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I'm currently studying Kittel's Solid State Physics and in his chapter on Phonon heat capacity, we need to first calculate the total energy $U$. Phonons have energy $E_n = (n+1/2)\hbar\omega$ and he first calculated the average energy $\langle E\rangle$ and using the Boltzmann factor, he showed: $$\langle E\rangle = \dfrac{1}{2}\hbar\omega+\hbar\omega\dfrac{1}{e^{\hbar\omega/k_BT}-1}=\dfrac{1}{2}\hbar\omega+\hbar\omega\langle n \rangle$$ so then we must have $$\langle n \rangle=\dfrac{1}{e^{\hbar\omega/k_BT}-1}$$ I recognise this as the Bose-Einstein but I'm surprised to see this as being interpreted as an average of the number of states. I always thought this was a probabilistic distribution and in fact, Kittel does seem to use this as a probability since he later writes: $$U=\int d\omega\ \hbar\omega D(\omega)\langle n \rangle$$ where $D(\omega)$ is the density of state. In this expression $D$ already accounts for the number of photons so $\langle n \rangle$ must be some probability weight? I'm sure something is flawed in my understanding so any help is appreciated!

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    $\begingroup$ <n> isn't an average number of states. It's the average number of phonons in a state with frequency $\omega$ at temperature T. $\endgroup$ – Samuel Weir Nov 1 '18 at 5:27
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I wouldn't say that $D\left(\omega\right)$ accounts for the number of phonons; I'd say it accounts for the number of vibrational (phonon) modes at a given energy. A phonon is a unit of energy in a vibrational mode. Then, $\left<n\right>$ is the average number of phonons in a given vibrational mode (what Samuel Weir said).

So the integrand is the energy of a phonon $\hbar \omega$ times the average number of those phonons in a mode $\left<n\right>$ times the density of modes at that energy $D\left(\omega\right)$.

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