Harmonic Oscillator

Our proof of the equipartition theorem depends crucially on the classical approximation. To see how quantum effects modify this result, let us examine a particularly simple system that we know how to analyze using both classical and quantum physics; namely, a simple harmonic oscillator. Consider a one-dimensional harmonic oscillator in equilibrium with a heat reservoir held at absolute temperature $T$. The energy of the oscillator is given by

$$\epsilon = \frac{p^{2}}{2\,m} + \frac{1}{2}\,\kappa\, x^{2},$$ (5.378)

where the first term on the right-hand side is the kinetic energy, involving the momentum, $p$, and the mass, $m$, and the second term is the potential energy, involving the displacement, $x$, and the force constant, $\kappa$. Each of these terms is quadratic in the respective variable. So, in the classical approximation, the equipartition theorem yields:

$$\frac{\left\langle p^{2}\right\rangle}{2\,m} = \frac{1}{2}\, k_B\,T,$$ (5.379)

$$\frac{1}{2}\,\kappa\left\langle x^{2}\right\rangle = \frac{1}{2} \,k_B\,T.$$ (5.380)

That is, the mean kinetic energy of the oscillator is equal to the mean potential energy, which equals $(1/2)\,k_B\,T$. It follows that the mean total energy is

$$\langle \epsilon\rangle = \frac{1}{2}\, k_B\,T + \frac{1}{2}\, k_B\,T = k_B\,T.$$ (5.381)

According to quantum mechanics (see Section 4.3.7), the energy levels of a harmonic oscillator are equally spaced, and satisfy

$$\epsilon_n = \left(\frac{1}{2}+n\right) \hbar \,\omega,$$ (5.382)

where $n$ is a non-negative integer, and

$$\omega = \sqrt{\frac{\kappa}{m}}.$$ (5.383)

Making use of the Boltzmann distribution, (5.329), the mean value of the quantum number $n$ for such an oscillator is

$$\langle n\rangle = \frac{\sum_{n=0,\infty}\exp[-(n+1/2)\,\hbar\,\... ...0,\infty} x^{\,n}}= x\,\frac{d}{dx}\!\left[\ln\sum_{n=0,\infty} x^{\,n}\right],$$ (5.384)

where $x=\exp(-\hbar\,\omega/k_B\,T)$. Now,

$$\sum_{n=0,\infty} x^{\,n} = \frac{1}{1-x},$$ (5.385)

so

$$\ln \sum_{n=0,\infty} x^{\,n}=-\ln(1-x),$$ (5.386)

and

$$x\,\frac{d}{dx}\!\left[\ln\sum_{n=0,\infty} x^{\,n}\right]= \frac{x}{1-x}.$$ (5.387)

Hence, we deduce that

$$\langle n\rangle = \frac{1}{\exp(\hbar\,\omega/k_B\,T)-1}.$$ (5.388)

Thus, the mean energy of the oscillator,

$$\langle \epsilon\rangle = \left(\frac{1}{2}+\langle n\rangle\right)\hbar\,\omega,$$ (5.389)

takes the form

$$\langle \epsilon\rangle =\left[\frac{1}{2}+\frac{1}{\exp(\hbar\,\omega/k_B\,T)-1}\right]\hbar\,\omega.$$ (5.390)

Consider the limit

$$\frac{\hbar\, \omega}{k_B\,T} \ll 1,$$ (5.391)

in which the thermal energy, $k_B\,T$, is large compared to the separation, $\hbar\,\omega$, between successive energy levels. In this limit,

$$\exp\left(\frac{\hbar\,\omega}{k_B\,T}\right) \simeq 1 + \frac{\hbar\, \omega}{ k_B\,T},$$ (5.392)

so

$$\langle \epsilon\rangle \simeq \left(\frac{1}{2}+\frac{k_B\,T}{\h... ...t)\hbar\,\omega \simeq \left( \frac{k_B\,T}{\hbar\,\omega}\right)\hbar\,\omega,$$ (5.393)

giving

$$\langle \epsilon \rangle \simeq k_B\,T.$$ (5.394)

Thus, the classical result, (5.381), holds whenever the thermal energy greatly exceeds the typical spacing between quantum energy levels.

Consider the limit

$$\frac{\hbar \,\omega}{k_B\,T} \gg 1,$$ (5.395)

in which the thermal energy is small compared to the separation between the energy levels. In this limit,

$$\exp\left(\frac{\hbar\,\omega}{k_B\,T}\right) \gg 1,$$ (5.396)

and so

$$\langle \epsilon\rangle \simeq \left[\frac{1}{2}+ \exp\left(-\fra... ...omega}{k_B\,T}\right)\right]\hbar\, \omega \simeq \frac{1}{2} \,\hbar \,\omega.$$ (5.397)

Thus, if the thermal energy is much less than the spacing between quantum states then the mean energy approaches that of the ground state. Clearly, the equipartition theorem is only valid in the former limit, where $k_B\,T \gg \hbar\, \omega$, and the oscillator possess sufficient thermal energy to explore many of its possible quantum states.