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1 vote
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Understanding operator valued distributions

This is a common abuse of notation when using distributions in Physics in general, so not something specific to operator-valued ones. Given a test function $f(x)$ you can always define a distribution ...
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Wick rotation in Peskin and Schroeder's QFT

My first question is why $\ell^0 \equiv i \ell_E^0$ imply this counterclockwise rotation. The integration variable $\ell_E^0$ is supposed to be real. If $\ell_E^0$ runs from $-\infty$ to $\infty$ ...
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-1 votes

Wick rotation in Peskin and Schroeder's QFT

The euclidean variabile is on the $x$-axis of the Gauss plane, so the imaginary variable is on the vertical axis. And about the scalar product, you are in Minkowski spacetime, so there is not the ...
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How is Lorentz invariance of $S$-matrix related to vanishing of Hamiltonian density commutator at spacelike separations?

The correct statement is that IF ${\cal H}(x)$ is scalar and $[{\cal H}(x),{\cal H}(x')]=0$ for $x,x'$ spacelike related THEN the $S$ matrix is invariant under the unitary representation $U_{(\Lambda, ...
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Peskin and Schroeder's QFT book page 289

User Zack has already answered OP's first part. Concerning the equivalence between the interaction picture (4.31) and the path integral formulation in the Heisenberg picture (9.18), note that P&S ...
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-2 votes

How is Lorentz invariance of $S$-matrix related to vanishing of Hamiltonian density commutator at spacelike separations?

I think I can devise an explanation, but I am not 100% sure that it is correct. But I will try anyway and if there is a hole in my line of reasoning, someone will shout out to me hopefully. Here goes.....
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Understanding graviton propagators in Minkowski space

Taking your expression (6) for the propagator $$\Pi_{\mu\nu\rho \sigma} = \frac{1}{2 k^2} \left( \theta_{\mu\rho} \theta_{\nu \sigma} + \theta_{\mu\sigma} \theta_{\nu \rho} - \theta_{\mu\nu} \theta_{\...
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1 vote
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Peskin and Schroeder's QFT book page 289

In the path integral formalism, $\phi$ is not an operator, it is an integration variable. In other words, inside the integral $\int D\phi$, $\phi$ is just an ordinary classical field. So there’s no ...
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How does quantum field theory connect to dynamics of particles?

In QFT all processes are considered to be happening inside the "black boxes", and the theory is aiming at predicting the outputs of the black box from given inputs. Creation and annihilation ...
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What exactly does it mean by gauge-invariant "operators"?

The whole point of the quantization of gauge theories is that we don't really want a realization of the gauge symmetry in the quantum theory because the gauge symmetry is unphysical (see also this ...
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2 votes
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Two-point correlation function in functional integral

Perhaps it would be more symmetric if we write the denominator $\langle\Omega|\Omega\rangle=1$ on the left-hand sides explicitly. The denominators on the right-hand sides are important, and cannot in ...
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0 votes

Two identical output particles in decay and scattering process in quantum field theory

Let's me finish this post myself. First, we need to figure out where causes the symmetry factor $S$, it comes from integration in momentum space: (for example, Peskin and Schroeder's book, page 106, ...
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Do quarks oscillate via Charged Weak Interaction (CKM Matrix)?

Yes, this type of oscillation is possible. Your specific example is forbidden because there are no charmed baryon states whose masses overlap with the nucleon, so the $u\to c$ diagram you’ve drawn ...
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1 vote
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Handling the $\nabla \phi$ term in the Hamiltonian in a path integral

Yes, your reasoning is correct. To spell it out a bit more, the gradient $\nabla$ is an operator on position space, but doesn't act on Hilbert space. The field $\hat{\phi}(x)$ is an operator-valued ...
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2 votes
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Linking of a sphere with a Wilson line

Fact of life: In $d$ a dimensions, a $p$-dimensional closed manifold can link with a $q$ dimensional closed manifold if $p+q+1=d$. (cf. wiki: linking number) $\newcommand{\S}{\mathbb{S}}\newcommand{\R}...
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What are the sources of infinities in (unrenormalized) quantum field theory calculations?

Is there any consensus in the physics community about what at least some of the infinities sources are? There is a consensus in the broad sense, that the best theory we have is incomplete. This has ...
2 votes

Topological and non-topological defects?

