17

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 \...


15

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 ...


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 $....


13

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 ...


8

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 : ...


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

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 ...


6

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 ...


5

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.)


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

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 ...


5

Perhaps it would be more clear if the authors used the following notation: $$\begin{align}\oint_{|z-w|=\varepsilon}& \mathrm{d}z~{\cal R} a(z)b(w) \cr ~=~& \oint_{|z|=|w|+\varepsilon} \mathrm{d}z~ a(z)b(w)\cr ~-~& \oint_{|z|=|w|-\varepsilon} \mathrm{d} z ~b(w)a(z)\tag{6.15a} \cr ~=~&A(|w|\!+\!\varepsilon) b(w)-b(w)A(|w|\!-\!\varepsilon) \cr ...


5

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 ...


5

Section III of this classic illustrates the method. I'll bypass the subtle math and cut to the chase for your specific example, taking the trivial case of ξ real... you do general things to your satisfaction, yourself, or check the ref in the comment of @ZeroTheHero above. This is an identity between exponentials of operators. In Lie group theory, ...


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

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

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

This two definitions are not equal, and lead to different expressions for more complicated fields (composite ones) OPEs. This two definitions uses the same regularization (point-splitting regularization) but different subtraction schemes. The one adopted by Polchinski is given by subtracting the contractions between the fields, divergent and finite terms are ...


4

Various comments to the post (v3): One may speculate that seemingly unordered operators are in practice always ordered wrt. some order. - As long as the fields $\phi_i=\phi_i^{(+)}+\phi_i^{(-)}$ are linear in creation and annihilation operators, this should not be a problem. - Isserlis' theorem is related to the path-integral formulation of Wick's theorem,...


4

I'm having trouble trying to understand what the symmetry factor of a Feynman diagram really is. [...] I understand it has nothing to do with the number of contractions that lead to the same topology (Wick's theorem), nor with the $4!$ factor of the Lagrangian, nor with the $n!$ factor from the Dyson series expansion. This seems to be a rather deep ...


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_{...


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