New answers tagged

1

I am the first author of the Wikipedia page of Matsubara frequency and the maker of the summation table therein. The result of your summation is $$\frac{1}{2} \eta \left(-\frac{\left(\xi _1-\xi _2+\xi _3\right) n_{\eta }\left(\xi _2-\xi _3\right)}{\xi _3 \left(\xi _2^2-2 \left(\xi _3+\xi _4\right) \xi _2+\xi _3^2+\xi _4^2-\xi _5^2+2 \xi _3 \xi ...


0

I googled the LSZ reduction formula, but many of the are almost the same as Srednicki's except this one http://porthos.ist.utl.pt/~romao/homepage/publications/Lectures/Lectures-TCA-c2.pdf by Jorge Romao. He didn't insert a T directly in the expression of S matrix element like (5.13), so his final result contains a "disconnected term", plus (5.15). He stated ...


0

I have the same question as you. I do not know why time ordering operator is so powerful. It's simple from (5.13) to (5.14), if we regard T acts according to the t labelling a and a dag, that is a() and adag(). But I don't know why this T can also respond to the t labelling phi() which appears in (5.15)


0

Note that under an infinitesimal change in the metric of the form $g \to g + \delta g$ the action changes to $$ \delta S = \int T^{ab} \delta g_{ab} $$ Now, under Weyl transformations we have $$ g_{ab} \to e^{2\omega} g_{ab} \qquad \implies \qquad \delta g_{ab} = 2 \omega g_{ab} $$ For Weyl transformations $\omega$ is completely arbitrary. If we consider a ...


0

There is a general expression for the matrix representation of $N$ Grassmann numbers (called the Clifford-Jordan-Wigner representation), and it is intimately related to the matrix representation of the Euclidean $\gamma$-matrices in $D=2N$ dimensions ($\gamma$-matrices in $D=2N+1$ and Lorentzian $\gamma$-matrices are simple to find from these). I will simply ...


0

In the following paper Professor Pierre-Marie Robitaille has argued that thermal emission is due to vibrations of nuclei within the lattice of a material, and hence also of a blackbody: Robitaille, P.M. On the validity of Kirchhoff’s Law of thermal emission. IEEE Trans. Plasma Sci., 2003, v. 31, no. 6, 1263–1267.


4

It's a good question and I don't think there is a simple way of seeing it from the action. I just checked Fradkin (section 7.7 and 7.8) and through an RG analysis he shows that any half-integer spin behaves the same as the spin-$\frac{1}{2}$ case. But for the latter he then refers to the exact Bethe ansatz solution to show it is gapless! However, I wanted ...


4

Ok so, let's start by defining the photon in the context of quantum theory. A photon is one of the quantum states of definite energy and momentum of the theory of Quantum Electrodynamics. In QED without fermions, a photon of momentum $k$ and spin $s$ is defined a the state $$ |\gamma, \vec{k},s\rangle=a^{\dagger}(\vec{k},s)\,|0\rangle $$ Where ...


2

I'm afraid I can't give you an answer based on Landau's book, as I haven't read it. If you are referring to Landau's Theoretical Physics, Volume 4, I have it in my library so feel free to ask for more specific information in the comments. I advice you to have a look at other references too, especially at an elementary level. Anyways, the photon's ...


0

Hey I had the same problem, but the key is that you have to take care about the following thing, if you write $$ \mathcal{M}^\dagger = v^\dagger \gamma^5 (\gamma^0)^\dagger u = - v^\dagger \gamma^0 \gamma^5 u = -\bar{v} \gamma^5 u $$ so taking the conjugate of a spinor don't give you automatically the "barred version" The same actually happens when you take ...


2

Suppose first that you deal with free field theories. You start from canonical commutation or anticommutation relations for coordinate $q(x, t)$ and momentum $p(x, t)$: $$ \tag 0 [q_{i}(\mathbf x , t), p_{j}(\mathbf y , t)]_{\pm} = i\delta (\mathbf x - \mathbf y)\delta_{ij}, \quad [q_{i}(\mathbf x , t),q_{j}(\mathbf y , t)]_{\pm} = [p_{i}(\mathbf x , t), ...


6

How to see it in canonical quantization: All operators $\mathcal{O}$ in a quantum theory fulfill the Heisenberg equations of motion $$ \frac{\mathrm{d}}{\mathrm{d}t}\mathcal{O}(t) = \mathrm{i}[H,\mathcal{O}(t)]$$ where $H$ is the Hamiltonian density and which is exactly the quantum version of the classical Hamiltonian equations of motion. So the fields, as ...


