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As Lubos Motl mentions in a comment, for all practical purposes, OP's sought-for eq. (1) is proved via Wick's Theorem. It is interesting to try to generalize Wick's Theorem and to try to minimize the number of assumptions that goes into it. Here we will outline one possible approach. I) Assume that a family $(\hat{A}_i)_{i\in I}$ of operators ...

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Essentially, what the Wick theorem tells you is that the moments of a multivariate gaussian distribution are determinate by the second moments; for instance, for a $3D$ gaussian in $(x,y,z)$ space, the quantity $$\langle xyzx \rangle$$ can be calculated in terms of $\langle xy\rangle$, $\langle xz \rangle$, $\langle xx\rangle$ and $\langle yz\rangle ... 7 If the operators$X_i$can be written as a sum of an annihilation and a creation part$^1$$$X_i~=~A_i + A^{\dagger}_i, \qquad i~\in ~I, \tag{1}$$ $$A_i|0\rangle~=~0, \qquad \langle 0 |A^{\dagger}_i~=~0, \qquad i~\in ~I,\tag{2}$$ where $$[A_i(t),A_j(t^{\prime})] ~=~ 0, \qquad [A^{\dagger}_i(t),A^{\dagger}_j(t^{\prime})] ~=~ 0, \qquad i,j~\in ... 6 I) It is true that operator ordering procedures are idempotent operations$$\tag{1} T(T(\ldots))~=~T(\ldots)\quad\text{and}\quad N(N(\ldots))~=~N(\ldots). $$But it is not true that the outermost ordering cancels the effect of the innermost ordering$$\tag{2} T(N(\ldots))~=~T(\ldots)\quad\text{and}\quad N(T(\ldots))~=~N(\ldots). \quad (\longleftarrow ... 6 Applying$\partial_1\overline{\partial}_1$, we have the following. First term:$$\partial_1\overline\partial_1 G_1 = -\pi\alpha' \eta^{\mu_1\mu_2} \delta^2 (z_1-z_2,\overline z_1-\overline z_2) X^{\mu_3} (z_3,\overline z_3) X^{\mu_4} (z_4,\overline z_4)$$$$+\text{ } {\rm permutations~of~indices~} (2,3,4).$$In the second term,$z_1$can be the logarithm or ... 5 The case with the linear term is obtained from the original one by a simple shift, i.e. the substitution $$x = X + A^{-1} B$$ Substitute it to the exponent in your more general integral: $$-\frac 12 x^T A x + B^T x = -\frac 12 (X^T+B^T A^{-1}) A(X+A^{-1}B)+B^T (X+A^{-1}B)=\dots$$ I used$A=A^T$. Now, all the terms that are schematically$BX$i.e. linear ... 4 Right, one is only supposed to put the sources$J=0$to zero after the very last$J$-differentiation has been performed. Figuratively speaking, short of writing out the calculation in full detail: Some of the$J$s downstairs can "couple" to the$J$s upstairs in the exponential. 4 We can use $$\begin{split} : F : : G: = \exp \left( - \frac{\alpha'}{2} \int d^2 z_1 d^2 z_2 \log|z_{12}|^2\frac{\delta }{\delta X_F^\mu(z_1, {\bar z}_1)} \frac{\delta }{\delta X_{G\mu}(z_2, {\bar z}_2)} \right) :F G: \end{split}$$ This gives \begin{split} : \frac{i}{\alpha'} \partial X^\mu(z) : : e^{i k \cdot ... 4 I think there is a typo in the first formula. Let me propose this (partial) answer for the$3$first formulae: Because$H(z)H(0) \sim -ln(z)$, we may write the OPE for any pair of operators$F(H), G(H)$functions of$H$(in analogy with formula$2.2.10$p.$39$vol$1$) $$:F::G: = e^{- \large \int dz_1 dz_2 ln z_{12} \frac{\partial}{\partial ... 4 Here we will outline a strategy to prove the sought-for operator identity (4) from the following definitions of what the commutator and the normal order of two mode operators \alpha_m and \alpha_n mean:$$ [\alpha_m, \alpha_n]~=~ \hbar m~\delta_{m+n}^0, \qquad\qquad(1) :\alpha_m \alpha_n:~=~\Theta(n-m) \alpha_m \alpha_n ~+~ \Theta(m-n) \alpha_n ... 3 The normal ordening is a way to say: ''we throw away the zero-point energy'' (since it becomes infinity and wa say we only look at energy-differences), or to put it in the words of A. Zee: ''Create before you annihalite''. The chronological ordening comes in when you calculate the Feynman propagator (also called the Green's function), which is basically the ... 3 You don't need to use that. You can simply do the cross-contractions by hand. Let's do that. Note that I only care about the the$\frac{1}{z^4}$term to evaluate the central charge. We have \begin{split} T(z) T(w) & =\left( : \partial_z b c(z): - \lambda \partial_z : b c (z): \right) \left( : \partial_w b c(w): - \lambda \partial_w : ... 3 The physical limit is$a\to 0$so the terms in the operator that are subleading in$a/L$go to zero and may be neglected. This is a different situation from computing various sums and integrals (in Green's functions and scattering amplitudes) whose leading terms in an expansion diverge. The leading divergent piece may be unphysical and get subtracted by ... 3 Lubos Motl has already provided a correct answer. This answer uses a different approach in the spirit of perturbation theory with$j$-sources: $$\int_{\mathbb{R}^n} \! d^nx ~f(x)~e^{-\frac{1}{2}x^TAx +j^Tx}$$ $$~=~f\left(\frac{\partial}{\partial j}\right) \int_{\mathbb{R}^n} \! d^nx ~e^{-\frac{1}{2}x^TAx +j^Tx}$$ $$~\stackrel{\begin{matrix}\text{Gauss.}\\ ... 3 In classical physics, quantities are ordinary, commuting c-numbers. The order in which we write terms in expressions is of no consequence. In quantum field theory (QFT), on the other hand, quantities are described by operators that, in general, don't commute. Classical physics is a low-energy approximation of quantum physics - the road from quantum to ... 