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Columbia UniversityDepartment of Physics

QUALIFYING EXAMINATION

Monday, January 13, 202010:00AM to 12:00PM

Classical PhysicsSection 1. Classical Mechanics

Two hours are permitted for the completion of this section of the examination. Choose4 problems out of the 5 included in this section. (You will not earn extra credit by doing anadditional problem). Apportion your time carefully.

Use separate answer booklet(s) for each question. Clearly mark on the answer booklet(s)which question you are answering (e.g., Section 1 (Classical Mechanics), Question 2, etc.).

Do NOT write your name on your answer booklets. Instead, clearly indicate your ExamLetter Code.

You may refer to the single handwritten note sheet on 812” × 11” paper (double-sided) you

have prepared on Classical Physics. The note sheet cannot leave the exam room once theexam has begun. This note sheet must be handed in at the end of today’s exam. Pleaseinclude your Exam Letter Code on your note sheet. No other extraneous papers or booksare permitted.

Simple calculators are permitted. However, the use of calculators for storing and/or recov-ering formulae or constants is NOT permitted.

Questions should be directed to the proctor.

Good Luck!

Section 1 of 12

1. A vertical circular wire loop with its center at the origin rotates about the z-axis witha constant angular velocity ω. A bead can move freely without friction along the loop.

a) At what positions can the bead remain at a constant angle along the loop? Theanswers depend on the value of ω; you must give the answers for all cases.

b) Which of these positions are stable equilibria and which are unstable one?

Section 1 of 12

2. Consider a train car of mass M1 able to move without friction in one dimension to whichis mounted a pendulum composed of point mass M2 suspended by a massless rod oflength `. A drum of mass M3 is placed on the car for part (b). The drum is symmetricabout an axis directed out of the plane of the diagram, has moment of inertia I aboutthis axis and radius R. It is free to roll without slipping on the top surface of the car.Solve both parts (a) and (b) in the approximation that the motion of the pendulum isin the small angle approximation.

(a) With only the car and pendulum present describe in words the two independentmodes of the system. What is the frequency of oscillation if the system starts withthe car and pendulum at rest but the pendulum makes a non-zero angle θ0 withthe vertical direction.

(b) Now include the rolling drum in the system and describe in words the three inde-pendent modes of the system. What is the frequency of oscillation if the systemstarts with the car, drum and pendulum at rest but the pendulum makes a non-zeroangle θ0 with the vertical direction.

Section 1 of 12

3. A ball is bouncing vertically and perfectly elastically in a standing elevator. The max-imum height of the bouncing ball is h0. The upward acceleration of the elevator thenchanges very slowly from 0 to g/8. Using adiabatic invariants, find the new maximumheight of the bouncing ball.

Section 1 of 12

4. There is a toy called a celt or rattleback (see figure for a representation of the spinningrattleback) which has the strange property that when placed on a table (with friction)and spun in one specific direction it slows down, then rattles and starts to spin in theopposite direction. When confronted with this behavior you might be tempted to saythat it appears as if the law of conservation of angular momentum is violated. The fullanalysis of this motion is complex, however it can at least be shown simply that thevertical component of the angular momentum is not conserved under a special conditionfor r, which is the vector from the center of mass to the point of contact with the table.What is that condition? Use the following notation:

vCM is the velocity of the center of mass;

r is the vector from the center of mass to the point of contact;

F is the net force exerted by the table on the rattleback at the point of contact of therattleback with the table;

L is the angular momentum of the rattleback;

M is the mass of the rattleback;

y is a unit vector in the upward vertical direction.

You may also find the following vector identity useful: A · (B×C) = (A×B) ·C.

Section 1 of 12

5. Consider the three following Lagrangians:

1. L = eλt m2x2

2. L = m2x2e2γx

3. L = xωx

tan−1(xωx

)− 1

2ln (x2 + ω2x2)

(a) For one of the Lagrangians write out the equation of motion for the correspondingsystem. For a second, different Lagrangian, obtain the Hamiltonian and write outthe Hamiltonian equations of motion.

(b) For one of the Lagrangians analyzed in part a, provide the general solution to theLagrange or Hamiltonian equation of motion for t > 0 in terms of relevant initialconditions at t = 0. Clearly indicate the initial values that you have assumed.

(c) For the third Lagrangian not analyzed in part a, identify as many conserved quan-tities for the physical system as you can find and indicate what continuous trans-formations of position and/or time the conserved quantities can be associated with.Ideally, show how the transformation affects the Lagrangian and how the conservedquantity follows from the form of the transformation.

Section 1 of 12

Question 1 Solution

The generic Lagrangian is

L = T − V =1

2M(r2 + r2θ2 + r2 sin2 θφ2

)−MgR cos θ.

Here we have r = R, r = 0, and φ = ω, so this becomes

L =1

2MR2

(θ2 + ω2 sin2 θ

)−MgR cos θ

=1

2MR2θ2 − Veff,

where Veff = MgR cos θ − 12MR2ω2 sin2 θ. The stationary points are the zeros of V ′eff(θ). We

compute

V ′eff(θ) = −MgR sin θ −MR2ω2 sin θ cos θ

= −MR2ω2 sin θ( g

Rω2+ cos θ

).

Thus the stationary points are at sin θ = 0 =⇒ θ = 0, π and cos θ = − g

Rω2(for

g

Rω2≤ 1).

Ifg

Rω2> 1, then there is a minimum of the effective potential at θ = π and a maximum

at θ = 0. Thus θ = π is a stable equilibrium point, and θ = 0 is unstable, as expected.

Ifg

Rω2< 1, there is a minimum at θ = cos−1

(− gRω2

)and maxima at θ = 0, π. Thus

θ = cos−1(− gRω2

)is a stable point, and θ = 0, π are unstable.

Question 2 Solution

(a) Describe the state of the system by two coordinates: θ, the angle that the pendulummakes with vertical direction, and x, the location of the train car. One mode is simpletranslation with x(t) = x0 + v0t and θ = 0. The second mode is oscillation with thelocation of the center of mass fixed. The frequency can be found by solving Newton’sequations for the motion of the pendulum and the car:

M1x = M2gθ, M2

(`θ + x

)= −M2gθ.

Eliminating x from the second equation, one finds

M2`θ = −M2

(1 +

M2

M1

)gθ,

giving an oscillation frequency

ω =

√g

`

(1 +

M2

M1

).

(b) Two modes are simple to describe: i) Simple translation of the entire system at constantvelocity with θ = 0 and ii) The drum rolls at constant velocity while the car andpendulum are at rest and θ = 0. The third, oscillatory mode can be obtained fromthe four equations for the acceleration of each of the three masses and the angularacceleration of the drum. Introducing the force F that the car exerts on the drum andthe angle φ through which the drum has rotated, we can write

M1x = M2θ − F M2

(`θ + x

)= −M2gθ

M3

(Rφ+ x

)= F Iφ = −FR.

Eliminating F , φ, and x from these equations leaves

M2θ = −M2g

(1 +

M2

M1 + M3

1+M3R2/I

),

which yields the frequency

ω =

√√√√g

`

(1 +

M2

M1 + M3

1+M3R2/I

).

Question 3 Solution

In the frame of the elevator, the energy of the ball is

E =p2

2m+mg′y,

where g′ = g + a is the effective gravitational acceleration, with a the acceleration of theelevator.

The adiabatic invariant is

J =

∮p dy = 2

∫ h

0

p dy.

Conservation of energy implies

E = mg′h =⇒ p = m√

2g′(h− y),

so

J = 2m

∫ h

0

dy√

2g′(h− y) =4√

2

3m√g′h3/2.

Thush

3/20

√g′0 = h

3/2f

√g′f .

Since g′0 = 0 and g′f = 98g, we find

hf =

(8

9

)1/3

h0 =2

32/3h0.

Question 4 Solution

The following basic equation pertains to the motion:

F−Mgy = MdvCM

dt.

So,

F = M

(dvCM

dt+ gy

),

and from the torque about the center of mass

dL

dt= r× F.

Substituting for F,

dL

dt= r×M

(dvCM

dt+ gy

)= M

(r× dvCM

dt+ gr× y

).

Now we find the vertical component of the angular momentum, which we want to show isnot conserved:

d

dt(y · L) = M y ·

(r× dvCM

dt+ gr× y

)= M y ·

(r× dvCM

dt

)+M y · (gr× y) .

The second term vanishes since the vectors are perpendicular to each other, and the firstterm can be rewritten using the given vector identity

d

dt(y · L) = M (y× r) · dvCM

dt.

This then tells us the condition we need for the vertical component of the angular momentumto not be conserved, namely when

y× r 6= 0,

in other words, when r does not lie along the vertical direction.

If you want more detail on this complicated motion (e.g. why does the rattleback changerotation when spun in one direction and not the other, i.e. exhibit chirality? It is not evidentfrom what we showed above), see the full analysis by Bondi in Proc. R. Soc. Lond. A 405,265-274 (1986).

Question 5 Solution

(a) The Lagrange equations of motion and general solutions for the three different La-grangians are:

1. L = eλt m2x2

d

dt

(eλtmx

)= 0→ meλt (λx+ x) = 0→ x = −λx.

