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The number operator is given by:

$$\hat{n}= a^{\dagger} a.$$

For a presentation, I have to derive the expectation value of the anticommutator of the bosonic operators $a$ and $a^{\dagger}$ :

$$\langle \{a , a^{\dagger} \} \rangle = \langle 2 \, \hat{n} + 1 \rangle $$

How can I do this?

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  • $\begingroup$ You need to specify the stae in which you are taking the expectation value. $\endgroup$
    – mike stone
    Dec 11, 2020 at 15:49
  • $\begingroup$ If the state is $\sum\rho_{mn}|m\rangle\langle n|$, the expectation is $\operatorname{Tr}(\rho\sum_k(2k+1)|k\rangle\langle k|)=\sum_k(2k+1)\rho_{kk}$. $\endgroup$
    – J.G.
    Dec 11, 2020 at 16:51
  • $\begingroup$ $\rho_{kk}$ should than be the probability of each state or? This would give $\rho_{kk}=\frac{\exp(- \beta \, k \, E)}{Z}= \frac{\exp(- \beta \, k \, E)}{\exp(-\sum_k \beta \, k \, E)}$ $\endgroup$
    – Phicalc
    Dec 11, 2020 at 17:07

1 Answer 1

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$[a,a^\dagger]=1$ gives $aa^\dagger=1+a^\dagger a$

So $\{a,a^\dagger\}=aa^\dagger+a^\dagger a=2a^\dagger a+1=2\hat{n}+1$

To calculate the expectation value $\langle 2\hat{n}+1 \rangle$ we have (take $\hbar=1$ )

\begin{align} \langle 2\hat{n}+1 \rangle&=\text{Tr}[\hat{\rho} (2\hat{n}+1)] \\ &=\text{Tr}[\frac{e^{-\beta \omega (\hat{n}+1/2)}}{\text{Tr}[e^{-\beta \omega (\hat{n}+1/2)}]} (2\hat{n}+1)] \\ \end{align} First let's compute $Z=\text{Tr}[e^{-\beta \omega (\hat{n}+1/2)}]=e^{-\beta \omega/2}\sum_{n}\langle n|e^{-\beta \omega \hat{n}}|n \rangle=e^{-\beta \omega/2}\sum_{n}e^{-\beta \omega n} $

Then \begin{align} \text{Tr}[\frac{e^{-\beta \omega (\hat{n}+1/2)}}{Z} (2\hat{n}+1)]&=1+2\frac{e^{-\beta \omega/2} \sum_{n}n e^{-n \beta \omega}}{Z} \\ &=1+2\frac{ \sum_{n}n e^{-n \beta \omega}}{\sum_{n}e^{-\beta \omega n}} \\ \langle 2\hat{n}+1 \rangle&=1+\frac{2}{e^{\beta \omega}-1} \end{align}

Hope this is helpful.

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  • $\begingroup$ Sorry if my question was a bit misleading. I wanted to know how derive the expectation value of $\langle 2 \hat{n} + 1 \rangle$ $\endgroup$
    – Phicalc
    Dec 11, 2020 at 16:43
  • $\begingroup$ what is the state you are considering? $\endgroup$ Dec 11, 2020 at 16:45
  • $\begingroup$ All possible states I guess. Where the probability of each state is given by $p_n = \frac{\exp \left(- \beta \, n \, E_n \right)} {Z} $ with $Z$ being the partition function. According to my professor, this should yield the fluctuation-dissipation theorem. $\endgroup$
    – Phicalc
    Dec 11, 2020 at 16:54
  • $\begingroup$ Maybe you can calculate that by density operator. i.e. expectation value of an observable A$<A>=Tr[\rho A]$ where $\rho=\frac{e^{-\beta H}}{Tr[e^{-\beta H}]} $ $\endgroup$ Dec 11, 2020 at 17:09
  • $\begingroup$ I have updated my answer and see if that will help you, though I don't see the relation to fluctuation-dissipation thm. $\endgroup$ Dec 11, 2020 at 17:51

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