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We have a Harmonic Oscillator in the Thermal state $\tau(\beta)$ which is defined

$$\tau(\beta) = \frac{e^{-\beta H}}{\mathrm{Tr}(e^{-\beta H})}$$

where $Z = \mathrm{Tr}(e^{-\beta H})$ is known as the partition function.

Now I was asked to calculate the average initial energy

$$E(\tau(\beta)) = \mathrm{Tr}(\tau H)$$

Now I did this (correctly) by doing:

$$E(\tau(\beta)) = (\frac{1}{Z}) \sum_n n \cdot \mathrm{Exp}[-n \beta]$$

Now I was asked to calculate the corresponding variance. I am supposed to obtain

$$V(\tau(\beta)) =(\frac{1}{Z}) \sum_n (n -\varepsilon)^2 \cdot \mathrm{Exp}[-n \beta] \tag{1}$$

where I am not 100% sure what $\varepsilon$ is. I used the formula

$$V(\rho) = \mathrm{Tr}(\rho H^2) - (\mathrm{Tr}(\rho H))^2$$

but am not able to understand how I am supposed to get eq. (1) with this. Can anyone help?

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    $\begingroup$ Are you sure your eq(1) is correct? I would have thought $V=\frac{1}{Z}\sum_n (n-\epsilon)^2\exp(-n\beta)$ (assuming $\epsilon$ is the expectation of the energy, and $\hbar\omega=1$) $\endgroup$ – By Symmetry Jan 14 at 15:37
  • $\begingroup$ Oh yes sorry! That's what I meant. Could you maybe give me a reference or tell my why my way to calc. it did not yield the result? $\endgroup$ – CatoMaths Jan 15 at 10:02
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    $\begingroup$ There are 2 standard expressions for the variance of a random variable $X$, $\mathrm{Var}(X) = \langle (X-\langle X\rangle)^2\rangle = \langle X^2\rangle - \langle X\rangle^2$. How to derive one from the other is in every introductory statistics book. The result you are aiming for is essentially a version of the first form and the formula you used is essentially a version of the second $\endgroup$ – By Symmetry Jan 15 at 12:01
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Use that $\rho=\exp[-\beta H]/Z$, and write $H$ in its eigenbasis: $H=\sum n|n\rangle\langle n|$.

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