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1

As other answers mention, it was originally (in QED) about getting a neutral vacuum. It is useful to go back to Schwinger's old version of QED, before Dyson's approach became accepted. See Pauli: Selected topics in field quantization. Pauli presents both ways of looking at it: 1) define the electric current as sum of two terms (p.20 [6.4]), such that the ...

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

The choice of normal ordering prescription $:~:$ is typically 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 ...

2

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 ...

2

Lubos Motl has already provided a correct answer. This answer uses a different approach in the spirit of perturbation theory with sources: $$\int \! d^nx ~f(x)~e^{-\frac{1}{2}x^TAx +j^Tx} ~=~f\left(\frac{\partial}{\partial j}\right) \int \! d^nx ~e^{-\frac{1}{2}x^TAx +j^Tx} ~~\stackrel{\begin{matrix}\text{Gauss.}\\ \text{int.}\end{matrix}}{=}~~C~ ... 4 The case with the linear term is obtained from the original one by a simple shift, i.e. the substitution$$ 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 ...

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