Discussion
The system consists of two indistinguishable particles, each of which can be in either of the two energy states, 0 or \(\varepsilon\). The system is in thermal equilibrium with a heat bath at temperature T. There is also an interaction energy term that increases the energy by an amount \(\delta\) if the two particles are in the same energy state. The average energy of the system is \(\langle E \rangle\) = ..
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.
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