−( xa+i2nπaL )

2

dx = a

∫ ∞+i2nπaL

−∞+i2nπaL

e−y2

dy = a

∫ ∞−∞

e−y2

dy = a√π. (43)

9In Swedish: kvadratkomplettering.10In the beautiful theory of complex analysis you learn how to overcome the many lim-

its in real-valued mathematical analysis. You also explain a lot of the weird trigonometricsubstitutions suggested for solving real valued integrals.

11

c) Q. What are the ”good” linear combinations of ψn and ψ−n for this problem?Show that with these states you get the first-order correction using eq.(6.9) inthe text book. (Note: this is the first order correction for nondegenerate pertur-bation theory.)

Sol:Let us start by clarify the goals of this problem:

1) Find the ”good” linear combinations of the degenerate states ψn and ψ−n.2) Using these good states, verify that the first order nondegenerate

energy correction gives the same answer as eq.(41).

In the case of a two-fold degeneracy the ”good” states must be of the form:

ψ = αψ0a + βψ0

b (44)

Eq.(6.22) in the text book gives:

αWaa + βWab = αE1 =⇒ β = αE1 −Waa

Wab

so in our case we obtain:

β+ = α+E1

+ −Waa

Wabβ− = α−

E1− −Waa

Wab(45)

Recall that we computed Waa,Wab, E1± in the previous problem (a=n, b=-n).

We will now use information about these to obtain the coefficients in eq.(45).

From eq.(36) and eq.(38) we see that:

E1± = Wnn ∓ |Wn,-n| (46)

which inserted into eq.(45) gives:

β+ =−|Wn,-n|Wn,-n

α+ β− =|Wn,-n|Wn,-n

α−

Using the result in eq.(40) we now obtain:

β+ = α+ β− = −α−. (47)

Finally, we are now able to give the ”good” linear combinations:

ψ+ = α+ψn+β+ψ-n = α+ψn−α+ψ-n =α−√L

(ei

2πnxL − e−i

2πnxL

)=

2α+√L

sin(2πnL x

)

ψ− = α−ψn+β−ψ-n = α−ψn+α−ψ-n =α−√L

(ei

2πnxL + e−i

2πnxL

)=

2α−√L

cos(2πnL x

)

12

To determine α± you can either compute the integrals or just note that sincewe added two orthonormal functions to create ψ± we must have α+ = α− = 1√

2.

Now we will attack the second part of the problem: calculate the firstorder energy corrections E1

± for nondegenerate states by using the good statesthat we have just found:

ψ+ =

√2

Lsin(2πnL x

)ψ− =

√2

Lcos(2πnL x

)(48)

We will in this solution do E1+ = 〈ψ+|H ′|ψ+〉 and leave the other case as an

exercise.

E1+ = 〈ψ+|H ′|ψ+〉 = 〈ψ+| − V0e−x

2/a2 |ψ+〉 = − 2

LV0

∫ L2

−L2

e−x2

a2 sin2(2πnxL

)dx

(49)using the hint from the previous problems:

∫ L2

−L2

e−x2

a2 sin2(2πnxL

)dx ≈

∫ ∞−∞

e−x2

a2 sin2(2πnxL

)dx. (50)

Using Euler’s formula we obtain:∫ ∞−∞

e−x2

a2 sin2(2πnxL

)=

∫ ∞−∞

e−x2

a21

4i2

(ei

4πnxL + e−

i4πnxL − 2

)dx

= −1

4

∫ ∞−∞

e−x2/a2ei4πnx/Ldx − 1

4

∫ ∞−∞

e−x2/a2e−i4πnx/Ldx +

1

2

∫ ∞−∞

e−x2/a2dx︸ ︷︷ ︸

a√π/2

.

(51)Like in the previous problem, the first two integrals are mean. Therefore wejust give the results11∫ ∞

−∞e−

x2

a2−i 4nπxL dx =

∫ ∞−∞

e−x2

a2+i 4nπxL dx = e

−(2nπaL

)2

a√π (52)

So combining eq.(52), eq.(51) and eq.(49) gives:

E1+ = −V0

La√π[1− e−( 2nπa

L )2]

(53)

which is in agreement with eq.(41), the result obtained using degenerate per-turbation theory. Now you should verify that you have understood the solutionby redoing the steps for E1

−.