-1

I can't comment on Hobson; I haven't read it. But I can comment on the ideas that you posted. This is not an easy subject to unravel. Without looking too hard you'll find seemingly endless discussions. In my opinion, it boils down to what you mean by the words "particle". If your picture of a particle is a little bit of "something" that has an ...


0

how can fields explain why, when you watch which slit the "particle" goes through, does the interference pattern disappear? a) Quantum field theory is a different mathematical tool and gives the same calculations as with simple first quantization calculations except it is just extremely more efficient in setting up solutions for complicated boundary ...


-3

After reading Art Hobson's article titled, "There are no particles, there are only fields" published in The American Journal of Physics in 2013, I'm wondering what other experts think of his main thesis I think there's some merit in it, but there's plenty wrong with it too. See his paper on the arXiv where you can read this: the Schroedinger field is a ...


2

Ok, I actually found the answer. As the question was up-voted, I'm going to write it down. I would argue as follows. Let's say that we want to define the cross section at an arbitrary renormalization scale. Then we put $$ \sigma=\sigma(s,M,a_{s}(M)) $$ as these are the only variables on which the cross section can explicitly depend (of course we are ...


1

Thanks for your help. I agree this was homework-like - should've posted it somewhere else. This is an equality $$\int^\infty_{-\infty}f(x)dx=\int^\infty_{-\infty}f(-x)dx$$ so I guess mathematically it doesn't matter the sign of your substitution.


2

The $a,a^\dagger$ act on the Fock space. If you write a random $\delta(\vec x - \vec y)$ is it neither an element of a Fock space nor an operator on it - the equation doesn't really make sense without further context. However, $\delta$ is just a distribution on functions of spacetime and not operator-valued itself, so the meaning of $a(x)\delta(x-y)$ is ...


2

First let us address "emty space". Empty space is a theoretical concept, a space where there is no matter and no energy. In our universe, no matter how far away one goes in space, it is not empty. It contains the cosmic microwave background radiation, cool photons, which is at a temperature of 2.7 K . Within quantum mechanics and elementary paricles, the ...


0

The universe is not EMPTY. It is governed by dark matter (80%)!.The universe is also filled with neutrino. THE universe wouldn't then be expanding because new matter particles is being created(however they again disappear). Think it like a balloon being blown, as more particle goes in, the balloon expands. In other words, expansion is directly proportional ...


0

Such phenomena as Casimir effect (especially dynamic version of the effect) and Lamb shift are typical manifestations of the quantum fluctuations. An existence of quantum fluctuations is really intuitive fact if you use an uncertainty principle, which tells us that particles can pop out of the vacuum during a very short time interval.


1

this paper might help.$^1$ It's written pedagogically and hence is easier to read. It goes on to discuss a lot more than just color decomposition too. $^1$ Scattering Amplitudes, Henriette Elvang, Yu-tin Huang.


1

If $\mathbf{A}$ does not depend on time it follows from Maxwell's equations that, excluding a linear growth of $\mathbf{E}$ in time, that it is also time-independent. We can try to quantise $\mathbf{A}$, \begin{equation*} \mathbf{A(x)=}\sum_{j}\int d\mathbf{k}\frac{1}{\sqrt{2k}}\{a_{\mathbf{k}% j}^{\ast ...


1

Can I simplify this to pick out of the summation only the mode that matches the transition energy $\hbar \omega_{21}$? This would be because photons are emitted/absorbed entirely anyway, so the energies have to match, and the photons which do not have the right energy simply won't interact with the atom That's exactly right. Let's see it in detail. ...


1

That $g^{-1}\mathrm{d} g$ is Liealgebra-valued for a Lie group-valued function $g$ has nothing to do with the particular model or with physics, it is true for all matrix groups. Write $g(x) = \exp(k(x))$, where $k(x)$ is now Lie algebra-valued and $\exp$ is the usual power series in the case of a matrix group. Then $\partial_\mu g = \partial_\mu k\exp(k)$ by ...


0

Regularization is a small deformation of the theory in a scale that is far from your concerns (see this). By the Renormalization Group we understand that at IR scale, some UV deformations are irrelevant. The analytical continuation like $n^{-s}$ when $s$ is small affects more large $n$ values than the small $n$. This deformation is an smooth one through all ...