3 The choice of normal ordering prescription :~: is usually adjusted to the choice of vacuum state |\Omega\rangle so that the bra-ket-sandwich of normal-ordered operators$$\langle \Omega|:\hat{\cal O}_1\ldots \hat{\cal O}_n : |\Omega\rangle~=~0 $$vanishes. The relation of normal ordering prescription to Wick theorem and other operator ordering ... 3 Wick's theorem tells us that$$ \mathcal{T}(\phi_1\dots\phi_N) =\ :\phi_1\dots\phi_N: + :\text{pairwise contractions}:$$where :\ : is normal ordering. Immediately from the definition of normal ordering (all annihilators to the right, all creators to the left), the expectation value of anything that is normal-ordered and not a constant vanishes because the ... 2 Hints: Note that radial ordering {\cal R} and normal-ordered/non-singular terms are implicitly implied in OP's eq. (2), see eq. (A) below. From (i) the OPEs$$ {\cal R} [\beta(z)\gamma(w)]~=~\frac{1}{z-w} ~+~ :\beta(z)\gamma(w):,\tag{A}  {\cal R} [\beta(z)\beta(w)]~=~ :\beta(z)\beta(w):, \qquad{\cal R} [\gamma(z)\gamma(w)]~=~:\gamma(z)\gamma(w):, ... 2 First of all, note that the radial operator ordering${\cal R}$is implicitly implied in many textbooks of CFT (e.g. Ref. 1). For instance, eq. (2.2.7) on p. 39 in Ref. 1 is discussing Wick's theorem between two operator ordering prescriptions. In this case between normal ordering$:~:$and radial ordering${\cal R}$. See also e.g. this Phys.SE post. The ... 2 Computing time-ordered products via Wick's theorem is fairly straightforward, schematically, $$\mathcal{T} \left\{ \phi_1 \phi_2 \dots \phi_n\right\} = \; : \sum \mathrm{all \;possible \; contractions}:$$ where colons denote normal-ordering, and for simplicity we have chosen a real scalar field, and the notation$\phi_n$denotes a field evaluated at the ... 2 Comment to the question (v2): Yes, the authors of Ref. 1 are cheating. They are not using Wick's theorem (although one in principle could do so). They know that the modified Bessel function$I_0$of first kind has an asymptotic series expansion in terms of a generalized hypergeometric function$^1$$$e^{\frac{1}{4g}} ... 2 In the particular case where f(\vec x) is a sum of expressions, where each expression has a total even power of the x^i like : f(x) = x_1x_2 + x_1^2 x_2 x_3 + ..., . we may present general expressions. we have :$$I(\Sigma)=\int d^{n} x\, e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}\tag{1} = \sqrt{(2 \pi)^ndet \Sigma}$$We have : ... 2 The main point is that in the context of the Dyson series it is by definition implicitly assumed that the time ordering operator T does not act/operate/re-order operators inside the interaction Hamiltonian H_I(t). In plain English: It does not 'see' what operators that H_I(t) consists of. It only re-orders between different appearances of the ... 2 An other way is to begin with (2.2.1.4)$$:e^{i k_1.X(z,\bar z)}:~ :e^{i k_2.X(0,0)}: ~\sim ~|z|^{\alpha' k_1.k_2} ~:e^{i (k_1 + k_2).X(0,0)}:$$Now, derive this expression relatively to k_1^\mu, then doing k_1=0, we get :$$:i X_\mu(z, \bar z):~:e^{i k_2.X(0,0)}: ~\sim(\alpha' (k_2)_{\mu} \ln|z| + i :X_\mu(0,0):):e^{i k_2.X(0,0)}:$$Now, we derive ... 2 In the operator^1 formulation, the vital assumption, which makes the standard Wick's theorem hold, is the assumption that the contractions are in the center of the pertinent operator algebra. This is often stated casually as the contractions should be c-numbers, meaning that the contractions should (super)commute with all pertinent operators. ^1 A ... 2 You can always use Wick's theorem when you're describing expectation values with respect to free field states (that is, non-interacting, like a single Slater determinant state). For example, in Hubbard-Stratonovich or (generally) Variational Mean-Field Theory (like Bogliubov-deGennes) you take the interaction terms and decouple them into a model that is ... 2 The first term in your second equation does not contain any singularities and is hence part of the "non-singular terms" at the end of the last expression. To find the final form you just need to perform the derivatives of the logarithm terms and Taylor expand the term :\!\partial X^\mu(z)\partial'X_\mu(z')\!: around z=z'. The singular contributions from ... 1 I was able to confirm both of your formulas, assuming that the operators are fermionic and normal ordered. Here is the general algorithm for generating a cumulant average of n variables (taken from page page 34 of G.W.Gardiner, Stochastic Methods: A Handbook for the Natural and Social Sciences, attributed to van Kampen):$$G_c(X_1,X_2,\ldots, ... 1 Normal-ordered non-singular terms in the OPE are still there in principle, but they are sometimes omitted (and therefore only implicitly implied) in the notation. This is because many important physical quantities only depend on the singular terms of the OPE. By the way, speaking of implicitly implied things, be aware that many authors don't write the ... 1 OP wrote (v1): In the beginning of proof of this theorem one says that permutation of fields in the left and right side of (1) doesn't change it. Answer: This is essentially due to the fact that operator ordering prescriptions (such as e.g. time ordering$T$or normal ordering$::$) are (graded) symmetric$\$ ...

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