First of all, let's set up the stage correctly. The setting will be a completely general $d$-dimensional Euclidean quantum field theory, so the answer will be applicable to string theory and ...
1 vote

Cluster decomposition $\stackrel{?}{=}$ Translation invariance

I completely agree with the answers already provided by hft and Andrew. Thanks. Here is an explicit counter-example (closely related to Andrew’s) but expressed in the language of ladder operators. ...
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Cluster decomposition $\stackrel{?}{=}$ Translation invariance

Cluster decomposition $\stackrel{?}{=}$ Translation invariance No, these are not the same. Conversely, starting with the requirement of translation invariance, I think we can deduce the above form ...
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4 votes
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Cluster decomposition $\stackrel{?}{=}$ Translation invariance

The following interaction is translation invariant but nonlocal \begin{equation} H_{\rm int} = \lambda \int d^3 x \int d^3 y \phi^2(\vec{x}) \phi^2(\vec{y}) \end{equation} An $S$-matrix computed with ...
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1 vote
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Peskin and Schroeder's QFT eq. (9.14): Gaussian momentum field integration of phase space path integral

Briefly, there are 2 issues: It is safest to Wick rotate $t_E=it_M$ to make the Gaussian integrals exponentially damped rather than oscillatory. (NB: Don't also Wick rotate the momentum field $\pi_M=...
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Energy and momentum conservation using Dirac delta function

Thanks for the answer, In scattering problems $$\sigma = \int |M^2| d^4 p_f 2\pi \delta^4(p_f -p_i) \delta(p_f^2 -m_0^2)$$ the $d^4p_f$ integration is all over phase space. So I can assume that ...
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6 votes
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Defining propagators as vacuum expectation values of products of field operators

It's simply because that commutator is proportional to identity operator, as your result imply. Considering a normalized vacuum state then $$ \left[ \hat{\phi}^+(x), \hat{\phi}^-(y) \right] = \text{...
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Contraction with external legs in $S$-matrix

I want answer my post myself. Actually, my analysis in post about the two cases is wrong!!! In both case (1) and (2), $$ \left\langle\mathbf{p}_1 \mathbf{k}_2\left|T\left\{\phi\left(x_1\right) \phi\...
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1 vote
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What does effective quantum field theory mean?

In theoretical physics, quantum field theory (QFT) is a theoretical framework that combines classical field theory, special relativity, and quantum mechanics. Italics mine. QFT can be used to model ...
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1 vote
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Dirichlet/Neumann conditions for the quantized string

The boundary conditions of a classical string theory are constraints that are an integral part of fully specifying the actual physical theory, not some sort of ad-hoc conditions you are allowed to ...
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4 votes
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$Z_n$ gauge theory from $U(1)$

You need to think about what "$B$ is a periodic scalar" means: When $B$ is $2\pi$-periodic, then since $V$ acts on $B$ as $B\mapsto B + \frac{2\pi}{n}$, you necessarily have $V^n = 1$ since $...
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3 votes

Is momentum space "less physical" than position space?

I would concede that position-space is often more "intuitive" or "less abstract" than momentum-space. But as for "more or less physical"--I would use the word "...
12 votes
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Are particles merely "relatively stable" patterns that can appear on their respective fields?

There is no such thing as "looking at a quantum field", and particles aren't just "relatively stable patterns". You're thinking about this with a classical intuition (that there ...
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Are particles merely "relatively stable" patterns that can appear on their respective fields?

Am I mistaken to think that then particles merely refer to "relatively stable" patterns that can appear on these fields? In the field theoretical standard model of particle physics, the ...
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2 votes

Energy and momentum conservation using Dirac delta function

The scattering matrix elements you calculate in QFT have to be integrated over various phase space measures to yield measurable results. In this process all delta functions are expended and you end up ...
3 votes
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Motivation of Grassmann fields in the Faddeev-Popov method for free Gluon fields

Well, one argument goes as follows: The path integral $Z$ for a gauge theory needs gauge-fixing, typically a product of Dirac delta distributions: $\prod_{x,a} \delta(G^a(x))$. To make the path ...
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3 votes
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Measure of Functional Integral in Path Integral Formulation