11As a fun exercise, prove the second step. You don’t need any complex analysis for that.

13

6.9

Q. Consider a quantum system with just three linearly independent states. Sup-pose the Hamiltonian in matrix form is:

H = V0

1− ε 0 00 1 ε0 ε 2

(54)

where V0 is a constant and ε << 1.

a)

Q. Write down the eigenvectors and eigenvalues of the unperturbed Hamiltonian(ε = 0).

Sol:This is very straightforward. If ε = 0 then:

H0 = V0

1 0 00 1 00 0 2

(55)

and we immediately see that the eigenvectors are:

|α1〉 =

100

|α2〉 =

010

|α3〉 =

001

(56)

with the three eigenvalues:

λ1 = V0 λ2 = V0 λ3 = 2V0 (57)

b)

Q. Solve for the exact eigenvalues of H. Expand each of them as a power seriesin ε up to second order.

Sol:This is something that we have done a lot. Start by setting up the eigenvaluematrix equation:

H|α〉 = λ|α〉

V0

1− ε 0 00 1 ε0 ε 2

abc

= λ

abc

=⇒

∣∣∣∣∣∣V0(1− ε)− λ 0 0

0 V0 − λ V0ε0 V0ε 2V0 − λ

∣∣∣∣∣∣ = 0

14

=⇒ (V0(1− ε)− λ)(V0 − λ)(2V0 − λ)− V0ε · V0ε · (V0(1− ε)− λ) = 0

(V0(1− ε)− λ) [(V0 − λ)(2V0 − λ)− V0ε · V0ε] = 0

(V0(1− ε)− λ)[λ2 − 3V0λ+ V 2

0 (2− ε2)]

= 0

which has the solutions:

(V0(1− ε)− λ) = 0 ⇔ λ1 = V0(1− ε)

λ2−3V0λ+V 20 (2−ε2) ⇔ λ2 =

V02

(3−

√1 + 4ε2

)λ3 =

V02

(3 +

√1 + 4ε2

)expanding

√1 + 4ε2 ≈ 1 + 2ε2 + . . . gives:

λ1 ≈ V0 − εV0 (58)

λ2 ≈ V0 − ε2V0 (59)

λ3 ≈ 2V0 + ε2V0 (60)

N.B: we chose λ2 = V0

2

(3−√

1 + 4ε2)

out of convenience. We could have cho-sen the other square root instead if we wanted to.

c)

Q. Use first- and second-order nondegenerate perturbation theory to find theapproximate eigenvalue for the state that grows out of the nondegenerate eigen-vector of H0. Compare this to the exact value obatined in (a).

Sol:For nondegenerate perturbation theory we get the first order correction from:

E1n = 〈α0

n|H ′|α0n〉 (61)

and since our perturbation (on matrix form) is:

H′ = εV0

−1 0 00 0 10 1 0

(62)

and the state that is nondegenerate is the one with the unperturbed eigenvalue2V0 we find:

E13 = 〈α0

3|H ′|α03〉 =

(0 0 1

)εV0

−1 0 00 0 10 1 0

001

= 0 (63)

15

(which means that there is no first-order energy correction due to the pertur-bation).

The second-order correction is given by:

E23 =

∑m6=3

|〈α0m|H ′|α0

3〉|2

E03 − E0

m

=|〈α0

1|H ′|α03〉|2

E03 − E0

1

+|〈α0

2|H ′|α03〉|2

E03 − E0

2

(64)

and computing the inner products:

〈α01|H ′|α0

3〉 =(1 0 0

)εV0

−1 0 00 0 10 1 0

001

= 0 (65)

〈α02|H ′|α0

3〉 =(0 1 0

)εV0

−1 0 00 0 10 1 0

001

= εV0 (66)

gives:

E23 =|〈α0

2|H ′|α03〉|2

E03 − E0

2

=ε2V 2

0

2V0 − V0= ε2V0 (67)

which is the same as the result you obtained in (a).

16

d)

Q. Use degenerate perturbation theory to find the first-order correction to thetwo initially degenerate eigenvalues. Compare to the exact results.

Sol:

First order degenerate perturbation theory gives the energy correction:

E1± =

1

2

[Waa +Wbb ±

√(Waa −Wbb)2 + 4|Wab|2

](68)

with Wab ≡ 〈α0a|H ′|α0

b〉.