The velocity, x has general solution x = Ce−λt and the displacement has generalsolution,

x = C1 + C2e−λt.

2. L = m2x2e2γx

d

dt

(mxe2γx

)− γmx2e2γx = 0→ me2γx

(x+ 2γx2 − γx2

)= 0→ x = −γx2

The velocity satisfies

1/x = C + γt→ x =1

C + γt.

Then, the position satisfies

x =

∫dt

1

C + γt=

1

γln (C + γt) + C ′.

3. L = xωx

tan−1(xωx

)− 1

2ln (x2 + ω2x2)

∂L

∂x=

1

ωxtan−1

(x

ωx

)+

x

ω2x2

(1 +

x2

ω2x2

)−1

− x

x2 + ω2x2

=1

ωxtan−1

(x

ωx

)+

x

ω2x2

(ω2x2

ω2x2 + x2

)− x

x2 + ω2x2

=1

ωxtan−1

(x

ωx

)∂L

∂x= − x

ωx2tan−1

(x

ωx

)− 1

1 + (x/ωx)2

x

ωx

x

ωx2− ω2x

x2 + ω2x2

= − x

ωx2tan−1

(x

ωx

)− 1

x.

The equation of motion is therefore

− x

ωx2tan−1

(x

ωx

)+

1

ωx

ω2x2

x2 + ω2x2

ωxx− ωx2

ω2x2+

x

ωx2tan−1

(x

ωx

)+

1

x= 0.

This simplifies toxx− x2

x2 + ω2x2+ 1 = 0 =⇒ x+ ω2x = 0,

which has general solution of either form:

x = C cos (ωt− φ), or x = A cos (ωt) +B sin (ωt) .

(b) The Hamiltonian, H(x, p) can be obtained using the Legendre transformation,H = px − L. Evaluating the Hamiltonian and the corresponding equations of motionfor the three Lagrangians

1. L = eλt m2x2 The generalized momentum is

p =∂L

∂x= eλtmx→ x =

p

me−λt.

Then,

H =p2

me−λt − m

2eλt[ pme−λt

]2

=p2

me−λt

(1− 1

2

)=

p2

2me−λt.

The Hamiltonian equations of motion are, then,

x =∂H

∂p=

p

me−λt, (1)

p = −∂H∂x

= 0. (2)

Then, since p is a constant, the general solution can be written,

x = C − p

mλe−λt.

2. L = m2x2e2γx The generalized momentum is

p =∂L

∂x= mxe2γx → x =

p

me−2γx,

so

H =p2

me−2γx − m

2e2γx

[ pme−2γx

]2

=p2

me−2γx

(1− 1

2

)=

p2

2me−2γx.

Then, the Hamiltonian equations of motion are:

x =∂H

∂p=

p

me−2γx (3)

p = −∂H∂x

=p2

2m(−2γ) e−2γx = −2γH. (4)



Monday, January 13, 20202:00PM to 4:00PMClassical Physics

Section 2. Electricity, Magnetism & Electrodynamics


Use separate answer booklet(s) for each question. Clearly mark on the answer booklet(s)which question you are answering (e.g., Section 2 (Electricity etc.), Question 2, etc.).






Good Luck!

Section 2 Page 1 of 11

1. A plane wave is normally incident on a perfectly reflecting mirror. A glass photographicplate is placed on the mirror and forms a small angle α to the mirror. The photographicemulsion is nearly transparent. But when later developed, a striped pattern is founddue to the action of the wave. Predict the spacing of the stripes. Ignore reflections orattenuation due to the photographic plate.


2. Consider a spherical capacitor with a fixed radius a for the outer spherical shell andvacuum between the shells. If the electric field at the surface of the inner sphericalshell cannot exceed a value E0, for what radius b of the inner shell is the stored energymaximized, and how much energy can be stored?


3. Imagine that an ideal magnetic dipole m is located at the origin of an inertial systemS ′ that moves with speed v in the x direction with respect to inertial system S. In S ′,the vector potential can be expressed as

A =µ0

4π

m× r′

|r′|2,

and the scalar potential is zero. Find the scalar potential V in S (make certain allelements of the answer are expressed for an observer in S).


4. A spherical shell with radius R and uniform surface charge density σ spins with angularfrequency ω around a diameter. Find the magnetic field at its center.


5. Consider a magnetic field B with energy density UB = 10−16mec2/r3e where me is electron

mass and re is the classical electron radius. An electron is injected with velocity v0 ⊥ B;v0 << c. How many circular orbits will it make around B before its kinetic energyE0 = mev

20/2 is reduced to E1 = 10−2E0?


Question 1 Solution

For constructive interference, we have

2h

λ(2π) + π = 2πn =⇒ h =

2n− 1

4λ

Since h = d sinα,

d =(2n− 1)

4

λ

sinα

Thus the stripes are regularly spaced a distance λ/ sinα apart. The first one is at distanceλ/4 sinα from the point of contact.

Question 2 Solution

For convenience, let b = ka, with 0 < k < 1. If E0 is the field at radius ka, then for r > ka,

E = E0(ka)2

r2. Energy stored in the field is

U =ε02

∫E2 dV

=ε02

∫ a

ka

(E0

(ka)2

r2

)2

4πr2 dr

= 2πε0k4a4E2

0

∫ a

ka

1

r2dr

= 2πε0k4a4E2

0

(1

ka− 1

a

)= 2πε0a

3E20

(k3 − k4

).

We take the derivative to find the maximum:

3k2 − 4k3 = 0 =⇒ k =3

4.

Thus the maximum stored energy is achieved at b = 34a. The stored energy is

U = 2πε0a3E2

0

((3

4

)3

−(

3

4

)4)

=27

128πε0a

3E20 .

Alternatively, we can solve this problem using the capacitance of the sphere,

C = 4πε0ab

a− b= 4πε0a

k

1− k.

If we let Q be the charge on the inner sphere, the electric field at the inner surface is

E =Q

4πε0(ka)2=⇒ Q = 4πε0k

2a2E0.

Thus the energy is

U =Q2

2C=

(4πε0k2a2E0)

2

8πε0ak

1−k= 2πε0a

3E20

(k3 − k4

),

agreeing with our expression above.

Question 3 Solution

Start by applying the Lorentz transformation V = γ (V ′ + vA′x), realizing that V ′ = 0. So

V = γvµ0

4π

myz −mzy

(r′)3.

We still need to transform r′, which is

(r′)2 = γ2(R2 − v2

c2R2 sin2 θ

),

where R is the vector in S from the (instantaneous) location of the dipole to the point ofobservation and θ is the angle between R and x. And thus we have

V = γvµ0

4π

myz −mzy

γ3R3(1− v2

c2sin2 θ

)3/2=µ0

4π

v · (m×R)(

1− v2

c2

)R3(1− v2

c2sin2 θ

)3/2 .

Question 4 Solution

We divide the sphere into rings centered on the z axis. These rings have width dw = Rdθ.The speed of each ring is v = ωR sin θ. Focusing on one ring at angle θ, the charge passinga given point in time dt is

dq = σ(dw)(vdt) = σωR2 sin θ dθ dt.

Thus the ring current is

I =dq

dt= σωR2 sin θ dθ.

The Biot-Savart field dB produced by a small piece of the ring with length d` has magnitude

dB =µ0

4π

I d`

R2.

Integrating over the whole ring, the horizontal components cancel, with the vertical compo-nents (dB sin θ) add. The length of the ring is ` = 2πR sin θ. Thus the contribution to thenet field due to a ring of width dθ at angle θ is

Bring = zµ0

4π

I`

R2sin θ

= zµ0

4π

(σωR2 sin θ dθ) (2πR sin θ)

R2sin θ

= z1

2µ0σωR sin3 θ dθ.

Integrating over θ from 0 to π, and doing so by rewriting sin3 θ = sin θ (1− cos2 θ), yields

B = z1

2µ0σωR

∫ π

0

sin θ(1− cos2 θ

)dθ

= z1

2µ0σωR

(1

3cos3 θ − cos θ

) ∣∣∣∣π0

= z2

3µ0σωR.

Question 5 Solution

The electron with velocity v ⊥ B experiences acceleration due to the Lorentz force:

a =−ev×B

mec.

It rotates around B with Larmor radius r found from a = v2/r. The rotation period is

T =2πr

v=

2πv

a= 2π

mec

eB.

The acceleration a gives the second time derivative of the dipole moment |d| = ea, and theelectron loses energy due to the dipole radiation with rate given by the Larmor formula:

dE

dt= −2d2

3c3= −2e4v2B2

3m2ec

5= −4e4B2

3m3ec

5E =⇒ ln

E0

E(t)=

4e4B2t

3m3ec

5.

The energy is reduced to E1 after time

t1 =3m3

ec5

4e4B2lnE0

E1

.

Using UB = B2/8π and re = e2/mec2 one finds the number of Larmor orbits

N1 =t1T

=3m2

ec4

8πe3BlnE0

E1

=3 ln(E0/E1)

(8π)3/2 (r3eUB/mec2)1/2=

3 ln 100

10−8(8π)3/2≈ 1.1× 107.



