DOCTORAL GENERAL EXAMINATION WRITTEN EXAM | WITH...

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DOCTORAL GENERAL EXAMINATION

WRITTEN EXAM — WITH SOLUTIONS

September 4, 2015

DURATION: 75 MINUTES PER SECTION



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WRITTEN EXAM - CLASSICAL MECHANICS — WITH SOLUTIONS

September 4, 2015

DURATION: 75 MINUTES

1. This examination has two problems. Read both problems carefully be-fore making your choice. Submit ONLY one problem. IF YOU SUBMITMORE THAN ONE PROBLEM FROM A SECTION, BOTH WILL BEGRADED, AND THE PROBLEM WITH THE LOWER SCORE WILLBE COUNTED.

2. If you decide at this time not to take this section of the exam, pleaseinform the faculty proctor. ONCE YOU BEGIN THE EXAM, IT WILLBE COUNTED.

3. Calculators may not be used.

4. No books or reference materials may be used.

Classical Mechanics 1: Striking the sphere

A hard sphere of radius R is fixed at the center of a spherically symmetric potential U(r).The potential U(r) declines monotonically to zero as r −→∞.

A beam of test particles is aimed at the sphere from a great distance. All the particlesare moving parallel to one another with speed v � c. The number density of particles is nparticles/cm3, and each particle has a mass m.

(a) (4 pts) Calculate the cross-section for striking the sphere, in terms of U(R), v, and m.

(b) (2 pts) Is it possible for the cross-section to vanish? If so, under what conditions?

(c) (4 pts) Now take U(r) = −GMm/r, the gravitational potential between the sphere ofmass M and the test particle of mass m.

Whenever a particle strikes the sphere, the particle sticks to the sphere and increasesthe sphere’s mass but without appreciably changing the radius. Calculate the timerequired for the sphere’s mass to increase from an initial value of Mi to a final valueof Mf .

1

• Solutions by J. Winn (August 2015). An earlier version of this problem also appeared onthe Fall 1992 exam.

(a) The figure below illustrates the case of a repulsive potential, although the result is thesame for an attractive potential. Consider the particles with impact parameter b justlarge enough for the point of closest approach to equal R. All particles with smallerimpact parameters will collide with the sphere.

To figure out b in terms of given quantities, use the conservation of energy,

1

2mv2 =

1

2mv20 + U(R), (1)

as well as the conservation of angular momentum,

mvb = mv0R, (2)

giving v0 = v(b/R). Combining these results,

1

2mv2 =

1

2mv2

(b

R

)2+ U(R) (3)

1 =

(b

R

)2+

2U(R)

mv2(4)

b2 = R2[1− 2U(R)

mv2

]. (5)

The cross-section for hitting the sphere is therefore

σ = πb2 = πR2[1− 2U(R)

mv2

]. (6)

2

(b) From our previous answer we see that for the cross-section to vanish, U(R) must bepositive. Specifically, σ vanishes when

1

2mv2 = U(R), (7)

and the particles with b = 0 have just enough energy to touch the sphere.

(c) In time dt, the number of particles hitting the sphere is nvσdt. Therefore

dM

dt= mnvσ (8)

dM

dt= mnvπR2

[1 +

2GMm

mv2R

](9)

dM

dt= K

[1 +

M

M0

], (10)

where we have defined K ≡ mnvπR2 and M0 ≡ v2R/2G. Now we may integrate:

dM

1 +M/M0= Kdt (11)∫ Mf

Mi

dM

1 +M/M0= Kt (12)

M0 ln (M0 +Mf )−M0 ln (M0 +Mi) = Kt (13)

t =M0K

ln

(M0 +MfM0 +Mi

)(14)

t =v

2πGmnRln

(v2R/2G+Mfv2R/2G+Mi

). (15)

3

Classical Mechanics 2: Particle sliding on a rotating circular wire

A pointlike particle of mass m is constrained to move on a circular wire of radius R. Theparticle can slide without friction. The circular wire spins with constant angular speed ωabout the vertical diameter. The force of gravity mg is acting on the particle.

(a) (2 pts) Write the Lagrangian for the system using θ as the generalized coordinate.Identify an effective potential V (θ).

