Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Quantum anharmonic oscillator

Posted by peeterjoot on March 13, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Question: Quantum anharmonic oscillator ([1] pr 3.30)

The energy levels of a quantum-mechanical, one-dimensional, anharmonic oscillator may be approximated as

\begin{aligned}\epsilon_n = \left( n + \frac{1}{{2}} \right) \hbar \omega - x \left( n + \frac{1}{{2}} \right)^2 \hbar \omega\qquad n = 0, 1, 2, \cdots\end{aligned} \hspace{\stretch{1}}(1.0.1)

The parameter x, usually \ll 1, represents the degree of anharmonicity. Show that, to the first order in x and the fourth order in u \equiv \hbar \omega/k_{\mathrm{B}} T, the specific heat of a system of N such oscillators is given by

\begin{aligned}C = N k_{\mathrm{B}} \left( \left( 1 - \frac{1}{{12}} u^2 + \frac{1}{{240}} u^4 \right) + 4 x \left( \frac{1}{{u}} + \frac{1}{{80}} u^3 \right) \right)\end{aligned} \hspace{\stretch{1}}(1.0.2)


We can expand the partition function in a first order Taylor series about x = 0, then evaluate the sums

\begin{aligned}Z_1 &= \sum_{n = 0}^\infty \exp\left( -\beta \left( n + \frac{1}{{2}} \right) \hbar \omega + \beta x \left( n + \frac{1}{{2}} \right)^2 \hbar \omega \right) \\ &= \sum_{n = 0}^\infty \exp\left( - \left( n + \frac{1}{{2}} \right) u + x \left( n + \frac{1}{{2}} \right)^2 u \right)\approx\sum_{n = 0}^\infty e^{ - \left( n + \frac{1}{{2}} \right) u }\left( 1 + x u \left( n + \frac{1}{{2}} \right)^2 \right).\end{aligned} \hspace{\stretch{1}}(1.0.3)

The quadratic sum can be evaluated indirectly as it can be expressed as a derivative

\begin{aligned}Z_1 &= \left( 1 + x u \frac{d^2}{du^2} \right)\sum_{n = 0}^\infty e^{ - \left( n + \frac{1}{{2}} \right) u } \\ &= \left( 1 + x u \frac{d^2}{du^2} \right)e^{-u/2}\sum_{n = 0}^\infty e^{ - n u } \\ &= \left( 1 + x u \frac{d^2}{du^2} \right)e^{-u/2}\frac{ 1 }{1 - e^{-u} } \\ &= \left( 1 + x u \frac{d^2}{du^2} \right)\frac{ 1 }{e^{u/2} - e^{-u/2} } \\ &= \left( 1 + x u \frac{d^2}{du^2} \right)\frac{ 1 }{2 \sinh(u/2)}\end{aligned} \hspace{\stretch{1}}(1.0.4)

Finally, evaluation of the derivatives gives us

\begin{aligned}Z_1=\frac{ 1 }{\sinh(u/2)}\left( 1 + x u \frac{2 \coth^2(u/2) - 1}{8} \right).\end{aligned} \hspace{\stretch{1}}(1.0.5)

Now we’d like to compute the specific heat in terms of derivatives of u. First, for the average energy

\begin{aligned}\left\langle{{H}}\right\rangle &= -N \frac{\partial {}}{\partial {\beta}} \ln Z_1 \\ &= -N \hbar \omega \frac{\partial {}}{\partial {u}} \ln Z_1.\end{aligned} \hspace{\stretch{1}}(1.0.6)

The specific heat is

\begin{aligned}C_{\mathrm{V}} &= \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {T}} \\ &= \frac{\partial {u}}{\partial {T}} \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {u}} \\ &= \frac{\hbar \omega}{k_{\mathrm{B}}} \frac{\partial {(1/T)}}{\partial {T}} \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {u}} \\ &= -\frac{\hbar \omega}{k_{\mathrm{B}} T^2} \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {u}} \\ &= -\frac{k_{\mathrm{B}} u^2}{\hbar \omega} \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {u}},\end{aligned} \hspace{\stretch{1}}(1.0.7)


\begin{aligned}C_{\mathrm{V}} = N k_{\mathrm{B}} u^2 \frac{\partial^2 {{}}}{\partial {{u}}^2} \ln Z_1.\end{aligned} \hspace{\stretch{1}}(1.0.8)

Actually computing that is messy algebra (See \nbref{pathria_3_30.nb}), and the result isn’t particularily interesting looking. The plot fig. 1.1 is interesting though and shows negative heat capacities near zero and a funny little jog near C_{\mathrm{V}} = 0.

Fig 1.1: Quantum anharmonic heat capacity


Also confirmed in the Mathematica notebook is equation eq. 1.0.2, which follows by first doing a first order series expansion in x, then a subsequent series expansion in u.


[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: