GATE 2021 PH – Question 47
Two atoms can occupy three quantum states with energies $0,\epsilon,2\epsilon$, in equilibrium at $T=(k_B\beta)^{-1}$. Match CD (classical distinguishable), CI (classical indistinguishable), FD (Fermi–Dirac) and BE (Bose–Einstein) statistics to the partition functions: $$Z_1=e^{-\beta\epsilon}+e^{-2\beta\epsilon}+e^{-3\beta\epsilon},$$ $$Z_2=1+e^{-\beta\epsilon}+2e^{-2\beta\epsilon}+e^{-3\beta\epsilon}+e^{-4\beta\epsilon},$$ $$Z_3=1+2e^{-\beta\epsilon}+3e^{-2\beta\epsilon}+2e^{-3\beta\epsilon}+e^{-4\beta\epsilon},$$ $$Z_4=\tfrac12+e^{-\beta\epsilon}+\tfrac32e^{-2\beta\epsilon}+e^{-3\beta\epsilon}+\tfrac12e^{-4\beta\epsilon}.$$
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Correct answer: (C) CD: Z3, CI: Z4, FD: Z1, BE: Z2.
Explanation
Put $u=e^{-\beta\epsilon}$. For distinguishable classical particles, $Z_{CD}=(1+u+u^2)^2=1+2u+3u^2+2u^3+u^4=Z_3$. The classical indistinguishable correction divides this by $2!$, giving $Z_4$. Fermions occupy two different states, so $Z_{FD}=u+u^2+u^3=Z_1$. Boson occupation pairs give energies $0,\epsilon,2\epsilon,2\epsilon,3\epsilon,4\epsilon$, hence $Z_{BE}=1+u+2u^2+u^3+u^4=Z_2$. The matching is C.