(b) (2 pts) Write down the Euler Lagrange equation and derive the equation of motion interms of m, g, R, ω, and θ. (Do not solve it.)

(c) (2 pts) Find constant values θi, i = 1, 2, ..., for which θ(t) = θi is a stationary solutionof the equation of motion. Express your answer in terms of ω,R, and g. Do all solutionsexist for all values of ω?

(d) (4 pts) Consider now small displacements from each of the equilibrium values θi iden-tified in part (c). Determine, as a function of ω whether such displacements lead tostable or unstable oscillations. If the small-amplitude oscillation is stable, determinethe corresponding oscillation frequency Ωi. Summarize your results in a graph whereyou plot the θi as functions of ω and label the various parts of the curves as ”stable”or ”unstable”.

4

• Solutions by Markus Klute (August 2015) An earlier version of this problem also appearedon the Spring 2009 exam.

(a) The velocity ~v of the particle is

~v = R θ̇ ~eθ + ωR sin θ ~eφ . (1)

The Lagrangian L is then given by

L = T − V = 12m(R θ̇ ~eθ + ωR sin θ ~eφ

)2−mgR (1− cos θ) . (2)

Simplifying and dropping constants

L =1

2mR2 θ̇2 +

1

2mω2R2 sin2 θ +mgR cos θ =

1

2mR2 θ̇2 − V (θ) , (3)

with

V (θ) = −12mω2R2 sin2 θ −mgR cos θ . (4)

(b) The Euler Lagrange equationd

dt

(∂L

∂θ̇

)=∂L

∂θ(5)

gives1

2mR22θ̈ = −mgR sin θ +mω2R2 sin θ cos θ . (6)

Canceling out constants results in

θ̈ = − gR

sin θ + ω2 sin θ cos θ . (7)

(c) If θ is constant then θ̈ = 0, so from the part (b) of this problem we get

sin θ(ω2 cos θ − g

R

)= 0 . (8)

There are three solutions

θ1 = 0 , (9)

θ2 = cos−1( gRω2

), (10)

θ3 = π . (11)

The solutions θ1 and θ3 exist for all values of ω. The solution θ2 exists for

ω2 >g

R≡ ω20 . (12)

5

(d) From the equation of motion we find

mR2θ̈ =∂L

∂θ= −∂V

∂θ. (13)

Around the equilibrium points θi we write θ = θi + � and from∂V∂θ|θi we find

mR2 �̈ = −∂2V

∂θ2

∣∣∣∣θi

�. (14)

The oscillation frequency Ωi is therefore

Ω2i =1

mR2∂2V

∂θ2

∣∣∣∣θi

. (15)

Using the potential from part (a) gives

∂2V

∂θ2= −mω2R2(cos2 θ − sin2 θ) +mgR cos θ . (16)

We therefore findΩ2i = −ω2(2 cos2 θi − 1) + ω20 cos θi . (17)

For θ1 = 0 we findΩ2i = ω

20 − ω2 . (18)

This point is stable for ω < ω0 and unstable for ω > ω0.

For θ2 satisfying cos θ2 =ω20ω2

, we find

Ω2i = −ω2(

2ω40ω4− 1)

+ω40ω2

= ω2 − ω40

ω2. (19)

This is stable for ω > ω0.

For θ3 = π we findΩ2i = −ω20 − ω2 < 0 , (20)

which is always unstable.

6



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WRITTEN EXAM - ELECTRICITY AND MAGNETISM — WITH SOLUTIONS

September 4, 2015






Electromagnetism 1: Magnetic mirror

When a charged particle orbits around magnetic field lines while also drifting along thefield into a region of higher field strength, then the particle experiences a force that reducesthe component of velocity parallel to the field. This force slows the motion along the fieldlines and may reverse it. This is the basis of a “magnetic mirror,” illustrated below.

In this problem we will investigate this phenomenon using a cylindrical coordinate systemin which the z-axis is the symmetry axis, r denotes the cylindrical radius (the distance fromthe z-axis), and φ is the azimuthal angle measured from the x-axis, as illustrated below.

