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I'm referencing this thesis which should be open-access.

In Appendix D.1 "High temperature expansion in general", the author writes the high temperature expansion in the following way:

$$ \begin{align*} \langle \hat{O} \rangle &= \frac{\sum_i \langle i| \hat{O} e^{-\beta \hat{H}} |i\rangle}{\sum_i \langle i| e^{-\beta \hat{H}} |i\rangle} \\ &= \beta^0 \Bigl[ \frac{1}{\Theta} \mathrm{Tr}(\hat{O})\Bigr] -\beta^1 \Bigl[ \frac{1}{\Theta} \mathrm{Tr}(\hat{O}\hat{H}) - \frac{1}{\Theta^2} \mathrm{Tr}(\hat{O})\mathrm{Tr}(\hat{H}) \Bigr] \\ &\qquad + \beta^2 \Bigl[ \frac{1}{2}\frac{1}{\Theta} \mathrm{Tr}(\hat{O}\hat{H}^2) - \frac{1}{\Theta^2} \mathrm{Tr}(\hat{OH})\mathrm{Tr}(\hat{H}) \\ &\qquad\quad - \frac{1}{2}\frac{1}{\Theta^2} \mathrm{Tr}(\hat{O})\mathrm{Tr}(\hat{H}^2) + \frac{1}{\Theta^3} \mathrm{Tr}(\hat{O})\mathrm{Tr}(\hat{H})^2 \Bigr] + \mathcal{O}(\beta^3) \end{align*} $$ where $\hat{O}$ is some operator and trace is over multiparticle states $|i\rangle$; $\Theta \equiv \mathrm{Tr}(I)$ is the dimension of the problem.


My question is: How did they do this expansion? (My attempt:) Clearly there has been an expansion of the exponential in the numerator in terms of $\beta$, $$ \sum_i \langle i| \hat{O} e^{-\beta \hat{H}} |i\rangle = \sum_m \frac{(-\beta)^m}{m!} \sum_i \langle i| \hat{O} \hat{H}^m |i\rangle \tag{1} $$

but I'm not sure 1) how or where $\Theta$ comes from, and also 2) why there is a split of the traced terms: $\mathrm{Tr}(\hat{O}\hat{H})$, $\mathrm{Tr}(\hat{O})\mathrm{Tr}(\hat{H})$ in $\beta^1$ for example. And also 3) how to formally divide out the denominator $\sum_i \langle i| e^{-\beta \hat{H}} |i\rangle$, like after substituting (1) back into the original equation:

$$ \frac{\sum_i \langle i| \hat{O} \hat{H}^m |i\rangle}{\sum_i \langle i| e^{-\beta \hat{H}} |i\rangle} = \text{terms for each $\beta^m$} $$

Can someone enlighten me on this?

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To first order in $\beta$, the numerator reads $$\eqalign{ \sum_i<i|Oe^{-\beta H}|i> &=\sum_i <i|O|i>-\beta\sum_i<i|OH|i>\cr &={\rm Tr}\ \!O-\beta\ \!{\rm Tr}\ \!OH\cr }$$ while the denominator is $$\eqalign{ \sum_i<i|e^{-\beta H}|i> &=\sum_i <i|i>-\beta\sum_i<i|H|i>\cr &=\Theta-\beta\ \!{\rm Tr}\ \!H\cr &=\Theta\Big(1-{\beta\over\Theta}\ \!{\rm Tr}\ \!H\Big)\cr }$$ Since $\beta$ is small, the inverse is (to first order) $${1\over\Theta}\Big(1-{\beta\over\Theta}\ \!{\rm Tr}\ \!H\Big)^{-1}={1\over\Theta}\Big(1+{\beta\over\Theta}\ \!{\rm Tr}\ \!H\Big)$$ The average is finally to first order in $\beta$ $$\eqalign{ \langle O\rangle&={1\over\Theta}\Big({\rm Tr}\ \!O-\beta\ \!{\rm Tr}\ \!OH\Big)\Big(1+{\beta\over\Theta}\ \!{\rm Tr}\ \!H\Big)\cr &={1\over\Theta}{\rm Tr}\ \!O -{\beta\over\Theta}{\rm Tr}\ \!OH +{\beta\over\Theta^2}{\rm Tr}\ \!O\ {\rm Tr}\ \!H\cr }$$ I let you extend the calculation to higher orders.

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  • $\begingroup$ There's a typo for your denominator line, but otherwise I get the gist of the method. Thank you! ($\sum_i<i|e^{-\beta H}|i>$, not $\sum_i<i|Oe^{-\beta H}|i>$.) $\endgroup$ – ksgj1 Nov 7 '18 at 2:36
  • $\begingroup$ Thanks! I have updated the answer. I am happy to have been helpful! $\endgroup$ – Christophe Nov 7 '18 at 19:55

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