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5 votes

Renormalization in quantum field theory by discretizing space (but not time)

Does this work (that is, are all the regularized integrals associated with Feynman diagrams actually finite in this scheme) or is there some problem I'm not seeing? At least perturbatively (which is ...
pseudo-goldstone's user avatar
0 votes

In QFT when performing path integral, why don’t we divide it by the volume of Poincaré group, as what we did for gauge group?

We should consider the generating function $Z[J]$ rather that just the partition function $Z = Z[0]$, because it computes the correlation functions which contain the physical content of the theory. If ...
SolubleFish's user avatar
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2 votes
Accepted

Time ordered correlator from path integral: equation of motion?

Let $P(\Phi)$ be a polynomial in a set of field variables $\Phi= \{\phi_1(x), \phi_2(x),\ldots\}$, and consider a correlation function $$ \langle{P(\Phi)}\rangle= \frac 1 Z \int d[\Phi] P(\Phi)e^{-S[\...
mike stone's user avatar
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-1 votes

Problem with Bohr Frequency in Quantized Radiation - Matter interaction

The energy variation as per Dirac's Time-Dependent Perturbation Theory would be no other than field-as-atom's relativistic confounding of radiation gauge measurement with interference on the part of ...
Quantologist's user avatar
0 votes

Operator-state correspondence in QFT

A small enhancement to the answer above, also meant as a answer to the comment of @LoganM. In general you can do the same steps, you can do the coordinate transformation to $z,\bar{z}$, and then look ...
Qwe90909 's user avatar
4 votes

Wilsonian RG in QFT: what is the difference between renormalized and bare couplings?

A main idea of the Wilsonian renormalization group (RG) flow is, as OP writes, that the renormalized coupling constants $g$ become the UV bare coupling constants $g_0$ when the renormalization scale $\...
Qmechanic's user avatar
  • 201k
4 votes

Doppler shift of a single photon

If your photon has a wave vector $\vec k$ in the lab frame, the it has a covariant 4-vector in the lab: $$ k^{\mu} = (\omega/c, \vec k) = (k, \vec k) $$ where I have used the dispersion relation: $$ \...
JEB's user avatar
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3 votes

Doppler shift of a single photon

A single photon is a standard electromagnetic wave, just one of the minimum amplitude. The Doppler shift is the same as for electromagnetic waves of higher amplitudes.
Jerrold Franklin's user avatar
6 votes
Accepted

Are Weinberg's soft theorems relevant when making predictions about collider physics?

Gravitons have never been observed, much less graviton scattering, so the graviton soft theorem is not relevant for particle physics experiments. The graviton soft theorem has been shown to be related ...
Andrew's user avatar
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0 votes

Can Special Relativity's Effect on Proton Diameter Explain Magnetic Fields?

I don't have time to watch the video, but I'm assuming that we are dealing with two frames here: Frame 1: the protons are stationary, the electrons are in motion, and the wire is overall neutral ...
Brian Bi's user avatar
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0 votes

Confusion with Weinberg's QFT book, Volume I, Equation 2.5.3 (one-particle states as irreps of Poincare group)

Here is my answer. Feel free to correct me. Under an arbitrary proper, orthochronous Lorentz transformation, $\Lambda$, takes us from $p\to p^\prime=\Lambda p$. However, we have to consider the set of ...
Solidification's user avatar
0 votes

Confusion with Weinberg's QFT book, Volume I, Equation 2.5.3 (one-particle states as irreps of Poincare group)

Weinberg proves directly before the equations you mention that $$P^\mu U(\Lambda)\Psi_{p, \sigma} = \Lambda^\mu{_\rho}p^\rho U(\Lambda)\Psi_{p, \sigma}.$$ In particular, $U(\Lambda)\Psi_{p, \sigma}$ ...
Silly Goose's user avatar
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1 vote

Question on Majorana Path Integral

As Prahar stated in the comments, this is analogous to integration on the complex plane. More simply, if you are integrating on $\mathbb{R}^2$, you pick coordinates $(u, v)$ and construct a measure $\...
pseudo-goldstone's user avatar
2 votes

A free parameter when switching from $\phi$ to $a$

It should be fine as long as you put the right normalization in $\pi$. If one takes $$\pi(x) = i\sqrt{\frac{m_x}{2}} (a_x^\dagger - a_x)$$ then everything works out. In general, you can always add a ...
pseudo-goldstone's user avatar
2 votes
Accepted

Expanding the generating functional $W[J]$ for connected diagrams as a power series in $\hbar$

OP is essentially asking about the $\hbar$/loop-expansion for the generating functional $W_c[J]$ of connected diagrams, i.e. that the power of $\hbar$ in a diagram is given by the number of loops. ...
Qmechanic's user avatar
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1 vote
Accepted

Why are the corrections to the effective Lagrangian (Wilsonian renormalization) given by connected diagrams only?

