Your Answer : (Not Answered)

Correct Answer : A

Second sound is the wave motion of entropy and temperature.

In the two-fluid model of He II, the liquid is a mixture of a superfluid component, which carries no entropy, and a normal component, which carries all of it. In ordinary (first) sound both components move together, giving a pressure and density wave. In second sound they move in opposite directions with no net mass flow, so the total density and pressure stay almost constant while the local entropy, and with it the temperature, oscillates. It is a temperature wave, excited by a heater rather than a loudspeaker.

The pressure and density waves describe first sound, and the motion of the normal component alone is not second sound, which needs the counterflow of both components.

Your Answer : (Not Answered)

Correct Answer : C

The keyed probability is \(\left\{\frac{10!}{4!6!}\left[\frac{1}{2}\right]^{10}\right\}^{2}\).

Each walker takes 10 independent steps, each left or right with probability 1/2. The probability that one walker takes 4 steps one way and 6 the other is binomial:

\[ \begin{aligned} p &= \frac{10!}{4!\,6!}\Big(\frac{1}{2}\Big)^{10} \\ &= \frac{210}{1024} \end{aligned} \]

The walkers are independent, so the probability that both make that same choice, and so stand together on the same site after 10 steps, is the product \(p \times p = p^{2}\).

A single factor \(p\) is the probability for one walker only, and a cube would apply to three walkers. Squaring only the binomial coefficient while squaring the whole bracket again, as in the last option, double-counts the factors.

Your Answer : (Not Answered)

Correct Answer : B

The fluctuation is proportional to \(1/\sqrt{N}\).

In the canonical ensemble, the mean square energy fluctuation is \(\langle(\Delta E)^{2}\rangle = kT^{2}C_{V}\). Both \(C_{V}\) and \(\langle E \rangle\) are proportional to \(N\), so the relative fluctuation is:

\[ \begin{aligned} \frac{\Delta E}{\langle E \rangle} &= \frac{\sqrt{kT^{2}C_{V}}}{\langle E \rangle} \propto \frac{\sqrt{N}}{N} \\ &= \frac{1}{\sqrt{N}} \end{aligned} \]

This is why the energy of a macroscopic system is sharply defined. 1/N is too fast a fall-off, and \(1/(3\sqrt{N})\) adds an arbitrary factor. The absolute fluctuation grows like \(\sqrt{N}\), not \(N\).

Your Answer : (Not Answered)

Correct Answer : A

The Maxwell-Boltzmann distribution holds for distinguishable particles.

Maxwell-Boltzmann statistics counts particles as identifiable individuals, with no limit on how many can occupy a state. It applies to classical systems and, as a limit, to dilute gases at high temperature where the average occupancy of each state is much less than one.

Indistinguishable quantum particles obey Fermi-Dirac statistics if they have half-integral spin (fermions, subject to the exclusion principle) and Bose-Einstein statistics if they have integral spin (bosons). So the options naming all indistinguishable particles, or particles of a given spin, describe the quantum distributions instead.

Your Answer : (Not Answered)

Correct Answer : C

The number of photons increases by a factor of 8.

The equilibrium photon number density in a cavity is obtained by integrating the Planck distribution:

\[ \begin{aligned} n &= \frac{2\zeta(3)}{\pi^{2}}\left(\frac{kT}{\hbar c}\right)^{3} \\ &\propto T^{3} \end{aligned} \]

Doubling T multiplies the number by \(2^{3} = 8\). The energy density goes as \(T^{4}\) and would rise 16 times; 2, 6 and 9 do not come from either law.

Your Answer : (Not Answered)

Correct Answer : C

The number of photons increases by a factor of 8.

Integrating the Planck distribution over all frequencies gives the photon number density in a blackbody cavity:

\[ \begin{aligned} n &= \frac{2\zeta(3)}{\pi^{2}}\left(\frac{k_{B}T}{\hbar c}\right)^{3} \\ &\propto T^{3} \end{aligned} \]

Doubling T multiplies n by \(2^{3} = 8\).

