# Normal ordering of bosonic commuators

1. Oct 29, 2014

### Idoubt

I am trying to understand wick's theorem and normal ordering mostly from Peskin and Schroeder. Now I have this problem with how normal ordering is defined. It seems to me that if you take the normal ordering of a commutator it should always be zero.

Here is what I understand normal ordering to be. If there is some operator $\hat{A}$ which is a product of bosonic operators $a_p$ and $a_p^{\dagger}$ , then normal ordering of of $\hat{A}$ is $N(\hat{A})$ where the creation operators are all moved to the left and the annihilation operators are all moved to the right. This essentially would mean that inside a normal ordering all operators can be commuted. For example,
$$N(a_pa_k^{\dagger}) = N(a_k^{\dagger}a_p) = a_k^{\dagger}a_p$$

Now taking the normal ordering of a commutator,
$$N([a_p,a_k^{\dagger}]) = N(a_pa_k^{\dagger} - a_k^{\dagger}a_p) = N(a_pa_k^{\dagger}) -N(a_k^{\dagger}a_p) = 0$$

But if I had used the fact that the commutator of $[a_p,a_k^{\dagger}] = \delta^{(3)}(p-k)$ which is a c-number, then

$$N([a_p,a_k^{\dagger}]) = N(\delta^{(3)}(p-k)) = \delta^{(3)}(p-k)$$

So there is a contradiction. Can someone explain to me what is the right way to think about normal ordering?

2. Oct 29, 2014

### bapowell

Normal ordering is not linear:
$$N(a a^\dagger) = N(1 + a^\dagger a) = N(1) + N(a^\dagger a) = 1 + a^\dagger a \neq a^\dagger a$$