Free Boson propagator and normal ordering

Question

let's take a 2d free boson CFT as in page 27 of Schellekens lecture notes: I have the following facts:

1. I can expand the (holomorphic part of the) field as $$ φ(z) = \hat q + i\hat p\log(z) + ∑_{n≠0}\frac{α_n}{n}z^{-n}; $$
2. The non-vanishing commutation relation are $$ [\hat q,\hat p]=i; \qquad\qquad\qquad [α_n,α_m]=nδ_{m+n,0}; $$
3. $\hat p$ and $\hat q$ are hermitian operators (Schellekens says real);
4. Vacuum $|0⟩$ is defined this way $$ α_n|0⟩=0=\hat p|0⟩\qquad (n>0); $$
5. Since $[\hat p,e^{k\hat q}]= e^{k\hat q}$ I can define momentum states $$ |k;0⟩ = e^{k\hat q}|0⟩, $$ consequently I can regard $\hat q$ as a creation operator. In fact in the above lecture notes he writes

We define normalordering of $p$ and $q$ in such a way that $p$ is always to the right of $q$.

NOW

I want to compute the propagator as he does in section 2.5. If I do not use normal order there is a term $$ ⟨0|\hat q^2|0⟩, $$

when I expand $φ(z)φ(w)$ and I take the $q$-$q$ part, which I'm not able to evaluate; in particular

1. He uses normal ordering procedure trating $\hat q$ as a creation operator so that $$ ⟨0|\hat q =0, $$ and he get the right result;
2. If it's true, being hermitian, I would also have $\hat q|0⟩=0$, but this seems to contradict the fact that $\hat q$ is a creation operator as expressed in point 5.

I thought that hermitian operator couldn't be creation or annihilation ones: again due to the hermiticity I would have $$ 0=⟨0|\hat q \hat p|0⟩ = ⟨0|[\hat q,\hat p]|0⟩ + ⟨0|\hat p\hat q|0⟩ = i + ⟨0|\hat p\hat q|0⟩=i $$

Where am I wrong? How can I compute propagator just using the usual commutator trick (i.e. $⟨0|\hat p\hat q|0⟩ = ⟨0|\hat q\hat p|0⟩ + ⟨0|[\hat p,\hat q]|0⟩$)?

Answer 1

1. Note first of all that the oscillator modes $\hat{\alpha}_{n\neq 0}$ are irrelevant for OP's question, so let's get rid of them for simplicity. Then besides the identity ${\bf 1}$, we only have two independent operators $\hat{q}$ and $\hat{p} \sim \hat{\alpha}_0$, i.e. the standard Heisenberg algebra with non-zero CCRs $$[\hat{q},\hat{p}]~=~i\hbar~{\bf 1}.\tag{1}$$

2. Now the choices of bra vacuum state $\langle 0|$, ket vacuum state $|0\rangle$, and normal ordering $:~:$ are not unique, but has to fulfill some general requirements, cf. my Phys.SE answer here. However, $$\langle 0|^{\dagger}~=~|0\rangle \qquad\qquad (\longleftarrow\text{In general wrong!} )\tag{2}$$ is not a requirement. Each set of consistent choices constitutes a so-called picture. There exist bijective conversion formulas$^1$ between different pictures. There are a priori no reason why only one picture should exist. See also e.g. this Phys.SE post.

3. For the standard Wick normal-order $:~:$ the conditions $$ \langle 0| \hat{a}^{\dagger}~~=~~0~~=~~\hat{a}|0\rangle , \qquad\qquad \langle 0|0\rangle~~=~~1 .\tag{3}$$ is a natural choice to fulfill the general requirements. In this case eq. (2) holds. Here the creation/annihilation operators are defined as $$ \hat{a}~\equiv~\frac{1}{\sqrt{2}}(\hat{q}+i\hat{p}), \qquad \hat{a}^{\dagger}~\equiv~\frac{1}{\sqrt{2}}(\hat{q}-i\hat{p}), \qquad [\hat{a},\hat{a}^{\dagger}]~=~\hbar~{\bf 1}.\tag{4} $$

4. Another choice is $\hat{p}\hat{q}$-ordering $:~:$. The conditions $$ \langle 0| \hat{p}~~=~~0~~=~~\hat{q}|0\rangle , \qquad\qquad \langle 0|0\rangle~~=~~1, \tag{5}$$ is a natural choice to fulfill the general requirements.

5. Schelleken's normal-order $:~:$ is known as $\hat{q}\hat{p}$-ordering. He chooses this to have a simple description of the operator $\hat{p} \sim \hat{\alpha}_0$. The conditions $$ \langle 0| \hat{q}~~=~~0~~=~~\hat{p}|0\rangle , \qquad\qquad \langle 0|0\rangle~~=~~1,\tag{6}$$ is a natural choice to fulfill the general requirements.

6. With the conditions (6), the bra vacuum state $\langle 0|$ becomes a position eigenstate with $q=0$, and the ket vacuum state $|0\rangle$ becomes a momentum eigenstate with $p=0$. In particular in this case eq. (2) cannot hold.

7. It is straightforward to develop a pertinent notion of coherent states in the framework of $\hat{q}\hat{p}$-ordering. The coherent bra states $$\langle q|~:=~ \langle 0|\exp\left( \frac{iq\hat{p}}{\hbar}\right), \qquad \langle q|\hat{q}~\stackrel{(1)+(6)}{=}~ \langle q|q, \tag{7} $$ become position states; and the coherent ket states $$|p\rangle~:=~ \exp\left( \frac{ip\hat{q}}{\hbar}\right)|0\rangle, \qquad \hat{p}|p\rangle~\stackrel{(1)+(6)}{=}~p|p\rangle,\tag{8} $$ become momentum states.

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$^1$ E.g. one may show that the ket vacuum state (6) in the $\hat{q}\hat{p}$̂ -ordering is a squeezed state $$|p\!=\!0\rangle~~\propto~~ \exp\left( \frac{(\hat{a}^{\dagger})^2}{2\hbar}\right)|a\!=\!0\rangle$$ wrt. the standard Fock space ket vacuum state (3).