Partition Function Questions and Answers

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 \(\mu x\).

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 \(\omega = 2k_{B}T/\hbar\).

Answer: A

The answer is \(\frac{k_{B}}{2\varepsilon}\ln[\frac{N-E/\varepsilon}{N+E/\varepsilon}]\).

From \(N = N_{+} + N_{-}\) and \(E = \varepsilon(N_{+} - N_{-})\), \(N_{\pm} = \frac{1}{2}(N \pm \frac{E}{\varepsilon})\). The entropy, with Stirling's formula, is

\[ \begin{aligned} S = k_{B}[&N\ln N - N_{+}\ln N_{+} \\ &- N_{-}\ln N_{-}] \end{aligned} \]

Since \(\frac{dN_{\pm}}{dE} = \pm\frac{1}{2\varepsilon}\):

\[ \begin{aligned} \frac{1}{T} = \frac{\partial S}{\partial E} &= \frac{k_{B}}{2\varepsilon}(\ln N_{-} - \ln N_{+}) \\ &= \frac{k_{B}}{2\varepsilon}\ln\frac{N - E/\varepsilon}{N + E/\varepsilon} \end{aligned} \]

Option (D) misses the factor \(\frac{1}{2}\) from \(\frac{dN_{\pm}}{dE}\), and the \(\ln N\) options do not depend on \(E\), so they cannot give a temperature that changes with energy. For \(E \gt 0\) the result is negative, the familiar negative temperature of a population inversion.

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.

Answer: B

The average energy is \(\frac{\epsilon}{e^{\epsilon/kT} + 1}\).

With levels 0 and \(\epsilon\), the partition function is \(Z = 1 + e^{-\epsilon/kT}\). The mean energy is the Boltzmann-weighted average:

\[ \begin{aligned} \langle E\rangle &= \frac{0\cdot 1 + \epsilon\, e^{-\epsilon/kT}}{1 + e^{-\epsilon/kT}} \\ &= \frac{\epsilon}{e^{\epsilon/kT} + 1} \end{aligned} \]

Check the limits: at T = 0 it gives 0, and at high T it tends to \(\epsilon/2\), as both levels become equally occupied. The form with \(-1\) in the denominator is the mean energy of a harmonic oscillator mode (Planck's formula), which has infinitely many levels. The forms with \(e^{-\epsilon/kT}\) give the wrong limits, and the one with \(-1\) even becomes negative.

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.

Answer: A

The partition function is \(z = 2e^{0} + 3e^{-1}\).

The partition function sums the Boltzmann factor of each level, weighted by its degeneracy: \(z = \sum g_{i}e^{-E_{i}/k_{B}T}\). At T = 1 K, \(k_{B}T = 1.3806 \times 10^{-23}\) J, exactly \(E_{1}\), so \(E_{1}/k_{B}T = 1\):

\[ \begin{aligned} z &= 2e^{-0} + 3e^{-1} \\ &= 2 + 3e^{-1} \approx 3.10 \end{aligned} \]

The options with 6 multiply the two degeneracies together, which would be the rule for combining two independent systems, not for summing over the levels of one system.

Answer: B

The two levels are equally occupied when T is infinite.

In thermal equilibrium the ratio of occupations is given by the Boltzmann factor:

\[ \frac{N_{\varepsilon}}{N_{0}} = e^{-\varepsilon/k_{B}T} \]

For any \(\varepsilon \gt 0\) this ratio is less than 1 at every finite temperature, and it reaches 1 only in the limit \(T \rightarrow \infty\). At T = 0 all particles are in the lower level, and at very low temperature nearly all are. At a very high but finite temperature the populations approach equality without reaching it, which is why 'very high' is not exact.

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.

Answer: C

The average energy is \(E = \frac{\hbar v}{e^{\hbar v/kT} - 1}\).

Planck assumed the oscillator energy is quantised in steps of \(h\nu\) (the paper writes \(\hbar v\) for the quantum). Weighting each level \(nh\nu\) with the Boltzmann factor gives the mean energy:

\[ \begin{aligned} \langle E \rangle &= \frac{\sum nh\nu\,e^{-nh\nu/kT}}{\sum e^{-nh\nu/kT}} \\ &= \frac{h\nu}{e^{h\nu/kT} - 1} \end{aligned} \]

A single quantum or \(n\) quanta is the energy of a particular state, not the thermal average, and \(mc^{2}\) is rest energy. At high temperature the expression tends to \(kT\), the classical value.