Expectation value of creation-/ annihilation operator

Question

Within second quantization say we have an arbitrary many-particle state. What is then the expectation value of a creation-/ annihilation Operator in the case of the particles being fermions/bosons?

I am reading my Professors lecture notes on second quantization and at one point I think he uses, that in the fermionic case said expectation values vanish, but for bosons don't, but I can't find a way to prove it to myself.

Edit: After some research I reckon it has something to do with coherent states?

Answer 1

Let $\hat{a}^{\dagger}$ and $\hat{a}$ be the creation and annihilation operators respectively. Then what follows is a lifting from the usual bosonic algebra, viz:

Then \begin{align} \langle n|\hat{a}^{\dagger}|n \rangle \,&\propto\, \langle n | n+1 \rangle \\ &=0 \\ \langle n|\hat{a} \hat{a}^{\dagger}|n \rangle &= \langle n|(1 + \hat{a}^{\dagger}\hat{a})| n \rangle \\ &= \langle n|n \rangle + \langle n|\hat{a}^{\dagger}\hat{a})| n \rangle \\&= 1 + n \end{align} One can use the commutation relationd for these but it can get tedious, I prefer to utilise the ladder operator method.