Wednesday, January 15, 202010:00AM to 12:00PM

Modern PhysicsSection 1. Quantum Mechanics


Use separate answer booklet(s) for each question. Clearly mark on the answer booklet(s)which question you are answering (e.g., Section 1 (Classical Mechanics), Question 2, etc.).






Good Luck!


1. Consider the two-dimensional simple harmonic oscillator with Hamiltonian

H0 =p2x2m

+p2y2m

+1

2k(x2 + y2).

(a) Write down the general equation for the energy eigenvalues of this Hamiltonian.

(b) Write down the energy eigenvalues for the ground state and the first excited state.What are the degeneracies of these states, if any?

(c) Write down the energy eigenfunctions for the ground state and the first excitedstate.

A perturbation is added to the above Hamiltonian of the form H ′ = bxy, where b << 1.

(d) Find the energy eigenvalues of the perturbed Hamiltonian H0 + H ′ for the groundand the first two excited states.

(e) Sketch the energy level diagram for the ground and first two excited states of theperturbed Hamiltonian.


2. Consider a particle constrained to move in two dimensions (xy-plane) on a circular ring.The only variable of the system is the azimuthal coordinate angle φ. The state of thesystem is described by a wave function Ψ(φ), which must satisfy the symmetry

Ψ(φ+ 2π) = Ψ(φ).

The wave function must also be normalized, i.e.,∫ 2π

0

|Ψ(φ)|2 dφ = 1.

Suppose initially that there is no potential applied to the particle. The Hamiltonian forthe system thus consists only of the kinetic energy of the particle, which can be written

H0 =L2z

2K,

where the operator Lz = −i~ ddφ

and K is a constant.

(a) Calculate the eigenvalues and normalized eigenfunctions of the Hamiltonian H0.Which of the energy levels are degenerate?

Now suppose that the particle also experiences a potential, which can be treated as asmall perturbation to the overall Hamiltonian:

H ′ = −λ cos(2φ),

where λ is a constant.

(b) Considering the perturbed Hamiltonian, calculate the lowest non-vanishing correc-tion to the ground state energy due to H ′. The following integral may be useful:

1

2π

∫ 2π

0

einφ cos(mφ) dφ =1

2(δm,n + δm,−n) ,

where δi,j is the Kronecker δ-function.

(c) Calculate the ground state wave function to first order in λ.


3. Consider a hydrogen atom. The spin-orbit interaction is written as

Hso = AS · L,

where S is the spin of the electron, L is the orbital angular momentum and A =e2/2m2c2r3.

(a) Describe in words the origin of the spin-orbit interaction.

(b) Construct the basis of wavefunctions/eigenstates that diagonalizes Hso.

(c) Obtain the spin-orbit interaction energies for hydrogen in the state with radialquantum number n = 2. You may express your results in terms of the matrixelements of A without explicitly evaluating those matrix elements.


4. A 2× 2 matrix is parameterized as ρ = 12(A1 + σ ·B), where 1 is the identity matrix,

A is a real number, B is 3-dimensional vector of real numbers, and σ represents the 3Pauli spin matrices:

σ1 = σx =

(0 11 0

), σ2 = σy =

(0 −ii 0

), σ3 = σz =

(1 00 −1

)(a) Find the conditions on A and B if this matrix is a valid density matrix for a pure

state. You may find this identity (written using Einstein summation convention)useful:

σiσj = δij1 + iεijkσk.

(b) Assuming the 2-state system is spin-1/2, and the matrix is in the z-representation,find the values of A and B that maximize the expectation value of 〈Sy〉, the y-component of the spin operator.

(c) What conditions on A are B are needed to represent the density matrix for a mixedstate?


5. Consider a particle of mass m moving in the three-dimensional potential

V (r, θ, φ) = V0δ(r − a),

where V0 is a constant. Suppose the angular momentum ` = 0. Then the Schrodingerequation for the radial wavefunction is(

d2

dr2+

2

r

d

dr+

2m

~2(E − V0(.r − a)

))R(r) = 0.

In terms of k =√

2mE~2 , give

(a) the solution to the Schrodinger equation when 0 < r < a,

(b) the solution to the Schrodinger equation when r > a.

(c) Give the matching condition for parts (a) and (b) at r = a.

(d) If one writes R(r) = h(2)0 (kr) + e2iδh

(1)0 (kr) for r > a, can you evaluate e2iδ? What

is δ when V0 is very large? (Note the difference between the Dirac delta functionand the phase shift, δ.)

The solutions to the differential equation(d2

dz2+

2

z

d

dz+ 1− `(`+ 1)

z2

)f(z) = 0,

are the spherical Bessel functions j`(z) and n`(z), or equivalently, the spherical Hankel

functions h(1)` (z) and h

(2)` (z). For ` = 0, we have

j0(z) =sin(z)

z, n0(z) = −cos(z)

z, h

(1)0 (z) =

eiz

iz, h

(2)0 (z) = −e

−iz

iz.


Question 1 Solution

(a) This is just two independent one-dimensional simple harmonic oscillator, so the generalsolution is just Enx,ny = Enx + Eny = (nx + ny + 1)~ω, with ω2 = k/m.

(b) The energies are Eground = E0,0 = ~ω and E1 = E01 = E10 = 2~ω. The first excitedstate is doubly degenerate.

(c) The wavefunction for the ground state is ψ0 = ψ0(x)ψ0(y). The wavefunction forthe first excited state is ψ1 = ψ0(x)ψ1(y) or ψ1(x)ψ0(y). The ψnx(x) and ψny(y) arejust the solutions to the one-dimensional simple harmonic oscillator. Recalling thatψ0 ∝ e−β

2x2/2 and ψ1 ∝ xe−β2x2/2, where β2 = mω/~. Normalizing, we obtain

ψ0(x) =

(β

π1/2

)1/2

e−β2x2/2

ψ1(x) =

(2β3

π1/2

)1/2

xe−β2x2/2,

and the same for the y direction.

(d) The ground state energy is unchanged, Eground = ~ω, since

〈00|xy|00〉 =

∫ ∞−∞

ψ0(x)xψ0(x) dx

∫ ∞−∞

ψ0(y)yψ0(y) dy = 0

by parity.

Computing the energy of the first excited state involves finding the eigenvalues of theperturbing Hamiltonian, which in this basis is

H ′ = b

(〈10|xy|10〉〈10|xy|01〉〈01|xy|10〉〈01|xy|01〉

).

The diagonal elements are (by parity)

〈1|x|1〉〈0|y|0〉 = 0, 〈0|x|0〉〈1|y|1〉 = 0.

The off-diagonal elements obey

〈01|xy|10〉 = 〈10|xy|01〉 = (〈0|x|1〉)2 .

Thus we only need to calculate∫ ∞−∞

ψ0(x)xψ1(x) dx =1√2β.

Thus

H ′ = b

(0 1/2β2

1/2β2 0

).

Aside: This is way easier using

x =1√2β

(ax + a†x

)y =

1√2β

(ay + a†y

),

where

a†x |00〉 = |10〉 , a†y |00〉 = |01〉 , ax |10〉 = |00〉 , ay |01〉 = |00〉 , etc.

Then right away 〈01|xy|10〉 = 1/2β2 by inspection.

To find the eigenvalues of H ′, we solve

det

(−λ b/2β2

b/2β2 −λ

)= 0 =⇒ λ = ± b

2β2.

Thus the perturbed energies are ~ω and 2~ω ± b/2β2.

(e) We find

2~ω + b/2β2

2~ω − b/2β2

~ω

Question 2 Solution

(a) The eigenvalues of H0 = − ~2

2K

d2

dφ2are En =

m2~2

2K, where m is an integer. The

normalized eigenfunctions are

Ψm(φ) =1√2πeimφ.

All the energy levels except m = 0 are doubly degenerate, since m2 = (−m)2.

(b) The first-order correction to the ground state is

E(1)0 =

⟨Ψ

(0)0 (φ)

∣∣∣H ′∣∣∣Ψ(0)0 (φ)

⟩= − λ

2π

∫ 2π

0

cos(2φ) dφ = 0.

The second-order correction is

E(2)0 =

∑m 6=0

∣∣∣⟨Ψ(0)0

∣∣∣H ′∣∣∣Ψ(0)m

⟩∣∣∣2E

(0)0 − E

(0)m

=∑m 6=0

∣∣∣− λ2π

∫ 2π

0eimφ cos(2φ) dφ

∣∣∣2−m2~2/2K

= − λ2K

2π2~2∑m 6=0

1

m2

∣∣∣∣∫ 2π

0

eimφ cos(2φ) dφ

∣∣∣∣2= −λ

2K

2~2∑m6=0

1

m2|δ2,m + δ2,−m|2

= −λ2K

4~2.

