# Why is there a sum over $\mathbf{k}$ in Debye's calculation

In Einstein's calculation for the specific heat of solids, the expression for the average energy in 1d is $$\langle E\rangle = \hbar\omega\left(n_B(\beta\hbar\omega)+\frac{1}{2}\right)$$ In the book I'm reading, the author says:

Debye decided that the oscillation modes of a solid were waves with frequencies ω(k) = v|k| with v the sound velocity—and for each k there should be three possible oscillation modes, one for each direction of motion. Thus he wrote an expression entirely analogous to Einstein’s expression $$\langle E\rangle = 3\sum_{\mathbf{k}}\hbar\omega (\mathbf{k})\left(n_B\left(\beta\hbar\omega(\mathbf{k})\right)+\frac{1}{2}\right)$$

And then he goes on to do some calculations and arrive at the specific heat expression. My problem is that I have no idea why there is that summation over k. Summing over the k's means summing over all the possible wavelengths, right? But why should that be? Shouldn't that equation be true for some sort of average wavelength to fit the average energy without the sum?

• The modes are independent each others, exactly as happens to photons, so, each one has its Planck distribution. Therefore, you have $3\sum_{\bf k}\langle E({\bf k})\rangle$.
• Suppose there are 3 oscillators in the whole solid, and there are two possible values of k, $k_1$ and $k_2$ (just for the sake of argument). Then just for one oscillator, it may oscillate at frequencies corresponding to wavevectors $k_1$ or $k_2$, And each wavevector has three possible oscillation modes, so the average energy of one oscillator should be $\frac{3\langle E(k_1)\rangle+3\langle E(k_2)\rangle}{6}$ right? And the average energy of the whole solid should be the sum of all the combination of k for the three atoms and then divide that by the number of possible combinations? Nov 18, 2019 at 16:36