Partition Function Questions and Answers

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.

Answer: A

The average energy is the Boltzmann-weighted average of \(\delta\), \(2\varepsilon + \delta\) and \(\varepsilon\), with the both-in-0 state entering as \(\delta e^{-\beta\delta}\).

For two indistinguishable particles (counted as bosons) there are three distinct states of the system:

• both in state 0: energy \(0 + 0 + \delta = \delta\);
• one in each state: energy \(\varepsilon\), no interaction term;
• both in state \(\varepsilon\): energy \(2\varepsilon + \delta\).

The partition function is \(Z = e^{-\beta\delta} + e^{-\beta\varepsilon} + e^{-\beta(2\varepsilon + \delta)}\), and \(\langle E \rangle = \sum E_{i}e^{-\beta E_{i}}/Z\), which is the keyed expression.

The option with \((\delta + \varepsilon)\) gives the both-in-0 state the wrong energy. The option without Boltzmann factors in the numerator is not a weighted average, and \((2\delta + 3\varepsilon)/3\) is a plain average with no temperature dependence.

Answer: B

The correct statement is \(\langle M \rangle = \frac{N\mu^{2}}{k_{B}T}B_{z}\) with \(\chi \propto 1/T\).

For spin-1/2 moments the two energies are \(\mp\mu B_{z}\), and

\[ \begin{aligned} \langle M \rangle &= N\mu\tanh\frac{\mu B_{z}}{k_{B}T} \\ &\approx \frac{N\mu^{2}B_{z}}{k_{B}T} \end{aligned} \]

in the limit \(\mu B_{z} \ll k_{B}T\) (low field, high temperature), using \(\tanh x \approx x\). The magnetisation is linear in the field, and the susceptibility \(\chi = \partial M/\partial B\) varies as \(1/T\): Curie's law.

\(\chi \propto T\) contradicts the Curie law and goes with the inverted expression in the last option. The option with \(N\mu B_{z}^{2}\) is not linear in the field and has the wrong dimensions.