You may find it useful to remember

~∇ · ~A = 1r

∂

∂r(rAr) +

1

r

∂Aφ∂φ

+∂Az∂z

. (1)

1

Consider a magnetic field ~B which is axially symmetric around the z-axis. The z-component is

Bz(r, z) = B0 +B′0z, (2)

where B′0 is a constant.

We inject a particle with mass m and charge q > 0 at point A, located at x = r0 andy = z = 0. The particle’s initial velocity has both horizontal and vertical components:

~v = −vh0ŷ + vz0ẑ, (3)

where vh0 =qmcB0r0 and vz0 � vh0. Both components of the velocity are non-relativistic.

(a) (1 pt) Calculate the radial component of the magnetic field, Br(r, z).

(b) (1 pt) Show that throughout the subsequent motion,

v2(t) = v2h0 + v2z0. (4)

Now assume that during each orbit around the z-axis, the horizontal velocity is nearlyin the −φ̂ direction and the change in radius ∆r of the orbit is negligible compared to theinstantaneous radius r.

(c) (1 pt) What is the radius r of the orbit as a function of vφ, Bz and physical constants?

(d) (2 pts) Find an equation for the z-dependence of vh (the horizontal speed) as theparticle drifts in the z-direction. One way to do so is to write the equation of motionfor the φ̂ component of the velocity, and then use vzdt = dz.

(e) (3 pts) Using your result from part (c), integrate your equation from part (d) to showthat the horizontal speed varies with z as

vh(z)

vh0=

√Bz(z)

B0. (5)

(f) (2 pts) Find the value of z for which vz = 0. This is the reflection point of themagnetic mirror, where the spiraling particle stops its upward motion and starts movingdownward.

Express your answer entirely in terms of B0, B′0, vh0 and vz0.

2

• Solutions by J. Winn (August 2015), based on a version of this problem that appeared inthe Spring 1986 exam.

(a) Requiring the divergence of ~B to vanish,

1

r

∂

∂r(rBr) +

∂Bz∂z

= 0 (6)

∂

∂r(rBr) = −B′0r (7)

rBr = −r2

2B′0 + constant. (8)

Note that the constant of integration must be zero, for Br to remain finite at r = 0.Therefore

Br = −rB′0

2. (9)

(b) The only force on the particle is the Lorentz force, which cannot change the kineticenergy of the particle:

~F =q

c~v × ~B (10)

dEKdt

= ~F · ~v = 0 (11)

EK =1

2mv2 =

1

2m(v2h + v

2z) = constant (12)

from which the desired result follows.

(c) Since the motion is nearly circular, we have

mv2hr

=qvhBzc

, (13)

which gives

r =mvhc

qBz. (14)

(d) The φ component of the Lorentz equation of motion is

Fφ = mdvφdt

=q

c(vzBr − vrBz) , (15)

and since we are assuming a nearly circular orbit, vr ≈ 0. Thus

dvφdt

=q

mcvzBr = −

q

mcvzrB′0

2. (16)

Dividing both sides by vz, and using vh = −vφ (the particle circulates in the −φdirection) we obtain

dvhvzdt

=dvhdz

=q

mc

rB′02. (17)

3

(e) Replace r by mvφc/qBz in the preceding equation, giving

dvhdz

=q

mc

mvhcB′0

2qBz(18)

1

vh

dvhdz

=1

2

B′0Bz. (19)

We integrate to find vh(z): ∫ vhvh0

dvhvh

=1

2B′0

∫ z0

dz

B0 +B′0z(20)

ln

(vhvh0

)=

1

2ln

[Bz(z)

B0

](21)

vh(z)

vh0=

√Bz(z)

B0. (22)

(f) Denote by zr the height of the reflection point, at which vz(zr) = 0. From part (b),

[vh(zr)]2 = v2h0 + v

2z0. (23)

Now we make use of our result from part (e), by writing[vh(zr)

vh0

]2= 1 +

(vz0vh0

)2(24)

B0 +B′0zr

B0= 1 +

(vz0vh0

)2(25)

allowing us to solve for zr:

zr =

(B0B′0

)(vz0vh0

)2. (26)

4

Electromagnetism 2: Electromagnetic waves in a plasma

A plane electromagnetic wave of angular frequency ω propagates in a uniform plasmawith electron density Ne. The plasma is locally neutral, ρ = 0. The electromagnetic wavegenerates periodic currents within the plasma that, as the problem will show, modify theindex of refraction of the medium compared to that of the vacuum.

