Question to a calculation rule -> Dirac formalism

[Lindsayyyy]

Hi everyone

Homework Statement

I have a question: Am I allowed to do the following, where the a's are operators and the alphas are quantum states.

$$\langle \alpha \mid a^\dagger a^\dagger a a \mid \alpha \rangle = \langle \alpha \mid a^\dagger a^\dagger \mid \alpha \rangle \langle \alpha \mid a a \mid \alpha \rangle$$


Well, I think I am allowed to, but I'm not sure how to verify this.

Thanks for your help

 

[vela]

No, you can't. For that to be true, you'd have to have $\lvert \alpha \rangle \langle \alpha \rvert = 1$, which is almost certainly not the case.

 

[Lindsayyyy]

No, you can't. For that to be true, you'd have to have $\lvert \alpha \rangle \langle \alpha \rvert = 1$, which is almost certainly not the case.

and when the alpha is a coherent state?

 

[dextercioby]

Not even then, it would require the annihilation operator to be a self-adjoint operator (moreover a projector), which is not the case.

 

[Lindsayyyy]

Ok thank you.

my problem is the following: I want to simplyfy the following expression

$$\langle \alpha \mid a^\dagger a^\dagger a a + a^\dagger a \mid \alpha \rangle$$

I'm very uncertain about calculation rules when it comes to operators and states

is the expression above the same as?:
$$\langle \alpha \mid (a^\dagger a^\dagger a a \mid \alpha \rangle + a^\dagger a \mid \alpha \rangle)$$

or doesn't that help at all?

I think the aim here is to get an expressin which only has the alpha (as an eigen value) in it. The task is about calculating the expectation value

 

[Lindsayyyy]

$$\langle \alpha \mid a^\dagger a^\dagger a a + a^\dagger a \mid \alpha \rangle = \langle \alpha \mid a^\dagger a^\dagger a a \mid \alpha \rangle + \langle \alpha \mid a^\dagger a \mid \alpha \rangle <br /> = \alpha \alpha^* \langle \alpha \mid a^\dagger a \mid \alpha \rangle +|\alpha|^2 <br /> = |\alpha|^4 +|\alpha|^2$$

sorry for making a new reply, I tried to edit, but didn't come out in latex format.
To get to this solution I used that

$$a\mid \alpha \rangle = \alpha \mid \alpha \rangle$$

and the equivalent for the bra vector

 

[dextercioby]

It looks OK to me.