GATE 2026 PH – Question 42
Two identical particles with a fixed total energy $E=2\hbar\omega$ are in thermal equilibrium in a one-dimensional harmonic oscillator potential $\frac12m\omega^2x^2$. Let the entropy of the particles be denoted by $S_F$ if they are fermions with spin $1/2$ ($s_z=\pm\hbar/2$) and $S_B$ if they are bosons with spin 0. Then, which of the following options is correct? ($k_B$ is the Boltzmann constant)
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Correct answer: (B) $S_F=2k_B\ln2,\ S_B=0$
Explanation
The two particles are in a 1-D harmonic oscillator with total energy $E=2\hbar\omega$. Each particle has an energy $\left(n+\tfrac12\right)\hbar\omega$, so
$$\left(n_1+\tfrac12\right)+\left(n_2+\tfrac12\right)=2\;\Rightarrow\;n_1+n_2=1 .$$
The only occupation is one particle in $n=0$ and the other in $n=1$.
**Bosons (spin 0).** The wave function must be symmetric under exchange. With two different orbitals there is exactly **one** symmetric state. So $\Omega_B=1$ and
$$S_B=k_B\ln1=0 .$$
**Fermions (spin ½).** The total wave function must be antisymmetric. With the two particles in different orbitals ($n=0$ and $n=1$):
- spatial symmetric $\times$ spin singlet: 1 state,
- spatial antisymmetric $\times$ spin triplet: 3 states.
That gives $\Omega_F=4$ states, so
$$S_F=k_B\ln4=2k_B\ln2 .$$
Answer: $S_F=2k_B\ln2$ and $S_B=0$ (option B).