Derivation of time ordered 2d scalar propagator I am trying to derive the following result (formula 3.7a from the book "Basic Concepts of String Theory")
$$\langle X^\mu_L(\bar{z})\, X^\nu_L(\bar{w}) \rangle = \frac{1}{4}\alpha'\eta^{\mu\nu}\ln\bar{z} -\frac{1}{2}\alpha'\eta^{\mu\nu}\ln(\bar{z}-\bar{w}) \quad(1) $$
where $\bar{z} = e^{2\pi i(\tau +\sigma)/l}$ and $\bar{w} = e^{2\pi i(\tau' +\sigma')/l}$ with $(\tau,\sigma)\sim(\tau,\sigma + l)$ parametrizing the closed string world sheet.
The propagator is defined as:
$$\langle X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma') \rangle = \mathcal{T}[X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma')] - :X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma'):\quad (2)$$
where $\mathcal{T}[\dots]$ denotes time-ordering and $:\dots:$ denotes normal-ordering.
I'm trying to derive (1) from the following mode expansion
$$X^\mu_L(\tau+\sigma)=\frac{1}{2}x^\mu + \frac{\pi\alpha'}{l}p^\mu(\tau+\sigma)+i\sqrt{\frac{\alpha'}{2}}\sum_{n>0}
\left(\frac{1}{n}\bar{\alpha}^\mu_n e^{-\frac{2\pi}{l} in(\tau+\sigma)} 
-\frac{1}{n}(\bar{\alpha}^\mu_n)^\dagger e^{\frac{2\pi}{l} in(\tau+\sigma)} \right) \quad(3) $$
with the commutation relation
\begin{align}
[\bar{\alpha}^\mu_m, (\bar{\alpha}^\nu_n)^\dagger] &= m\delta_{m,n}\eta^{\mu\nu}, \quad (m,n > 0)\\
[x^\mu, p^\nu] &= i\eta^{\mu\nu}
\end{align}
and we define $:p^\nu x^\mu:=x^\mu p^\nu$.
From the definition (2) and using (3) with the commutation relations, I was able to show that for $\tau > \tau'$,
\begin{align}
\langle X^\mu_L(\bar{z})\, X^\nu_L(\bar{w}) \rangle &= -i\frac{\pi\alpha'}{2l}\eta^{\mu\nu}(\tau+\sigma) + \frac{\alpha'}{2}\eta^{\mu\nu}\sum_{n>0}\frac{1}{n}\frac{\bar{w}^n}{\bar{z}^n} \\
&=\frac{1}{4}\alpha'\eta^{\mu\nu}\ln\bar{z} -\frac{1}{2}\alpha'\eta^{\mu\nu}\ln(\bar{z}-\bar{w})
\end{align}
which agrees with (1). But the same derivation would lead to that for $\tau <\tau'$,
$$\langle X^\mu_L(\bar{z})\, X^\nu_L(\bar{w}) \rangle = \frac{1}{4}\alpha'\eta^{\mu\nu}\ln\bar{w} -\frac{1}{2}\alpha'\eta^{\mu\nu}\ln(\bar{w}-\bar{z})$$
I wonder if there is any subtlety in my derivation which I have neglected or what would be the correct way to derive (1) from the expansion (3).
 A: Maybe I'm missing something but you can get as far as
\begin{align}
\langle X^\mu_L(\bar{z})\, X^\nu_L(\bar{w}) \rangle &= -i\frac{\pi\alpha'}{2l}\eta^{\mu\nu}(\tau+\sigma) + \frac{\alpha'}{2}\eta^{\mu\nu}\sum_{n>0}\frac{1}{n}\frac{\bar{w}^n}{\bar{z}^n}
\end{align}
without assuming anything about the ordering yet. Then there are two cases.

*

*If $\tau^\prime - \tau$ has a positive imaginary part, the propagator above is the expression you're trying to derive.


*If $\tau^\prime - \tau$ has a negative imaginary part, the sum in the propagator above doesn't even converge.
This looks like a version of the statement that only time ordered correlators make sense in Euclidean CFT. This also seems to be the conclusion from
$$\langle X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma') \rangle = \mathcal{T}[X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma')] - :X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma'):$$
if we take the vacuum expectation value of both sides.
Update
I would say the plane co-ordinates $(z, \bar{z})$ are not more or less meaningful than the cylinder co-ordinates $(\sigma, \tau)$. There is some semantics going on about what we call Euclidean but I think I see what we're doing differently now.

*

*You're actually computing $\mathcal{T}[X^\mu(\tau, \sigma)\, X^\nu(\tau',\sigma')]$ which means you're treating equation (2) as an equation which should be used.


*I'm just shoving $X^\mu_L(\bar{z}) X^\nu_L(\bar{w})$ between two vacuum states, as CFT people always do, which means I'm treating equation (2) as just something that we can keep in mind for later if the two-point function needs to be interpreted.
The second approach might be more subtle than I realized due to the $:p^\nu x^\mu: = x^\mu p^\nu$ convention for zero modes. But I agree that, when taking the first approach, I get your two expressions with $\ln(\bar{z} - \bar{w})$ and $\ln(\bar{w} - \bar{z})$. So that supports the expression in the previous question with the two Heaviside functions.
If this happened for vertex operators, I would be more worried. But $X_L$ and $X_R$ are not genuine CFT operators so maybe their correlators are more sensitive to how you quantize the theory.
