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I think I have a basic understanding of what degeneracies are: Two or more different states of a quantum mechanical system are said to be degenerate if they give the same value of energy upon measurement (thanks for stating it very clearly in the comments).

I am trying to now apply this concept to a problem:

A system possess three energy levels $E_1=\varepsilon$, $E_2=2\varepsilon$, $E_3=3\varepsilon$, with degeneracies $g(E_1)=g(E_3)=1, g(E_2)=2$. Find the heat capacity of the system.

My attempt: from a beginner's perspective I know to apply the partition function for a system with degeneracies as the first step in order to be able to obtain more information about the system. Thus,

$$\beta=\frac{1}{kT}$$ $$Z=\sum_i g_ie^{-\beta \varepsilon_i} \ $$ $$Z=g_ie^{-\beta \varepsilon} + 2g_ie^{-\beta 2\varepsilon} + g_ie^{-\beta 3\varepsilon}$$

But I don't quite understand. Also, the hint at the back of the book simply says to take the zero of the energy scale at $E_1=0$ and then proceeds to give the final answer as: $$C=2k\frac{x^2e^x}{(e^x+1)^2}$$

But this isn't really helpful in understanding the concept. What do the degeneracies of a system do to the solution of the partition function?

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  • $\begingroup$ @AnubhavGoel oops, that's what I meant to write! I'll edit. But then what about the rest of the problem? How does this affect solving the partition function? $\endgroup$ – whatwhatwhat Feb 9 '16 at 3:15
  • $\begingroup$ @AnubhavGoel it's ok bro, I'm right there with you :'( $\endgroup$ – whatwhatwhat Feb 9 '16 at 3:17
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    $\begingroup$ I'm writing an answer now :-) $\endgroup$ – Robbie Feb 9 '16 at 3:19
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I'll show you how the book derives that solution.

First, we start with the definition for energy fluctuation: $$ \langle(\Delta E)^2\rangle =\frac{\partial^2}{\partial \beta^2}\ln(Z) $$ where $Z$ is the Canonical Partition Function. The heat capacity, $C$, can be defined as: $$ C=\frac{1}{k_BT^2}\langle (\Delta E)^2\rangle $$

Next, let's calculate $Z$ for your question: \begin{align} Z&=\sum_i g_i e^{-\beta \epsilon_i}\\ &=1\cdot e^{-\beta \epsilon}+2\cdot e^{-2\beta \epsilon}+1\cdot e^{-3\beta \epsilon} \end{align} Where the degeneracies are the factors in front of each term. The little "trick" now is to factorise this, because later we plan on taking the logarithm and so we want to write this expression as a product of factors. \begin{align} Z&=e^{-\beta \epsilon}(1+2\cdot e^{-\beta \epsilon}+e^{-2\beta \epsilon})\\ &=e^{-\beta \epsilon}(e^{-\beta \epsilon}+1)(e^{-\beta \epsilon}+1)\\ &=e^{-\beta \epsilon}(e^{-\beta \epsilon}+1)^2 \end{align}

Okay, now that we have the canonical partition function written as a product (rather than sum) we can take the natural logarithm. \begin{align} \ln(Z)&=\ln\left(e^{-\beta \epsilon}(e^{-\beta \epsilon}+1)^2\right)\\ &=\ln\left(e^{-\beta \epsilon}\right)+\ln\left((e^{-\beta \epsilon}+1)^2\right)\\ &=-\beta\epsilon+2\ln\left(e^{-\beta \epsilon}+1\right)\\ \end{align}

Next step is to calculate the energy fluctuation by taking the second derivative of this function: \begin{align} \langle(\Delta E)^2\rangle&=\frac{\partial^2}{\partial \beta^2}\ln(Z)\\ &=\frac{\partial^2}{\partial \beta^2}\left(-\beta\epsilon+2\ln\left(e^{-\beta \epsilon}+1\right)\right)\\ &=\frac{\partial}{\partial \beta}\left(-\epsilon +\frac{2(-\epsilon e^{-\beta\epsilon})}{e^{-\beta\epsilon}+1}\right)\\ &=\frac{\partial}{\partial \beta}\left(-\epsilon +\frac{-2\epsilon}{1+e^{\beta\epsilon}}\right)\\ &=-(-1)2\epsilon(\epsilon e^{\beta\epsilon})\left(1+e^{\beta\epsilon}\right)^{-2}\\ &=\frac{2\epsilon^2e^{\beta\epsilon}}{\left(1+e^{\beta\epsilon}\right)^2} \end{align} The final step is to calculate the heat capacity using the definition above: \begin{align} C&=\frac{1}{k_BT^2} \langle(\Delta E)^2\rangle\\ &=\frac{1}{k_BT^2}\frac{2\epsilon^2e^{\beta\epsilon}}{\left(1+e^{\beta\epsilon}\right)^2}\\ &=\beta^2k_B\frac{2\epsilon^2e^{\beta\epsilon}}{\left(1+e^{\beta\epsilon}\right)^2}\\ &=k_B\frac{2(\beta\epsilon)^2e^{\beta\epsilon}}{\left(1+e^{\beta\epsilon}\right)^2} \end{align} Hence, with $x=\beta\epsilon$ this can be written as: \begin{align} C=k_B\frac{2x^2 e^{x}}{\left(1+e^{x}\right)^2} \end{align} If any steps don't make sense let me know and I'll explain them!

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  • $\begingroup$ Thanks for that @user36790, I never can remember all those LaTeX symbols! $\endgroup$ – Robbie Feb 9 '16 at 3:44
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    $\begingroup$ Quick reference on MathJax: meta.math.stackexchange.com/q/5020 $\endgroup$ – user36790 Feb 9 '16 at 3:47
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    $\begingroup$ Wait, in the "Next step is to calculate the energy fluctuation by taking the second derivative of this function:" shouldn't line 3 be: $\frac{\partial}{\partial \beta}\left(-\epsilon +\frac{2(-\epsilon e^{-\beta\epsilon})}{e^{-\beta\epsilon}+1}\right) $ (in other words, the right term should be positive?). Derivative of ln(x) = 1/x*chain rule. The chain rule here makes the term negative, but you have 2 negative signs. $\endgroup$ – whatwhatwhat Feb 9 '16 at 3:53
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    $\begingroup$ @whatwhatwhat Yes, they are equivalent expressions. You can see the full derivation in puccini.che.pitt.edu/~karlj/Classes/CHE2101/Shell/… (starting on page 6). $\endgroup$ – Robbie Feb 9 '16 at 4:09
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    $\begingroup$ Robbie, I understood all of this very well!! Thanks!!!!!1 $\endgroup$ – whatwhatwhat Feb 9 '16 at 4:27

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