The factor 16 is the Stefan-Boltzmann result for the energy density, which goes as \(T^{4}\): the photons become both more numerous and more energetic on average. Factors of 2 and 4 would correspond to linear and quadratic dependences that do not apply to blackbody radiation.

Your Answer : (Not Answered)

Correct Answer : D

On the key, there are 3 accessible microstates.

For indistinguishable particles a microstate is fixed by how many particles occupy each level, not by which particle is where. With no limit on the number per level (Bose counting), we need three energies from {0, 1, 2, 3} in units of \(\varepsilon\) adding to 4:

\[ \begin{aligned} 4 &= 3 + 1 + 0 \\ 4 &= 2 + 2 + 0 \\ 4 &= 2 + 1 + 1 \end{aligned} \]

(4 + 0 + 0 is impossible because the highest level is \(3\varepsilon\).) That gives 3 microstates. One microstate would be the answer for fermions, which forbid the doubly occupied levels in the second and third sets. Four would come from counting an arrangement twice; distinguishable particles would give many more, since each set could be permuted.

Your Answer : (Not Answered)

Correct Answer : A

The magnetic moment is proportional to the difference between the population of the two states.

In a two-level system, the states have moments \(+\mu\) and \(-\mu\) along \(\vec{B}\). With fractional populations \(n_{1}\) and \(n_{2}\), the net moment is \(\mu(n_{1} - n_{2})\), so it is set by the population difference, which the Boltzmann factor makes \(\tanh(\mu B/kT)\).

The sum of the fractional populations is always 1 and carries no information. The product and the ratio do not give the net moment directly.

Your Answer : (Not Answered)

Correct Answer : A

The constant is \(C = -J^{2}\).

The four configurations have energy -J (both spins equal, two states) or +J (opposite, two states). The partition function and mean energy are

\[ \begin{aligned} Z &= 2e^{\beta J} + 2e^{-\beta J} = 4\cosh\beta J \\ \langle E\rangle &= -\frac{\partial\ln Z}{\partial\beta} = -J\tanh\beta J \end{aligned} \]

At high temperature \(\beta J\) is small and \(\tanh\beta J \approx \beta J\), so

\[ \langle E\rangle \approx -\frac{J^{2}}{k_{B}T} \]

The negative sign reflects the slight preference for the lower-energy aligned states. \(+J^{2}\) has the wrong sign, and \(-2J^{2}\) or \(4J^{2}\) would follow from miscounting the states.

Your Answer : (Not Answered)

Correct Answer : A

The probability is 1/18.

Each step has 6 equally likely directions, so there are \(6^{3} = 216\) three-step walks. To return in exactly three steps, the walker must go round a small triangle: on a triangular lattice, two steps at 120 degrees to each other followed by the third side.

The first step can be any of 6 directions; for each, there are 2 ways to complete a triangle back to the start (turning left or right). So there are \(6 \times 2 = 12\) returning walks:

\[ P = \frac{12}{216} = \frac{1}{18} \]

1/216 counts just one walk, and 1/36 and 1/12 use the wrong number of closing paths.

Your Answer : (Not Answered)

Correct Answer : D

The fluctuation is \(\Delta M = \frac{1}{\sqrt{N}\sinh(\mu H/KT)}\). (The paper prints the square of \(\langle M\rangle\) as a trailing 2.)

Each moment points along or against the field with probabilities \(\propto e^{\pm x}\), where \(x = \mu H/KT\). For one moment \(\langle m\rangle = \mu\tanh x\) and \(\langle m^{2}\rangle = \mu^{2}\). For N independent moments the means and variances add:

\[ \begin{aligned} \langle M\rangle &= N\mu\tanh x \\ \langle M^{2}\rangle - \langle M\rangle^{2} &= N\mu^{2}(1 - \tanh^{2}x) = \frac{N\mu^{2}}{\cosh^{2}x} \end{aligned} \]

\[ \begin{aligned} \Delta M &= \frac{\sqrt{N}\mu/\cosh x}{N\mu\tanh x} \\ &= \frac{1}{\sqrt{N}\sinh x} \end{aligned} \]

The cosh and tanh options keep only part of this ratio, and \(1/\sqrt{N}\) drops the field and temperature dependence altogether, although it shows the general rule that relative fluctuations fall as \(1/\sqrt{N}\).

