# Normal ordering of operators: Commutator or no commutator [duplicate]

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My exercise for a quantum optics course tells me to "find a normal ordering" for an Operator $$\hat{O}(\hat{a},\hat{a}^\dagger)$$, which is given as a (rather complicated) string of $$\hat{a}$$'s and $$\hat{a}^\dagger$$'s in random order.

To me, it is not clear, whether the tasks wants me to apply the normal ordering operator $$:\ :$$, which acting on an operator simply pulls all daggers to the right while neglecting the commutator rule or whether to use the proper commutation to create a normally ordered expression.

But that is not directly the question. When I open a textbook or my lecture note, when it is time to introduce the $$P$$, $$Q$$ and $$W$$ functions, they normally start by looking at some operator $$\hat{O}$$, then defining $$:\hat{O}: = \hat{O}^{(N)}$$. For example $$:\hat{n}^2:= \left(\hat{a}^\dagger\right)^2\hat{a}^2$$. Then an expectation value is calculated: $$\left\langle:\hat{n}^2:\right\rangle = \int d^2\alpha\ P(\alpha)\left|\alpha\right|^4$$

My problem is, that I don't see why one would use the normal ordering operator, because clearly the normally ordered operator is different from the unordered operator and will have a different expectation value.

And to extend that question: Can I calculate the expectation value for any normally ordered operator using the $$P$$ function? If I wanted the expectation value for $$\hat{n}^2$$ instead of $$\left\langle:\hat{n}^2:\right\rangle$$, would I have to bring the operator into normal order using the commutator relation?

## marked as duplicate by Qmechanic♦May 19 at 16:47

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• For your first question: normal ordering is not a well-defined function on operators, which probably accounts for your confusion. To impose it, you don't use the commutators. See the related question here. – knzhou May 19 at 16:46
• Possible duplicates: physics.stackexchange.com/q/323801/2451 , physics.stackexchange.com/q/345898/2451 and links therein. – Qmechanic May 19 at 16:48
• To reopen this post (v1), consider to only ask 1 subquestion per post, and remove duplicate parts. – Qmechanic May 19 at 16:49
• @Qmechanic While these posts are surely helpful and take away my confusion, they don't actually answer the questions phrased in this way. They only focus on the $:\ :$ itself. – HerpDerpington May 19 at 16:58
• 1) You are correct that $:n^2:$ and $n^2$ are different operators and have different expectation values. 2) normal ordered expressions are nice because it is very easy to calculate their expectation values (using the coherent state basis) 3) we bring up the whole idea of normal ordered operators because in the Glauber theory of photodetection we see that statistics of the (classical) measured photocurrent (such as the variance or two-time correlation function) are naturally expressed in terms of normal ordered strings of (quantum) creation and annihilation operators. – jgerber May 19 at 17:05