# Computing correlation function $\langle e^{i\beta \phi(x)}e^{-i\beta\phi(0)}\rangle$ for massless scalar field $\phi$

I am currently reading Shankar's "Bosonization: How to make it work for you in condensed matter" (http://inspirehep.net/record/408901/). In page 9, I am stuck with computing the correlation function in (39).

The part of the article is as follows:

My question is about deriving (40). Following the guide given in few lines below the article, I found out $$:e^A: = :e^{A^++A^-}:=e^{A^+}e^{A^-}$$ but how can I proceed to (40). Moreover, if I just assume (40), I cannot understand (42). Letting $$A=i\beta\phi(x), \quad B=i\beta\phi(0)$$ and using (40) gives $$e^{i\beta\phi(x)}+e^{-i\beta\phi(0)}=:ie^{i\beta(\phi(x)-\phi(0))}: \exp\left(\beta^2\langle \phi(x)\phi(0)-\frac{\phi(x)^2+\phi(0)^2}{2}\rangle\right).$$ How can I match this with (42)?

First of all there is a typo on the left-hand side of eq. (40) in Ref. 1. It should read $$T(e^Ae^B)~=~\ldots.\tag{40'}$$ Here the time-order $$T$$ is often not explicitly written in the notation. E.g. there is also an implicitly written time-order $$G_{\beta}~\equiv~\langle T\left(\ldots \right)\rangle\tag{39'}$$ inside the correlator function (39). Under the time-order symbol the operator ordering doesn't matter so we can write eq. (40') as $$T(e^C)~=~:e^C:~ e^{\frac{1}{2}\langle C^2 \rangle},\tag{40"}$$ where $$C~=~A+B.$$ Eq. (40") is a standard form of Wick's theorem, see e.g. my related Phys.SE answer here.
Equation 40 should read (a correct version can be found can be found in Shankars book) $$e^Ae^B = :e^{A+B}: e^{\left\langle AB + \frac{1}{2}\left(A^2+B^2\right)\right\rangle}$$ The procedure is essentiall laid out in the paragraphs under eq.(40); given $$A$$ and $$B$$ commute with $$[A,B]$$, we have eq.(41) (this can be derived as a special case of the Baker-Campbell-Haausdorff formula) \begin{align} e^{A+B} &= e^A e^B e^{-\frac{1}{2}[A,B]}\\ &= e^B e^A e^{\frac{1}{2}[A,B]}\;. \end{align} We can now write $$A = A^+ + A^-$$ interms of creation and destruction parts (and similarly $$B$$) and use the above expression to reorder terms until we arrive at the normal ordered expression $$:e^{A+B}:$$ multiplied by a bunch of commutators.
Finally we note that \begin{align} \langle C D \rangle &= \langle (C^++C^-)(D^++D^-) \rangle\\ &=\langle C^-D^+\rangle \\ &= [C^-,D^+] \end{align} and use this to write the commutators in terms of the vacuum expectation values to obtain the final result.