Number operator in the ground state

[pleasehelpmeno]

Homework Statement

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$^{3}$ $\textbf{k}$

 

[G01]

Homework Statement

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$^{3}$ $\textbf{k}$

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.

 

[pleasehelpmeno]

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

 

[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.

 

[pleasehelpmeno]

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 canceled as there are also two degrees of freedom.

 

[pleasehelpmeno]

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.

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

 

[G01]

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

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.

 

[pleasehelpmeno]

yeah thought so thx