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##\textbf{1)}##

My current understanding is that unlike Einstein's model, he views a 1D solid as a chain of atoms similar not unlike a 1D chain of masses coupled with springs. Thus a chain of N oscillators has N vibrational modes in 1D and 3N in 3D. This means that we have taken the normal mode frequencies for longitudinal oscillations to be the same as those the transverse oscillations. Is this really the case (classically and quantum mechanically) or was the assumption made for simplicity?

##\textbf{2)}##

The energy for a mode of frequency ##\omega## is given by ##\hbar \omega (n_b(\beta \hbar \omega) + \frac{1}{2})##, with ##n_b## being the Bose-Einstein distribution, and telling us the average level of excitation for that particular mode. It was then implied that we can take the entire mode to be in itself a a) simple harmonic oscillator with b) the energy given above. What does he mean by a) and b)? Also with regards to 2), why does a chain of atoms have the same expression for energy as a single quantum harmonic oscillator? In classical mechanics we would have to sum the energy contributions of individual masses (and spring constants) to get the total energy of the system.

##\textbf{3)}##

It was then stated that we could also view each mode as a boson of frequency ##\omega## occupied on average ##n_b## times. Why do we take a mode, which really is the vibration of a chain of atoms, as a single particle? Or was this meant as an analogy rather than a literal interpretation? Also, what does it mean for a boson to be "occupied" ##n_b## times?

##\textbf{4)}## Later the concept of the density of states ##g(\omega)## was introduced. Am I right in saying that ##g(\omega) \ d\omega## can be interpreted as the number of chains of atoms in the considered solid occupying vibrational modes between frequencies ##\omega## and ##\omega + d\omega##?

PS I have not studied Statistical Mechanics (that comes in the future) and thus am unfamiliar with the Bose-Einstein distribution and properties of Bosons.