As we obtain a reciprocal lattice for a given crystal we see that discrete values of wavevectors are allowed but a phonon wavevector spectrum is a continuum. Is there a relation between reciprocal wavevectors and phonon wavevector?
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Question: Are you completely sure that the phonon wavevector spectrum is continuous? For any finite body, phonon modes are (nominally) quantized in much the same fashion as the fermion seas of electrons in metals. It's quasi-continuous, sure, but a true continuum spectrum for phonons is necessarily a bit of an abstraction. |
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Suppose we take a classical approach and model the crystal lattice as a series of coupled simple harmonic oscillators. if the number of individual lattice sites $N$ is large (but not infinite), we may assume periodic boundary conditions, which result in the quantization of allowed wavevectors. Using Ashcroft and Mermin pg 430 as a reference, the above explanation is shown in more detail below. For the 1D spring model, We can write the potential energy as $$ U^{harm} = \frac{1}{2} K \sum_n [u(na) - u([n+1]a)]^2, $$ Where a is the lattice spacing, and u(na) denotes the equilibrium location of the nth lattice site. This allows us to arrive at the following force equation: $$ M \ddot{u}(na) = - \frac{\partial U^{harm}}{\partial u(na)} = -K [ 2u(na) - u([n-1]a)-u([n+1]a)]. $$ Assuming that $N$ is large, we introduce the following Born-von Karman boundary conditions $$ u([N+1]a) = u(a), $$ $$ u(0) = u(Na). $$ The guess solution for the above differential equation is as follows: $$ u(na,t) = e^{i(kna-wt)}. $$ Incorporating boundary conditions into the guess solution gives us the requirement that $$ k = \frac{2 \pi}{a} \frac{n}{N}, $$ where $n$ is an integer. Thus, we have quantized the allowed phonon modes in the crystal. |
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