QUALIFYING EXAMINATION, Part 1

1:00 pm – 5:00 pm, Thursday September 3, 2015

Attempt all parts of all four problems.

Please begin your answer to each problem on a separate sheet, write your 3 digit codeand the problem number on each sheet, and then number and staple together the sheetsfor each problem. Each problem is worth 100 points; partial credit will be given.

Calculators and cell phones may NOT be used.

1

Problem 1: Mathematical Methods

(a) The formula

sinx = x∞∏n=1

(1− x2

n2a2

)is known to work for a particular value of a.

(i) (10 points) Find this value of a by requiring that the right-hand side of the formulareproduces all the zeros of sinx.

(ii) (10 points) Evaluate∞∏n=1

(1− 1

4n2

).

(iii) (15 points) Evaluate∞∏

n=2

(1− 1

n2

).

(b) (30 points) Starting with Cauchy’s theorem

f(z) =1

2πi

∮C

f(z′)dz′

z′ − z,

where C is a contour in the complex plane which encloses z and within which f(z′) isanalytic, derive the formula for df(z)/dz as a contour integral.

Evaluate the integral ∮|z|=3.141

eazdz

z2(z − 1).

(c) A classical particle of mass m = 1 moves in a potential

V (x, y) = x2 + 2xy − 2y2. (1)

(i) (20 points) Determine the stationary point of the potential (i.e., the equilibrium pointof the particle). Is it a maximum, minimum or saddle point?

(ii) (15 points) Determine at least one direction along which the particle may be dis-placed from the stationary point so that it will execute stable oscillations. What is thecorresponding frequency ω for this oscillation?

2

Problem 2: Classical Mechanics

Consider an N -pendulum, the generalization of the double pendulum (see Fig. 1).Each link consists of a massless string of length l ending in a bob of mass m. Thebottommost mass is denoted by n = 1 and the topmost mass is n = N . The angulardisplacement of mass n from the vertical is denoted by θn and is assumed to be small.

θn+1

θn

l

θn-1

xn

xn-1

Figure 1: N-pendulum

(a) (10 points) Find an expression for sin θn in terms of the horizontal displacements xnand xn+1 of the pendulum bobs from their equilibrium positions.

(b) (25 points) Write the (linear) equation of motion for each displacement xn in termsof other displacements.

Hint: in the limit of small θn, you can assume that the tension Tn in the n-th string (justabove mass n) is a constant Tn = nmg, independent of the displacements.

(c) (25 points) Assuming a harmonic solution xn ∝ sin(ωt + δ) with the same frequencyand phase for all xn, derive a recurrence relation for xn+1 in terms of xn−1 and xn.

(d) (40 points) Find the normal frequencies of the double pendulum (N = 2) using therecurrence relations in (c).

Hint: The x-displacement is zero at the top hanging point of the pendulum.

3

Problem 3: Electromagnetism I

An infinitely long hollow (non-conducting) circular cylinder of radius R is fixed at apotential V = V0 sinϕ (see Fig. 2).

V sin0 z

R

ϕ

ϕ

Figure 2: Hollow cylinder with a potential V = V0 sinϕ.

(a) (15 points) Using cylindrical coordinates r, ϕ, z where the z axis is taken along thesymmetry axis of the cylinder, argue that the electrostatic potential V is independent ofz and must satisfy V (r,−ϕ) = −V (r, ϕ).

(b) (20 points) Find the electrostatic potential Vin(r, ϕ) inside the cylinder (r ≤ R).

(c) (20 points) Find the electrostatic potential Vout(r, ϕ) outside the cylinder (r ≥ R).

(d) (25 points) Calculate the surface charge density σ on the cylinder as a function of ϕ.

(e) (20 points) Find the local capacitance of the device per unit length as a function ofthe angle ϕ, i.e., calculate C = dQ/dV where dQ/dϕ is the charge per unit angle (andper unit length of the cylinder) and dV/dϕ is the voltage per unit angle.

Hint: the solution of Laplace’s equation in cylindrical coordinates which is independentof z has the general form

V (r, ϕ) = a0 + b0 ln r +∞∑n=1

anrn sin(nϕ+ αn) +

∞∑n=1

bnr−n sin(nϕ+ βn) ,

where an, bn, αn, βn are constants.

