# Using the Debye model, show that the contribution of the zero point energy to the lattice vibrational energy is given by $𝑈=(9/8)𝑁𝑘_B𝜃_𝐷$

Using the Debye model, show that the contribution of the zero point energy to the lattice vibrational energy is given by $$𝑈={9\over8}𝑁𝑘_B𝜃_𝐷$$.

I know that The zero point energy is $$E={1\over2}\hbar\omega$$ and $$U=9𝑁𝑘_BT({T\over𝜃_𝐷})^3\int_0^{x_D}{x^3\over e^x-1}dx$$ where $$x_D={𝜃_𝐷\over T}$$. However I don't really know what to do from this, given that it's zero point energy, does that mean I need to make approximations with $$T\ll 𝜃_𝐷$$? Any help is appreciated.

• Have you tried getting a value for the integral? – Aaron Stevens Oct 5 '18 at 3:40
• The integral as it stands is unreasonably complex for the class level so I assume I need to make an approximation. When I approximate $e^x-1=x+1$, the indefinite integral works out to be $-ln(x+1)+{x^3\over 3} -{x^2\over 2}+x$ – Henry Mullen Oct 5 '18 at 7:23

Here's a crack at an answer.

Total energy may be written as the sum of the energies over all phonon modes, indexed by wavevector, $$\textbf{k}$$, and polarization index, $$\textit{p}$$:

$$U_{lat} = \sum_{\textbf{k}}\sum_{\textit{p}}U_{\textbf{k},\textit{p}} = \sum_{\textbf{k}}\sum_{\textit{p}} \langle n_{\textbf{k},\textit{p}} \rangle\hbar\omega_{\textbf{k},\textit{p}}$$

But since we are only interested in the case where all the atoms are in the ground state, this becomes,

$$U_{lat} = \sum_{\textbf{k}}\sum_{\textit{p}} {1\over 2} \hbar\omega_{\textbf{k},\textit{p}}$$

Suppose the crystal has $$D_{\textit{p}}(\omega)d\omega$$ modes of a give polarization $$\textit{p}$$ in the frequency range $$\omega$$ to $$\omega + d\omega$$, then the energy is now,

$$U = \sum_{\textit{p}} \int {1\over2}\hbar\omega D_{\textit{p}}(\omega)d\omega$$

We can now apply the Debye Model. The density of states is,

$$D(\omega) = \frac{V\omega^2}{2\pi^2v^3}$$

and if there are N primitive cells the total number of acoustic modes is $$N$$. The cutoff frequency is then,

$$\omega^3_D = \frac{6\pi^2v^3N}{V} \tag{1}$$

We define the Debye temperature as,

$$\theta_D^3 = \frac{6\hbar^3v^3\pi^2N}{k_B^3V} \tag{2}$$

We can assume that the phonon velocity $$v$$ is independent of polarization, so we can multiply by a factor of 3. Putting all this together we integrate from 0 to the cutoff frequency to give us:

\begin{align} U &= \left({3\over2}\right)\cdot \left( \frac{\hbar V}{2\pi^2v^3} \right) \int_0^{\omega_D} \omega^3 d\omega\\\\&= \left({3\over2}\right)\cdot \left( \frac{\hbar V}{2\pi^2v^3} \right)\cdot\left( {1\over4}\right)\omega_D^4 \end{align}

By playing around with (1) and (2), we can see that $$\omega_D = \frac{\theta_D k_B}{\hbar}$$, which gives us,

$$U = \left({3\over2}\right)\cdot \left( \frac{\hbar V}{2\pi^2v^3} \right)\cdot\left( {1\over4}\right)\cdot\left( \frac{\theta_D^4 k_B^4}{\hbar^4} \right)$$

We'll rearrange this slightly to get,

$$U = \left( {3\over8} \right)\cdot \left( \frac{V}{2\pi^2v^3\hbar^3} \right) \theta_D^4 k_B^4 \tag{3}$$

And by playing around with Eq (2) again, we can see that,

$$\frac{V}{2\pi^2v^3\hbar^3} = \frac{3N}{k_B^3\theta_D^3}$$

And from here substitute the above in Eq (3) to get,

$$U = \left( {3\over8} \right)\cdot \left( \frac{3N}{k_B^3\theta_D^3} \right) \theta_D^4 k_B^4$$

Which reduces to,

$$U = {9\over8}Nk_B\theta_D$$

This gives us what we want. The only step that I'm a little uncertain on is effectively letting $$\langle n \rangle = {1\over2}$$. So if anybody could shed some light on this, it would be much appreciated.