# Mean energy in the Debye calculation?

I am reading 'The Oxford Solid State Basics' by S.H.Simon page 11 on which it say's that the mean energy in the Debye calculation of the specific heat capacity is: $$\left<E\right>=3 \sum_{\vec k} \hbar \omega(\vec k)\left(n_B(\beta \hbar \omega(\vec k))+\frac{1}{2}\right)$$ where $n_B(\beta \hbar \omega(\vec k))=1/(e^{\beta \hbar \omega}-1)$. However a minimal explanation is given of where it this comes from, and looking in other sources I can't seem to find any explanation. In short my question is: Where does this expression come from?

Additional Information

Here is my main concern: The above expression is indicating that the mean energy of an oscillator with wave vector $\vec k$ (for one polarization) is: $$\left<E_{\vec k}\right>=\hbar \omega(\vec k)\left(n_B(\beta \hbar \omega(\vec k))+\frac{1}{2}\right)$$ for an simple harmonic oscillator the mean energy is: $$\left<E_{SHO}\right>=\hbar \omega\left(n_B(\beta \hbar \omega)+\frac{1}{2}\right)$$ which is found by a summation over all $n$ of the Boltzmann factor with energy $\hbar \omega\left(n+\frac{1}{2}\right)$, but for each $n$ the wavenumber $k$ takes a different value, so the summation is really a mean over the wavenumber $k$. The book draws an anology between $\left<E_{\vec k}\right>$ and $\left<E_{SHO}\right>$ so I cannot see how $\left<E_{\vec k}\right>$ can be a function of $\vec K$.

• Because phonon energies are a function of momentum, so you have to average over all crystal momenta. – Jon Custer Oct 7 '16 at 15:38
• @JonCuster Please could you expand on this comment. Thanks – Quantum spaghettification Oct 10 '16 at 14:53