Daily Archives: Sat, 4 February 2017

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.