(c) The first-order correction to the ground state wavefunction is

∣∣∣Ψ(1)0

⟩=∑m6=0

⟨Ψ

(0)m

∣∣∣H ′∣∣∣Ψ(0)0

⟩E

(0)0 − E

(0)m

∣∣Ψ(0)m

⟩=

λ/2

2~2/K

∣∣∣Ψ(0)2

⟩+

λ/2

2~2/K

∣∣∣Ψ(0)−2

⟩=λK

4~21√2π

(e2iφ + e−2iφ

)(from part (b))

=λK

2√

2π~2cos(2φ).

Question 3 Solution

(a) In the rest frame of the electron the proton is moving with velocity −v, which producesa magnetic field B ∼ L. The coupling of this field to the intrinsic magnetic momentfrom S is the spin-orbit interaction Hso.

(b) We rewrite the spin-orbit interaction Hso = AS · L as

Hso =A

2

(J2 − L2 − S2

),

where J = S + L is the total angular momentum. The states with quantum numbersJ , L, and S make this Hamiltonian diagonal. The states can be written as |j,m, l, s〉,where m is the z-component of J . In this basis the matrix elements of Hso are

〈Hso〉 = 〈A(r)〉nl~2(j(j + 1)− l(l + 1)− 3

4

).

(c) For n = 2, the allowed values of l are 0 or 1. The allowed values of j are 1∓ 1/2. Thevalue j = 1/2 is obtained from l = 0, 1. There are two terms. The value j = 3/2 isobtained from l = 1 only. There is one term. There are also two matrix elements:

〈A(r)〉2,0 = A0, 〈A(r)〉21 = A1.

The spin-orbit interaction energies are then obtained with the above equations.

Question 4 Solution

(a) A valid density matrix requires Tr [ρ] = 1 =⇒ A = 1 (since Pauli matrices aretraceless). For a pure state, it must also satisfy ρ2 = ρ so

ρ =1

2(1 + σ ·B) =⇒ ρ2 =

[1

2(A1 + σ ·B)

]2=

1

2· 1

2

[1 + 2σ ·B + (σ ·B)2

].

Using

(σ ·B)2 = (σiBi)(σjB

j) = δijBiBj

1 + iεijkBiBjσk = B ·B1 + i(B×B) · σ = B ·B1,

we get

ρ2 =1

2· 1

2(1 + 2~σ ·B + (B ·B)1) , (1)

which requires B ·B = 1 to satisfy ρ2 = ρ.

(b) We find

〈Sy〉 = Tr[ρSy

]=

1

2Tr

[(1 + σ ·B)

~2σy

]=

~2· 1

2Tr [(σxBx + σyBy + σzBz)σy] ,

but σxσy = iσz, which is traceless, and similarly for σzσy, while σyσy = 1, so B =

(0, 1, 0) maximizes 〈Sy〉 to ~2, which makes physical sense.

(c) For a mixed state, we must still have Tr [ρ] = 1, so we still require A = 1 (of course).But a mixed state is signified by Tr [ρ2] < Tr [ρ]; using results from part (a) we see thatthe necessary condition is |B|2 < 1.

Those seeking more information should look at the Wikipedia article on the Blochsphere.

https://en.wikipedia.org/wiki/Bloch_sphere

https://en.wikipedia.org/wiki/Bloch_sphere

Question 5 Solution

(a) For 0 < r < a, the differential equation is(d2

dr2+

2

r

d

dr+

2mE

~2

)R(r) = 0,

which we can rewrite as (d2

d(kr)2+

2

kr

d

d(kr)+ 1

)R(kr) = 0.

Thus the solution is R(kr) = Cj0(kr). We do not include the n0(kr) solution becauseit is not regular at r = 0.

(b) For r > a, we have the same differential equation, but we find

R(kr) = C1h(1)0 (kr) + C2h

(2)0 (kr).

(c) The matching conditions are

Cj0(ka) = C1h(1)0 (ka) + C2h

(2)0 (ka)

and ∫ a+ε

a−εdr

(d2

dr2+

2

r

d

dr+ k2 − 2mV0

~2δ(r − a)

)R(r) = 0.

The second condition implies

C1h(1)0

′(ka) + C2h

(2)0

′(ka)− Cj′0(ka) =

2mV0C

~2kj0(ka).

(d) Using part (c), we eliminate C, finding

C1h(1)0

′(ka)+C2h

(2)0

′(ka)−j

′0(ka)

j0(ka)

(C1h

(1)0 (ka) + C2h

(2)0 (ka)

)=

2mV0~2k

(C1h

(1)0 (ka) + C2h

(2)0 (ka)

).

Thus

e2iδ =C2

C1

= −h(1)0

′(ka)−

(j′0(ka)

j0(ka)+ 2mV0

~2k

)h(1)0 (ka)

h(2)0

′(ka)−

(j′0(ka)

j0(ka)+ 2mV0

~2k

)h(2)0 (ka)

.

For V0 →∞,

e2iδ = −h(1)0 (ka)

h(2)0 (ka)

= e2ika,

so δ = ka.



Wednesday, January 15, 20202:00PM to 4:00PM

Modern PhysicsSection 4. Relativity and Applied Quantum Mechanics


Use separate answer booklet(s) for each question. Clearly mark on the answer booklet(s)which question you are answering (e.g., Section 4 (Relativity and Applied Quantum Me-chanics), Question 2, etc.).



have prepared on Modern Physics. The note sheet cannot leave the exam room once theexam has begun. This note sheet must be handed in at the end of today’s exam. Pleaseinclude your Exam Letter Code on your note sheet. No other extraneous papers or booksare permitted.



Good Luck!


1. The deuteron is a bound state of a neutron and a proton. The deuteron ground statebinding energy is EB = −2.22 MeV. Model the proton-neutron potential as a squarewell of unknown depth V0 and range a = 2.0 fm, and assume that the ground state is ans-wave.

(a) Using the fact the the deuteron is a very weakly bound state, make a rough estimatefor V0. You may take the neutron and proton masses to be equal with Mc2 =940 MeV. You may find it useful to use ~c = 197.3 MeV-fm.

(b) Improve your estimate by taking into account the boundary conditions for theradial wave-function at r = a to find an approximate solution to the resultingtranscendental equation, keeping in mind that the deuteron is loosely bound.


2. A laser beam has photon number density n. Find the photon density n′ in a framemoving along the beam with velocity v.


3. Consider the following simplified model of neutrino oscillation, involving only a two-levelsystem:

We begin by assuming that there are two unique neutrino mass states, represented by|ν1〉 and |ν2〉, each with mass m1 and m2, respectively, and that each is an eigenstateof the Hamiltonian in the two-dimensional Hilbert Space spanned by these two basisvectors, i.e.:

H |ν1〉 = E1 |ν1〉 ,

H |ν2〉 = E2 |ν2〉 ,

and 〈νi|νj〉 = δij.

We also assume that when neutrinos are produced in a weak interaction they are alwaysproduced as so-called “flavor eigenstates,” defined in terms of the mass states as follows:

|νe〉 = cos θ |ν1〉+ sin θ |ν2〉 ,

referred to as an electron neutrino, and

|νµ〉 = − sin θ |ν1〉+ cos θ |ν2〉 ,

referred to as a muon neutrino.

Given a neutrino is produced at time t = 0 as a pure muon neutrino eigenstate, calculatethe probability that, after some time t > 0 measured in the lab frame, this neutrino wouldbe detectable (through a weak interaction) as an electron neutrino.

When is this probability maximal?


4. A particle of charge q and rest mass m0 is accelerated from rest by a uniform electricfield E = V0

Lx, which is generated by applying a potential difference V0 between two

parallel metal plates separated by a distance L.

(a) What is the kinetic energy of the particle when it reaches the second plate? Recallthat dE = F · dr.

(b) What is the particle’s Lorentz factor γ?

Suppose that the particle passes through a hole in the second plate into a region wherethe electric field E = 0, but there is a uniform magnetic field B = B0z. The particle isobserved to move in a circular arc of radius r.

(c) Derive an expression for r in terms of B0, V0, c, and the mass-to-charge ratio m0/q.

(d) Compute E and B in the rest frame of the particle immediately after it emergesfrom the hole, expressed in terms of B0 and the dimensionless velocity β.


5. A stream of protons traveling at velocity v = 0.9900c are directed into a two slit ex-periment where the slit separation is d = 4.00× 10−9m. An interference pattern resultson the viewing screen. What is the angle between the center of the pattern and thesecond minimum? You can assume the distance between the screen with the slits andthe viewing screen is much larger than any other relevant distance.


Question 1 Solution

(a) For s-wave states, and a two-body system with reduced mass µ, the Schrodinger equa-tion becomes

− ~2

2µ

1

r∂2r (rψ) + V (r)ψ = Eψ

which implies u(r) = rψ(r) satisfies a simple one-dimensional Schrodinger equation.The boundary condition at r = 0 requires u(r) ∝ sin k0r for r < a, with k20 = 2µ

~2 (V0 +E). (In this expression the well-depth is −V0, that is, V0 is a positive number while Eis negative for a bound state.)

If the state is very weakly bound, the wave function barely ”turns over” at r = a, thatis, k0a ≈ π

2, which implies

V0 ≈~2

2µ

( π2a

)2− E = 27.8 MeV.

