23

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{...


13

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 $....


12

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 \...


10

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 ...


9

Short explanation: Polchinski's eq. (1) is not a formula that transforms no normal order into normal order: The expression ${\cal F}$ on the right-hand side of eq. (1) is implicitly assumed to be radially ordered. In fact, eq. (1) is a Wick theorem for changing radial order into normal order, cf. e.g. this Phys.SE post. Longer explanation: When dealing ...


7

The choice of normal ordering prescription $:~:$ is usually adjusted to the choices of bra vacuum state $\langle \Omega|$ and ket vacuum state $|\Omega\rangle$, so that $$\langle \Omega|:\hat{\cal O}_1\ldots \hat{\cal O}_{n>0} : |\Omega\rangle~~=~~0 , \qquad\qquad \langle \Omega|\Omega\rangle~~=~~1 .$$ The relation of normal ordering prescription to ...


6

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 ...


6

We can use \begin{equation} \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} \end{equation} This gives \begin{equation} \begin{split} : \frac{i}{\alpha'} \partial X^\mu(z) : : e^{i k \cdot X(...


6

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{equation} \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 : ...


5

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 ...


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 ...


5

Applying normal ordering twice is not a valid operation. In fact, "normal ordering" is not a proper operator at all, and only defined on a single product of operators. It is undefined when acting on sums of operators, since trying to extend it linearly fails: $$ :a a^\dagger: = : a^\dagger a + 1 : = :a^\dagger a: + :1: = a^\dagger a + 1 \neq a^\dagger a = :a ...


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.


4

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.)


4

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 ...


4

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 ...


4

In your example there's two contractions giving two terms. Its amplitude is $$ \left< \psi^{\dagger}_a(x) \psi^b(y) \phi(z) \right> = -i \lambda \int d^4 s \, \Delta_M(z - s) \Delta_m(y-s)^b_{\;c} \Delta_m(x-s)^c_{\;a} $$ $$ - i \lambda \int d^4 s \Delta_M(x - y) ^b_{\;a} \Delta_M(0)^c_{\;c} \Delta_m(s-x) + \mathcal{O}\left(\lambda^2\right). $$ How ...


4

Wick's theorem is a very important theorem in Quantum Field Theory used to reduce products of creation and annihilation operators to sums of products of pairs of these operators. While I cannot give you a complete course here, I'll just state some of the basic ideas so you can get the idea on what's happening and not get lost in the mathematics (don't ban me ...


4

There cannot be such an expression because the covariance matrix and displacement don't contain the relative phase information. This is easy to see, since they are computed from the reduced density matrix (which does not depend on the phase of the state). A way around this can be to include a third reference state and $\vert\chi\rangle$ and consider the ...


3

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 ...


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 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

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 ...


3

I) Recall first the $\phi\phi$-Operator Product Expansion (OPE): $$\tag{A} {\cal R}\left\{\phi(z,\bar{z})\phi(w,\bar{w})\right\} ~-~: \phi(z,\bar{z})\phi(w,\bar{w}): ~=~C(z,\bar{z};w,\bar{w}) ~{\bf 1}, $$ where the contraction is assumed to be a $c$-number: $$\tag{B} C(z,\bar{z};w,\bar{w})~=~ \langle 0 | {\cal R}\left\{\phi(z,\bar{z})\phi(w,\bar{w})\right\...


3

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_{...


3

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.


3

The normal-ordered operator is defined as the difference between the operator and the ground state expectation value of that operator, e.g. $$[\hat{O}] \equiv \hat{O}-\langle \hat{O} \rangle$$ so the ground state expectation value of the normal-ordered operator should be zero. In your case of free fermi sea, if $k<k_F$, $$\begin{align} [c^{\dagger}_k ...


3

I) Definition. Given two operator ordering prescriptions, denoted by, say, $T$ and $::$, the corresponding (generalized) contraction $$C(\hat{A},\hat{B})~=~ T(\hat{A}\hat{B})~-~:\hat{A}\hat{B}:$$ of two operators $\hat{A}$ and $\hat{B}$ is the difference in ordering prescriptions. II) In applications, a contraction $C(\hat{A},\hat{B})~\propto~{\bf 1}$ ...


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