# Thermodynamics of harmonic oscillators – classical and quantum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.5, Exercise 7.5.4.

One application of harmonic oscillator theory is in the behaviour of crystals as a function of temperature. A reasonable model of a crystal is of a number of atoms that vibrate as harmonic oscillators. From statistical mechanics, the probability ${P\left(i\right)}$ of finding a system in a state ${i}$ is given by the Boltzmann formula

$\displaystyle P\left(i\right)=\frac{e^{-\beta E\left(i\right)}}{Z} \ \ \ \ \ (1)$

where ${\beta=1/kT}$, with ${k}$ being Boltzmann’s constant and ${T}$ the absolute temperature, and ${Z}$ is the partition function

$\displaystyle Z=\sum_{i}e^{-\beta E\left(i\right)} \ \ \ \ \ (2)$

The thermal average energy of the system is then

 $\displaystyle \bar{E}$ $\displaystyle =$ $\displaystyle \sum_{i}E\left(i\right)P\left(i\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sum_{i}E\left(i\right)e^{-\beta E\left(i\right)}}{Z}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial\left(\ln Z\right)}{\partial\beta} \ \ \ \ \ (5)$

For a classical harmonic oscillator, the energy is a continuous function of the position ${x}$ and momentum ${p}$:

$\displaystyle E_{cl}=\frac{p^{2}}{2m}+\frac{1}{2}m\omega^{2}x^{2} \ \ \ \ \ (6)$

The classical partition function is then

 $\displaystyle Z_{cl}$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\beta p^{2}/2m}e^{-\beta m\omega^{2}x^{2}/2}dp\;dx\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}e^{-\beta p^{2}/2m}dp\int_{-\infty}^{\infty}e^{-\beta m\omega^{2}x^{2}/2}dx\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{2\pi}{\beta m\omega^{2}}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi}{\omega\beta} \ \ \ \ \ (10)$

Where we used the standard formula for Gaussian integrals to get the third line. The average classical energy is, from 5

$\displaystyle \bar{E}_{cl}=-\frac{\partial\left(\ln Z_{cl}\right)}{\partial\beta}=\frac{1}{\beta}=kT \ \ \ \ \ (11)$

The average energy of a classical oscillator thus depends only on the temperature, and not on the frequency ${\omega}$.

For a quantum oscillator, the energies are quantized with values of

$\displaystyle E\left(n\right)=\hbar\omega\left(n+\frac{1}{2}\right) \ \ \ \ \ (12)$

The quantum partition function is therefore

$\displaystyle Z_{qu}=e^{-\beta\hbar\omega/2}\sum_{n=0}^{\infty}e^{-\beta\hbar\omega n} \ \ \ \ \ (13)$

The sum is a geometric series, so we can use the standard result for ${\left|x\right|<1}$:

$\displaystyle \sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x} \ \ \ \ \ (14)$

This gives

$\displaystyle Z_{qu}=\frac{e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}} \ \ \ \ \ (15)$

The mean quantum energy is again found from 5, although this time the derivative is a bit messier, so is most easily done using Maple. However, by hand, you’d get

 $\displaystyle \bar{E}_{qu}$ $\displaystyle =$ $\displaystyle -\frac{\partial\left(\ln Z_{qu}\right)}{\partial\beta}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1-e^{-\beta\hbar\omega}}{e^{-\beta\hbar\omega/2}}\left[-\frac{1}{2}\frac{\hbar\omega e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}}-\frac{\hbar\omega e^{-\beta\hbar\omega/2}e^{-\beta\hbar\omega}}{\left(1-e^{-\beta\hbar\omega}\right)^{2}}\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar\omega}{2}\left(\frac{1+e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar\omega}{2}\left(\frac{1-e^{-\beta\hbar\omega}+2e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\omega\left(\frac{1}{2}+\frac{1}{e^{\beta\hbar\omega}-1}\right) \ \ \ \ \ (20)$

The average energy is the ground state energy ${\hbar\omega/2}$ plus a quantity that increases with increasing temperature (decreasing ${\beta}$). For small ${\beta}$ we have

