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Harmonic Oscillators

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

$\displaystyle E = \frac{p^{ 2}}{2 m} + \frac{1}{2} \kappa  x^{ 2},$ (7.136)

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:

$\displaystyle \frac{\overline{p^{ 2}}}{2 m}$ $\displaystyle = \frac{1}{2}  k T,$ (7.137)
$\displaystyle \frac{1}{2} \kappa  \overline{x^{ 2}}$ $\displaystyle = \frac{1}{2}  k T.$ (7.138)

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

$\displaystyle \overline{E} = \frac{1}{2}  k T + \frac{1}{2}  k T = k T.$ (7.139)

According to quantum mechanics, the energy levels of a harmonic oscillator are equally spaced, and satisfy

$\displaystyle E_n = \left(n + \frac{1}{2}\right) \hbar  \omega,$ (7.140)

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

$\displaystyle \omega = \sqrt{\frac{\kappa}{m}}.$ (7.141)

(See Section C.11.) The partition function for such an oscillator is given by

$\displaystyle Z = \sum_{n=0,\infty} \exp(-\beta  E_n) = \exp\left(-\frac{1}{2}...
...eta  \hbar  \omega\right) \sum_{n=0,\infty} \exp(- n \beta  \hbar \omega).$ (7.142)

Now,

$\displaystyle \sum_{n=0,\infty} \exp(- n \beta  \hbar  \omega) = 1 + \exp(-\beta \hbar \omega) + \exp(-2 \beta \hbar \omega) + \cdots$ (7.143)

is simply the sum of an infinite geometric series, and can be evaluated immediately to give

$\displaystyle \sum_{n=0,\infty} \exp(- n \beta  \hbar  \omega) = \frac{1}{1-\exp(-\beta \hbar  \omega)}.$ (7.144)

(See Exercise 1.) Thus, the partition function takes the form

$\displaystyle Z = \frac{ \exp[-(1/2) \beta \hbar \omega]}{1-\exp(-\beta \hbar \omega)},$ (7.145)

and

$\displaystyle \ln Z = - \frac{1}{2} \beta \hbar \omega -\ln \left[1- \exp(-\beta \hbar \omega)\right].$ (7.146)

The mean energy of the oscillator is given by [see Equation (7.35)]

$\displaystyle \overline{E} = - \frac{\partial}{\partial \beta} \ln Z = - \left[...
...(-\beta \hbar \omega) \hbar \omega} {1-\exp(-\beta \hbar \omega)}\right],$ (7.147)

or

$\displaystyle \overline{E} = \hbar  \omega \left[ \frac{1}{2} + \frac{1}{\exp( \beta  \hbar  \omega)-1} \right].$ (7.148)

Consider the limit

$\displaystyle \beta \hbar \omega = \frac{\hbar  \omega}{k T} \ll 1,$ (7.149)

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

$\displaystyle \exp( \beta \hbar \omega) \simeq 1 + \beta  \hbar  \omega,$ (7.150)

so

$\displaystyle \overline{E} \simeq \hbar \omega\left[\frac{1}{2} + \frac{1}{\be...
...omega}\right] \simeq \hbar \omega\left[ \frac{1}{\beta \hbar \omega}\right],$ (7.151)

giving

$\displaystyle \overline{E} \simeq \frac{1}{\beta} = k T.$ (7.152)

Thus, the classical result, (7.139), holds whenever the thermal energy greatly exceeds the typical spacing between quantum energy levels.

Consider the limit

$\displaystyle \beta \hbar \omega = \frac{\hbar  \omega}{k T} \gg 1,$ (7.153)

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

$\displaystyle \exp( \beta \hbar \omega) \gg 1,$ (7.154)

and so

$\displaystyle \overline{E} \simeq \hbar \omega \left[\frac{1}{2} + \exp(-\beta \hbar \omega)\right] \simeq \frac{1}{2}  \hbar  \omega.$ (7.155)

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


next up previous
Next: Specific Heats Up: Applications of Statistical Thermodynamics Previous: Equipartition Theorem
Richard Fitzpatrick 2016-01-25