Inner Product Between States of Multiple Particles

[Wledig]

$$<p_1 p_2|p_A p_B> = \sqrt{2E_1 2E_2 2E_A 2E_B}<0|a_1 a_2 a_{A}^{\dagger} a_{B}^{\dagger} |0>$$ $$=2E_A2E_B(2\pi)^6(\delta^{(3)}(p_A-p_1)\delta{(3)}(p_B-p_2) + \delta^{(3)}(p_A-p_2)\delta^{(3)}(p_B-p_1))$$

The identity above seemed easy, until I tried to prove it. I figured I could work this out backwards opening all the commutators:

$$[a_A,a_{1}^{\dagger}]\cdot[a_B,a_{2}^{\dagger}]+[a_A,a_{2}^{\dagger}]\cdot[a_B,a_{1}^{\dagger}]$$

My idea was that this would reduce to $a_1 a_2 a_{A}^{\dagger} a_{B}^{\dagger}$, but that just didn't happen. What am I overlooking here?

 

[vanhees71]

You should use the commutation relations to bring one annihilation operator to the very right, leading to 0, because it annihilates the vacuum state. The only building block you really need is that (obviously in your convention of normalization of the momentum states),
$$[\hat{a}_1,\hat{a}_2^{\dagger}]=(2 \pi)^3 \delta^{(3)}(\vec{p}_1-\vec{p}_2).$$

 

[Wledig]

Ok, I tried it again. This time I tried doing it directly using the commutator relation to open the $a_2a_{A}^{\dagger}$ term, but I seem to be getting a sign error somehow:
$$a_1((2\pi)^3\delta^3(p_2-p_A)+a_{A}^{\dagger}a_2)a_{B}^{\dagger} $$
$$=a_1 a_{B}^{\dagger}(2\pi)^3\delta^3(p_2-p_A)+ a_1a_{A}^{\dagger}a_2a_{B}^{\dagger}$$
$$=((2\pi)^3\delta^3(p_1-p_B)+\underbrace{a_{B}^{\dagger}a_1}_0)((2\pi)^3\delta^3(p_2-p_A))+a_1a_{A}^{\dagger}((2\pi)^3\delta^3(p_2-p_B)+\underbrace{a_{B}^{\dagger}a_2}_0)$$

$$=(2\pi)^6(\delta^3(p_1-p_B)\delta^3(p_2-p_A)+\delta^3(p_2-p_A)+\delta^3(p_1-p_A)\delta^3(p_2-p_B))$$

 

[vanhees71]

I don't see, where you get the 2nd term in the bracket from, and then there's everything fine.

 

[Wledig]

I did the same thing that I had done with $a_2a_{A}^{\dagger}$, but this time for $a_2 a_{B}^{\dagger}$:

$$a_2 a_{B}^{\dagger} = [a_2,a_{B}^{\dagger}] + a_{B}^{\dagger}a_2$$

I did this to $a_1 a_{B}^{\dagger}$ and $a_1 a_{A}^{\dagger}$ too. The result is pretty close to what I wanted, but not the same. I should somehow be getting commutator terms like $[a_{A},a_{2}^{\dagger}]$ to account for $\delta^{3}(p_A-p_2)$ for example.

 

[Wledig]

Nevermind, you're right. It's the same thing as the expression I wanted, since: $$\delta^3(p_A-p_2)=\delta^3(-(p_2-p_A) )= \frac{\delta^3(p_2-p_A)}{|-1|}=\delta^3(p_2-p_A)$$
Everything is fine.

 

[vanhees71]

Note that this entire exercise is formalized in "Wick's theorem" and finally written in an ingenious notation as Feynman diagrams!