Our degenerate states are a=1 and b=2 so we obtain:

W11 =(1 0 0

)εV0

−1 0 00 0 10 1 0

100

= −εV0

W12 =(1 0 0

)εV0

−1 0 00 0 10 1 0

010

= 0

W22 =(0 1 0

)εV0

−1 0 00 0 10 1 0

010

= 0

Insert these results into eq.(68) to obtain:

E1± =

1

2

[−εV0 ±

√(εV0 − 0)2 + 0

]=

{0 for E1

+

−εV0 for E1−

(69)

Is this consistent with the results in (b)? Yes, because there we found:

E11 = V0λ− εV0 (70)

E12 = V0λ− ε2V0 (71)

and in this problem we found that one state gets forced down by −εV0 whilethe other state is unaffected in first order of ε.

17

6.32

Q. Suppose the Hamiltonian H for a particular quantum system is a function ofsome parameter λ; let En(λ) and ψn(λ) be the eigenvalues and eigenfunctionsof H(λ). The Feynman-Hellmann theorem states that

∂En∂λ

=

⟨ψn

∣∣∣∂H∂λ

∣∣∣ψn⟩ (72)

assuming either that En is nondegenerate or if degenerate that the ψn’s are the”good” linear combinations of the degenerate eigenfunctions.

a)Q. Prove the Feynman-Hellmann theorem. Hint: use eq.(6.9) in the text bookfor this.

Sol:There are at least two ways to prove this theorem. We first present the sug-gested way by the problem statement and then the original way Feynman didit as an alternative.

Method 1Since H is a function of λ everything changes according to λ. To organize ourthoughts, let us first define a fixed point λ = λ0 and then observe a change fromthis fixed point:

H(λ0 + dλ) = H(λ0) + V (λ0 + dλ) = H0 +H ′(λ0 + dλ) (73)

the purpose of this is to reconnect with out notion of perturbation. Whateverchanges our Hamiltonian is due to the last term, which is the perturbation.

We can identify this perturbation in the spirit of a true physicist by using apower series expansion of the new Hamiltonian:

H(λ0 + dλ) = H0 +∂H(λ0)

∂λdλ+O(dλ2) + . . . (74)

everything after H0 in eq.(74) would thus be our perturbation. We shall how-ever soon see that we only need to consider the first term.

Now consider the left hand side of eq.(72). (Our aim is to use valid manipula-tions that end up with the right hand side of eq.(72)).

For our fixed point λ = λ0 the left hand side can be written as:

∂En(λ0)

∂λ= limdλ→0

En(λ0 + dλ)− En(λ0)

dλ= limdλ→0

E1n(λ0 + dλ)

dλ(75)

where the last step is valid since En(λ0 + dλ) = En(λ0) + E(1)n (λ0 + dλ) for

small enough dλ.

18

According to eq.(6.9) in the text book (first order nondegenerate energy correc-tion):

E1n(λ0 + dλ) = 〈ψ0

n|H ′(λ0 + dλ)|ψ0n〉. (76)

so eq.(75) now becomes:

∂En(λ0)

∂λ= limdλ→0

〈ψ0n|H ′(λ0 + dλ)|ψ0

n〉dλ

=

⟨ψ0n

∣∣∣ limdλ→0

H ′(λ0 + dλ)

dλ

∣∣∣ψ0n

⟩(77)

where the last step is valid because dλ is just a scalar. Notice that it is reassur-ing that our result resembles eq.(72).

Proceeding now with the right hand side of eq.(77) we recall the power seriesexpansion in eq.(74):

H(λ0 + dλ) = H0 +∂H(λ0)

∂λdλ+O(dλ2) + . . . = H0 +H ′(λ0 + dλ)

From this we see that:

limdλ→0

H ′(λ0 + dλ)

dλ=

∂H(λ0)∂λ dλ

dλ=∂H(λ0)

∂λ(78)

which gives us a new form of eq.(77):

∂En(λ0)

∂λ=

⟨ψ0n

∣∣∣∂H(λ0)

∂λ

∣∣∣ψ0n

⟩(79)

Now we are almost done. Recall that ψ0n are the unperturbed eigenfunctions

and can be written as ψ0n = ψn(λ0). So eq.(79) has a more general form:

∂En(λ0)

∂λ=

⟨ψn(λ0)

∣∣∣∂H(λ0)

∂λ

∣∣∣ψn(λ0)