Assume the plasma has no resistivity and neglect radiation pressure effects as well ascurrents due to the ions.

(a) (3 pts) Relate the current ~J(~r, t) in the plasma to the wave’s electric field ~E(~r, t) orderivatives thereof. Assume magnetic forces can be neglected.

(b) (3 pts) Write down the appropriate Maxwell equations and derive the wave equation.Find the dispersion relation ω(k) and the lowest frequency electromagnetic wave thatcan propagate the plasma.

(c) (2 pts) Find the phase and group velocity for electromagnetic waves in the plasma.Compare those velocities with c, the speed of light in vacuum.

(d) (1 pt) Find the index of refraction n of the plasma as a function of frequency.

(e) (1 pt) If a plane electromagnetic wave is incident on a plane interface between a vacuumand the plasma, what is the critical angle for total reflection, measured from the normalto the interface?

5

• Solutions by Markus Klute (August 2015) An earlier version of this problem also appearedon the Fall 2010 exam.

(a) As stated in the problem, we ignore the ions and focus on the electrons. The motionof an electron at location ~r is

m~̈s = −e ~E(~r, t) , (1)where ~s is the small displacement vector of the electron, lying along the direction ofthe electric field. Because of the harmonic nature of the wave, the solution for ~s willalso be harmonic with angular frequency ω so that ~̈s = −ω2~s. We thus get

~s(~r, t) =e

mω2~E(~r, t) . (2)

It follows that the velocity of the electron is given by

~v(~r, t) =e

mω2∂ ~E(~r, t)

∂t. (3)

The overall current density in the plasma at ~r is given by

~J = −eNev(~r, t) = −e2Nemω2

∂ ~E(~r, t)

∂t. (4)

(b) In Gaussian units the appropriate Maxwell equations are

~∇ · ~B = 0 (5)~∇ · ~E = 4πρ = 0 (6)

~∇× ~B = 4πc~J +

1

c

∂ ~E

∂t(7)

~∇× ~E = −1c

∂ ~B

∂t(8)

Note that ρ is taken to be zero due to the condition of local charge neutrality. We nowuse the results for ~J to simplify Maxwell’s equations for the plasma.

~∇× ~E = −1c

∂ ~B

∂t(9)

~∇× ~B = −4πe2Ne

mcω2∂ ~E

∂t+

1

c

∂ ~E

∂t(10)

=1

c

(1− 4πe

2Nemω2

)∂ ~E

∂t(11)

We can combine these two equations, while making use of ~∇ · ~E = 0, to find the waveequation in the plasma

∇2 ~E =(

1− 4πe2Ne

mω2

)1

c2∂2 ~E

∂t2. (12)

6

We can write this as

∇2 ~E =(

1−ω2pω2

)1

c2∂2 ~E

∂t2, (13)

where ωp is the plasma frequency

ω2p ≡[

4πe2Nem

]cgs

or

[e2Ne�0m

]mks

. (14)

The wave equation admits to plane wave solutions with dispersion relation

k2 =

(1−

ω2pω2

)ω2

c2, (15)

or equivalently,

ω =√ω2p + k

2c2 . (16)

Since k has no real solution for ω < ωp, we conclude that the propagating waves musthave a frequency above the plasma frequency.

(c) The phase and group velocities are given by

vp =ω(k)

kand vg =

dω

dk. (17)

We thus find

vp =c√

1− ω2p

ω2

and vg =kc2

ω= c

√1−

ω2pω2

. (18)

The group velocity is less than c and the phase velocity larger than c.

(d) The index of refraction n is defined from the phase velocity as vp =cn. We thus find

n =

√1−

ω2pω2

. (19)

Note that n < 1.