$\sqrt{\frac{m}{2\pi ih \Delta t}}$ is the famous Feynman fudge factor. It can be derived from the Gaussian momentum integrations in the corresponding Hamiltonian phase space path integral, see e.g. ...
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7 votes
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Dimensional analysis of quantized Klein-Gordon Field

I know that the creation and annihilation operator need to have dimension 1, since they related with one-particle state. I don't know what rule you think you're using, but this is wrong. Each ladder ...
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Motivation of Grassmann fields in the Faddeev-Popov method for free Gluon fields

Main idea Indeed as @JeanbaptisteRoux pointed out in a comment, the motivation is that you can think of the determinant of an operator to a positive power as coming out of a fermionic integral. To ...
1 vote

QFT- Computing matrix elements when momentum is NOT conserved

Momentum is not conserved when the lagrangian density depends explicitly on $x$, that is when the interaction vertices are no longer constant but functions of spacetime points. This typically happen ...
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33 votes
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Is momentum space "less physical" than position space?

Your question can be equivalently phrased as whether position space is more physical than momentum space. In a sense, yes. One of the basic facts about our universe (as far as we can tell) is that it ...
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Is momentum space "less physical" than position space?

Well, to begin with, the universe we happen to live and move around in is position space- and we build & run our experiments in position space too. So if one wanted to compare the results of a ...
1 vote

Does the vacuum really have infinite energy density?

In reality the vacuum energy is zero. It must be so to be in line with the measured flat geometry of space-time of the universe. Dark Energy can be explained by the cosmological constant, which is ...
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Definition of symmetry factor $p$ in Feynmans diagrams symmetry factor in Coleman's "Introduction to Many-Body Physics"

For a Feynman diagram with $n$ vertices, Coleman's notation for a permutation $\pi\in S_n$ is the image $(\pi(1) \pi(2)\ldots \pi(n))$; the notation should not be read as cycle notation for a cyclic ...
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3 votes

Ultraviolet Power Counting in the Four Photon Vertex

The same argument is made in Ref. 1: On p. 319 it is argued that when we Taylor expand an amplitude in external momenta, the $n$th Taylor coefficient has its superficial degree of divergence (SDOD) ...
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Wick's Theorem Example from Stoof & Gubbels

It is allowed, it is just that $\langle\phi^*\phi^*\rangle=0$!
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2 votes
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Peskin and Schroeder's QFT eq. (7.88)

In $d=2N$, and in $d=2N+1$, dimensions the trace of the gamma-matrix space identity matrix is ${\rm tr}({\mathbb I}_{\rm spinors})$ is $2^N$.
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1 vote
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Inonu-Wigner contraction in Weinberg Volume I

This is necessary since $$[\underbrace{K_i}_{{\cal O}(v^{-1})},\underbrace{P_j}_{{\cal O}(mv)}]=i\underbrace{M}_{{\cal O}(m)}\delta_{ij}.$$
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Why does the Dirac Lagrangian not already use operators (instead of canonical quantization)?

Note that the Lagrangian density, the Hamiltonian density and the equations of motion all can be deduced from each other via the Legrende transformation for fields and the Euler - Lagrange equation ...
3 votes

Why standard model uses Lie groups like $SU(2)$ and not $SL(2,\mathbb{C})$?

A gauge transformation acts on the wave function and does not transform it in space and time but in an abstract internal space. The corresponding gauge covariant derivative has Lorentz indices that ...
1 vote

Why does the Dirac Lagrangian not already use operators (instead of canonical quantization)?

In quantization the starting point is a classical theory. On one hand, in a classical Hamiltonian formulation, there is a (super) Poisson bracket $\{\cdot,\cdot\}$, which we can formally replace with ...
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1 vote

Continuum limit of lattice field theory

This isn't really specific to lattice theory - doing stuff in "imaginary time" means you're doing Euclidean QFT. This isn't so much "relativistic" or "non-relativistic" ...
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3 votes

Propagators for a generic Lagrangian density

Formally speaking, if the Hessian matrix of the quadratic action is non-degenerate, the free propagator is given by the inverse matrix, cf. e.g. my related Phys.SE answer here. Concerning OP's last ...
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