(b) To refine this estimate, we must take into account the wave function for r > a, whichgoes as e−qr with q2 = −2µ

~2 E. Matching u(r and u′(r) at r = a leads to

tan k0a = −k0q,

which is the transcendental expression mentioned in the statement of the problem.

To obtain an approximate solution, use the fact that k0a is only slightly greater thanπ2

(again, the “loosely bound” condition) to express k0a = π2

+ δ to rewrite the aboveequation as

cot δ =k0q.

We expect δ << 1, leading to the approximate solution δ ≈ qk0

, so

k0a =π

2+ δ =

π

2+

q

k0,

which is a quadratic equation for k0 with solutions

k0 =π2±√

(π2)2 + 4aq

2a.

We take the + solution since that gives us π2

+ a small correction. Numerically, thisresults in k0 = 1.82

a, which then gives

V0 ≈~2

2µ

(1.82

a

)2

− E = 36.7 MeV,

very close to the value of 36.5 MeV obtained by exact (numerical) solution to the tran-scendental equation. This differs significantly from the zeroth order estimate becausethe π

2≈ 1.57 → 1.82 enters as the square in the expression for the potential depth,

even though the wave-function does indeed “barely turn over”, as shown in the belowfigure:

Question 2 Solution

Solution using 4-flux

Choose the x-axis along the beam. The four-flux of photon number is F µ = cn(1, 1, 0, 0).The corresponding flux in the moving frame F µ′ = cn′(1, 1, 0, 0) is related to F µ by Lorentztransformation:

F 0′ = γF 0 − γβF 1 = γ(1− β)cn =⇒ n′ = γ(1− β)n,

where β = v/c and γ = (1− β2)−1/2.

Solution using worldlines

Let S be the beam cross section. Consider a piece of the beam of length l = xA − xB. Itcontains N = nlS photons. As the beam propagates with the speed of light c, the boundariesof the piece, xA and xB, also move with c. Their worldlines in the (ct, x) plane satisfy

xA = ctA, xB = ctB − l, (1)

(the reference time t = 0 was chosen when xA = 0). When the same piece is viewed in themoving frame, its cross section is A′ = A (the Lorentz boost along x does not touch y,z),and its length is changed to l′. Then

n′l′A′ = n lA = N = const. =⇒ n′

n=l

l′.

It remains to find l′. For a pair of events (ctA, xA) and (ctB, xB), Lorentz transformation ofthe 4-vector connecting the events gives

c(tA − tB) = γc(t′A − t′B) + γβ(x′A − x′B) (2)

xA − xB = γ(x′A − x′B) + γβc(t′A − t′B). (3)

The meaning of length l′ is l′ = x′A − x′B at t′A = t′B, so Eqs. (2) and (3) give

c(tA − tB) = γβl′

xA − xB = γl′.

Using (xA − xB)− c(tA − tB) = l (see Eq. 1), one finds

γ l′ − γβ l′ = l =⇒ l = γ(1− β)l′ =⇒ n′ = γ(1− β)n.

Question 3 Solution

At t = 0, the initial state is given as |νµ〉 = cos θ |ν1〉− sin θ |ν2〉. After some time t, we knoweach mass state will evolve according to:

ψ(t) = e−iHt/~ψ(0) = e−iEt/~ψ(0).

Therefore, the time-evolved neutrino state is given by:

|νµ(t)〉 = − sin θe−iE1t/~ |ν1〉+ cos θe−iE2t/~ |ν2〉 .

The probability that this state will be detectable as an electron neutrino is represented bythe quantum-mechanical overlap of the time-evolved state and the definite electron neutrinostate:

Pµ→e = |〈νe|νµ(t)〉|2

=∣∣(cos θ 〈ν1|+ sin θ 〈ν2|)

(− sin θ |ν1〉 e−iE1t/~ + cos θ |ν2〉 e−iE2t/~

)∣∣2=∣∣− sin θ cos θe−iE1t/~ + cos θ sin θe−iE2t/~

∣∣2= sin2 θ cos2 θ

∣∣e−iE2t/~ − e−iE1t/~∣∣2

= sin2 θ cos2 θ(1 + 1− eiE2t/~e−iE1t/~ − e−iE2t/~eiE1t/~

)= sin2 θ cos2 θ (2− 2 cos (E2t/~− E1t/~))

= 4 sin2 θ cos2 θ sin2

(E2 − E1

~t

)= sin2(2θ) sin2

(E2 − E1

~t

).

The probability oscillates with a frequency of (E2 − E1)/(2π~), and reaches its maximumwhenever t = (2n+ 1)π

2~/(E2 − E1).

Question 4 Solution

(a) We first note that

dE = F · dr = qE · dr =qV0L

dx.

Thus the kinetic energy of the particle when it reaches the second plate is

K =

∫ E

0

dE ′ =

∫ L

0

qV0L

dx = qV0.

(b) We find

K = (γ − 1)m0c2 = qV0 =⇒ γ = 1 +

qV0m0c2

.

(c) Since the acceleration is perpendicular to the velocity, the relativistic version of New-ton’s second law yields

qv×B = γm0a

qvB0 (x× z) = γm0ω2rr.

Here r = −y, so we have

qωrB0 = γm0ω2r =⇒ ω =

qB0

γm0

.

We then find that r is

r =v

ω=βc

ω

=βcγm0

qB0

=m0c

qB0

γ√

1− γ−2

=m0c

qB0

(1 +

qV0m0c2

)√1− 1

(1 + qV0m0c2

)2(from part (b))

=1

B0

√2V0

m0

q+V 20

c2(No need to simplify this far.)

(d) The fields in the lab frame are

E = 0, B = B0z.

Thus the fields in the rest frame of the particle are

B′ = γB0z =B0√

1− β2z, E′ = γ(−βB0)y = − βB0√

1− β2y.

Question 5 Solution

We let L >> d be the distance from the slits to the viewing screen and let y be the distanceup the screen, as in the figure.

The distances the protons travel from slits 1 and 2 to the screen are

`1 =

√L2 +

(y +

d

2

)2

≈ L+1

2L

(y +

d

2

)2

`2 =

√L2 +

(y − d

2

)2

≈ L+1

2L

(y − d

2

)2

The path length difference is thus

`1 − `2 ≈dy

L.

For the second minimum, we set this equal to 3λ/2, yielding

y =3

2

λL

d.

The deBroglie wavelength is

λ =h

p=

h

γmv=

4.1356× 10−15 eV · s0.99c√1−0.992 (938× 106 eV/c2)

= 1.88× 10−16 m.

Thus the angle is

θ ≈ tan θ =y

L=

3λ

2d= 7.06× 10−8 rad.



Friday, January 17, 202010:00AM to 12:00PM

General PhysicsSection 5. Thermodynamics and Statistical Mechanics


Use separate answer booklet(s) for each question. Clearly mark on the answer booklet(s)which question you are answering (e.g., Section 5 (Thermodynamics and Statistical Mechan-ics), Question 2, etc.).






Good Luck!


1. Consider a gas contained in volume V at temperature T . The gas is composed of Ndistinguishable particles of zero rest mass, so that the energy E and momentum p of theparticle are related by E = pc. The number of single-particle energy states in the rangep to p + dp is 4πV p2 dp/h3. Find the equation of state and the internal energy of thegas, and compare with an ordinary gas.


2. Consider a mole of helium gas in container of size one liter at room temperature. Assumethe helium is an ideal, noninteracting gas in the classical limit. The single-particle

partition function for an atom is given by Z1 = V(mkBT2π~2

)3/2.

(a) Please compute the free energy of the gas (use Stirling’s approximation).

(b) Now compute the chemical potential.

(c) Now assume that this container is connected to a large reservoir of helium gas, whichhas the same pressure and temperature. The container is free to exchange particleswith the reservoir. Consider an ideal gas of particles of mass M in a volume V inthermal equilibrium with a reservoir with which it can exchange particles. Please

compute the fluctuation in particle number

√〈N(t)−N〉2

Nwhere the brackets indicate

a long time average.

(d) Assuming the number density is the same as in parts (a) and (b), how big shouldthe container have been so that the fluctuation in particles defined in (c) is 10%?(Work this out numerically.)


3. For an elastic filament it is found that, at a finite range in temperature, a displacementx requires a force (tension)

T = ax+ bT + cTx

where a, b, and c are constants. Furthermore, the heat capacity C of the filament atconstant displacement is proportional to temperature, i.e.

Cx = A(x)T.

(a) Use an appropriate Maxwell relation to calculate the partial derivative of theentropy, S, with respect to displacement at fixed temperature T, i.e. calculate(∂S/∂x)T .

(b) Show that the coefficient A in the equation for specific heat, above, has to in factbe independent of x, i.e.

dA

dx= 0.

(c) Give the expression for S(T, x) assuming S(0, 0) = S0.

(d) Calculate the heat capacity at constant tension, i.e. CT as a function of displace-ment and temperature.


4.

