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)

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    $\begingroup$ Have you fleshed out your thinking with N =2, just two coupled harmonic oscillators? What do you see in the limit of vanishing mutual coupling? How do you define a mode? $\endgroup$ Commented Dec 10, 2023 at 14:48
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    $\begingroup$ Those N atoms are bound in a crystal, not isolated from each other. $\endgroup$
    – Jon Custer
    Commented Dec 10, 2023 at 15:19
  • $\begingroup$ " it is oscillating at a different frequency $\omega_k$. Is that why?" Indeed, that's why. The ground state energy is $\hbar ( \omega_0+ \omega_1+ \omega_2+...+\omega_{N-1})/2$, and does not collapse to your expression unless you switch off the couplings and trivialize the mode resolution. $\endgroup$ Commented Dec 10, 2023 at 16:29
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    $\begingroup$ This might help, but your misconception is ignoring that the hamiltonian is the sum of those of all decoupled modes, whose zero-point energies you sum, for the ground state. You may not jam all particles into the same mode, which is a linear combination of them. Do you want an illustration for N=2 ? $\endgroup$ Commented Dec 11, 2023 at 14:22
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    $\begingroup$ I don't know how explicit Kittel's book is on lattice vibrations, but neat expositions of phonons (what photons are to the EM field, phonons are to a lattice) can be found in the classic text of Fetter & Walecka. $\endgroup$ Commented Dec 12, 2023 at 16:13

1 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.

  • $\begingroup$ @CosmasnZachos Thank you so much for your help ... U very precisely cleared my doubt ... And many thanks for the resources ... Unfortunately, for some medical emergencies I'm not able to go through those... But I'll definitely go through once I'm alright... Thanks again for putting this effort... I'm sure u did this amidst ur busy schedules $\endgroup$ Commented Dec 13, 2023 at 17:26

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