Question

Two identical particles with a fixed total energy \(E = 2\hbar\omega\) are in thermal equilibrium in a one-dimensional harmonic oscillator potential \(\frac{1}{2}m\omega^2x^2\). Let the entropy of the particles be denoted by \(S_F\) if they are fermions with spin \(\frac{1}{2}\) (\(S_z = \pm\frac{\hbar}{2}\)) and by \(S_B\) if they are bosons with spin 0. Then, which of the following options is correct? (\(k_B\) is the Boltzmann constant)

• A. \(S_F = k_B\ln 2,\ S_B = k_B\ln 2\)
• B. \(S_F = 2k_B\ln 2,\ S_B = 0\)
• C. \(S_F = 4k_B\ln 2,\ S_B = 0\)
• D. \(S_F = 2k_B\ln 2,\ S_B = k_B\ln 2\)

Hint:

Total energy fixes \(n_1+n_2=1\), so the two particles must sit in levels 0 and 1. Count the allowed spin/label combinations for fermions and for bosons separately, then use \(S=k_B\ln\Omega\).

Answer 1

The Correct Option is B

Step 1: Find which single-particle levels can be occupied.
The single particle energies are $\varepsilon_n = (n+\frac12)\hbar\omega$, so two particles with total energy $E=2\hbar\omega$ need $n_1+n_2=1$. The only non-negative integer pair is $\{0,1\}$: one particle in the ground level, one in the first excited level. No pair with $n_1=n_2$ works, since that would need $n_1+n_2$ to be even, and 1 is odd.

Step 2: List the fermion microstates directly.
Label each single-particle state by (level, spin). The particle in level 0 can carry spin $+\frac12$ or $-\frac12$, and independently, the particle in level 1 can also carry spin $+\frac12$ or $-\frac12$. Listing all four combinations:
(0,up)(1,up), (0,up)(1,down), (0,down)(1,up), (0,down)(1,down).
Since the two particles already sit in different spatial levels, exchange antisymmetry never removes any of these four combinations, they are all valid two-fermion states. So $\Omega_F=4$, giving
\[ S_F = k_B \ln 4 = 2 k_B \ln 2 \]

Step 3: List the boson microstates directly.
Spin-0 bosons carry no spin label, so the only information is which levels are occupied: one particle in level 0, one in level 1. For identical particles, swapping "which particle" is in which level is not a new state, it is the same physical configuration counted once. So there is exactly one microstate, $\Omega_B=1$, giving
\[ S_B = k_B\ln 1 = 0 \]

Step 4: Match against the options.
Only the pair $S_F=2k_B\ln2$ and $S_B=0$ appears together, which is option (B). Any option with $S_B\neq0$ is wrong because the identical bosons have just one way to arrange themselves here, and $4k_B\ln2$ overshoots the fermion count, which stops at 4 combinations, not 16.

Final Answer:
Direct enumeration of the allowed (level, spin) combinations gives $\Omega_F=4$ and $\Omega_B=1$. \[ \boxed{S_F = 2k_B\ln 2,\ S_B = 0} \]