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

Question

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?

Frank · Accepted Answer

$[a,a^dagger]=1$ gives $aa^dagger=1+a^dagger a$
So ${a,a^dagger}=aa^dagger+a^dagger a=2a^dagger a+1=2hat{n}+1$
To calculate the expectation value $langle 2hat{n}+1 rangle$ we have (take $hbar=1$ )
begin{align}
langle 2hat{n}+1 rangle&=text{Tr}[hat{rho} (2hat{n}+1)] 
&=text{Tr}[frac{e^{-beta omega (hat{n}+1/2)}}{text{Tr}[e^{-beta omega (hat{n}+1/2)}]} (2hat{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} (2hat{n}+1)]&=1+2frac{e^{-beta omega/2} sum_{n}n e^{-n beta omega}}{Z} 
&=1+2frac{ sum_{n}n e^{-n beta omega}}{sum_{n}e^{-beta omega n}} 
langle 2hat{n}+1 rangle&=1+frac{2}{e^{beta omega}-1}
end{align}
Hope this is helpful.

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

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