# Why do we need the quantization for lattice vibration?

I've been reading the Wikipedia article on phonon. So, my understanding is what they get is the discrete energy levels of vibration from quantization. But the discrete energy level is not only the property of the quantum system but also the property of classical harmonic oscillator.

And if they can describe the vibration with the classical harmonic oscillator model, why do they need to introduce the so-called second quantization for lattice vibration?

Do they get anything new which we cannot obtain from the classical harmonic oscillator?

The comment below and the answer by @Vadim mention that the classical harmonic oscillator has a continuous energy spectrum. I add some reference of Wikipedia article stating different idea:

From Wikipedia, Phonon article:

In the article, the displacement of atom positions are modeled as

$$u_n = \sum_{Nak/2\pi=1}^n Q_k \exp(ikna)$$

and the discrete $$k$$ values leads to the discrete normal modes.

For the second reference, I link the Quantum harmonic oscillator article:

The quantity $$k_n$$ will turn out to be the wave number of the phonon, i.e. $$2\pi$$ divided by the wavelength. It takes on quantized values, because the number of atoms is finite.

I extracted the quote in the section just before imposing the commutation relations and so before quantization.

Their point seems that the atoms are placed in discrete positions inside the finite size matter and the discreteness leads to the discrete wavelength solutions.

• You wrote "the discrete energy level is not only the property of the quantum system but also the property of classical harmonic oscillator". What do you mean? The classical harmonic oscillator has a continuous energy, not discrete levels. – HicHaecHoc Jul 16 '20 at 7:40
• @HicHaecHoc I'm referring to the Wikipedia article I linked. In the classical treatment section, it derives the normal modes as a solution with a discrete Fourier transform. – hbadger19042 Jul 16 '20 at 7:51
• Well, the normal modes are an ensemble of classical harmonic oscillators. In the quantum treatment they are replaced by quantum harmonic oscillator. Are you aware of the differences between a classical harmonic oscillator and a quantum one? It would be useful to know to correctly aim an answer. – HicHaecHoc Jul 16 '20 at 8:18
• @HicHaecHoc I think I know. The Quantum oscillator is to impose the commutator relationship between normal coordinate and conjugate momentum. – hbadger19042 Jul 16 '20 at 8:27
• @HicHaecHoc I updated the question including more reference on the discreteness in the classical harmonic oscillator in the lattice. – hbadger19042 Jul 16 '20 at 10:37

## 1 Answer

The classical oscillator does not have discrete levels, its energy is $$E=\frac{p^2}{2m} + \frac{m\omega^2x^2}{2},$$ which can take any value greater or equal to zero. On the other hand, for a quantum oscillator only the energy values $$E_n = \hbar\omega\left(n+\frac{1}{2}\right)$$ are possible.

Whether to use classical or quantum description for a physical system is not a matter of our choice - rather we choose the description that is more consistent with the real world. Quantum mechanics describes the real world physical phenomena better than classical one, although in some problems quantum effects may be neglected and the classical description suffice. In case of phonons quantum description is necessary, e.g., to obtain the expressions for specific heat that are consistent with experiments. On the other hand, sound propagation in solids is mostly described using classical elasticity.

Finally, in case of wave phenomena, like electromagnetic wave or phonons, the formalism called second quantization, which is in fact first quantization!

Update
In the reference (added later to the question) the wave numbers $$k_n$$ and the corresponding frequencies $$\omega_n=c_{ph}k_n$$ refer to different oscillators. In other words, the oscillations are possible only with these frequencies, but the energy of oscillations at any particular frequency can still be arbitrary (if the oscillators are classical). While such "quantization" due to the number of atoms and the finite size of a system is typical for wave phenomena, it is not really a quantum effect, but simply a buzz word used instead of saying discreteness.

One should note however that mathematically the quantum quantization and discreteness of spectrum arise in the same way, since in quantum description particles are described by waves, whose spectra may become discrete when the motion is constrained.

• I added some references explaining the classical harmonic oscillator with discrete wavelength solutions for lattice. (Let me know if I misinterpreted it.) – hbadger19042 Jul 16 '20 at 10:39
• @Kevin In the reference that you give the wave numbers $k_n$ and the corresponding frequencies $\omega_n=c_{ph}k_n$ refer to different oscillators. In other words, the oscillations are possible only with this frequencies, but the energy of oscillations at any particular frequency can still be arbitrary (if the oscillator is classical). While such "quantization" due to the number of atoms and the finite size of a system is typical for wave phenomena, but it is not really a quantum effect, but simply a buzz word used instead of saying discreteness. – Roger Vadim Jul 16 '20 at 11:13
• I see. The amplitude of the vibration can be arbitrary and so the energy can be also. I was too focused in frequency. – hbadger19042 Jul 16 '20 at 11:18
• @Vadim this explanation in the comment is a useful addition to the answer. It'd be better to have it in the answer body so that potential comments cleanup doesn't remove it. – Ruslan Jul 16 '20 at 11:23