Why are phonons quantized? I'm a third year physics major taking an introductory subject on condensed matter physics. I'm trying to understand the concept of phonons in a 1D lattice.
I understand the classical treatment of (1D) lattice vibrations (wikipedia has a derivation https://en.wikipedia.org/wiki/Phonon#Classical_treatment) leading to equations for the Fourier coefficients that have the same form as the equations you get from decoupled harmonic oscillators.
It doesn't say it in the wikipedia article, but the next step my lecturer took was to say
"because the equations we get are the same as for non-interacting harmonic oscillators, we can quantise the energy of the normal modes (the phonons) like non-interacting harmonic oscillators – ie. as E=(n+1/2)ħω". This is the step I don't understand.
I understand that for an isolated harmonic oscillator, the energy eignevalues are E=(n+1/2)ħω (I've done the derivation of this from Schrodinger's equation in a previous class). That means that the quantization of energy in that problem arises from Schrodinger's equation, right? But surely Schrodinger's equation doesn't apply to the Fourier coefficients? Then why should the Fourier coefficients satisfying the same equations as decoupled harmonic oscillators lead to the same quantization?
 A: I agree that the standard derivation of phonon energy quantization is kind of annoying. If you find it unsatisfactory, you can do a "quantum from the start" derivation and it's not really any harder. Doing it that way makes it more clear where the quantization comes from. For example, see J. M. Ziman's book Electrons and Phonons (hopefully available thru your school's library) section 1.3. I think that section should be digestible for an upper-level undergrad, but I make no guarantees.
Ziman even writes

The conventional way with this problem is to start with the classical solution of the equations of motion, and introduce the minimum of 'quantization' at the last possible moment. Accounts along these lines are available in all the standard books. An alternative derivation, which is quantum mechanical throughout, is not any more difficult than the classical analysis, and introduces in a natural manner several important and useful techniques.

EDIT: I'll add that what's basically going on is a change of variables from the position/momentum of each atom to their Fourier transforms. Because the position/momentum were good quantum variables (with proper commutation relations), the same is true for their Fourier transforms. So, you can solve the problem with the new variables. You've probably seen change of variables in, say, spin 1/2 systems. E.g. You can take a Hamiltonian and states in the +/-x basis and rewrite them in +/-z and everything is fine. The same thing is happening for photons; it's just less obvious because the transformation is more complicated and the standard way of handling phonons only introduces quantization after the change of basis, which makes it hard to see what's going on.
A: "...But surely the Schrodinger's equation doesn't apply to the Fourier coefficients, right?"
To begin, I have to say that I personally always found to the word "quantization" as very confusing and not exactly explanatory as to what kind of action it refers to.
Now coming to your question that I copied above, the answer is in fact yes you can (sort of). The thing that you should be aware of is that the very term "quantization" is in fact a hail from the heroic early days of Quantum Theory when people like Niels Bohr where doing things noting short of wizardry (Correspondence Principle, Complementarity Principle, etc) to somehow squeeze the Physics they grew up with into the Procrustean bed of the contemporary experimental evidence.
It is then that the protocol for the "quantization of a theory" has crystallized. In short, the algorithm was/is as follows:

*

*Formulate your problem in classical terms as a Hamiltonian theory just the way Goldstein, Landau&Lifshits vol.1, or any other Classical Mechanics textbook of your liking, prescribes. Namely, you have to find:

1-1) a set of (classical) variables $\{ Q_i \}$ that fully describe your system. These variables dubbed "generalized coordinates" do not necessarily have to look like "coordinates". They can be anything but the necessary requirement is that knowing them you should always be able to tell how your system looks like $\textit{now}$. For example the $3N$ coordinates of $N$ particles or any other (linearly independent) linear combination of them (hence the "generalized" quality of the generalized coordinates).
1-2) Construct the corresponding classical Lagrangian in terms of generalized coordinates $\{ Q_i \}$ and their time derivatives ("generalized speeds") $\{ \dot{Q}_i \}$.
1-3) Using the found Lagrangian $L(\{ Q_i \}, \{ \dot{Q}_i \})$ find "generalized momenta" $\{ P_i\}$ according to $P_i \equiv \frac{\partial L}{\partial \dot{Q}_i}$.
This was all classical so far. Here comes the magic of "quantization":


*Up till now all $Q_i$'a and $P_i$'s were nothing but a bunch of number. Now you say that you forgo that, and say you have no idea anymore as to what they are. The only thing you $\textit{postulate}$ you know is that

$$Q_i P_j - P_j Q_i = i\hbar \delta_{ij}$$
This is called Heisenberg commutation relations and this is the central element, the keystone of Quantum Theory. To dispell any possible misunderstanding, regardless of how Heisenberg arrived at them, they do not follow from anywhere. Quite in contrary, everything follows from it.


*This is the last and the hardest step. You have to $\textit{guess}$ what ever could those $Q_i$'a and $P_i$'s be. For example, Mr. Born had an idea that they could be matrices. Why not? Matrices don't commute, hence the Heisenberg commutation relations are satisfied. By consistently following this $\textit{representation}$, Heisenberg has constructed his Matrix Mechanics.
Schoedinger (or rather Dirac) went another way and noticed that Heisenberg commutation can be obtained if one thinks of $P$'s and $Q$'s as some sort of calculus operation performed of continuous functions. Hence Schoedinger's Wave Mechanics (you can try to find your own representation if you want, with some other algebraic objects. Who knows, maybe it will open new vistas in Physics, as Feynman's approach did).

