In solving the Classical model for lattice dynamics [Rossler pag 38] we find that the lattice admits $$d\cdot N\cdot r = \#modes$$ where

$d=$dimension of the problem

$N=$ number of atoms

$r=$ number of atoms in the primitive cell

So, classically we have that only a finite number of modes (for a finite lattice) is possible.

When we quantize the Hamiltonian we get the hamiltonian of an HO

$$\hat{H}=\Sigma_s\Sigma_\vec{q}\hbar\omega_s(\vec{q})\left(a^\dagger_{s\vec{q}} a_{s\vec{q}}+\frac{1}{2}\right)$$

where $s$ is a branch index.

So, in the Quantum case it seems that we can have an infinite number of phonons.

Am I missing something? Since the creation of a phonon corresponds physically to excite a vibration mode in the lattice, it seem me strange that we can create an infinite number of phonons in QM while in classical mechanics we can have only a finite number of excitations.

  • 1
    $\begingroup$ In both cases number of modes is the same finite number and there is infinity of different states possible. What is different in the classical model is that energy of a mode is not limited to multiples of $\hbar \omega$. $\endgroup$ Jul 30, 2015 at 20:39

1 Answer 1


Ok, I got the point: while classically I can (ideally) vary the energy of an oscillation mode by varying its amplitude from 0 to $\infty$, in QM I can vary the energy from $\frac{\hbar \omega}{2}$ to $\infty$ per steps of $\hbar\omega$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.