Example 3: Phonons in a Solid Lattice

Having computed the partition function of a harmonic oscillator, we now compute the partition function of the normal modes of a solid at low temperature. According to the harmonic approximation, the Hamiltonian of the system is

$$\hat{H} =\sum_{\alpha=1}^{DN} \hat{h}_{\alpha},$$ (176)

where DN is the number of normal modes, with D the dimensionality of the lattice and $\hat{h}_{\alpha}$ is the Hamiltonian of a harmonic oscillator with a frequency $\omega_{\alpha}$ and eigenvalues

$$E_n(\alpha)=\hbar \omega_{\alpha}(\frac{1}{2} + n_{\alpha}),$$ (177)

with $n_{\alpha}=1,2 ...$ An arbitrary vibrational state $\xi$ of the lattice can be specified by the DN normal mode frequencies $\omega_{\alpha}$ and vibrational quantum numbers $n_{\alpha}$. The energy of such state is

$$E_{\xi}=\sum_{\alpha=1}^{DN}[n_{\alpha}\hbar \omega_{\alpha} + \frac{\hbar}{2}\omega_{\alpha}].$$ (178)

The canonical partition function for the lattice is

$$Z(\beta, N)=\sum_{n_1} \sum_{n_2}\sum_{n_3}... \Bigg ( -\beta \sum_{\alpha=1}^{DN} n_{\alpha} \hbar \omega_{\alpha} +\frac{\hbar}{2} \omega_{\alpha} \Bigg),$$ (179)

which according to Eq.(174) becomes,

$$Z(\beta, N) = \prod_{\alpha}^{DN} \frac{e^{-\beta \frac{\hbar \om... ...{\hbar \omega_{\alpha}}{2}} - e^{-\beta \frac{\hbar \omega_{\alpha}}{2}})^{-1},$$ (180)

and

$$Z = -\sum_{\alpha=1}^{DN} (e^{\beta \frac{\hbar \omega_{\alpha}}{2}} -e^{-\beta \frac{\hbar \omega_{\alpha}}{2}}),$$ (181)

or in the continuous representation,

$$Z= -\int_0^{\infty} d\omega g(\omega) (e^{\beta \frac{\hbar \omega}{2}}-e^{-\beta \frac{\hbar \omega}{2}})$$ (182)

where $g(\omega)$ is the density of states --i.e., the number of vibrational states with frequencies between $\omega$ and $\omega + d\omega$.