# Tag Info

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Here is a simple Feynman diagram. electron electron elastic scattering if time is the y axis,( electron positron elastics scattering if time is the x axis) Real particles are the incoming and the outgoing that can be measured in an experiment in the lab. The exchanged photon is called virtual. The difference between real particles and virtual ...

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The reason for many contradictory statements regarding the nature of virtual particles is that they are often invoked for heuristical explanations of phenomena that arise within the framework of quantum field theory. One then tries to further justify those explanations by attributing certain properties to virtual particles they do not actually possess. ...

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As showed by Dirac and others QFT is a disjoint theory of RQM; as a consequence, QFT cannot solve the problems of RQM. You must want to revise the section "8.3 Does QFT solve the problems of relativistic QM?" of the Phys. Found. paper Quantum mechanics: Myths and facts. This is the conclusion: Thus, instead of saying that QFT solves the problems of ...

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The isospin $I_3$ of the antineutron is +1/2.

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I'll try to point out things I'm seem that are not fine. You are making confusion between $k$ and $\kappa$. There is no problem with your convention for fourier transform. It's only a little unconventional in this kind of problem (based on the books I used) You have to integrate! Note that the integral you found has $d^3{\bf k} dk'$ you must solve it. The ...

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The only dimension-five operators allowed by the SM are neutrino masses, $(HL_i)(HL_j)$. So we mostly talk about dimension-six operators because for almost any question they're the first higher-dimension operators that can appear.

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Threshold corrections is a term that appears when you discuss effective field theories (EFTs). An EFT is an approximation of a full theory which is valid at low energies, ie below some threshold. Let $A_{\mbox{eff}}$ be any amplitude as calculated in the EFT and $A_{\mbox{full}}$ the amplitude for the same process calculated in the full theory. The ...

0

From Goldstone theorem we know that $\langle0|J^\mu|\pi\rangle$ isn't zero, that's all.

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The logs that you get at one loop are scheme independent pretty much like the beta function at one loop is. There is a nice and neat discussion about it on the second volume of Weinberg.

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The consequence if the vacuum (i.e. the VEV of the scalar field) was not $SU(2)_L\times U(1)_Y$ charged is that it will not break this symmetry and also the consequence if the vacuum was $SU(3)$ charged is that it will break this symmetry which is believed to be an exact symmetry in the Standard Model (SM). More precisely, for your first question, check my ...

