# Question to a calculation rule -> Dirac formalism

1. Nov 12, 2012

### Lindsayyyy

Hi everyone

1. The problem statement, all variables and given/known data

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.

2. Nov 12, 2012

### vela

Staff Emeritus
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.

3. Nov 12, 2012

### Lindsayyyy

and when the alpha is a coherent state?

4. Nov 12, 2012

### dextercioby

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

5. Nov 12, 2012

### 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

Last edited: Nov 12, 2012
6. Nov 12, 2012

### Lindsayyyy

$$\langle \alpha \mid a^\dagger a^\dagger a a + a^\dagger a \mid \alpha \rangle = \langle \alpha \mid a^\dagger a^\dagger a a \mid \alpha \rangle + \langle \alpha \mid a^\dagger a \mid \alpha \rangle = \alpha \alpha^* \langle \alpha \mid a^\dagger a \mid \alpha \rangle +|\alpha|^2 = |\alpha|^4 +|\alpha|^2$$

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

7. Nov 12, 2012

### dextercioby

It looks OK to me.