(e) A plane wave incident on a planar boundary with a medium having an effective indexof refraction n is subject to Snell’s law

nvac sin θinc = nplasma sin θrefr , (20)

where the angles are measured with respect to the direction normal to the interface.For our problem, this results in

sin θinc = n sin θrefr . (21)

Since the index of refraction n of the medium is less than unity the refracted anglereaches 90◦ before the incident angle reaches 90◦. Thus, the critical incident angle,after which the incident radiation is totally reflected, is

sin θcritical = n . (22)

7



Room 4-315 Fax: (617) 258-8319


WRITTEN EXAM - STATISTICAL MECHANICS — WITH SOLUTIONS

September 4, 2015






Statistical Mechanics 1: Potts Model

The q-state Potts model generalizes the Ising model. There is a variable σi ∈ {1, 2, . . . , q}at each lattice site. The Hamiltonian is given by a sum over nearest neighbors:

HPotts = −3J

2

∑〈i,j〉

δσi,σj . (1)

There are N lattice sites. Assume J > 0. For parts (a) and (b), q ≥ 2 is arbitrary, and for(c) and (d), we assume q = 3.

(a) (1 pt) What is the entropy of the system at T = 0?

(b) (3 pts) For this part only, suppose the N sites are arranged on a line with open bound-ary conditions. Write down the (exact) free energy.

(c) (1 pt) For the q = 3 case, show that the model is equivalent to

H = −J∑〈i,j〉

~si · ~sj (2)

with each ~si restricted to take values in the set

~si ∈{(

10

),

(−1/2√

3/2

),

(−1/2−√

3/2

)}(3)

(d) (5 pts) Let ~m denote the mean-field magnetization∑

i ~si/N . Suppose that each sitein the lattice interacts with z other sites. Use a mean-field approximation to derivean expression for the free energy. (Your answer should be left in terms of a solutionto a transcendental equation.) In other words, replace the true nearest-neighbor inter-actions with an approximation in which every site interacts with every other site withan interaction strength rescaled appropriately. Is there a first-order (i.e. discontinuousin m) phase transition?

[Hint: it may be helpful to consider ~m of the form (m, 0).]

1

• Solutions by Aram Harrow (August 2015)

(a) kB ln(q)

(b)

Z =∑

σ1,...,σN

Mσ1,σ2Mσ2,σ3 · · ·MσN−1,σN , (4)

where Mi,i = e−β3J/2 and Mi,j = 1 for i 6= j. Let I denote the q × q identity matrix

and u = 1/√q, where 1 is the all-ones vector. Then

M = quuT + (e−β3J/2 − 1)I,

and so MN = (e−β3J/2 + q − 1)NuuT + (e−β3J/2 − 1)N(I − uuT ). We compute

Z =∑i,j

(MN)i,j = quTMNu = q(e−β3J/2 + q − 1)N ,

and soF = −kBT (ln(q) +N ln(e−β3J/2 + q − 1)). (5)

(c) The inner product ~si · ~sj is either 1 (when i = j) or −1/2 (when i 6= j). This yieldsthe same Hamiltonian up to an overall additive constant.

(d) We can replace the sum over 〈i, j〉 with zN

times the sum over all pairs i, j. Then weget

E = −J zNN2 ~m · ~m = −JzN |~m|2.

If ~m = (m, 0) then the energy is E = −JzNm2.The entropy in this case can be determined from the populations N1, N2, N3. LetNi = Npi. We can determine p1, p2, p3 by solving the linear system of equations

p1 −p2 + p3

2= m (6)

p2 − p3 = 0 (7)p1 + p2 + p3 = 1 (8)

(9)

which have solution p1 =1+2m

3, p2 = p3 =

1−m3

. The entropy is

S = NkB

3∑i=1

pi ln(1/pi) =NkB

3

((1 + 2m) ln

3

1 + 2m+

2(1−m)3

ln3

1−m

). (10)

Thus the free energy is

F = E − TS = N(−Jzm2 − kBT

3

((1 + 2m) ln

3

1 + 2m+

2(1−m)3

ln3

1−m

)).

(11)

2

To calculate the derivative, observe that

~∇S(p1, p2, p3) =3∑i=1

(− ln(pi)− 1)ei and∂~p

∂m=

1

3

2−1−1

, (12)so that dS

dm= NkB

23

ln 1−m1+2m

.

