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Unformatted text preview: Quiz 3 Solutions 1 Fermionic Photons (a) The partition function for the mode ω m (for a single polarization) is given by summing exp(- s /τ ) over all states s . In this case, because we’re treating the photons as fermions, there can be no more than one photon per state. Thus there are only two states: one with zero photons and one with one photon, and so: Z = 1 + e- ¯ hω m /τ Then the average number of photons in this mode is just: < N > = 1 Z (0) + (1) e- ¯ hω m /τ = e- ¯ hω m /τ 1 + e- ¯ hω m /τ = 1 1 + e ¯ hω m /τ We need to multiply this by 2 to account for the two independent polarizations: < N m > = 2 1 + e ¯ hω m /τ Note this is basically a derivation of the Fermi-Dirac distribution (for zero chemical potential). Although we hadn’t covered this by the time of the quiz, this derivation uses only material from chapter 3. (b) The average energy of the total system is just the sum of the contribution from each mode, which is: < U > = m ¯ hω m < N m > Using part (a) we get:...
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