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


*

*I can expand the (holomorphic part of the) field as $$ φ(z) = \hat q +
    i\hat p\log(z) + ∑_{n≠0}\frac{α_n}{n}z^{-n}; $$

*The non-vanishing commutation relation are $$ [\hat q,\hat p]=i;
    \qquad\qquad\qquad [α_n,α_m]=nδ_{m+n,0}; $$

*$\hat p$ and $\hat q$ are hermitian operators (Schellekens says real);

*Vacuum $|0⟩$ is defined this way
$$
α_n|0⟩=0=\hat p|0⟩\qquad (n>0);
$$

*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 


*

*He uses normal ordering procedure trating $\hat q$ as a creation operator so that 
$$
⟨0|\hat q =0,
$$
and he get the right result;

*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. $⟨0|\hat p\hat q|0⟩ = ⟨0|\hat q\hat p|0⟩ + ⟨0|[\hat p,\hat q]|0⟩$)? 
 A: *

*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
$$[\hat{q},\hat{p}]~=~i\hbar~{\bf 1}.\tag{1}$$

*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.

*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 
$$ \hat{a}~\equiv~\frac{1}{\sqrt{2}}(\hat{q}+i\hat{p}), \qquad  \hat{a}^{\dagger}~\equiv~\frac{1}{\sqrt{2}}(\hat{q}-i\hat{p}), \qquad  [\hat{a},\hat{a}^{\dagger}]~=~\hbar~{\bf 1}.\tag{4} $$

*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.

*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.

*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.

*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.  
--
$^1$ 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).
