Transition Amplitude in Free Field Theory In my course notes for quantum field theory, there is a section on calculating the probability of a 2 particle state making a transition to the 1 particle state:
$ =<\vec{k}_{out} | \vec{k}_{1} \vec{k}_{2}>$
$=<0|a^{\dagger} _{k _{out}} a _{k _{1}}a _{k _{2}}|0>$
$=<0|[a^{\dagger} _{k _{out}} a _{k _{1}}]a _{k _{2}}|0>$ + $<0|a _{k _{1}} a^{\dagger} _{k _{out}}a _{k _{2}}|0>$
= $0$
I can see how the 3rd step leads to $0$, but I do not understand how the 2nd step leads to the 3rd step. 
 A: Edit: The convention  I used is apparently different than yours, although the method is still the same. What notes are you using?
We use the commutation relations
$$
[a_p,a^\dagger_q] = (2\pi)^3\delta^{(3)}(\vec{p}-\vec{q}), [a_p,a_q]=[a^\dagger_p,a^\dagger_q
]=0
$$
Now we start with $$
\langle k|k_1k_2\rangle=\langle 0|a_k a^\dagger_{k_1}a^\dagger_{k_2}|0\rangle
$$
Now we use the fact that $a_k a^\dagger_{k_1}=a^\dagger_{k_1}a_k+(2\pi)^3\delta^{(3)}({k}-{k_1})$.
This gives
$$
\langle 0|a_k a^\dagger_{k_1}a^\dagger_{k_2}|0\rangle=\langle 0|\left(a^\dagger_{k_1}a_k+(2\pi)^3\delta^{(3)}({k}-{k_1})\right)a^\dagger_{k_2}|0\rangle
$$
$$
=\langle0|a^\dagger_{k_1}a_ka^\dagger_{k_2}|0\rangle+(2\pi)^3\delta^{(3)}(k-k_1)\langle0|a^\dagger_{k_2}|0\rangle
$$
we use the commutation relations again
$$
=\langle0|a^\dagger_{k_1}\left(a^\dagger_{k_2}a_k+(2\pi)^3\delta^{(3)}(k-k_2)\right)|0\rangle+(2\pi)^3\delta^{(3)}(k-k_1)\langle0|a^\dagger_{k_2}|0\rangle
$$
$$
=0+(2\pi)^3\delta^{(3)}(k-k_2)\langle0|a^\dagger_{k_1}|0\rangle
+(2\pi)^3\delta^{(3)}(k-k_1)\langle0|a^\dagger_{k_2}|0\rangle=0$$
