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Consider an Einstein solid. Each oscillator has quantized energy $E = n\hbar\omega$, where $n \geq 0$ is an integer. How can I compute the average energy and the specific heat at constant volume of an Einstein solid?

Ultimately, I want to show that the average energy expression obeys the equipartition theorem in the high temperature limit.


I know that the average energy is defined by

$$\overline{E} = \frac{1}{Z}\sum_{s}E(s)e^{-\beta E(s)}.$$

So, I can substitute my expression for energy into what I have above. Does this just mean that the average energy is given by this:

$$\sum_{n = 0}^{\infty}n\hbar\omega e^{-\beta n \hbar \omega}.$$

Is there any way to compute this summation? I'm not so sure if this is even correct. All help is appreciated.

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1 Answer 1

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Consider this summation first \begin{align} \sum_{n=0}^{\infty} e^{-nx} \; = \; \frac{1}{1 - e^{-x}} \end{align} where $x =\beta \hbar \omega > 0$. Note that it is a geometric series with common ratio $e^{-x} < 1$.

Now, take derivative with respect to $x$ so that we can "pull" the integer $n$ down :). This trick is very common in statistical mechanics. \begin{align} \frac{d}{dx} \sum_{n=0}^{\infty} e^{-nx} \; &= \; \frac{d}{dx} \Big( \frac{1}{1 - e^{-x}} \Big) \\ \sum_{n=0}^{\infty} \frac{d}{dx} \Big(e^{-nx} \Big) \; &= \; \frac{d}{dx} \Big( \frac{1}{1 - e^{-x}} \Big) \\ \sum_{n=1}^{\infty} n e^{-nx} \; &= \; -\frac{e^{-x}}{(1-e^{-x})^{2}} \end{align}

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  • $\begingroup$ But, using your notation, shouldn't our sum become $\sum_{n = 1}^{\infty} nx/\beta \cdot e^{-nx}$? $\endgroup$
    – Julie
    Commented Oct 1, 2018 at 8:14
  • $\begingroup$ @Julia, you are right. But $\beta$ is just a constant, just "throw" it to the right hand side. $\endgroup$
    – K_inverse
    Commented Oct 1, 2018 at 8:48
  • $\begingroup$ @Julia, I let $x = \beta \hbar \omega$ for convenience. You can have a different definition of $x$ as long as you feel comfortable. Cheers :) $\endgroup$
    – K_inverse
    Commented Oct 1, 2018 at 8:58

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