The Wilsonian effective action is defined as $$\begin{align} \exp&\left\{-\frac{1}{\hbar}W_c[J^H,\phi_L] \right\} \cr ~:=~& \int_{\Lambda_L\leq |k|\leq \Lambda_H} \! {\cal D}\frac{\phi_H}{\...
Qmechanic's user avatar
  • 201k
2 votes

Transmitting data via quantum fields

All fields in our universe are limited to the speed of light for propagation. If you want long-range communication, you need a massless excitation. For it to be practical, you need an easy & cheap ...
niels nielsen's user avatar
0 votes

Weinberg QFT Vol 1, charge renormalization

One thing we said is wrong: When we rewrite the term $$ q_B \overline{\psi}^B A^B_\mu \gamma^\mu \psi^B = Z_1 q_R \overline{\psi}^R A^R_\mu \gamma^\mu \psi^R $$ where $A_B = \sqrt{Z_3}A_R$ and $A_R$ ...
zixuan feng's user avatar
1 vote
Accepted

Does matter in the outside universe affect Hawking radiation?

Hawking radiation emits all sorts of particles, including charged particles. Adding a big charge to the outside of the black hole would not alter, in principle, the particles emitted by the black hole,...
Níckolas Alves's user avatar
0 votes

In which systems does Planck's constant apply? Is everything thus quantized?

Most physical interactions we know of are currently described by quantum mechanical theories. At present, there are multiple candidates for a quantum theory of space and time, which would also be a ...
alanf's user avatar
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1 vote

Does the density matrix still holds for infinite subsytems?

I think the issue here is the assumption here is the assumption that there is always a well defined maximally mixed state. If we take the quantum states in the computational basis as a binary ...
By Symmetry's user avatar
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1 vote
Accepted

In which systems does Planck's constant apply? Is everything thus quantized?

Of course, at the classical or macroscopic level, not all things are quantized, e.g. the energy of a ball rolling downhill or the amount of soup a person can eat in the morning. However, at the ...
Albertus Magnus's user avatar
4 votes

What, specifically is meant by "particles are popping in and out of existence all the time?"

For a less technical answer, things like vacuum fluctuations and virtual particles and so on are purely artefacts of a mathematical trick we use to simplify the problem. When we want to know what ...
Nullius in Verba's user avatar
4 votes

What, specifically is meant by "particles are popping in and out of existence all the time?"

Thank you for this question. I've been waiting for an opportunity to address this issue. It is an important valid question. The short answer is: there are no particles popping in and out of existence. ...
flippiefanus's user avatar
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0 votes

How to obtain the explicit form of Green's function of the Klein-Gordon equation?

I was interested in this problem, specifically in computing the retarded Green's function, for which only the $\tau^2>0$ sector survives. One way you can do this is by first breaking down the ...
Sara R.'s user avatar
  • 31
1 vote

Why reasonable observables are made of an even number of fermion fields?

One explanation is that on one hand an observable should be a quantity/operator that could take a value in a measurement in the form of (a tuple of) ordinary numbers. On the other hand (an odd product ...
Qmechanic's user avatar
  • 201k
1 vote

If "borrowing energy for a short time" interpretation of HUP is wrong, then how are the ranges of fundamental forces explained?

I think the reason for resistance to the notion of "borrowing energy" is that in the Feynman method the energy and momentum is strictly conserved at every stage: in every vertex and in every ...
Andrew Steane's user avatar
1 vote
Accepted

Charge renormalization choice in QED

This is common practice whenever dimensional regularization is used. Simply consider the dimensions of the various terms in $d$ space-time dimensions. As the action $S= \int d^d x \, \mathcal{L}$ is ...
Hyperon's user avatar
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3 votes

How can a QFT field act on particle states in Fock space?

The two sentences : Excitations of the quantum fields are particles and The quantum fields acts on the multi-particles states in the Fock space both encapsulate the same idea because excitations ...
SolubleFish's user avatar
  • 5,538
2 votes

How can a QFT field act on particle states in Fock space?

In both quantum mechanics (QM) and quantum field theory (QFT) Heisenberg picture observables give us information about the evolution of measurable quantities. Observables don't directly represent the ...
alanf's user avatar
  • 7,293
2 votes
Accepted

Making sense of stationary phase method for the path integral

You want to compute the generating functional $$Z[J]=\int [d\phi] e^{\frac{i}{\hbar} (S[\phi]+\phi \cdot J)}, \quad Z[0]=1, \quad \phi\cdot J:=\int\! dx \, \phi(x) J(x)$$ in the quasiclassical ...
Hyperon's user avatar
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1 vote

Making sense of stationary phase method for the path integral

Your understanding seems fine (if we can deal with these questions it will be perfect). To obtain (2) the picture I was taught is to see $\phi$ as a big vector with values in each point of space-time. ...
Jos Bergervoet's user avatar
7 votes

How can a QFT field act on particle states in Fock space?

You had encountered exactly the same situation already in elementary quantum mechanics, but maybe you did not recognize the analogies. Consider a single harmonic oscillator with angular frequency $\...
Hyperon's user avatar
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5 votes

How can a QFT field act on particle states in Fock space?

