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## Hot answers tagged wick-theorem

10

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 $\hat{A}_i\in{... 9 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 $.... 9 If the operators$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 $$[A_i(t),A_j(t^{\prime})] ~=~ 0, \qquad [A^{\dagger}_i(t),A^{\dagger}_j(t^{\prime})] ~=~ 0, \qquad i,j~\in ~I,\... 7 I) It is true that operator ordering procedures are idempotent operations$$\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 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 ... 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 ... 5 Eq. 2.A.30 is a somewhat non-trivial identity for the ground-state$|\psi_0\rangle$which only uses the ground-state property that$\eta_q|\psi_0\rangle =0$. (Of course, as the OP has noted,$c_i|\psi_0\rangle \neq 0$.) What we need to show is that $$I=\langle \psi_0 | (c_j+c_j^\dagger)(c_i+c_i^\dagger)|\psi_0\rangle =\delta_{ij}.$$ Using Eq. 2.A.37a, we ... 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 X(... 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: $$[\alpha_m, \alpha_n]~=~ \hbar m~\delta_{m+n}^0, \qquad\qquad(1)$$ $$:\alpha_m \alpha_n:~=~\Theta(n-m) \alpha_m \alpha_n ~+~ \Theta(m-n) \alpha_n \... 4 I think there is a typo in the first formula. Let me propose this (partial) answer for the 3 first formulae: Because H(z)H(0) \sim -ln(z), we may write the OPE for any pair of operators F(H), G(H) functions of H (in analogy with formula 2.2.10 p.39 vol 1)$$:F::G: = e^{- \large \int dz_1 dz_2 ln z_{12} \frac{\partial}{\partial H(z_1)}\frac{\... 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. 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 Lubos Motl has already provided a correct answer. This answer uses a different approach in the spirit of perturbation theory with$j$-sources: $$\int_{\mathbb{R}^n} \! d^nx ~f(x)~e^{-\frac{1}{2}x^TAx +j^Tx}$$ $$~=~f\left(\frac{\partial}{\partial j}\right) \int_{\mathbb{R}^n} \! d^nx ~e^{-\frac{1}{2}x^TAx +j^Tx}$$ $$~\stackrel{\begin{matrix}\text{Gauss.}\\ \... 3 The physical limit is a\to 0 so the terms in the operator that are subleading in a/L go to zero and may be neglected. This is a different situation from computing various sums and integrals (in Green's functions and scattering amplitudes) whose leading terms in an expansion diverge. The leading divergent piece may be unphysical and get subtracted by ... 3 The normal ordening is a way to say: ''we throw away the zero-point energy'' (since it becomes infinity and wa say we only look at energy-differences), or to put it in the words of A. Zee: ''Create before you annihalite''. The chronological ordening comes in when you calculate the Feynman propagator (also called the Green's function), which is basically the ... 3 You don't need to use that. You can simply do the cross-contractions by hand. Let's do that. Note that I only care about the the \frac{1}{z^4} term to evaluate the central charge. We have \begin{split} T(z) T(w) & =\left( : \partial_z b c(z): - \lambda \partial_z : b c (z): \right) \left( : \partial_w b c(w): - \lambda \partial_w : ... 3 In the operator^1 formulation, the vital assumption, which makes the standard Wick's theorem hold, is the assumption that the contractions are in the center of the pertinent operator algebra. This is often stated casually as the contractions should be c-numbers, meaning that the contractions should (super)commute with all pertinent operators. ^1 A ... 3 Wick's theorem tells us that$$ \mathcal{T}(\phi_1\dots\phi_N) =\ :\phi_1\dots\phi_N: + :\text{pairwise contractions}:$$where :\ : is normal ordering. Immediately from the definition of normal ordering (all annihilators to the right, all creators to the left), the expectation value of anything that is normal-ordered and not a constant vanishes because the ... 2 You can always use Wick's theorem when you're describing expectation values with respect to free field states (that is, non-interacting, like a single Slater determinant state). For example, in Hubbard-Stratonovich or (generally) Variational Mean-Field Theory (like Bogliubov-deGennes) you take the interaction terms and decouple them into a model that is ... 2 The first term in your second equation does not contain any singularities and is hence part of the "non-singular terms" at the end of the last expression. To find the final form you just need to perform the derivatives of the logarithm terms and Taylor expand the term :\!\partial X^\mu(z)\partial'X_\mu(z')\!: around z=z'. The singular contributions from ... 2 In the particular case where f(\vec x) is a sum of expressions, where each expression has a total even power of the x^i like : f(x) = x_1x_2 + x_1^2 x_2 x_3 + ..., . we may present general expressions. we have :$$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 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 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 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 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 normal ordering is a valid operation provided one can undo it by an appropriate choice of counterterms (of existing couplings or field renormalisations). (How this is done in practice is explained here: http://arxiv.org/abs/1512.02604.) 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$$$e^{\frac{1}{4g}} \frac{Z(g)}{\sqrt{2\pi}}~=~\frac{1}{\... 2 Hints: Note that radial ordering {\cal R} and normal-ordered/non-singular terms are implicitly implied in OP's eq. (2), see eq. (A) below. From (i) the OPEs$$ {\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 I think the subscript$\text{IN}$refers to operators in interaction picture rather than "incoming" fields (what are those anyway?). There exists a way to morph the second expression into some interaction-picture operators sandwiched between the free theory vacuum: $$\langle \Omega|T[\phi(x_1)\phi(x_2)...\phi(x_n)]|\Omega\rangle = \langle \Omega|T[\phi_{... 2 This is explained in my Phys.SE answer here. In a nutshell, under appropriate assumptions, one may show that$$ \text{Contract}[\phi(x)\phi(y)]~=~D_F(x-y) ~{\bf 1},$$where${\bf 1}\$ is the identity operator.

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