(a) The Bose Einstein Condensation temperature TBEC is the temperature at which amacroscopic number of bosons in a gas enter the ground state. At this temperature,the thermal wave length (or spread of the wave function) of a boson becomes com-parable to the average distance between adjacent bosons. Use this fact to estimateTBEC for an ideal monoatomic, non-relativistic Bose gas system in three dimensionswhich has a particle density, n (bosons per volume) and the boson mass, m. It isnot necessary to derive TBEC rigorously using the partition function and Riemannzeta function. We will, however, accept such an answer if it is correctly done.

(b) For spin-1/2 fermions in 3 dimensions, one can calculate the Fermi temperature TFfrom the Fermi energy εF = kBTF for non-interacting Fermi gas with density n andmass of each fermion m. Show that TBEC and TF has the same dependence on nand m. This can be done either by using the uncertainty principle or by calculatingεF for free Fermions in a 3-dimensional box.

(c) When one resorts to a rigorous calculation, TBEC and TF are comparable within afactor of 2. Explain why this is so with qualitative arguments. We are not askingyou to perform a rigorous calculation of these numbers, but rather remember whatTBEC and TF stand for and explain why they are comparable.

(d) Describe one phenomenon in the real world which can be attributed to the Bose Ein-stein Condensation, and describe an experimental method to measure the transitiontemperature.


5. A mixed stated in quantum mechanics is described by a density matrix rather than by avector in the Hilbert space. The density matrix, ρ, is an operator that allows to computeexpectation values of observables through

〈O〉 = Tr(ρ O). (1)

The trace ‘Tr’ is defined as

Tr(. . .)

=∑n

〈n| · · · |n〉 , (2)

where the sum is taken over any complete orthonormal set of states |n〉n.

Consider then a 1-D harmonic oscillator with proper frequency ω, at finite temperatureT = 1/β. The corresponding density matrix (setting kB = 1) is

ρ =e−βH

Tr(e−βH

) , (3)

where H is the Hamiltonian. Notice that Tr(ρ) = 1.

(a) Compute the average energy:E ≡ 〈H〉 . (4)

(b) Compute the so-called von Neumann entropy:

S ≡ −Tr(ρ log ρ

). (5)

(c) For E and S computed above, discuss the low-temperature (T E0) and high-temperature (T E0) limits.

(d) For the high-temperature limit of item 3 above, consider changing the temperatureby an infinitesimal amount dT . Verify the thermodynamical identity

dE = TdS . (6)

[Hint: To compute traces, use the orthonormal basis made up of energy eigenstates. Inthis basis, all traces relevant for the problem reduce to geometric series or derivativesthereof (with respect to β). Also, by using directly the expression (3), you can relate thecomputation of S in part (b) to that of E in part (a).]


6. Fermions with peculiar dispersion

One (unproven) theory says that matter in the core of a neutron star can be viewed asa gas of fermions with the excitation spectrum

ε±(k) = ± ~2

2M(k − kF )2

where the − pertains to k < kF and the + to k > kF .

Suppose that at T = 0 all of the states with energy less than 0 are filled and all of thestates with energy greater than zero are empty.

(a) What is the chemical potential at T = 0?

(b) Assuming that the density of states in reciprocal space is constant around |k| = kF ,i.e. approximating

∫d3k → k2F

∫dq with q = k−kF and neglecting any temperature

dependence of the chemical potential find the temperature dependence of the meanenergy of the system E(T )− E(T = 0).

(c) Using the same approximations as in (b), find the specific heat and give the leadingtemperature dependence as T → 0.


Question 1 Solution

The partition function for N particles is

Z =

∫dV1 · · · dVn

d3p1 . . . d3pN

h3Ne−E/kBT .

Substituting the given density of states gives

Z =

(4πV

h3

∫ ∞0

dp e−pc/kBTp2)N

=

(8πV

h3

(kBT

c

)3)N

.

The free energy is F = −kBT lnZ, and dF = −p dV − SdT , so

p = −(∂F

∂V

)T

= kBT∂ lnZ

∂V=NkBT

V.

The internal energy is

U = − ∂ lnZ

∂(1/kBT )= 3NkBT.

For an ordinary gas, we have p = NkBT/V and U = 32NkBT , so the pressure is the same,

but the energy is off by a factor of 2.

Question 2 Solution

(a) The N -particle partition function is given by Z = ZN1 /N !. The free energy of an ideal

gas is F = −kBT ln(Z). Plug and chug to get

F (N, V, T ) = −NkBT −NkBT ln

(V

Nλ3T

).

with λT =

√2π~2

mkBT.

(b) The chemical potential is

µ =∂F

∂N= −kBT ln

V

Nλ3T.

(c) Examination of the formula for the grand canonical partition function shows that

⟨(N − 〈N〉)2

⟩= T 2 ∂

2

∂µ2ln Ω

= T∂N

∂µ

= T

(∂µ

∂N

)−1.

From the result of (b) we evaluate this as

T

(∂µ

∂N

)−1= N,

so the relative number fluctuation is just 1√N

or 1√nV

.

(d) From the answer to (c), 1√nV

= 0.1 or V = 100n

= 166 nm3.

Question 3 Solution

(a) If F is the free energy then(∂S

∂x

)T

= − ∂2F

∂x∂T= − ∂

∂T

∂F

∂x=∂T∂T

= b+ cx

(b) From the relation C = T (∂S/∂T ) we find A(x) = ∂S/∂T so ∂A(x)/∂x = ∂2S/∂T∂xand taking the derivative with respect to T of the answer in (a) shows the result is 0

(c) Because the cross derivative ∂2S/∂x∂T = 0 we have

S(x, T ) = Sx(x) + ST (T ),

and integrating the results of the previous problems

S(x, T ) = S(0, 0)AT + bx+1

2cx2.

(d) The change in tension is

dT = (a+ cT )dx+ (b+ cx)dT

so if dT = 0 then dxdT

= − b+cxa+cT

and

CT = T∂S

∂T T= T

(∂S

∂T

)x

+ T

(∂S

∂x

)T

dx

dT= AT − T (b+ cx)2

a+ cT.

Question 4 Solution

Examples of possible solutions:

(a) At temperature T , p2/2m ∼ kBT =⇒ p =√

2mkBT . Let λ = h/p by the ther-mal wavelength. In a BEC, λ becomes comparable to the inter-boson distance n−1/3.Therefore, p2 = h2 × n2/3 = 2mkBTBEC, so

TBEC ∼h2

2kB

n2/3

m.

An exact calculation gives

TBEC =1

2πζ(3/2)2/3h2

kB

n2/3

m≈ 0.084

h2

kB

n2/3

m.

(b) Suppose you divide the volume V into a small sub-cube of the edge size ∆x. In eachsub-cube, we can put two fermions (spin up and down). Thus N = 2V/(∆x)3 where Nis the total number of fermions. Thus ∆x = 21/3(V/N)1/3. By the uncertainty principle,∆p∆x ∼ ~. Then

(∆p)2

2m∼ εFermi ∼ ~22−2/32−1

n2/3

m.

So

TFermi ∼1

25/3

~2

kB

n2/3

m.

An exact calculation for a free electron gas in three dimensions yields

TFermi =(3π2)2/3

2

~2

kB

n2/3

m.

(c) As you can see above, exact results for TBEC and TFermi are comparable within a factor 2.This can be understood if you remember that both energy scales represent a boundarybetween quantum mechanical (boson or fermion) particles and higher energy classicalparticles. Below these energy scales, bosons and fermions form quantum degenerateliquid. In the classical gas at higher temperatures, distinction between bosons andfermions becomes unnecessary.

(d) Examples of Bose-Einstein condensates and methods to measure the condensation tem-perate:

(a) Superfluid Helium-4: Superfluidity can be verified by vanishing viscosity measuredby torsion oscillator, critical opalescence, lambda-shaped peak of the specific heat,and more.

(b) Superconductivity: BCS superconductors undergo a special case of BEC at thecritical temperature Tc, where formation of bosons from fermions (pair formation)and condensation of newly formed bosons occur simultaneously. Critical temper-ature can be measured as zero resistivity, diamagnetism due to Meissner effect,specific heat anomaly, and many other phenomena.

(c) BEC of cold atoms: BEC of dilute cold boson gas can be observed from the velocitydistribution peaking at zero momentum of the gas below TBEC.

