What's the expectation value of the normally ordered commutator between annihilation and creation operator, $:[a,a^\dagger]:$?

Question

According to the commutation relation of annihilation and creation operators,

$$[a,a^{\dagger}]=1. \tag{1}$$

I would like to calculate the vacuum expectation value of the normal order of this commutator. We claim that since this commutator is just a c-number, then there is no effect from the normal ordering. We therefore get

$$\langle 0|(:[a,a ^\dagger]:)|0 \rangle=1. \tag{2}$$

However, if we expand the commutator first and do normal order later, we will get something like

$$\langle0|(:[a,a^\dagger]:)|0\rangle=\langle0|(:aa^\dagger-a^\dagger a:)|0\rangle=\langle0|(a^\dagger a-a^\dagger a)|0\rangle=0.\tag{3}$$ Which is contradicting with itself.

Answer 1

It is probably best to try to avoid writing commutators $[\cdot,\cdot]$ inside the normal-order argument $:\ldots:$, because it is likely to create confusion. However OP wants to know what would happen, and that's of course a fair question.

The logic goes as follows: Since operators$^1$ inside the normal-order argument $:\ldots:$ do (super)commute, e.g.

$$ :[a,a^{\dagger}]:~=~0, \tag{A}$$

it is implicitly assumed that one does not apply CCR relations, such as eq. (1), inside the normal order argument.

Hence OP's eq. (3) is formally correct, while eq. (2) is not.

See also e.g. this and this Phys.SE post for similar notational issues with operator orderings.

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$^1$ The 'operators' inside the normal-order argument are strictly speaking symbols, not operators.

Answer 2

The normal order symbol is a linear function in the sense that $${:} z_1d_1+z_2d_2 {:}= z_1{:} d_1 {:} + z_2 {:} d_2 {:}\quad \text{for} \quad z_1,\,z_2\in \mathbb{C}\quad \text{and}\quad d_1,\,d_2 \in \mathcal{A}$$ so it is totally fine to use the commutator inside the columns. The crucial point is to specify what is the domain on which the normal order is defined as a function, that is the algebra $\mathcal{A}$. This is the free algebra, which is the set of all possible linear combinations of formal finite products of creation and annihilation operators. In particular $a a^\dagger - a^\dagger a \neq 1$ in $\mathcal{A}$ because the canonical commutation relation $[a,a^\dagger]=1$ does not apply inside $\mathcal{A}$, that is inside the columns. This is a more mathematical restatement of the answer by Qmechanic. See also my answer to this other post where an axiomatic definition of normal order is provided.

Answer 3

The commutation relation between creation and annihilation operator in a quantum field theory is, $[a(t,x),a^{\dagger}(t,x')] = (2\pi) \delta(x-x')$. If you use this commutation you will see that the first equation (1) will give you infinite value. Then normal ordering ensures that you get a finite result.