Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Midterm II reflection, take II, with approximate anharmonic oscillator solution

Posted by peeterjoot on March 11, 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: Perturbation of classical harmonic oscillator (2013 midterm II p2)

Consider a single particle perturbation of a classical simple harmonic oscillator Hamiltonian

\begin{aligned}H = \frac{1}{{2}} m \omega^2 \left( {x^2 + y^2} \right) + \frac{1}{{2 m}} \left( {p_x^2 + p_y^2} \right) + a x^4 + by^6\end{aligned} \hspace{\stretch{1}}(1.0.12)

Calculate the canonical partition function, mean energy and specific heat of this system.

This problem can be attempted in two ways, the first of which was how I did it on the midterm, differentiating under the integral sign, leaving the integrals in exact form, but not evaluated explicitly in any way.

That solution was posted previously.

Alternately, by Taylor expanding around c = 0 and d = 0 with those as the variables in the Taylor expansion (as now done in the Pathria 3.29 problem), we can form a solution in short order. Given my low midterm mark, it seems very likely that this was what was expected.

Performing a two variable Taylor expansion of Z, about (c, d) = (0, 0) we have

\begin{aligned}Z \approx\frac{2 \pi m}{\beta}\int dx dye^{- \beta m \omega^2 x^2/2}e^{- \beta m \omega^2 y^2/2}\left( 1 - \beta a x^4 - \beta b y^6  \right)=\frac{2 \pi m}{\beta}\frac{ 2 \pi}{\beta m \omega^2}\left( 1 - \beta a \frac{3!!}{(\beta m \omega^2)^2} - \beta b \frac{5!!}{(\beta m \omega^2)^3}  \right),\end{aligned} \hspace{\stretch{1}}(1.0.22)

or

\begin{aligned}\boxed{Z \approx\frac{(2 \pi/\omega)^2}{\beta^2}\left( 1 - \frac{3 a }{\beta (m \omega^2)^2} - \frac{15 b }{\beta^2 (m \omega^2)^3}  \right).}\end{aligned} \hspace{\stretch{1}}(1.0.23)

Now we can calculate the average energy

\begin{aligned}\left\langle{{H}}\right\rangle = - \frac{\partial {}}{\partial {\beta}}\ln Z= - \frac{\partial {}}{\partial {\beta}}\left( -2 \ln \beta + \ln \left( 1 - \frac{3 a }{\beta (m \omega^2)^2} - \frac{15 b }{\beta^2 (m \omega^2)^3}  \right)  \right)=\frac{2 \beta}-\frac{    \frac{3 a }{\beta^2 (m \omega^2)^2}+    \frac{30 b }{\beta^3 (m \omega^2)^3}}{   1    - \frac{3 a }{\beta (m \omega^2)^2}   - \frac{15 b }{\beta^2 (m \omega^2)^3}}.\end{aligned} \hspace{\stretch{1}}(1.0.24)

Dropping the c, d terms of the denominator above, we have

\begin{aligned}\boxed{\left\langle{{H}}\right\rangle=\frac{2 \beta}-    \frac{3 a }{\beta^2 (m \omega^2)^2}-    \frac{30 b }{\beta^3 (m \omega^2)^3}.}\end{aligned} \hspace{\stretch{1}}(1.0.25)

The heat capacity follows immediately

\begin{aligned}\boxed{C_{\mathrm{V}} = \frac{1}{{k_{\mathrm{B}}}} \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {T}}= 2 - \frac{6 a k_{\mathrm{B}} T}{(m \omega^2)^2} - \frac{90 k_{\mathrm{B}}^2 T^2 b }{(m \omega^2)^3}.}\end{aligned} \hspace{\stretch{1}}(1.0.26)

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