4

Problem 4: Electromagnetism II

An infinitely long infinitesimally thin and massless cylindrical non-conducting tube ofradius R carries a charge with uniform surface charge density of σ (charge/unit area).The tube rotates around its symmetry axis with a fixed angular velocity ω in free space(see Fig. 3). The formulas below are given in both Gaussian and SI units.

ω

R

z

Figure 3: Rotating cylindrical tube.

(a) (10 points) Calculate the electric field ~E inside and outside the cylinder.

(b) (15 points) Calculate the magnetic field ~B inside and outside the cylinder.

Hint: treat the system as a solenoid.

(c) (20 points) Next consider the case where the cylinder’s angular velocity ω is notconstant, but given by ω = αt, starting with ω = 0 at t = 0. What is the magnetic fieldinside the cylinder? Calculate the electric field (including its direction) just inside thecylinder.

Hint: use Faraday’s law of induction to calculate the electric field due to the changingmagnetic field.

(d) (20 points) Calculate the Poynting vector

~S =

{c4π

( ~E × ~B) (Gaussian units)1µ0

( ~E × ~B) (SI units)

just outside and just inside the cylinder.

(e) (35 points) Integrate the Poynting vector over the inner surface of the cylinder todetermine the rate of energy flow per unit length into the cylinder’s interior. Compare

5

this to the rate of change of the total field energy within the cylinder per unit length. Arethese rates the same? Explain why.

Hint: the energy density u in an electromagnetic field is given by

u =

{18π

(E2 +B2) (Gaussian units)12

(ε0E

2 + 1µ0B2)

(SI units).

6

N2\

€x'dlÈ;{ËUri

R

, v,¡=u'ff+u, iH-rH

å3 v'a.=",(} W-#)."(#-#)*" |($oau-ä),'v=I*o('#).##.#

Explicit Forms of Vector Operations

Let er, €2, €¡ b€ orthogonal unit vectors associated srith the coordinate directions

specified in the headings on the left, and A1, Az, A¡ be the corresponding

components of A. Then

vú=e, ffi*",ff*^ffiv.¡,:ðrA'+**&ôxt' ôxz- ðxg

:

v x a : e, (H-*)*' (#-#) -* (*-*)v'1,:ffi*ffi*ffi

vú=e' #*",1#** ^håËo .o=+$ tr a,l*ffiS rrt" oA,)*^hq#

vxÀ:er **[*n*oAt-H].., [^h #-] $ t,",r]*", + [#

('ot-#]

u',þ :+ *(t #).r..|6 * (""' #).*- #[No,",t ", ] $ (r S)=i $ t+il

.Ea9(Düìè

AND d FUNCTIONS

Note: A square-root sign is to be understood ovet elter|cocfficicnt, e.g., for -8/15 read -JW'

35. CLEBS CH-G ORDAN COEFFICIENTS, SPHERICAL HARMONIC S,

2x!/2

t- xt/+L +L/21

3/2

+r -r/2o +r/2

3/2 L/2-112 +Ll2I

4: \Æ"* o 2

"l =-l*sir.i,eiö

2x]-

L/3 2/32'/3 -r/3

2+

o -r/2-r +r/2

3

X1

3/2 r/2-L/2 -r/2

+1

F1 +11 1

32+2 +2I

2

+2

w:'/T^G"",'r-å)

",] : -ly- si,'l,eoso eió

"î :i'{*

"i,.2 s.2nó

0

+1

2/3 rl37/3 -2/3

+1 0

0+1

/3 2/3/3 -r/3

2

+1

-t -r/2

1

+1

L/2 r/2t/2 -t/2

+2 -1+1 0

0+1

3lZ-3/2

x1

321+1 +1 +1

35. Clebsch-Gordan coefficients 1

Yl^ = ?r) Ytr.

1/1sB/rs2/s

+t- -L00

-1 +1

2ll,2/,:

1

2L0000

r/3 3/s7/6 -3/ro

-L/2 rho

Ll6 1/z r/32/3 o-r/3r/6 -r/2 L/3

3/2xt

+z -Ll2+L +L/2

s/2 3/2-3/2+3/2

+al2 +Ll 1

+1 -100

-1 +1

rls 4/!4ls -Ll!

sl2

321000

-1 0

+3/2 O

+L/2 +1.