(d) Other possible case includes BEC of neutron star, BEC of excitons in semicon-ductors, and possible BEC of pre-formed pairs in underdoped regions of high-Tccuprate and other unconventional superconductors. One can refer to literatureson how to measure TBEC in these systems

Question 5 Solution

For the one-dimensional Harmonic oscillator, the energy eigenstates |n〉n=0,1,2,... are suchthat

H |n〉 = E0

(n+

1

2

)|n〉 , n = 0, 1, 2, . . . ,

where E0 = ~ω is the energy gap between two adjacent states. Since ρ is a function of Halone, ρ is also diagonal in this basis:

ρ |n〉 = ρn |n〉 ,

with eigenvalues

ρn =e−βE0(n+1/2)

∞∑m=0

e−βE0(m+1/2)

The sum at the denominator is a geometric series,

e−βE0/2

∞∑m=0

(e−βE0

)m=

e−βE0/2

1− e−βE0

and so the eigenvalues of ρ are

ρn =(1− e−βE0

)e−βE0n

(a) We have

E = Tr(ρH)

=(1− e−βE0

)E0

∞∑n=0

(e−βE0n

)(n+

1

2

)(1)

=(1− e−βE0

)E0

(− 1

E0

∂

∂β+

1

2

) ∞∑n=0

(e−βE0

)n(2)

=1

2E0 −

(1− e−βE0

) ∂∂β

1

1− e−βE0(3)

= E0

(1

2+

e−βE0

1− e−βE0

)(4)

(b) Given the definition of the thermal density matrix in the problem, we have

S = −Tr (ρ log ρ)

= β Tr(ρH)

+ Tr (ρ) log(

Tr(e−βH

))= βE + log

(e−βE0/2

1− e−βE0

)= β

(E − 1

2E0

)− log

(1− e−βE0

)= βE0

e−βE0

1− e−βE0− log

(1− e−βE0

).

(c) At low temperatures, we have e−βE0 1, and so

E =1

2E0

(1 +O

(e−βE0

)), S = O

(βE0 e

−βE0)<< 1.

This makes sense: at very low temperatures compared to the gap, the probability of be-ing in an excited state is exponentially small. So, up to exponentially small corrections,the system is in the ground state, with energy E0/2 and zero entropy.

At high temperatures, we have e−βE0 ' 1− βE0, and so:

E ≈ 1

β= T, S ≈ − log(βE0) = log

T

E0

>> 1

The energy is consistent with equipartition, while the entropy is very high, and of theright order of magnitude: all states up to E ∼ T are equally populated; there areN ∼ E/E0 ∼ T/E0 of them, and the entropy is S ∼ logN .

(d) For an infinitesimal temperature variation dT , from above we get

dE ' dT , dS ' dT

T,

which is consistent with dE = TdS.

Question 6 Solution

(a) The chemical potential at T = 0 is zero.

(b) The internal energy is (making the indicated approximation and noting that ε− = −ε+)

E =

∫dk

ε(k)

eε(k)T + 1

≈ k2F2π2

∫ ∞0

dqε+(q)

eε+(q)

T + 1− ε+(q)

e−ε+(q)

T + 1

or

E(T )− E(0) =k2F2π2

∫ ∞0

dq ε+(q)

(1− tanh

ε+(q)

2T

).

(c) Define u = ~2q24MT

so q =√

4MTu~2 and dq =

√MT~2

du√u

so

E(T )− E(0) =k2Fπ2

(M

~2

) 32

T32

∫ ∞0

du√u (1− tanhu) ,

so the specific heat is C =∂E

∂T∝ T

12 .



Friday, January 17, 20202:00PM to 4:00PM

General PhysicsSection 6. Various Topics


Use separate answer booklet(s) for each question. Clearly mark on the answer booklet(s)which question you are answering (e.g., Section 6 (General Physics), Question 2, etc.).






Good Luck!


1. In answering the questions below, make reasonable estimates for any needed parame-ters/quantities not already provided.

Consider an idealized Sun and Earth as blackbodies in otherwise empty space. TheSun has a surface temperature TS = 6000 K, and heat transfer processes on the Earthare effective enough to keep the Earths surface temperature uniform. The radius of theEarth is RE = 6.4×106 m, the radius of the Sun is RS = 7.0×108 m, and the Earth-Sundistance is d = 1.5× 1011 m. The mass of the Sun is MS = 2.0× 1030 kg.

(a) Find the temperature of the Earth.

(b) Find the radiation force on the Earth.

(c) Compare these results with those for an interplanetary granule in the form of aspherical, perfectly conducting black body with a radius R = 0.1 cm, moving in acircular orbit around the Sun at a radius equal to the Earth-Sun distance d.

(d) At what distance from the Sun would a metallic particle melt if its melting tem-perature Tm = 1550 K?

(e) For what size particle would the radiation force calculated in part (c) be equal tothe gravitational force from the Sun at a distance d?

Note: You will need the values of the Stefan-Boltzmann constant σ and the gravitationalconstant G.


2. A small bit of solid matter is released at the top of out atmosphere. It is pulled downby gravity through three atmospheric gas layers, described below, before reaching theearth’s surface. What is its characteristic fall velocity while passing through each layer?

Each of the three layers has its own characteristic gas mass density ρ of mass m moleculesmoving about with a randomized thermal velocity v. These molecules scatter elasticallyon neighbors with a cross-section σ.

The bit of matter has a fixed mass M and radius R. In a vacuum it would have an earthdirected acceleration g. Because of the drag force on the bit from its passage throughthe local atmosphere with downward speed V , its downward acceleration is essentiallycanceled.

In layer I, the first it passes through, the air molecules’ mean free path for scatteringλ >> R and V is supersonic V > v. In the second, layer II, with large ρ, λ >> R butV < v. In the third, layer III, λ < R ansd V << v. The drag force on the falling bitbecomes proportional to RV . (Stokes’ Law is appropriate.)

What is the characteristic V , as a function of R, g, M , ρ, v, σ, and m in each layer?


3. When it rains, is it better to run fast or to walk slowly (assuming you don’t want to gettoo wet)? Motivate your answer with an actual computation. In particular, you shouldcompute the total amount of water hitting you as you move from A to B, as a function,among other things, of your speed. You can adopt several simplifications:

1. You are a parallelepiped, gliding smoothly at constant (horizontal) speed v.

2. The density of rain drops is so high that you can treat rain as a continuum, withconstant mass density ρrain, moving vertically downwards at constant speed vrain.

3. The angle at which rain hits you does not matter, nor does its speed relative toyou: you are perfectly absorbent.

4. Ignore, of course, relativistic effects.


4. Consider normal modes of vibration of a linear chain of identical atoms of mass Min the harmonic approximation. Assume that coupling between atoms exist only fornearest neighbor atoms. These couplings are represented by single equal springs withforce constant K. Normal modes have wave vector k and frequency ω. Also assume thatthe chain has N atoms and length L = Na.

(a) Write the equation of motion for displacements u of normal modes and obtainthe normal mode frequencies ω(k) as function of the magnitude of the wavevector|k| = k.

(b) Longitudinal normal modes have u ‖ k and transverse modes have u ⊥ k. Howmany longitudinal modes and transverse modes exist for each value of wave vector.Why are modes with wave vectors k and −k degenerate

(c) Show that for long wavelengths the equation of motion reduces to a continuumelastic wave equation. What is the speed of sound?

(d) Use the cyclic boundary conditions to obtain the density of states of modes. Howmany modes are in the first Brillouin zone of the linear chain?

(e) In a Debye model ω(k) is approximated by a linear dispersion ω(k) = csk, wherecs is the speed of sound in the chain. Obtain the specific heat due to longitudinalvibrations of the chain of identical atoms in the Debye approximation and obtainthe expression for the heat capacity (specific heat at constant volume) in the limitof T → 0.


5. Please read the entire question before beginning and note that there are equations at theend of the problem. The hints are there to benefit you but you can still do the problemin any way you choose.

In this part of the problem, we will explore both the Poisson distribution and the normaldistribution and their relation. All given equations can be used without proof - in allother cases you must show your work!

The Poisson distribution (form is given at the bottom of the page) is a discrete probabilitydistribution defined for integers k ≥ 0 and mean λ > 0 that expresses the probability ofa given number of events occurring in a fixed interval of time and/or space. In physics,we often have an expectation for a given rate of some type of event so we use the Poissondistribution to describe the probability of a given number of events occurring in a giventime interval assuming that our expectation is the rate multiplied by our given timeinterval. The Poisson distribution is a limit of the Binomial distribution with p→ 0 andN →∞.

Below is an image of the Poisson distribution for different mean values.

(a) Show that the mean (expectation value) and the variance of the Poisson distributionare both equal to λ.

Hint 1: Use may use the exponential summation given below.

Hint 2: Remember that σ2x = 〈x2〉 − 〈x〉2.

(b) Show that in the limit of λ >> 1 and k ∼ λ (k on the order of λ) that the Poissondistribution can be approximated by the normal distribution.

Hint 1: Let x = k = λ(1 + δ) where we know δ << 1 since k ∼ λ.

Hint 2: You will need to use Stirling’s Approximation (see equations below).

Hint 3: To approximate (1 + x)ax where x << 1, we can first use the naturallogarithm to get it in a more convenient form and then exponentiate the result.

Dark matter experiments, in their simplest form, are counting experiments. You have anexpectation for your background over some period of time T . These experiments then


take data over this time period T and look at the total of number of events to determinewhether they see a statistically significant excess of events or not (this excess of coursebeing from dark matter interactions!).

Consider two detectors. In detector A we expect 1 background event in a single yearand in detector B we expect 1000 background events in a year. We turn on our detectorsand after one year we find that detector A counted a total of 11 events and detector Bsaw a total of 1010 events. Assume that we know our background perfectly.

(c) What is the probability that each detector’s number of events is from background?

(d) If we can safely reject the background hypothesis, then it is possible that we canclaim a direct detection of dark matter. In which detector is it less likely that thesignal can be explained by background alone? Explain your reasoning.

