# Debye model and reciprocal space

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## Main Question or Discussion Point

Hi everyone, I need a little help understanding how periodic reciprocal space applies to the Debye model for solids. Many thanks in advance!

If we start with the general derivation of a dispersion relation for a 1D system, with atoms coupled by springs, one gets the following relation
$$\omega = 2\sqrt{\frac{k}{m}} |\sin \big( \frac{ka}{2} \big)|$$
whereby it is possible to show that the function is periodic with $\omega (k + 2\pi/a) = \omega (k)$.

However if we consider Debye's model, which comes as a result of taking the long wavelength (and small k) limit, we get
$$\omega = v_s k$$
where $v_s$ is the speed of sound.

$\textbf{Question}$: Does the concept of a periodic $k$ still apply here? Also, is the number of allowed modes for a chain of N masses still equal to N?

For example if I were to calculate energy contributions by individual frequencies $\omega (k)$ using the following equation
$$\langle E \rangle _k= \hbar \omega(k) \big( n_b (\beta \hbar \omega (k)) + 1/2 \big)$$
clearly energy for the $k=0$ mode differs from that of the $k = 2\pi /a$ mode. Should I pick $k = 0$ for my calculations or $k = 2 \pi /a$?

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Hi,

the Debye model assumes a linear dispersion relation, as you say. This model "forgets" the oscillators (like coupled springs); it treats the problem as if the solid was a box with phonons inside. The mean number of phonons excited at temperature T is given by Planck's distribution (Bose-Einstein with $\mu = 0$ ). So, since there is not a lattice of atoms now (only phonons in a box), you don't have the periodicity.

Hi,

the Debye model assumes a linear dispersion relation, as you say. This model "forgets" the oscillators (like coupled springs); it treats the problem as if the solid was a box with phonons inside. The mean number of phonons excited at temperature T is given by Planck's distribution (Bose-Einstein with $\mu = 0$ ). So, since there is not a lattice of atoms now (only phonons in a box), you don't have the periodicity.
Debye also postulated that a chain of N atoms (with motion in 1d) should only have N modes. Should I decide to find the expectation value of energy for this chain using
$$\langle E \rangle = \sum_{k} \hbar \omega(k) \big( n_b (\beta \hbar \omega (k)) + 1/2 \big)$$
how do I choose the N wavenumbers? I assume that the range to "pick" from is $0<k\leq 2\pi/a$ with a $k$-space interval of $\frac{2\pi}{Na}$? I'm also assuming that we ignore $k=0$ since by $\omega = vk$ it simply means that all the masses don't vibrate anyway. Is this right?

Thanks for assisting!

If you want to know the mean energy of your crystal (or chain, in this case), you have to sum over all k's which are accesibles given the temperature $T$ of the chain. For each mode ($\vec{k},$ p) (p is the branch) the mean energy is

$\langle E_p (k) \rangle = [ \langle n_p (k) \rangle+1/2 ] \hbar \omega_p (k)$

so the total mean energy is

$\displaystyle \langle E \rangle = \sum_{k, p} \langle E_p (k) \rangle$

Since the wavenumber here is a continuous variable, you have to replace the sum by an integral weighted with the density of states. In 1D this density is $D (\omega) = L \pi /c$, where c is the speed of sound ($\omega = c k$) and L the length of the chain. Finally, the calculation you have to do is

$\displaystyle \langle E \rangle = \int _0^{\omega_D} (\langle n \left( \omega) \rangle+1/2 \right) \hbar \omega \:D(\omega) d\omega$

where $\displaystyle \langle n (\omega) \rangle = \frac{1}{e^{\frac{\hbar \omega}{k T}}-1}$ is the ocupation number for phonons (Planck's distribution).

Since the wavenumber here is a continuous variable, you have to replace the sum by an integral weighted with the density of states. In 1D this density is $D (\omega) = L \pi /c$, where c is the speed of sound ($\omega = c k$) and L the length of the chain. Finally, the calculation you have to do is

$\displaystyle \langle E \rangle = \int _0^{\omega_D} (\langle n \left( \omega) \rangle+1/2 \right) \hbar \omega \:D(\omega) d\omega$

where $\displaystyle \langle n (\omega) \rangle = \frac{1}{e^{\frac{\hbar \omega}{k T}}-1}$ is the ocupation number for phonons (Planck's distribution).
I understand that as the number of atoms (N) becomes large, the wavenumber becomes continuous and we can do integrals. However if N is not large, (perhaps a chain of 2 atoms for example), does what I said here still apply?
I assume that the range to "pick" from is $0<k\leq 2\pi/a$ with a $k$-space interval of $\frac{2\pi}{Na}$? I'm also assuming that we ignore $k=0$ since by $\omega = vk$ it simply means that all the masses don't vibrate anyway.

I think not. Remember that in the Debye model phonons can have any wavenumber from 0 to $k_D = \omega _D /c$. Any state can be occupied if the temperature is high enough. Maybe you are mixing concepts: classical treatment of a chain of atoms and the Debye model, which is based on quantum statistical mechanics.

I'm not sure you can apply the Debye model to a chain of 2 atoms, or 5 or 20. It is a macroscopic model and uses statistical mechanics, and you can't make statistics with 2 atoms.