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 $$a_{p_4}^{\dagger}a_{p_2}^{\dagger}a_{p_3}a_{p_1}?$$

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

The normal-ordering will always put annihilation operators on the right and creation operators on the left like you said. Creation and annihilation operators on each side will appear in the same order they were in before the application of the normal-ordering operator. So this case you have

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

In the case of bosons you should note that it doesn’t make a difference since

$$[a_{p_1}, a_{p_2}] = [a_{p_1}^{\dagger}, a_{p_2}^{\dagger}] = 0 $$


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