# Number operator in the ground state

1. Jan 25, 2013

1. The problem statement, all variables and given/known data
Why does <0|$\frac{1}{(2\pi)^3}$∫ $\hat{a}^{\dagger}(t,r)$ $\hat{a}(t,r)$ d$^{3}$ $\textbf{k}$ |0> = $\frac{1}{\pi^2}∫$|β|^2 k^2 dk.

Where $\hat{a}$ and $\hat{a}^{\dagger}$ and its conjugate are bogulobov transformations given by:

$\hat{a}$(t,k) = $\alpha$(t)a(k) + β(t)$b^{\dagger}$(-k).

In the ground state a|0> =0 etc.

I am fairly certain it is some sort of table integral but i am not sure and want to prove it, any help or suggestions would be appreciated. I have taken the conjugate of the aforementioned a and multiplied it though but I don't understand why d$^{3}$ $\textbf{k}$ becomes k^2 dk. and how the pi factor changes, i.e. i get

$\frac{1}{(2\pi)^3}$∫ $b^{\dagger}$(-k)b(-k)|β|^2 d$^{3}$ $\textbf{k}$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 26, 2013

### G01

The change of integration measure is just the standard shift to spherical coordinates in k-space:
$$d^3k=k^2 dk d\Omega$$
Integrating over the angular integral $d\Omega$ is what changes the factors of Pi.

3. Jan 26, 2013

excellent thankyou, why do the operators bb^\dagger disappear is it just that in the ground state they become equal to 1?

4. Jan 26, 2013

### G01

Edit: I just caught your mistake. Check your work again. You should have $<0|b b^{\dagger}|0>$ not $<0|b^{\dagger}b|0>$ as you have in your first post.

Normal order all your operators and you should see how the operators drop out of the expression.

Last edited: Jan 26, 2013
5. Jan 26, 2013

sorry, $\hat{a}$(t,r) and its conjugate are bogulobov transofrmations of the fermionic creation and annihilation operators a(k) and b(-k).

The other bogulobv transformation although i didnt think it was needed is:

$b^{\dagger}$(t,k)=-β*(t)a(k)+ $\alpha$* $b^{\dagger}$(k).

My method was to take the * of $\hat{a}$(t,r) and multiply it with $\hat{a}$(t,r) and then cancel out using the ground state rules and that <0| $a^{\dagger}$(t,r) =0. but it still leaves b $b^{\dagger}$(-r). In the integral, one of the 2's is cancelled as there are also two degrees of freedom.

6. Jan 26, 2013

Thanks I didn't notice that, will that mean that it just goes to 1?

7. Jan 26, 2013

### G01

Work it out. Use the anticommutation relations to move operators such that you end up with the annihilation operators to the right (this is called normal ordering). You should see that you are left with a term that goes to zero and another that goes to 1.

8. Jan 26, 2013