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## Main Question or Discussion Point

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

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?

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?