To summarize you can quantize anything. Just make sure this something is part of a set of variables describing the state of of the system on the classical level, and that you can find "momenta" corresponding to this something. Once you have them, just postulate the Heisenberg commutation relation between them and find a suitable representation.
P.S. When I say "anything", I really mean anything. Take the predator-prey (Lotka-Volterra) equation. The "coordinates" would be the species' populations. The corresponding dynamics can be formulated in terms of a Hamiltonian (e.g. https://www.math.tecnico.ulisboa.pt/~rfern/Meus-papers/HTML/equadiff.pdf), then you can "quantize" it too. How useful it is? Probably not at all. But my point was that Fourier components are not the most absurd objects to quantize at all.
$\textbf{EDIT}$: Q: "...why should the Fourier coefficients satisfying the same equations as decoupled harmonic oscillators lead to the same quantization?" A: Since you are free to choose objects for $Q_i$ and $P_i$ as long as they commute according to Heisenberg, it is often convenient to "steal" them from another problem someone has solved before. Solutions don't smell. If two systems have equivalent Hamiltonians, so do their spectra.
A: One way to look at this that might be helpful in making it more intuitive is that even starting from the classical treatment you end up with position and momentum variables (one for each atom). Since we started with a model consisting of coupled springs and masses the classical equations of motion are linear with n-degrees of freedom.
So now instead of thinking of it as a bunch of individual springs connecting masses you just think of it as one super high dimensional spring with n-degrees of freedom.  Taking the first step into Quantum Mechanics is just saying "These things are Quantum Mechanical in nature and thus the position and momentum for each degree of freedom collectively obey known commutation relationships." Specifically ones you can find here:
https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator#N-dimensional_isotropic_harmonic_oscillator
The next step is tricky but the math is worked out on page 152 here:
https://curtwittig.com/wp-content/uploads/IV.-pp.121-214.pdf
Which basically says that if the individual site specific position and momentum operators obey the n-dimensional Quantum Harmonic Oscillators commutation relations then so do their Fourier transformed position and momentum operators. These Fourier transformed operators apply to the entire collection because they are in k-space and they also have to obey any Quantum Mechanics directly linked to the form of their commutation relationships. To put it more succinctly if we assume the vibrations of individual atoms in the lattice are quantized with a simple harmonic potential then so too must be all collective oscillations because that's how the math of linear algebra and the Fourier Transform work out.
I guess one way to look at it is that a collection of quantized coupled linear oscillators will always give rise to quantized collective oscillations and only quantized oscillations (this could and probably does hold for other systems but I'm sticking with standard assumptions here).
A: The surfaces of equal energy for our system are hyper-ellipsoids: for two particles each is an ellipse described by something like $x_1^2+(x_1-x_2)^2+x_2^2=\mathrm{const},$ for three particles it's an ellipsoid described by $x_1^2+(x_1-x_2)^2+(x_2-x_3)^2+x_3^2=\mathrm{const},$ etc.
The switch from the displacements $x_i$ of particles to the "displacements" of the eigenmodes is just a rotation in the configuration space: a rotation that aligns the axes of the hyper-ellipsoid to the coordinate axes. This rotation can be done both in classical and in quantum descriptions. The kinetic energy (be it a function in classical description or an operator in the quantum case) is invariant with respect to this rotation.
After we apply this change of coordinates, we get a new Schrödinger equation that indeed looks like that of a set of non-interacting harmonic oscillators, each with its own frequency. This lets us separate the variables and solve the equation for each eigenmode independently.
A bit more details on the derivation can be seen in my answer here.
A: You are correct that one needs to be careful when simply taking the classical equations of motion $\ddot u_n = -2 C u_n + C(u_{n+1} + u_{n-1})$ and saying "this looks like harmonic oscillator then QM is also a harmonic oscillator" but the idea is sound, in principal.
The spatial Fourier transformation is just a linear change of basis of the spatial degrees of freedom
$$ u_k = \sum_n e^{i k n} u_n$$
by itself, it is valid quantum mechanically as it is valid classically. It is just a different representation of the degrees of freedom.
So if we started from a classical Lagrangian / classical Hamiltonian for the spatial $u_n$ and ended up with a harmonic oscillator like equations of motion for $u_k$, and we want to make the claim "the quantized degrees of freedom $u_k$, which we take as the linear superposition of the different quantized spatial $u_n$, follow the quantum harmonic oscillator energy levels etc." the underlying assumption that we make is the following: The classical Lagrangian / Hamiltonian describing $u_k$ is the harmonic oscillator one, and we got it from a direct Fourier transformation of the classical Lagrangian / Hamiltonian describing $u_n$. This is valid assumption.
Assuming this is true (and you can verify that it is true), then we can start from said classical Hamiltonian, quantize $u_n$ and the associated momenta $p_n$, do the spatial Fourier transform which is valid both classically and quantum mechanically, and get the Hamiltonain for $u_k$, $p_k$ that is a quantum harmonic oscillator.
So to sum up, if we assume that we can deduce the classical Hamiltonian from the classical equations of motion, we can do the step of quantization without any problem. The spatial Fourier transform is just a change of basis and doesn't obstruct the quantization process.
A: Making the matters a bit simpler, phonons are not quantized, but their interactions with the lattices are quantized because of the discrete nature of those lattices. Phonons themselves are as continuous as it could get. Letting that sink and pausing to ponder for a while gives a solid foundation to move on to grasp all the resulting concepts.