2

(An answer from a student also studying Weinberg's book) Basically I'll be repeating what @Robin Ekman has said in his comment, but with some clarifications. First of all, here Weinberg is talking about the states of a single massive particle. That is, $\Psi_{k,~\sigma}$ and $\Psi_{k',~\sigma'}$ are two possible momentum eigenstates of this particle, and ...


0

1) The S-matrix must be Lorentz Covariant, rather than Lorentz Invariant. That is, if $\alpha$ and $\beta$ the in and out states, they must BOTH transform as the corresponding free-particle states (free particle state $\ne$ in/out state). $S_{\alpha,\beta} = \langle \beta | \alpha \rangle = \sum c(\alpha,\alpha') c(\beta,\beta') \ S_{\alpha',\beta'} $ (1). ...


1

I don't completely understand the question. How a scalar transforms is completely dictated by conformal symmetry. The transformation law is $$K_\mu \phi(x) = \big(\Delta x_\mu + x_\mu \, x_\nu \partial_\nu -x^2 \partial_\mu \big) \phi(x)$$ or if you wish $$\delta_K \phi(x) = a^\mu K_\mu \phi(x)$$ where $a^\mu$ are the infinitesimal parameters of the ...


3

Just one comment to the higgsss answer. Formally from the Wigner theorem we have that if there exist time shift symmetry, for which the scalar product of quantum mechanical rays is conserved, $$ \tag 1 |\langle \psi (t)|\kappa (t)\rangle| = |\langle \psi{'}(t+\tau)| \kappa{'}(t+\tau) \rangle|, $$ then the symmetry transformation acts on $|\psi\rangle$ as ...


3

(1) The operator $U(\Lambda,a)$ is a unitary "rotation" in the Hilbert space corresponding to an inhomogeneous Lorentz transformation of the spacetime coordinates. When $U(\Lambda,a)=\exp(iH\tau)$, it is an operator that adjusts the clock forward by $\tau$. Conceptually this is not a physical time evolution of the system. (2) A unitary rotation $U$ in the ...


6

Yes. Ordinary quantum field theory is as wrong as Newtonian gravity for not including GR effects. That is to say, it is a perfectly fine theory inside its domain of validity, which in this case means pretty much everything below the Planck scale, just as Newtonian mechanics is valid for speed much less than the relativistic scale (the speed of light). ...


1

Even if a theory is naively (classically) scale invariant (eg: the scalar theory with $\lambda \phi^4$ interaction therm), quantum mechanically, the 4-point scattering amplitude depends on the energy of the scattering particles (as can be shown by a one-loop computation. Tree level computations are the classical approximation). Suppose the scattering ...


0

The usual definition of normal ordered product is: $$:X^\mu(z,\bar z)X^\nu(w,\bar w): = X^\mu(z,\bar z) X^\nu(w, \bar w) - \langle X^\mu(z,\bar z) X^\nu(w, \bar w) \rangle $$ As you said, this is the regular part of the OPE, since only the divergent part of two operators gives non vanishing contribution to the correlator. Of course $$\langle ...


1

Disclaimer: Renormalization is a huge subject with many facets, such as, e.g. overlapping divergences of subgraphs, regularization, renormalization group, etc. Here we will only elaborate on OP's quote from Ref. 1. Ref. 1 is considering a Feynman diagram ${\cal F}(q_1, \ldots, q_E)$ in momentum Fourier space, with external 4-momenta $(q_1, \ldots, q_E)$, ...


1

Rather tautologically, a non-perturbative effect is one that is invisible to perturbation theory. An effect is invisible to perturbation theory exactly if it is in a non-analytic part of the partition function with respect to the coupling constant $g$. Observe that perturbation theory is essentially the Taylor expansion of the partition function $Z$ (or ...


2

1) As you know, $$ \tag 1 \theta\epsilon^{\mu \nu \alpha \beta}F_{\mu \nu}F_{\alpha \beta} = \theta\partial_{\mu}K^{\mu}, $$ where $K_{\mu}$ is the so-called Chern-Simons class. The Feynman diagrams method tells us that the term $(1)$ defines the diagram which corresponds to the two-photon (or two-three-four non-abelian bosons) vertex $V_{A}$ (where the ...