⟩(80)

Finally, notice that eq.(80) was obtained by taking a small step dλ from thestarting value λ0. Obviously, we could take another step dλ again, redoing allthe calculations on the last two pages, to obtain:

∂En(λ0 + dλ)

∂λ=

⟨ψn(λ0 + dλ)

∣∣∣∂H(λ0 + dλ)

∂λ

∣∣∣ψn(λ0 + dλ)

⟩. (81)

The last line of argument was our final piece to conclude the proof of:

∂En(λ)

∂λ=

⟨ψn(λ)

∣∣∣∂H(λ)

∂λ

∣∣∣ψn(λ)

⟩(82)

19

Method (2)In this alternative method we do not use perturbation theory but work withthe lft hand side of eq.(72) directly:

∂En∂λ

=∂

∂λ

[〈ψn|H|ψn〉

]=

⟨∂ψn∂λ

∣∣∣H∣∣∣ψn⟩+

⟨ψn

∣∣∣∂H∂λ

∣∣∣ψn⟩+

⟨ψn

∣∣∣H∣∣∣∂ψn∂λ

⟩=

⟨∂ψn∂λ

∣∣∣En∣∣∣ψn⟩+

⟨ψn

∣∣∣∂H∂λ

∣∣∣ψn⟩+

⟨∂ψn∂λ

∣∣∣H∣∣∣ψn⟩∗= En

⟨∂ψn∂λ

∣∣∣ψn⟩+

⟨ψn

∣∣∣∂H∂λ

∣∣∣ψn⟩+

⟨∂ψn∂λ

∣∣∣En∣∣∣ψn⟩∗= En

⟨∂ψn∂λ

∣∣∣ψn⟩+

⟨ψn

∣∣∣∂H∂λ

∣∣∣ψn⟩+ E∗n

⟨∂ψn∂λ

∣∣∣ψn⟩∗= En

⟨∂ψn∂λ

∣∣∣ψn⟩+

⟨ψn

∣∣∣∂H∂λ

∣∣∣ψn⟩+ E∗n

⟨ψn

∣∣∣∂ψn∂λ

⟩but remember now that the energies are real, so E∗n = En. Thus:

∂En∂λ

=

⟨ψn

∣∣∣∂H∂λ

∣∣∣ψn⟩+ En∂

∂λ[〈ψn|ψn〉]︸ ︷︷ ︸

0

(83)

where we have used that 〈ψn|ψn〉 = 1, so the last term is 0.

20

b)Q. Apply the theorem to the one-dimensional harmonic oscillator, (i) usingλ = ω (which yields a formula for 〈V 〉), (ii) using λ = ~ (which yields a for-mula for 〈T 〉) and (iii) using λ = m (which yields a relation between 〈V 〉 and〈T 〉). Compare your answers to problem 2.12 and the Virial theorem predictions.

Sol:Here we are going to use the formula we proved in the previous problem:

∂En∂λ

=

⟨ψn

∣∣∣∂H∂λ

∣∣∣ψn⟩for the harmonic oscillator. The Hamiltonian and the energies are given by:

H = − ~2

2m

d2

dx2+

1

2mω2x2 En = ~ω

(n+

1

2

)(84)

(i)Using this for λ = ω we get:

∂En∂ω

=

⟨ψn

∣∣∣∂H∂ω

∣∣∣ψn⟩∂En∂ω

=⟨ψn|mωx2|ψn

⟩=

⟨ψn

∣∣∣2Vω

∣∣∣ψn⟩From the energies we then see:

∂En∂ω

= ~(n+

1

2

)=Enω

and thus:En = 2 〈V 〉 (85)

which is familiar from problems 2.12 and 3.31.

(ii)Using λ = ~ instead we now get:

∂En∂~

=

⟨ψn

∣∣∣∂H∂~

∣∣∣ψn⟩∂En∂~

=

⟨ψn

∣∣∣− ~m

d2

dx2

∣∣∣ψn⟩ =

⟨ψn

∣∣∣2~T ∣∣∣ψn⟩

=2

~〈T 〉. (86)

From the expression for the energies you get (check this):

∂En∂~

=En~

so eq.(77) gives:

En = 2〈T 〉. (87)

Now you can do the third exercise (iii) to verify what you can get from (i) and(ii): 〈V 〉 = 〈T 〉.

21