2

-1

+L-L/z

LlS r/2!/5 o

Lls -rl2

slz 3/2+112 +3/2

3/2xL/2

s/27/2

1

-t

./2 r/2

./2 -L/2

m1 ^rlm1 mz I Coefficients

t

312+L/2

2/s 3/s3/s -2/s

2/5 3ls3ls -zls

JJMM

3/10-2/53/ro

-1 -t

2

-2

+312-1+L/2 0

-7/2 +7

0 -rl2-r +L/2

+3/2 -!/2+Ll2 +L/2

0-1-1 0

-2 +L

s/2 3/2 t/2tr/2 +L/2 +L/2

321-1 -1 -1

I

s/2

a;,o: ffirvY"

L/2

LlLo zls3/s LlLs

311-0 -8115

2/s a/2 a/Lo8/rs -L/6-31r0LlLs -L/3 3/s

+

3/2Ll2

3lS 2/52lS -3/s

1

L/4 314rl4 -al4

+r/2 -L/2-r/z +L/2

-L -L/2-2 +L/2

L/2-r/3

Ll6

5/z 3/2-3/2 -3/2

2r00

+r/z -L-t/2 o

-3/2 +L

4/s rlsals -4ls

rl2 u2v2 -a/2

-24

s/2 312 u2-L/2 -L/2 -L/2

32-2 -2

-L/2 -L/2-3/2 +u2

-inø

2/3 L/3

3/LO s/rs3/s -r/Is

LlLo -zls

-2 -t/2

s/2

2!-t -1

3/4 L/41-/4-3/4

?

-t

1

r/6-a/3

L/2

-312-112

-!/2-L-3/2 O

-2

5/2 3/2-312 -3123ls 2/s2/s -3/5

-3/2 -!

s/2-s l2

1

QUALIFYING EXAMINATION, Part 2

9:00 am – 1:00 pm, Friday September 4, 2015

Attempt all parts of all four problems.

Please begin your answer to each problem on a separate sheet, write your 3 digit codeand the problem number on each sheet, and then number and staple together the sheetsfor each problem. Each problem is worth 100 points; partial credit will be given.

Calculators and cell phones may NOT be used.

1

Problem 1: Quantum Mechanics I

An electron with charge e and mass me is placed in uniform magnetic field of strengthB pointing in the positive z-direction. The electron spin operator is ~S. The Hamiltonianfor this system is H = ωSz, where ω ≡ |e|B/mec.

At t = 0, the electron is in an eigenstate of ~S · n with eigenvalue ~/2, where n lies inthe x-z plane making an angle ϕ with the z-axis (see Fig. 1).

x

y

z

ϕn

Figure 1: A unit vector n in the x-z plane.

(a) (15 points) What is the unitary time-evolution operator U(t) ≡ U(t, 0) for this systemin terms of Sz and ω ?

(b) (20 points) Express the initial, normalized state of the system |t = 0〉 in terms of thenormalized spin up/down eigenkets |+〉 and |−〉 of Sz. Do this by acting on the state |+〉with the rotation operator of angle ϕ about the y axis: |t = 0〉 = Ry(ϕ)|+〉 = e−iSyϕ/~|+〉.In the spin 1/2 representation, Sy = (~/2)σy, where σy is a Pauli matrix.

(c) (30 points) Determine the state |t〉 of the system at later times t by acting on |t = 0〉with the time evolution operator U(t). What is the probability that a measurement of Sxat time t will yield the eigenvalue +~/2?

(d) (20 points) What is the expectation value of Sx as a function of time t?

(e) (15 points) What are the answers to parts (c) and (d) for ϕ = 0? Check your answersby a direct calculation without using your results in (c) and (d).

The Pauli matrices are given by

σx =

(0 11 0

)σy =

(0 −ii 0

)σz =

(1 00 −1

).

2

Problem 2: Quantum Mechanics II

The quantum Zeno effect is a quantum analog of the Zeno paradox. One of its versioncan be summarized as “a watched pot never boils.” Here we will use a two-level systemto demonstrate this behavior.

Consider a two-level system with two basis states identified as the “cold” state andthe “boiled” state

|ψcold〉 =

(10

)|ψboiled〉 =

(01

).

We assume that the system undergoes the following unitary time evolution

U (t+ ∆t, t) ≡ U (∆t) =

(cos(ν∆t/2) − sin(ν∆t/2)sin(ν∆t/2) cos(ν∆t/2)

).

At an initial time t = 0, the pot is prepared in the “cold” state |t = 0〉 = |ψcold〉. Thetime evolution can transform this “cold” state into the “boiled” state at a later time.