0

The general theory For each irreducible representation of the Poincare group with a given helicity $\pm s$, which is realized by relativistic creation-destruction field, $$ \hat{\psi}_{l}(x) = \sum_{\sigma = \pm s}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi )^{3}2\epsilon_{p}}}\left(u_{l}(\sigma , \mathbf p)\hat{a}(\sigma, \mathbf p)e^{-ipx} + v_{l}(\sigma ...


2

1) Suppose that you have two configurations (here I've used Coulomb gauge with euclidean time $\tau$): $$ \tag 0 A_{i}(x) = \begin{cases} 0 = U^{(0)}\partial_{i}(U^{(0)})^{-1}, \quad \tau = -\infty \\ U^{(1)}\partial_{i}(U^{(1)})^{-1}, \quad \tau = \infty\end{cases} $$ Such situation describes tunneling between vacua with topological charges $0$ and $1$. ...


1

You can have a look at Jean Zinn-Justin's book "Quantum Field Theory and Critical Phenomena", which is far from being fun to read, but has all the technical details you may want.


0

OP is basically asking (v4) the following. How the equation $$ \int\! \mathrm{d}^4x~ \left(i{\cal A}(x) + i\sum_r J_r(x) \langle F^r\left[x,\Phi\right] \rangle_J\right) ~=~0, \tag{4.5} $$ or equivalently, $$ \int\! \mathrm{d}^4x~ \left(i{\cal A}(x) + i\sum_r J_r(x) F^r\left[x,\frac{\delta}{i \delta J}\right]\right)Z[J] ~=~0, \tag{4.5'} $$ ...


4

The derivation in the given reference indeed seems confused and inconsistent. The crucial error seems to me that $$ S[\phi + \epsilon\delta\phi] = S[\phi]$$ is just not true for an infinitesimal symmetry. The definition of a symmetry is that $S[\phi']=S[\phi]$ (modulo boundary terms) for the finite transformation $\phi\mapsto \phi'$. Writing this ...


0

To calculate the decay width, you will need to first compute the scattering amplitudes for all processes contributing to the decay, to your desired order in perturbation theory. To compute the width for a two-body phase space, one has to evaluate the integral, $$\Gamma = \int \frac{d\Omega_{\mathrm{cm}}}{4\pi} \frac{1}{8\pi} \left( ...


0

There is a simple proof that the cancellation is impossible (at least unless you are willing to add to the classical a term proportional to $\hbar$), I am reformulating an answer by @Qmechanic in a simpler language: The anomaly, or the measure variation which is the typical source of the anomaly, contributes a term which is independent of $\hbar$, while any ...


4

Comments to the question (v2): Traditionally, the classical action $S$ sits in the Boltzmann factor $\exp\left[\frac{i}{\hbar} S\right]$ behind an inverse power of $\hbar$ in the path integral, while the path integral measure is independent of $\hbar$. In the conventional way of counting, we say that the Jacobian $J$ from the path integral measure is a ...


2

I don't know whether the OP will be satisfied this answers the original question, but I'd like to offer some context to all this. A free particle has uniform potential, so without loss of generality $V=0$. The Schrödinger equation then simplifies to $$\dot{\psi}=\frac{i\hbar}{2m}\nabla^2\psi\quad\left(1\right)$$ so ...


-1

It's interesting you are asking this question. Just few days ago an article was published in the Quanta Magazine discussing precisely this issue. There it goes beyond what you call “the usual quantum mechanics description of entanglement”. In the new theory, the quantum entanglement (QE) is applied not just across the space but also across the time. By ...


0

"Is this causality condition equivalent to Lorentz invariance?" No. Lorentz invariance ensures that points in spacetime that are spacelike (timelike) separated in one frame stay spacelike (timelike) in all inertial frames. Causality, in this context, is the notion that an event can not have effect at any spacelike separated point in spacetime. "Now the ...


10

What you call an operator theory is usually called the Heisenberg picture of quantum mechanics. What you call a wave function theory is usually called the Schroedinger picture of quantum mechanics. It is well-known that for every quantum mechanical model, the Heisenberg picture and the Schroedinger picture are fully equivalent through a dual description, ...


0

Okay, so there's two things to understand here: quantum field theory and quantum measurement. They are not very well-integrated at present; the best work in quantum measurement is done with "vanilla" quantum mechanics; the best work with quantum field theory takes a completely different tack from this and deals with Lagrangian densities and other such crazy ...



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