quantum mechanics - Normal ordering of the commutator between annihilation and creation operator

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

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.