When we observe whether the pot is “cold” or “boiled” at time t, we effectively per-form a measurement of the observable σz, with the +1 outcome for the “cold” state, andthe −1 outcome for the “boiled” state. The pot will be in the corresponding eigenstateof σz right after the measurement.

(a) (15 points) Suppose that no measurement is carried out. At what time tB does the potenter into the “boiled” state for the first time? In other words, find the smallest positivetB such that |tB〉 = |ψboiled〉.

(b) (30 points) Suppose that the pot is observed twice, first at time t = tB/2 and thenat t = tB. Our observations will change the evolution discontinuously. Compute theprobabilities associated with the four possible measurement outcomes, and complete thefollowing table.

1st measurement outcome 2nd measurement outcome Probability

“cold” “cold”“cold” “boiled”

“boiled” “cold”“boiled” “boiled”

(c) (30 points) Suppose that the pot is observed N times at t1 = tBN

, t2 = 2tBN, . . . , tN = tB.

Find the probability PZeno,N that all N measurements yield the “cold” state.

(d) (15 points) Find an approximation of PZeno,N for large values of N (N � 1) to thefirst order in 1/N .

(e) (10 points) Express the probability in (d) in terms of the time interval ∆t = tBN

andfind its value in the limit ∆t→ 0, i.e., when the pot is observed continuously.

3

Problem 3: Statistical Mechanics I

A molecular zipper is comprised of N links, each of which has two states: (i) thezipped-up state, which has energy 0 and degeneracy 1, and (ii) the unzipped state, whichhas energy ε (ε > 0) and degeneracy g (g > 1). This zipper can only unzip from its leftend, and a link cannot unzip unless all links to its left are already unzipped (see Fig. 2).In the following consider the limit N →∞.

Figure 2: Schematic of the molecular zipper discussed in parts (a)-(c). In this figure thereare m = 5 unzipped links. The dots indicate additional zipped links.

(a) (30 points) The molecular zipper’s partition function Z is finite at temperaturesT < TM . Calculate the partition function and determine the maximal temperature TMfor which your calculation is valid.

Hint: For x < 1,∞∑m=0

xm =1

1− x.

(b) (25 points) Calculate the average number of unzipped links 〈m〉 for T < TM .

(c) (20 points) For T < TM , what is the probability that one or more links of the zipperare unzipped?

4

(d) (25 points) A new, improved molecular zipper can unzip from both ends (see Fig. 3).A link cannot unzip unless all the links to its left or all the links to its right are alreadyunzipped. Determine the partition function Z2 and the mean number of unzipped links〈m2〉 for this new improved molecular zipper at T < TM . Assume that the number oflinks is infinite.

Figure 3: Schematic of the new, improved molecular zipper discussed in part (d). Thedots indicate additional zipped links.

5

Problem 4: Statistical Mechanics II

Consider a gas of N identical non-interacting spin-zero bosons in d dimensions, con-fined to a large volume V = Ld (you may assume periodic boundary conditions) and heldat temperature T . The dispersion relation, expressing the single-particle energy ε in termsof the magnitude of its momentum p = |~p|, is given by ε = aps where a and s are positivenumbers.

(a) (25 points) Find the single-particle density of states g(ε) as a function of energy.

Hint: in d dimensions dd~p = Sd pd−1dp where Sd is a constant.

(b) (25 points) Write an integral expression for the total number of bosons N in the limitof large volume V when the system is described by a chemical potential µ (assumingthere is no Bose-Einstein condensation). Evaluate the integral explicitly in terms of thefugacity z = eβµ and temperature T using the formula given below.

(c) (25 points) To determine whether Bose-Einstein condensation occurs, we have to ex-amine the expression for N in (b) in the limit µ → 0. Explain why this is the relevantlimit and determine the condition satisfied by s and d for which Bose-Einstein condensa-tion occurs. Check that this relation holds for the usual case of d = 3 and s = 2.

(d) (25 points) When Bose-Einstein condensation occurs, find for T < Tc the fraction〈n0〉/N of bosons in the single-particle ground state ~p = 0 as a function of T/Tc.

Useful relations: ∫ ∞0

dxxr−1

z−1ex − 1= Γ(r)Fr(z) ,

where Γ(r) the gamma function and Fr(z) is the function defined by its series expansionFr(z) =

∑∞n=1 z

n/nr.

Fr(1) diverges for r ≤ 1 and converges for r > 1.