Your Answer : (Not Answered)

Correct Answer : C

The number of ways is 0.

Identical fermions obey the Pauli exclusion principle: no two can occupy the same single-particle state. Three energy levels (each a single state) can hold at most three fermions, and even if each level held two of opposite spin, the limit would be six. Seven identical fermions therefore cannot be distributed in three levels at all.

The other numbers belong to other statistics: \(3^{7}\) counts distinguishable (Maxwell-Boltzmann) particles, and \(\frac{7!}{4!3!} = 35\) and 210 are combinatorial counts that ignore the exclusion principle.

Your Answer : (Not Answered)

Correct Answer : A

The minimum value is \( \frac{\mu_{1}T_{2} - \mu_{2}T_{1}}{T_{2} - T_{1}} \).

The whole system is closed, so \( dU_{2} = -dU_{1} \) and \( dN_{2} = -dN_{1} \), with volumes fixed. The total entropy change is

\[ \begin{aligned} dS = &\left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right) dU_{1} \\ - &\left(\frac{\mu_{1}}{T_{1}} - \frac{\mu_{2}}{T_{2}}\right) dN_{1} \end{aligned} \]

The second law requires \( dS \ge 0 \). Taking \( T_{2} \gt T_{1} \) and particles flowing into subsystem 1 (\( dN_{1} \gt 0 \)), divide through and multiply the fraction by \( T_{1}T_{2} \):

\[ \frac{dU_1}{dN_1}\ge\frac{\mu_1T_2-\mu_2T_1}{T_2-T_1} \]

The other three options change a sign in the numerator or the denominator, which no step of this derivation produces.

Your Answer : (Not Answered)

Correct Answer : C

The partition function is \( \frac{e^{\hbar\omega/k_{B}T}}{\left(e^{\hbar\omega/k_{B}T} - 1\right)^{2}} \).

The two quantum numbers are independent, so the sum factorises into two geometric series, with \( x = e^{-\hbar\omega/k_{B}T} \):

\[ \begin{aligned} Z &= x \sum_{n_{x}} x^{n_{x}} \sum_{n_{y}} x^{n_{y}} = \frac{x}{(1 - x)^{2}} \\ &= \frac{e^{\hbar\omega/k_{B}T}}{\left(e^{\hbar\omega/k_{B}T} - 1\right)^{2}} \end{aligned} \]

the last step multiplying top and bottom by \( x^{-2} \). The options with \( 2k_{B}T \) in the exponents halve the level spacing \( \hbar\omega \), and the option with \( +1 \) has the wrong sign in the denominator; none of them is the sum of these geometric series.

Your Answer : (Not Answered)

Correct Answer : C

The chemical potential at T = 0 is \(\frac{3}{2}\hbar\omega\).

⁴He atoms are bosons. For an ideal Bose gas the chemical potential must stay below the lowest single-particle energy, and as T falls to zero all the atoms condense into that ground state and \(\mu\) rises to meet it. The ground state of the 3D isotropic oscillator has zero-point energy \(\frac{1}{2}\hbar\omega\) in each of the three directions:

\[ \varepsilon_{0} = \tfrac{3}{2}\hbar\omega \]

So \(\mu \to \frac{3}{2}\hbar\omega\). Zero would be the answer only if energies were measured from the ground state, \(\frac{1}{2}\hbar\omega\) counts one direction only, and \(3\hbar\omega\) doubles the zero-point energy.

Your Answer : (Not Answered)

Correct Answer : B

The condition implies \(\dfrac{\partial\rho}{\partial t} \propto \{\rho, H\}\).