The field operators do not represent particles, they are operators that, if acting on a state (a vector in fock space which represents a state with or without particles), can change the state of the ...
Wolphram jonny's user avatar
7 votes

Feynman propagator as a sum over eigenfunctions

We start from the defining equation $$ (m^2+ \square_x -i \epsilon) G(x-x^\prime)=\delta^{(d)}(x-x^\prime) \tag{1} \label{1}$$ of the Green function of the Klein-Gordon operator in $d$ dimensions with ...
Hyperon's user avatar
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2 votes

Analytical continuation as regularization in Quantum Field Theory, the remaining questions

For your first question, analytical continuation is one way out of many ways to regulate a sum. As you pointed out, there is always an ambiguity in attributing a sum to a diverging series. Analytical ...
LPZ's user avatar
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0 votes
Accepted

Quantum effective action for Yang-Mills theories

Your action seems identical to the effective action in section 2.4.2 of David Tong's notes on gauge theory (modulo the sign in front of $[F_{\mu \nu}, \cdotp]$). It can be obtained using the ...
1 vote

Parity operator action on quantized Dirac field

I don't have my copy of the book to hand, but I assume that $P$ is there a linear operator on the Hilbert space. It does not commute with the creation and annihilation operators: $$ P\hat a_{\bf k} ...
mike stone's user avatar
  • 52.7k
0 votes

The contradiction between Gell-mann Low theorem and the identity of Møller operator $H\Omega_{+}=\Omega_{+}H_0$

Actually, the original Gell-Mann-Low Theorem(GLT) can be deduced by the Adiabatic Theorem(AT), which requires no degeneracy and discrete spectrum. However, in more general cases, say, degenerate ...
Sakana's user avatar
  • 21
1 vote

Is conservation of energy a local law in Quantum field theory?

Yes the stress energy tensor operator, also in curved spacetime, satisfies the local conservation equation in a distributional sense $$\nabla_a :\hat{T}^{ab}:_{\omega}(x)=0\:,\tag{1}$$ where the &...
Valter Moretti's user avatar
6 votes

Interpreting generating functional as sum of all diagrams

Perturbatively, one can formally argue that the partition function/path integral/functional integral/generating functional can be written as $$\begin{align} Z[J]~=~&\int {\cal D}\frac{\phi}{\sqrt{\...
Qmechanic's user avatar
  • 201k
3 votes
Accepted

Amputated connected 2-point function is inverse to connected 2-point function

TL;DR: Use the relations $$D_2~=~\int E_1A_2E_2\tag{1}$$ and $$E_1~=~D_2~=~E_2\tag{2}$$ to conclude OP's sought-for relation $$A_2~=~D_2^{-1}.\tag{3}$$ Eq. (2) can be argued in at least 2 ways: ...
Qmechanic's user avatar
  • 201k
7 votes
Accepted

Interpreting generating functional as sum of all diagrams

The generating functional by definition is an object that encodes the Green's functions in a series as $$Z[J]=\sum_{n=0}^\infty \dfrac{1}{n!}\int dx_1\cdots dx_n J(x_1)\cdots J(x_n) \langle \phi(x_1)\...
Gold's user avatar
  • 35.8k
0 votes

Are vacuum energy, zero point energy and vacuum fluctuations the same thing?

Vacuum energy and zero-point energy are the same thing. Vacuum fluctuations, however, do not exist. The vacuum state is completely stationary in time. The concept of vacuum fluctuations is from the ...
Jos Bergervoet's user avatar
2 votes

Are vacuum energy, zero point energy and vacuum fluctuations the same thing?

Terminology-wise, vacuum energy and zero-point energy are the same thing, though they are used in slightly different contexts. Zero-point energy applies to the residual energy of any quantum ...
user34722's user avatar
  • 1,753
0 votes

How does parity act on relativistic one-particle states?

I think your problem depends on whether the particle is massless or not, that is, the standard momenta associated to massless particle and massive particle are different. And $\sigma$ plays different ...
Ting-Kai Hsu's user avatar
1 vote
Accepted

Free acoustic phonon propagator

The reason why you get a weird pole at $\omega = 0$ in your first attempt to find the propagator of a free acoustic phonon is that the transversal oscillations which are always part of the ...
Gheorghita Victor-Basarab's user avatar
1 vote

Why does $S$-matrix theory end up being a covariant formalism when it is not obvious that it is?

The time ordered "$T$" product is formally Lorentz covariant because the fields commute at spacelike separation --- so "$t$" is in no way being singled out. I say "formally&...
mike stone's user avatar
  • 52.7k
1 vote

What particles are described by the Klein-Gordon Equation?

The Klein-Gordon equation is the field version of the special relativistic energy momentum relation. When one abandons the restriction that it should only be used for scalar fields it becomes clear ...
my2cts's user avatar
  • 23.9k
4 votes
Accepted

What particles are described by the Klein-Gordon Equation?

A real (hermitean) Klein-Gordon field describes chargeless spin $0$ bosons. Examples: $\pi^0$, Higgs. A complex (non-hermitean) Klein-Gordon field describes charged spin $0$ bosons, where the "...
Hyperon's user avatar
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