6

N2\

€x'dlÈ;{ËUri

R

, v,¡=u'ff+u, iH-rH

å3 v'a.=",(} W-#)."(#-#)*" |($oau-ä),'v=I*o('#).##.#

Explicit Forms of Vector Operations

Let er, €2, €¡ b€ orthogonal unit vectors associated srith the coordinate directions

specified in the headings on the left, and A1, Az, A¡ be the corresponding

components of A. Then

vú=e, ffi*",ff*^ffiv.¡,:ðrA'+**&ôxt' ôxz- ðxg

:

v x a : e, (H-*)*' (#-#) -* (*-*)v'1,:ffi*ffi*ffi

vú=e' #*",1#** ^håËo .o=+$ tr a,l*ffiS rrt" oA,)*^hq#

vxÀ:er **[*n*oAt-H].., [^h #-] $ t,",r]*", + [#

('ot-#]

u',þ :+ *(t #).r..|6 * (""' #).*- #[No,",t ", ] $ (r S)=i $ t+il

.Ea9(Düìè

AND d FUNCTIONS

Note: A square-root sign is to be understood ovet elter|cocfficicnt, e.g., for -8/15 read -JW'

35. CLEBS CH-G ORDAN COEFFICIENTS, SPHERICAL HARMONIC S,

2x!/2

t- xt/+L +L/21

3/2

+r -r/2o +r/2

3/2 L/2-112 +Ll2I

4: \Æ"* o 2

"l =-l*sir.i,eiö

2x]-

L/3 2/32'/3 -r/3

2+

o -r/2-r +r/2

3

X1

3/2 r/2-L/2 -r/2

+1

F1 +11 1

32+2 +2I

2

+2

w:'/T^G"",'r-å)

",] : -ly- si,'l,eoso eió

"î :i'{*

"i,.2 s.2nó

0

+1

2/3 rl37/3 -2/3

+1 0

0+1

/3 2/3/3 -r/3

2

+1

-t -r/2

1

+1

L/2 r/2t/2 -t/2

+2 -1+1 0

0+1

3lZ-3/2

x1

321+1 +1 +1

35. Clebsch-Gordan coefficients 1

Yl^ = ?r) Ytr.

1/1sB/rs2/s

+t- -L00

-1 +1

2ll,2/,:

1

2L0000

r/3 3/s7/6 -3/ro

-L/2 rho

Ll6 1/z r/32/3 o-r/3r/6 -r/2 L/3

3/2xt

+z -Ll2+L +L/2

s/2 3/2-3/2+3/2

+al2 +Ll 1

+1 -100

-1 +1

rls 4/!4ls -Ll!

sl2

321000

-1 0

+3/2 O

+L/2 +1.

2

-1

+L-L/z

LlS r/2!/5 o

Lls -rl2

slz 3/2+112 +3/2

3/2xL/2

s/27/2

1

-t

./2 r/2

./2 -L/2

m1 ^rlm1 mz I Coefficients

t

312+L/2

2/s 3/s3/s -2/s

2/5 3ls3ls -zls

JJMM

3/10-2/53/ro

-1 -t

2

-2

+312-1+L/2 0

-7/2 +7

0 -rl2-r +L/2

+3/2 -!/2+Ll2 +L/2

0-1-1 0

-2 +L

s/2 3/2 t/2tr/2 +L/2 +L/2

321-1 -1 -1

I

s/2

a;,o: ffirvY"

L/2

LlLo zls3/s LlLs

311-0 -8115

2/s a/2 a/Lo8/rs -L/6-31r0LlLs -L/3 3/s

+

3/2Ll2

3lS 2/52lS -3/s

1

L/4 314rl4 -al4

+r/2 -L/2-r/z +L/2

-L -L/2-2 +L/2

L/2-r/3

Ll6

5/z 3/2-3/2 -3/2

2r00

+r/z -L-t/2 o

-3/2 +L

4/s rlsals -4ls

rl2 u2v2 -a/2

-24

s/2 312 u2-L/2 -L/2 -L/2

32-2 -2

-L/2 -L/2-3/2 +u2

-inø

2/3 L/3

3/LO s/rs3/s -r/Is

LlLo -zls

-2 -t/2

s/2

2!-t -1

3/4 L/41-/4-3/4

?

-t

1

r/6-a/3

L/2

-312-112

-!/2-L-3/2 O

-2

5/2 3/2-312 -3123ls 2/s2/s -3/5

-3/2 -!

s/2-s l2

1