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We have $N$ oscillators and each of them is described by the Hamiltonian:

$$H = \frac{p^2}{2m} + \frac{Kq^4}{4} $$

I have to compute the average total energy $\langle E\rangle$ of the $N$ oscillators. But to do so, first I have to compute the one particle partition function and to do so I have to solve the following integral:

$$Z_1 (V,T) = \iint_{\mathbb{R}^2} e ^{-\beta \,H_1(p,q)}\,dp\,dq.$$

So in this case:

$$ Z_1 = \int_{-\infty}^{\infty} e^{-\beta\frac{p^2}{2m}}\,dp \int_{-\infty}^{\infty} e^{-\beta \frac{Kq^4}{4}}\,dq. $$

I know this integral can be solved by the Gauss method, knowing that:

$$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}.$$

For the first integral I got:

$$\int_{-\infty}^{\infty} e^{-\beta\frac{p^2}{2m}}\,dp = \sqrt{\frac{2m\pi}{\beta}}$$

I am having difficulties solving the second one:

$$\int_{-\infty}^{\infty} e^{-\beta \frac{Kq^4}{4}}\,dq.$$

I have tried to make a the change of variables: $q^4 = a^2,$ but this does not simplify the calculation. What method should I use?

Once you get $Z_1$:

$$Z_N = (Z_1)^N.$$

Then you just have to apply:

$$\langle E \rangle = -\frac{ \partial \log(Z_N)}{\partial \beta}.$$

ANSWER

$$\langle E \rangle = \frac{3N}{4\beta}.$$

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  • $\begingroup$ Wolfram Dev Platform gives an answer of $\Gamma(1/4)/(\sqrt{2}\sqrt[4]{\beta K}).$ $\endgroup$ – Adrian Keister Sep 28 '18 at 13:12
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    $\begingroup$ For any $n > 0\, \int_0^\infty e^{-x^n}dx = \Gamma(1+\frac{1}{n})$ - perhaps there is an application here given the spiltting of the half-line may be achievable. $\endgroup$ – Kevin Sep 28 '18 at 13:40
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    $\begingroup$ @JD_PM: I doubt that. Check the definitions in your textbook or notes. $\endgroup$ – Adrian Keister Sep 28 '18 at 13:44
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    $\begingroup$ @FelixMarin: It wouldn't surprise me if that was textbook-dependent. $\endgroup$ – Adrian Keister Sep 28 '18 at 19:50
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    $\begingroup$ @AdrianKeister That's right. $\endgroup$ – Felix Marin Sep 28 '18 at 19:57
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So, we have \begin{align*} Z_1&=\int_{-\infty}^{\infty} e^{-\beta\frac{p^2}{2m}}\,dp \int_{-\infty}^{\infty} e^{-\beta \frac{Kq^4}{4}}\,dq \\ &=\sqrt{\dfrac{2\pi m}{\beta}} \cdot \frac{\Gamma(1/4)}{\sqrt{2}\sqrt[4]{\beta K}} \\ &=\sqrt{\dfrac{\pi m}{\beta}} \cdot \frac{\Gamma(1/4)}{\sqrt[4]{\beta K}} \\ &=\frac{\sqrt{\pi m}\,\Gamma(1/4)}{\beta^{3/4}\,K^{1/4}} \\ &=\beta^{-3/4}\,\frac{\sqrt{\pi m}\,\Gamma(1/4)}{K^{1/4}}. \end{align*} Next, following your procedure, we do \begin{align*} Z_N&=\left(Z_1\right)^{\,N} \\ &=\left(\beta^{-3/4}\,\frac{\sqrt{\pi m}\,\Gamma(1/4)}{K^{1/4}}\right)^{\!N}, \\ \langle E\rangle&=-\frac{\partial}{\partial\beta} \, \ln\left[\left(\beta^{-3/4}\,\frac{\sqrt{\pi m}\,\Gamma(1/4)}{K^{1/4}}\right)^{\!N}\,\right] \\ &=-N\frac{\partial}{\partial\beta} \, \ln\left[\beta^{-3/4}\,\frac{\sqrt{\pi m}\,\Gamma(1/4)}{K^{1/4}}\right] \\ &=-N\frac{\partial}{\partial\beta} \left[-\frac34\ln(\beta)+\ln\left(\frac{\sqrt{\pi m}\,\Gamma(1/4)}{K^{1/4}}\right) \right] \\ &=\frac{3N}{4\beta}. \end{align*} Here you can see that the $\Gamma(1/4)$ just goes away because of the logarithm.

Lesson for you: Trust the intermediate results (at least temporarily), and keep going!

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    $\begingroup$ You are right, the definition of < E > has a negative sign! My bad. Now fixed. Thanks for your advise :) $\endgroup$ – JD_PM Sep 28 '18 at 14:21

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