The equilibrium value of m then satisfies

0 =1

N

dF

dm= −2Jzm− 2

3kBT ln

1−m1 + 2m

, (13)

or equivalently1−m1 + 2m

= exp(−3βJzm), (14)

where β ≡ 1/kBT .There is always a solution at m = 0 corresponding to the disordered state. Whetherthere is a solution at m > 0 depends on the value of βJz. To see this, we evaluate thesecond derivative of F , which is proportional to

−3βJz − 11−m

+2

1 + 2m.

At m = 0 this is 1− 3βJz which is negative for βJz > 1/3. When this occurs there isspontaneous order. This means that there is a first-order phase transition.

3

Statistical Mechanics 2: Magnetic susceptibility of a spinful Boltzmann gas

Consider a classical ideal gas of N spin-1/2 atoms moving in a container of volume V .In the presence of a weak external magnetic field H the energy of the nth such atom maybe taken to be

E(~pn, σn) =~p2n2m− γHσn (1)

Here σn = ±1 describes the two possible spin orientations of the atom and ~pn is the momen-tum of the atom. γ is a positive constant.

(a) (2 pts) Calculate the change in free energy due to the magnetic field.

(b) (3 pts) Calculate the average magnetization per atom 〈M〉 = 1Nγ〈∑

n σn〉.

(c) (3 pts) Calculate the variance in the magnetization 〈M2〉 − 〈M〉2.

(d) (2 pts) Use the above to calculate the magnetic susceptibility χ = d〈M〉dH

. How is thesusceptibility related to the variance?

4

• Solutions by Senthil Todadri (August 2015)

(a) Partition function Z is given by

Z =1

N !V N

(∫d3p

2π~∑σ=±1

e−β

(~p2

2m−γσH

))N(2)

The translational and spin degrees of freedom are decoupled. Therefore the totalpartition function factorizes as

Z = ZtransZs (3)

with Ztrans independent of H (the partition function of an ideal gas of spinless atoms),and

Zs = (2 cosh(βγH))N (4)

The free energy

F = Ftrans −N

βln(2 cosh(βγH)) (5)

(with Ftrans = − 1β lnZtrans).

(b) The average magnetization per atom is obtained as

〈M〉 = − 1N

∂F

∂H(6)

This gives〈M〉 = γtanh(βγH) (7)

(c) The variance is obtained from the partition function through

〈M2〉 − 〈M〉2 = 1N2β2

∂2Z

∂H2(8)

= − 1N2β

∂F

∂H(9)

=γ2

NSech2(βγH) (10)

(d) Differentiatiing 〈M〉 with respect to field we get the magnetic susceptibility

χ = βγ2Sech2(βγH) (11)

We also have

〈M2〉 − 〈M〉2 = kBTχN

(12)

5



Room 4-315 Fax: (617) 258-8319


WRITTEN EXAM - QUANTUM MECHANICS — WITH SOLUTIONS

September 4, 2015






Quantum Mechanics 1: A spin-1/2 particle in a magnetic field

A spin-1/2 particle interacts with a magnetic field via the Hamiltonian

H = ~σ · ~B, σx =(

0 11 0

), σy =

(0 −ii 0

), σz =

(1 00 −1

). (1)

(We absorb the other relevant constants into the definition of ~B.)

(a) (5 pts) Find the eigenvalues and eigenvectors of H. You may find it convenient to

write ~B as

~B =

B sin(θ) cos(φ)B sin(θ) sin(φ)B cos(θ)

.(b) (5 pts) Let ~B = B0ẑ. Suppose at time t = 0 the spin is initially pointing in the +x̂

direction. After time t = T , the spin is measured along the ŷ direction. What are thepossible outcomes and what are their probabilities?

1

• Solutions by Aram Harrow (August 2015)

(a) Observe that the Pauli matrices satisfy the multiplication rule {σi, σj} = 2δi,jI, where{X, Y } = XY + Y X is the anticommutator and I is the 2× 2 identity matrix. Fromthis we calculate

H2 =1

2{H,H} = 1

2

3∑i,j=1

BiBj{σi, σj} =∑i,j

BiBjδi,j = B2.

