If I understand correctly, in Debye’s theory we step away from the notion of individual motion of atoms, and so instead of modeling each individual atom as a harmonic oscillator (with U= \frac 12 kx^2, with x the displacement from equilibrium position), we consider each HO as some potential corresponding to a collective motion with a certain \vec k, the sum over all possible collective motions yields the total amount of modes (though I still don’t understand why it should equal 3N_{atoms}, as we do not consider individual atoms anymore).
From this, my questions follow. What do we consider to be our ‘x’ in case when our HO doesn’t represent a deviation from a physical position anymore?
What does n_B mean, corresponding to each HO? Is it the amount of atoms that correspond to a certain collective motion (with a certain \vec k)? If so, how does the amount of atoms in a collective motion make any sense? Collective motion implies all (or like a large group) atoms move alike. Perhaps a superposition of waves is meant that can be sustained by the present atoms (though the question then still remains why the amount of waves we encounter is exactly the number of atoms x 3).
Thank you for your question, Vlad.
First of all, I’d like to say that you would probably benefit from reading the corresponding chapter of a book that discusses this topic. Also, reading different books helps to get some perspective.
Now I will try to answer your questions. When describing motion in a solid, we use a displacement function \vec{u}(\vec{r},t), where \vec{u} is the distance an atom has displaced, and we use \vec{r} to indicate the static position of a given atom. If we go to 1D to make it easier, we have u(x,t) such that we could ask how is atom at x=a moving over time, which would be u(a,t).
When you look at the motion of such a collection of atoms, you figure out pretty quickly it is handy to look at normal modes, which are the natural solution for the equations of motion. So we move to reciprocal space, which is nothing but the Fourier transform of x (space). And if we ask how many modes there are in a solid with N atoms, that has a clear answer. For 1D, and if they can only move in one direction, you get N equations of motion (one per atom), so you will get N independent solutions. Those are your N states. If you had 2 polarizations or directions in which the atoms can move (say x and y), then you get 2N equations to solve (2 per atom, one in x, one in y, and N times, for all atoms). In reciprocal space, that results in N_{pol}N_{atoms} available k.
Once you do this, you don’t consider the amount of excitations in each atom, but the amount of excitations in the whole solid (that is the point of collective excitations), which is the difference with respect to Einstein’s model. And see that there is a certain number of atoms N_{atoms}, a certain number of modes N_{modes}, and a certain number of excitations. All those are different quantities. You can take 2 atoms (e.g. Hydrogen molecule) which only has 2 modes in 1D, and excite it 100 times, or infinite times (in principle). There is no limit but the energy you have available. So indeed the amount of waves you have is not limited by anything and is determined by the energy you put in the system, e.g., by temperature,e which you could compute by the Boltzmann distribution, which would give you the average amount of excitations in each mode. And we sometimes talk about how many modes can be excited at a given temperature. At high temperature, all modes are excited, which amount to N_{pol}N_{atoms}, but again, the number of excitations per mode will differ depending on the specific temperature.
Hope that helps 