The total time derivative of a phase-space function along the motion is its explicit time derivative plus its Poisson bracket with the Hamiltonian:

\[ \begin{aligned} \frac{d\rho}{dt} &= \frac{\partial\rho}{\partial t} + \{\rho, H\} = 0 \\ \frac{\partial\rho}{\partial t} &= -\{\rho, H\} \end{aligned} \]

This is Liouville's theorem: the phase-space density is constant along trajectories, and its local rate of change is fixed by the bracket with H. ∂ρ/∂t = 0 holds only for a stationary ensemble, which dρ/dt = 0 alone does not guarantee. Brackets with p·q/2 or q·q/2 involve functions that are not the Hamiltonian of the system and do not generate its motion.

Your Answer : (Not Answered)

Correct Answer : A

The correct dependence is (A): U starts flat at a non-zero value, rises, and levels off.

For a classical two-level system the occupation follows the Boltzmann factor, so

\[ \begin{aligned} U &= N\epsilon_{1} + N\Delta\,p_{2} \\ p_{2} &= \frac{e^{-\Delta/k_{B}T}}{1 + e^{-\Delta/k_{B}T}} \end{aligned} \]

At low T almost every particle is in the lower level, so \(U \approx N\epsilon_{1}\), and the rise is exponentially slow until \(k_{B}T\) approaches \(\Delta\). At high T the two levels become equally occupied, so U saturates at \(N(\epsilon_{1} + \Delta/2)\) instead of growing without limit.

(B) and (D) keep rising ever more steeply, as for a system with unlimited levels. (C) has U falling as T rises, which would mean a negative heat capacity.

Your Answer : (Not Answered)

Correct Answer : A

The partition function is \(\frac{e}{e^{2} - 1}\).

For a quantum oscillator with levels \(E_{n} = (n + \frac{1}{2})\hbar\omega\), the sum is geometric:

\[ \begin{aligned} Z &= \sum_{n=0}^{\infty} e^{-(n + 1/2)x} = \frac{e^{-x/2}}{1 - e^{-x}} \\ x &= \frac{\hbar\omega}{k_{B}T} = 2 \end{aligned} \]

So \(Z = e^{-1}/(1 - e^{-2})\); multiplying top and bottom by \(e^{2}\) gives \(e/(e^{2} - 1)\).

The \(+1\) in the denominator would come from an alternating series, which the oscillator does not have, and \(e/(e - 1)\) is what \(x = 1\) would give, forgetting the factor 2 in \(\omega = 2k_{B}T/\hbar\).

Your Answer : (Not Answered)

Correct Answer : A

The average moment is \(\frac{2\mu}{3}\left(\frac{\mu B}{k_{B}T}\right)\).

A spin-1 particle has three orientations, \(m = +1, 0, -1\), with energies \(-\mu B\), 0 and \(+\mu B\). Writing \(x = \mu B/k_{B}T\), the Boltzmann average is

\[ \begin{aligned} \langle \mu_{z} \rangle &= \mu\frac{e^{x} - e^{-x}}{1 + e^{x} + e^{-x}} \\ &= \mu\frac{2\sinh x}{1 + 2\cosh x} \end{aligned} \]

For \(x \ll 1\), \(\sinh x \approx x\) and \(\cosh x \approx 1\), so \(\langle \mu_{z} \rangle \approx \mu\frac{2x}{3}\). The factor 2/3 is the Curie-law factor \(J(J+1)/3\) for \(J = 1\). The other coefficients, \(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{3}{4}\), do not come from the three-level sum; a spin-1/2 particle, for comparison, would give \(\mu x\).

Your Answer : (Not Answered)

Correct Answer : C

The counts are (i) 64 and (ii) 20.

(i) Distinguishable particles: each of the 3 particles independently chooses one of 4 levels, so there are \(4^{3} = 64\) arrangements.

(ii) Identical bosons: only the occupation numbers matter, and any number may share a level. The number of ways to put 3 identical objects into 4 boxes is

\[ \binom{3 + 4 - 1}{3} = \binom{6}{3} = 20 \]

24 is \(4 \times 3 \times 2\), the count for distinguishable particles that may not share a level. 4 is \(\binom{4}{3}\), the count for fermions (at most one per level). 16 has no basis.