Thus H has eigenvalues in the set {±B}. Since trH = 0, H must have one eigenvalueequal to B and one equal to −B. To calculate these we need to solve

B

(cos(θ) sin(θ)e−iφ

sin(θ)eiφ − cos(θ)

)(xy

)= ±B

(xy

),

for x, y. Assume without loss of generality that x is real (since we can always add anoverall phase to make this true) and that sin(θ) 6= 0 (by continuity). Then from

±x = cos(θ)x+ sin(θ)e−iφy,

we find that e−iφy is also real. We can thus write x = cos(α) and y = sin(α)eiφ. Theeigenvalue equation then becomes

±(

cos(α)sin(α)eiφ

)=

(cos(θ) cos(α) + sin(θ) sin(α)

(sin(θ) cos(α)− cos(θ) sin(α))eiφ)

=

(cos(θ − α)

sin(θ − α)eiφ).

The +B eigenstate then has α = θ/2 while the −B eigenstate has α = π + θ/2, andso the corresponding eigenvectors are

|+B〉 =(

cos(θ/2)sin(θ/2)eiφ

)and |−B〉 =

(− sin(θ/2)cos(θ/2)eiφ

)(2)

(b) The state after time T is

e−iHT/~1√2

(11

)=

1√2

(e−iBT/~

eiBT/~

)=

1√2

(e−iω0T

eiω0T

),

where we have defined ω0 = B/~. The possible outcomes after a ŷ measurement are

±ŷ, corresponding to vectors 1√2

(1±i

). The corresponding probabilities are

Pr[±ŷ] =∣∣∣∣e−iω0T ± ieiω0T2

∣∣∣∣2 .These cases can be separately calculated as

Pr[+ŷ] =

∣∣∣∣e−iω0T+iπ4 + eiω0T−iπ42∣∣∣∣2 = sin2(ω0T − π/4) (3)

Pr[−ŷ] =∣∣∣∣e−iω0T−iπ4 + eiω0T+iπ42

∣∣∣∣2 = sin2(ω0T + π/4) (4)2

Quantum Mechanics 2: Two interacting fermions

Consider two identical fermions of mass m interacting with each other through an at-tractive harmonic potential. The Hamiltonian is

H =p212m

+p222m

+k

2(x1 − x2)2 (1)

x1,2 are the coordinates of the two fermions, and p1,2 are the conjugate momenta. Forsimplicity assume that the spins of both fermions are polarized in the same direction (e.g,in the ↑ direction) so that we may ignore the spin degree of freedom.

(a) (2 pts) State the restriction imposed by Fermi statistics on acceptable wave functionsψ(x1,x2) of this system.

(b) (2 pts) Rewrite H in terms of center-of-mass and relative coordinates.

(c) (3 pts) Ignoring the restriction imposed by Fermi statistics what is the ground stateenergy? What is the energy of the first excited bound state?

(d) (3 pts) Including the effects of Fermi statistics what is the ground state energy? Whatis the degeneracy of the ground state?

3

• Solutions by Senthil Todadri (August 2015)

(a) The wave function must be antisymmetric under exchange: ψ(x1,x2) = −ψ(x2,x1).

(b) Introduce center-of-mass R = x1+x22

, and relative coordinates x = x1 − x2. Thecorresponding conjugate momenta are P = p1 + p2 and p =

p1−p22

. In terms of thesethe Hamiltonian is

H =P2

4m+

p2

m+kx2

2(2)

(c) As P commutes with the rest of the Hamiltonian we can set fix it to determine theeigen-energies. The ground state and the lowest energy excited bound states will haveP = 0. The relative motion is described by a 3d simple harmonic oscillator at a

frequency ω =√

2km

.

Therefore ground state energy is 32~ω and first excited bound state is at 5

2~ω. The

ground state is singly degenerate and the excited bound state is triply degenerate.

(d) The antisymmetry condition means that the orbital angular momentum L of the rela-tive coordinate must be odd. In the ground state of the 3d oscillator, L = 0 and hencethis is not acceptable once we impose Fermi statistics. The first excited bound stateshave L = 1, and hence they are acceptable states. Thus for fermions, the ground stateenergy is 5

2~ω, and the ground state is three fold degenerate.

4

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