Discrete Expectation: 〈ka〉 =∑k

kap(k)

Exponential Sum:∞∑k=0

λk

k!= eλ

Expanding the Natural Log: ln(1 + x) =∞∑n=1

(−1)n+1xn

n

Poisson Distribution: p(k) =λke−λ

k!, λ > 0, k = 0, 1, 2, . . .

Normal Distribution: p(x) =1√

2πσ2e−(x−µ)

2/2σ2

Stirling Approximation: n! ≈√

2πn(ne

)n


6. Estimate the rms magnitude of the electric field associated with blackbody radiation inthermal equilibrium at room temperature. Note: σ = 5.7× 10−8 W/(m2·K4).


Question 1 Solution

(a) The total power radiated from the Sun is

PS = 4πR2SσεT

4S ,

where ε = 1 for a blackbody. Of this, the fraction that hits the Earth isπR2

E

4πd2, and in

equilibrium, with the Earth also radiating as a blackbody, then:

(4πR2

S

)σT 4

S

(πR2

E

4πd2

)=(4πR2

E

)σT 4

E,

so:

TE =

√RS

2dTS = 290 K.

(b) The radiation pressure is:

Prad =4σ

3cT 4S

(4πR2

S

4πd2

)= 7.1× 10−6 N/m2.

with the corresponding force on the Earth being:

FE = πR2EPrad = 9.2× 108 N.

(c) For the small granule, the temperature will be the same since it the same distance fromthe Sun as the Earth is. The radiation pressure on the granule is:

Fgr = πR2Prad = 2.2× 10−11 N.

(d) From the result of part (a), we can write the distance dm at which the granule will melt:

dm =1

2RS

(TSTm

)2

= 5.2× 109 m.

(e) Assume a spherical particle of mass m and radius r, then when the radiation andgravitational forces balance we have:

GMSm

d2= πr2Prad,

where m = 43πr3ρ, and we can assume a density of say ρ = 5.0× 103 kg/m3, then:

r =3

4

Pradd2

GMSρ= 1.8× 10−7 m.

Question 2 Solution

In layer I, we have

Mg ∼ R2V 2ρ =⇒ V ∼(Mg

ρR2

)1/2

.

In layer II, we have

Mg ∼ R2V vρ =⇒ V ∼ Mg

ρR2v.

In layer III, we have

Mg ∼ RλV vρ =⇒ V ∼ Mg

ρλvR.

Since λ ∼ mρσ

, we have

V ∼ Mgσ

vmR.

Question 3 Solution

Under the assumptions spelled out in the problem, it is better to run as fast as possible,v →∞.

Calling L the distance between A and B, the time it takes to cover that distance isT = L/v. Now call Stop the surface area of the top face of the parallelepiped (one’s head),and Sfront the surface area of the front face of the parallelepiped (one’s body). During T ,the mass of water that falls on one’s head is ρrainvrainStopT , which is proportional to T , andthus inversely proportional to v. On the other hand, the mass intercepted by one’s body isindependent of T and thus of v: it is simply ρrainSfrontL, the total mass of water containedin the ‘tube’ of cross section Sfront connecting A to B.

So, the total mass of rain intercepted (and thus absorbed) in moving from A to B atconstant speed v is

Mtot = ρrainL ·[Stop

vrainv

+ Sfront

].

As a function of v, this is schematically

Mtot ∼ const + 1/v,

which is minimized for v →∞.

Question 4 Solution

(a) The equation of motion is

Mu(n) = −k (2u(n)− u(n− 1)− u(n+ 1)) ,

where u(n) is the displacement of the atom at position n in the chain. The normalmode frequencies are

ω(k) = 2

√k

M

∣∣∣∣sin(1

2ka

)∣∣∣∣ =

√2K

M(1− cos ka).

(b) We have one longitudinal mode and two transverse modes. The modes with k and −kare degenerate because there is no dependence on the direction of propagation.

(c) Taking u(n) ∝ eikx(n), where x(n) = na. In the long wavelength limit, we have

u(n)− u(n− 1) ≈ ∂u(n− 1)

∂x(n− 1)ka

u(n)− u(n+ 1) ≈ −∂u(n+ 1)

∂x(n+ 1)ka.

The equation of motion is

M∂2u(n)

∂t2≈ −K

(∂u(n− 1)

∂x(n− 1)− ∂u(n+ 1)

∂x(n+ 1)

)ka.

We also have∂u(n− 1)

∂x(n− 1)− ∂u(n+ 1)

∂x(n+ 1)≈ ∂2u(n)

∂x(n)2ka.

Thus the equation of motion is

∂2u(n)

∂t2≈ K

M(ka)2

∂2u(n)

∂x(n)2.

It follows that the speed of sound is cs =√

KMka

(d) The chain has N atoms and length L = Na. The density of states is G(k) = L2π

. Wealso have dN = G(k) dk, and in the first Brillouin zone:∫ π/a

−π/aG(k) dk = N.

(e) The heat capacity at constant volume is

CV =L

π(kBT )

∫ xD

0

x

ex − 1dx,

where xD = ~cskD/kBT , where kD is the Debye wavevector. For T → 0,

CV ≈L

π(kBT )

∫ ∞0

x

ex − 1dx =

π

6L(kBT ).

Question 5 Solution

(a) We begin the problem by calculating the required expectation values: 〈k〉 and 〈k2〉

〈k〉 =∞∑k=0

kλke−λ

k!,⟨k2⟩

=∞∑k=0

k2λke−λ

k!.

We start with the expansion of exponential: eλ =∞∑k=0

λk

k!. Taking the derivative with

respect to λ on both sides yields

eλ =∞∑k=1

kλk−1

k!=⇒ λeλ =

∞∑k=1

kλk

k!.

Thus

〈λ〉 = e−λ∞∑k=0

kλk

k!= e−λλeλ = λ.

Now we repeat the derivative trick for λeλ =∞∑k=0

kλk

k!:

eλ + λeλ = eλ(1 + λ) =∞∑k=0

k2λk−1

k!=⇒ eλ

(λ+ λ2

)=∞∑k=0

k2λk

k!

Thus we find ⟨k2⟩

= e−λ∞∑k=0

k2λk

k!= e−λeλ

(λ+ λ2

)= λ+ λ2.

So the variance isσ2 =

⟨k2⟩− 〈k〉2 = λ.

(b) We start with p(k) =λke−λ

k!and want to get p(x) =

1√2πσ2

e−(x−µ)2/2σ2

in the limit of

λ→∞ and k ∼ λ. For k ∼ λ only care about points relatively close to the mean (nottails). Following the hints, we define x = k = λ(1 + δ) where δ is small. Thus

p(x) =λλ(1+δ)e−λ

(λ(1 + δ))!

=λλ(1+δ)e−λ√2πλ(1 + δ)

eλ(1+δ)λ−λ(1+δ)(1 + δ)−λ(1+δ) (Stirling’s approximation)

=eλδ√2πλ

(1 + δ)−λ−λδ−12 .

Following the hint about the logarithm, we find

ln(

(1 + δ)−λ−λδ−12

)=

(−λ− λδ − 1

2

)ln(1 + δ)

≈(−λ− λδ − 1

2

)(δ − 1

2δ2)

= −λδ − λδ2 − 1

2δ +

1

2λδ2 +

1

2λδ3 +

1

4δ2

≈ −λδ − 1

2λδ2.

Thus

p(x) ≈ eλδ√2πλ

e−λδe−λδ2/2

=1√2πλ

e−λδ2/2

=1√2πλ

e−(x−λ)2/2λ. (δ = x

λ− 1 = x−λ

λ)

From part (a), we know µ = 〈x〉 = λ and σ2 = λ, so we simply have a Gaussian in thelimit.

(c) The solution for this part comes from recognizing that we have a Poisson distributionfor the background. For detector A we find

p(k = 11;λ = 1) =1 · e−1

11!≈ 0,

so there is almost zero chance this is from the background. For detector B we can usethe Gaussian approximation to find

p(k = 1010;λ = 1000) =10001010e−1000

1010!≈ 1√

2π · 1000e−(1010−1000)

2/2·1000 = 0.12

(The Poisson calculation is numerically unstable.)

(d) The signal in detector A has only an extremely small chance of being from backgroundso it is very possible that is has seen dark matter!

Question 6 Solution

Assuming unity emissivity, the power radiated by a blackbody surface is

PBBR = σAT 4,

where T ≈ 290 K is the temperature and A is the surface area.

On the other hand, the power transported by an electric-field wave is given by the Poynt-ing vector ~S as

PP = ~S · ~A =1

cµ0

A〈E2〉 · 1

2· 1

2,

where c is the speed of light and µ0 is the permeability of free space. (The first factor of12

accounts for the fact that half of the BBR photons are directed toward the surface; thesecond factor of 1

2estimates the projection of the outward-going photons’ average direction

onto A: to be precise,∫ π/20

cos θ sin θ dθ = 12.)

Setting PBBR = PP, we find√〈E2〉 = 2T 2√cµ0σ ≈ 8 V/cm.

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