How is the energy of phonon modes $\left(n+\frac{1}{2}\right)\hbar \omega_k$ when each atom in the mode has $\left(p+\frac{1}{2}\right)\hbar \omega$?

Question

I'm having trouble understanding the quantisation of energy in normal modes of lattice vibrations. Stating Kittel :

The energy of a lattice vibration is quantized. The quantum of energy is called a phonon in analogy with the photon of the electromagnetic wave. The energy of an elastic mode of angular frequency $\omega_k$ is $$E_{n,k} = \left(n+\frac{1}{2}\right)\hbar \omega_k$$

I'm having trouble because a 'mode' consists of N atoms which execute simple harmonic motion. Each of them should be having their own quantised energy levels $\epsilon = \left(p+\frac{1}{2}\right)\hbar \omega_0$. If I assume all the atoms to be at the lowest energy possible, $\epsilon_0 = \frac{1}{2}\hbar \omega_0$, The mode should have a minimum energy of $\frac{N}{2}\hbar \omega_0$.

This is not in agreement with the energy of modes. I see that $\omega_k$ and $\omega_0$ are different, but not able to connect the dots.

Quantised energy states of simple harmonic oscillator depends on the natural frequency $\omega_0$ of the SHO. But when the SHO is a member of the normal mode, it is oscillating at a different frequency $\omega_k$. Is that why? (but even if a SHO is forced to oscillate at a frequency $\omega$, the quantised energy levels should still be $\epsilon_n = \left(n+\frac{1}{2}\right)\hbar \omega_0$)

Is it that the SHO quantisations don't hold when there are N of them connected and interacting? (but then I can go back... if the mode has a zero point energy of $\frac{1}{2}\hbar \omega_k$, the average energy of each oscillator in the mode would be $\frac{1}{2N}\hbar \omega_k$ which is much less than the SHO zero point energy. Feel like that can't happen)

Answer 1

The (linear) mode change of basis switches from N coupled oscillators to N uncoupled ones with different frequencies. You may not, then, jam all of the original oscillators ("particles") into just one mode, which is what you might be contemplating.

It's easiest to see this for N=2, $$ H= \frac{-\hbar^2(\partial_x^2+\partial_y ^2)}{2m} +\frac{m\omega^2}{2}(x-y)^2,\\ \hbox{where, defining } ~~~~u\equiv {x+y\over \sqrt 2}, v\equiv {-x+y\over \sqrt 2} ~~~~\leadsto \\ H= \frac{-\hbar^2(\partial_u^2+\partial_v ^2)}{2m} +\frac{m }{2}( \omega_u^2 u^2 + \omega_v^2v^2),\\ \omega_u= 0 ,~~~~~ \omega_v= \sqrt 2 \omega, $$ two now decoupled oscillators, "modes", the first (translation) with vanishing frequency, and the second (breather) with a higher frequency than the natural one.

The ground state, then, has zero-point energy $E= \frac{\hbar}{2}(\omega_u+ \omega_v)$, which you may easily extend to N>2 particles. You then see that these decoupled oscillators, modes, have their distinctively different 0-point energies add.

You cannot jam all N coupled particles into the lowest mode, and certainly not in one with the common natural frequency of the uncoupled oscillators: this is the very point of the natural mode resolution into independent oscillators.

But when the SHO is a member of the normal mode, it is oscillating at a different frequency $\omega_k$. Is that why? (But even if a SHO is forced to oscillate at a frequency $\omega$, the quantised energy levels should still be $\epsilon_n = \left(n+\frac{1}{2}\right)\hbar \omega_0$.)

No, as seen, each mode has energy $\epsilon_{k,n} = \hbar \omega_k\left(n+\frac{1}{2}\right) $, with ground state at $n=0$. It is a coherent superposition of particles, not a set of all of them.