# Normal ordering for several operators

I understand that if we consider a normal ordering of two operators $\mathcal{N}(a_{p_1}a_{p_2}^{\dagger})$, it will be $a_{p_2}^{\dagger}a_{p_1}$, where $a_{p_1}$ is an annihilation operator of a particle with a momentum $p_1$ and $a_{p_2}^{\dagger}$ is a creation operator of a particle with momentum $p_2$.

But what about normal ordering for several operators? For example, let's consider $\mathcal{N}(a_{p_1}a_{p_2}^{\dagger}a_{p_3}a_{p_4}^{\dagger}).$ Is it equal to $$a_{p_4}^{\dagger}a_{p_2}^{\dagger}a_{p_1}a_{p_3},$$ or $$a_{p_2}^{\dagger}a_{p_4}^{\dagger}a_{p_1}a_{p_3},$$ or $$a_{p_2}^{\dagger}a_{p_4}^{\dagger}a_{p_3}a_{p_1},$$ or $$a_{p_4}^{\dagger}a_{p_2}^{\dagger}a_{p_3}a_{p_1}?$$

• Normally, all of the above since the a's commute among themselves and so do the a daggers. – Abdelmalek Abdesselam Feb 22 '18 at 16:05

$$\mathcal{N}(a_{p_1}a_{p_2}^{\dagger}a_{p_3}a_{p_4}^{\dagger}) = a_{p_2}^{\dagger}a_{p_4}^{\dagger}a_{p_1}a_{p_3},$$
$$[a_{p_1}, a_{p_2}] = [a_{p_1}^{\dagger}, a_{p_2}^{\dagger}] = 0$$