- #1

dRic2

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*single*mode of oscillation of the lattice, that is something like ##u(x, t) = Ae^{i(kx-\omega t)} ## (in the classical limit). The energy corresponding to that mode should be ##E = \frac 1 2 \rho L^3 A^2 \omega^2 ##. If I equate this equation to ##E(k) = \hbar \omega(k) (N_{k} + \frac 1 2)## can I conclude that the number of phonons is

**exactly**##N_{k}## ?

I think the answer is no, but I'm not totally sure.

My reasoning is that, if ##N_{k}## is the exact umber of phonons, then the only possible way to describe the state of the lattice is

$$ | \psi > = |0, 0, 0, ..., N_{k}, 0, 0, ..., 0>$$

Isn't this in contradiction with the fact that phonons number is not conserved ? I am a bit confused about this... I can see that the particle number operator N does not commute with the phonon hamiltonian, but I don't know how to interpret this on empirical grounds.