Expectation value of the anticommutator of the bosonic creation and annihilation operator

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?

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

$$[a,a^\dagger]=1$$ 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}

• Sorry if my question was a bit misleading. I wanted to know how derive the expectation value of $\langle 2 \hat{n} + 1 \rangle$ Dec 11, 2020 at 16:43
• 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. Dec 11, 2020 at 16:54
• Maybe you can calculate that by density operator. i.e. expectation value of an observable A$<A>=Tr[\rho A]$ where $\rho=\frac{e^{-\beta H}}{Tr[e^{-\beta H}]}$ Dec 11, 2020 at 17:09