 $\displaystyle \bar{E}_{qu}$ $\displaystyle \rightarrow$ $\displaystyle \hbar\omega\left(\frac{1}{2}+\frac{1}{1+\beta\hbar\omega-1}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar\omega}{2}+\frac{1}{\beta}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle \rightarrow$ $\displaystyle kT \ \ \ \ \ (23)$

since as ${\beta\rightarrow0}$, ${\frac{1}{\beta}\gg\frac{\hbar\omega}{2}}$. Thus the quantum energy reduces to the classical energy 11 for high temperatures. The ‘high temperature’ condition is that

 $\displaystyle \frac{1}{\beta}$ $\displaystyle \gg$ $\displaystyle \frac{\hbar\omega}{2}\ \ \ \ \ (24)$ $\displaystyle T$ $\displaystyle \gg$ $\displaystyle \frac{\hbar\omega}{2k} \ \ \ \ \ (25)$

So far, we’ve considered the average behaviour of only one oscillator. Suppose we now have a 3-d crystal with ${N_{0}}$ atoms. Assuming small oscillations we can approximate its behaviour by a system of ${3N_{0}}$ decoupled oscillators. In the classical case, the average energy is found from 11:

$\displaystyle \bar{\mathcal{E}}_{cl}=3N_{0}\bar{E}_{cl}=3N_{0}kT \ \ \ \ \ (26)$

The heat capacity per atom is the amount of heat (energy) ${\Delta E}$ required to raise the temperature by ${\Delta T}$, so

$\displaystyle C_{cl}=\frac{1}{N_{0}}\frac{\partial\bar{\mathcal{E}}_{cl}}{\partial T}=3k \ \ \ \ \ (27)$

For the quantum system, we have from 20

 $\displaystyle \bar{\mathcal{E}}_{qu}$ $\displaystyle =$ $\displaystyle 3N_{0}\bar{E}_{qu}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3N_{0}\hbar\omega\left(\frac{1}{2}+\frac{1}{e^{\beta\hbar\omega}-1}\right) \ \ \ \ \ (29)$

The quantum heat capacity is therefore

 $\displaystyle C_{qu}$ $\displaystyle =$ $\displaystyle \frac{1}{N_{0}}\frac{\partial\bar{\mathcal{E}}_{qu}}{\partial T}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3\hbar\omega\frac{\partial}{\partial\beta}\left(\frac{1}{e^{\beta\hbar\omega}-1}\right)\frac{d\beta}{dT}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3\frac{\hbar^{2}\omega^{2}}{kT^{2}}\frac{e^{\hbar\omega/kT}}{\left(e^{\beta\hbar\omega}-1\right)^{2}} \ \ \ \ \ (32)$

We can define the Einstein temperature as

$\displaystyle \theta_{E}\equiv\frac{\hbar\omega}{k} \ \ \ \ \ (33)$

which gives the heat capacity as

$\displaystyle C_{qu}=3k\frac{\theta_{E}^{2}}{T^{2}}\frac{e^{\theta_{E}/T}}{\left(e^{\theta_{E}/T}-1\right)^{2}} \ \ \ \ \ (34)$

For large temperatures, the exponent ${\theta_{E}/T}$ becomes small, so we have

 $\displaystyle C_{qu}$ $\displaystyle \underset{T\gg\theta_{E}}{\longrightarrow}$ $\displaystyle 3k\frac{\theta_{E}^{2}}{T^{2}}\frac{1+\theta_{E}/T}{\left(1+\theta_{E}/T-1\right)^{2}}\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle \rightarrow$ $\displaystyle 3k \ \ \ \ \ (36)$

For low temperatures ${e^{\theta_{E}/T}\gg1}$ so we have

 $\displaystyle C_{qu}$ $\displaystyle \underset{T\ll\theta_{E}}{\longrightarrow}$ $\displaystyle 3k\frac{\theta_{E}^{2}}{T^{2}}\frac{e^{\theta_{E}/T}}{e^{2\theta_{E}/T}}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3k\frac{\theta_{E}^{2}}{T^{2}}e^{-\theta_{E}/T} \ \ \ \ \ (38)$

The heat capacity again reduces to the classical value for high temperatures. The observed behaviour at low temperatures is that ${C_{qu}\rightarrow T^{3}}$, so this simple model fails for very low temperatures. However, as is shown by Shankar’s figure 7.3 Einstein’s quantum model is actually quite good for all but the lowest temperatures.