# Number variance

1. Oct 22, 2007

### indigojoker

I need to find the number variance $$\langle (\Delta N ) ^2 \rangle$$ for the state $$| \beta \rangle = e^{\alpha a^{\dagger}-\alpha^{*} a}|1 \rangle$$

we know:
$$\langle (\Delta N ) ^2 \rangle$$
$$\langle a^{\dagger} a a^{\dagger} a \rangle$$
$$\langle a^{\dagger} (a^{\dagger}a +1) a \rangle$$
$$\langle a^{\dagger} a^{\dagger}a a+a^{\dagger} a \rangle$$
$$\langle \beta| a^{\dagger} (a^{\dagger}a +1) a |\beta \rangle$$

I know the relation (since this was derived):
$$[a^{\dagger},e^{\alpha a}]=-\alpha e^{\alpha a}$$
$$[a,e^{\alpha a^{\dagger}}]=\alpha e^{\alpha a^{\dagger}}$$

I could expand:
$$\langle 1|e^{\alpha a^{\dagger}-\alpha^{*} a} a^{\dagger} (a^{\dagger}a +1) a e^{\alpha a^{\dagger}-\alpha^{*} a}|1 \rangle$$

But i'm not sure how to apply the relation. any ideas would be appreciated.

2. Oct 22, 2007

### Physics Monkey

Hi indigojoker,

The first thing I would do is translate those commutator statements into something a little more concrete. It would be nice to know how a (or a^+) acts on your state. You might hope for a simple relationship since a annihilates the ground state and has a simple commutator with the operator that creates your state from the ground state.

3. Oct 23, 2007

### indigojoker

hmm, I'm not sure what you mean by something more concrete. Do you mean figuring out what:

$$a| \beta \rangle = ae^{\alpha a^{\dagger}-\alpha^{*} a}|1 \rangle$$
and
$$a^{\dagger}| \beta \rangle = a^{\dagger} e^{\alpha a^{\dagger}-\alpha^{*} a}|1 \rangle$$

is when simplified using:
$$[a^{\dagger},e^{\alpha a}]=a^{\dagger}e^{\alpha a}-e^{\alpha a} a^{\dagger}=-\alpha e^{\alpha a}$$
$$[a,e^{\alpha a^{\dagger}}]=a e^{\alpha a^{\dagger}}-e^{\alpha a^{\dagger}}a=\alpha e^{\alpha a^{\dagger}}$$

4. Oct 24, 2007

### Physics Monkey

Yes, that is what I had in mind. Here is a hint: since a annihilates the ground state, one can write [a,B]|ground> = (aB-Ba)|ground> = aB|ground> where B is anything!