Discussion
Two identical particles with a fixed total energy \(E = 2\hbar\omega\) are in thermal equilibrium in a one-dimensional harmonic oscillator potential \(\frac{1}{2}m\omega^{2}x^{2}\). Let the entropy of the particles be denoted by \(S_{F}\) if they are fermions with spin \(\frac{1}{2}\) \(\left(s_{z} = \pm\frac{\hbar}{2}\right)\) and \(S_{B}\) if they are bosons with spin 0. Then, which of the following options is correct?
(\(k_{B}\) is the Boltzmann constant)
Answer: B
The entropies are \(S_{F} = 2k_{B}\ln 2\) and \(S_{B} = 0\).
The single-particle levels are \((n + \frac{1}{2})\hbar\omega\). Two particles with total energy \(2\hbar\omega\) need \(n_{1} + n_{2} + 1 = 2\), so one particle is in n = 0 and the other in n = 1. The entropy is \(k_{B}\ln\Omega\), where \(\Omega\) counts the distinct many-particle states.
For spin-0 bosons the particles are identical and the occupation {0, 1} is a single state, so \(\Omega = 1\) and \(S_{B} = 0\). For spin-\(\frac{1}{2}\) fermions the two particles are in different levels, so each can have either spin: \(2 \times 2 = 4\) states, and
\[ S_{F} = k_{B}\ln 4 = 2k_{B}\ln 2 \]
Options with \(S_{B} = k_{B}\ln 2\) drop the zero-point energy, which would allow n = (0, 2) and (1, 1) for bosons. \(4k_{B}\ln 2\) would need 16 fermion states.
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