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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.

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)