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?


1 Answer 1


Debye's idea was to consider all the possible modes of the lattice, a more realistic hypothesis with respect to Einstein that considered just one instead. Then, each mode has its Planck distribution and so, your avearge is the sum of all the averages with respect to each mode. This is the reason why you get that sum. The factor 3 accounts for the possible oscillation modes that can be in three different directions, quite similar to the two polarization modes of photons.

  • $\begingroup$ Why is it the sum of all mode averages instead of being the average of all these mode averages? $\endgroup$ Nov 18, 2019 at 12:32
  • $\begingroup$ The modes are independent each others, exactly as happens to photons, so, each one has its Planck distribution. Therefore, you have $3\sum_{\bf k}\langle E({\bf k})\rangle$. $\endgroup$
    – Jon
    Nov 18, 2019 at 12:37
  • $\begingroup$ Sorry, but I've thought about it for quite some time and I still can't understand why they should be added. Even if the modes are independent, I don't see how that's relevant. $\endgroup$ Nov 18, 2019 at 16:27
  • $\begingroup$ 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? $\endgroup$ Nov 18, 2019 at 16:36
  • $\begingroup$ The average is intended on the quantum ensemble using the Planck distribution. What you are doing is meaningless. We are talking about quantized harmonic excitations of the lattice, normally called phonons, each one with its probability distribution for the single mode. $\endgroup$
    – Jon
    Nov 18, 2019 at 18:37

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