1

It is in fact a very simple matter if you use a different parametrization of the fields. Since we care about the Goldstone bosons only, just send $\lambda\rightarrow \infty$ so that the Higgslike state decouples. Moving to the following parametrization $$\phi_i(x)=U(x)\langle \phi_i\rangle \,,\qquad U(x)=e^{i \hat{T}^a \pi^a(x)}\,,\qquad ... 1 Like you said "A_\mu some dynamical U(1) gauge field that minimally couples to \phi". It means that the covariant derivative is :$$D_\mu \phi = \partial_\mu \phi + iqA_\mu\phi$$with q the U(1) charge of the scalar field. As a consequence, if \phi is not U(1) charged you will not have the second term in the covariant derivative and hence ... 5 I understand the statement in the following way: Pions, which are pseudo-goldstone bosons of chiral symmetry breaking, are described by the introduction of a unitary matrix U(x), defined as$$U(x)=\text{exp}\left(2i\pi^a(x)T^af_\pi^{-1}\right),$$where \pi^a is the pion field, f_\pi is the pion decay constant and T^a are the generators of the ... 2 Yes, e.g. all three Mandelstam variables$$ s~:=~(p_1+p_2)^2 ~\approx~ (m_1+m_2)^2 + \frac{m_1m_2}{2} ({\bf v}_1-{\bf v}_2)^2 ~>~0, t~:=~(p_1-p_3)^2~\approx~ (m_1-m_3)^2 - \frac{m_1m_3}{2} ({\bf v}_1-{\bf v}_3)^2 ~>~0, u~:=~(p_1-p_4)^2~\approx~ (m_1-m_4)^2 - \frac{m_1m_4}{2} ({\bf v}_1-{\bf v}_4)^2 ~>~0,$$are strictly positive in ... 1 Let me answer your questions, albeit slightly indirectly. We start with a local U(1) symmetry, i.e. a gauge symmetry, for a Lagrangian describing a scalar, \phi, and a gauge boson A_\mu. You write global. A global symmetry requries no gauge bosons, because its continuous parameter \epsilon\neq\epsilon(x) commutes with derivatives \partial_\mu. The ... 0 I will present the simplest example of beta functions arising in string theory, specifically within bosonic string theory. The states transform in the 24 \otimes 24 representation of SO(24), equivalent to three irreducible representations; schematically,$$(\mathrm{traceless \;symmetric} )\otimes (\mathrm{antisymmetric}) \otimes (\mathrm{singlet})$$To ... 1 So the key point to understanding this problem is to understand that it is the modes that contain information about the physical parameters of your photons (such as the momentum or angular momentum), and quantization is just a description of excitation of those modes. For instance the canonical quantization of the plane-wave expansion which you've ... 1 This is the basic idea of brane-world approaches to string theory. I don't know their current state. 2 Unification in physics is used differently in classical physics than in the quantum regimeof elementary particles. Unifying electricity and magnetism became necessary when functional measured relations appeared which connected the motion of charges with the magnetic field and the magnetic field with the motion of charges. The Biot-Savart law and Ampere's ... 4 The term "shell" originally derives from the non-relativistic version of the answer by @JamalS. In a non-relativistic theory, a free particle satisfies the following dispersion relation$$ E = \frac{ {\bf p}^2 }{ 2m } $$For a fixed energy a particle satisfies$$ {\bf p}_x^2 + {\bf p}_y^2 + {\bf p}_z^2 = 2 m E $$In momentum space, this is precisely the ... 8 A particle is said to be on-shell if it satisfies the relativistic dispersion relation,$$E^2 = p^2 +m^2$$in units wherein c=\hbar=1. If you graph it, you obtain a parabolic surface for massive particles, and a cone for massless particles, like a photon. This is known as the mass shell, it is quite literally a shell or surface. The momentum of a real ... 1 I have found an explanation. At the end of Section 22.2, working with Euclidean path integrals, Weinberg shows that$$-\frac{1}{32\pi^2} \int d^4x \epsilon^E_{ijkl}F_{\alpha ij} F_{\beta kl} \operatorname{tr} (t_\alpha t_\beta) = n_+ - n_-$$where t_\alpha are gauge group generators and n_\pm is the number of zero modes of the gauge covariant Dirac ... 1 The fields satisfy the wave equation. We can therefore write \begin{split} \phi(x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ a({\bf p}) e^{i p \cdot x} + b^\dagger({\bf p} ) e^{- i p \cdot x} \right] \\ \phi^\dagger (x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ b({\bf p}) e^{i p \cdot x} + ... 0 Even though you think of it as a single particle -- each of it's different properties like momentum, spin, etc (corresponding to each valid quantum number) sits in a Hilbert space of their own and the possible configurations of a particle sits in a tensor product of those Hilbert spaces.$$\mathcal{H_{particle}} = \mathcal{H_{momentum}} \otimes ...

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Since the 4-momentum must be conserved, there is no angular dependence in a 2-body decay in the CM frame. But If you are asking yourself what is the phase-space of such process, you should take a look at equation 4.72 of the book I mentioned in the comments: $$\Gamma(M\to 1 + 2)=\dfrac{|\vec{p_1}|}{32 \pi^2 M^2} \int \mathrm{d}\Omega ... 0 What you wrote is a lagrangian of a free theory. All free theories have infinite-dimensional symmetry, called higher-spin symmetry. The current you gave is a standard Noether current corresponding to one of the generators of the higher-spin symmetry. Usual space-time symmetry is a subalgebra of the higher-spin algebra. As Maldacena a Zhiboedov showed in 3d, ... 4 First, to be clear on what the graph is showing: as a function of the possible mass of the Higgs, it plots the fraction of Higgs bosons that will decay via each individual channel. Before we knew the mass of the Higgs boson, a plot like this one was useful for identifying the best channels to look at to detect the Higgs in various mass ranges. For example, ... 2 The Lagrangian density reads$$ \tag{1} {\cal L}~=~\frac{1}{2}\sum_{i=1}^2\left( \partial_{\mu}\phi^i \partial^{\mu}\phi^i -m^2\phi^i\phi^i\right). $$Consider an infinitesimal transformation$$\tag{2} \delta \phi^i~=~\epsilon Y^i_{\nu_2\ldots \nu_n},$$with generators$$\tag{3} Y^1_{\nu_2\ldots \nu_n}~:=~ ...

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In the case of a particle decaying to two identical particles, this is true. There is no angle-dependence in the scattering amplitude, so the integration is indeed trivial. When calculating scattering cross sections, this is in general not true: the scattering amplitude can depend on the phase space angle.

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I guess, it's true, although I never heard of it. Let's denote $S_F^{-1}(x)=(i\gamma_\mu\partial^\mu-m)(x)$ First let's compute $[S_F^{-1}(−x)]^\dagger$. Assuming that $(\partial_\mu)^\dagger=\partial_\mu$ we get $$[S_F^{-1}(−x)]^\dagger=(-i\gamma_\mu^\dagger\partial^\mu-m)(-x)=(i\gamma_\mu^\dagger\partial^\mu-m)(x)$$ Using $\gamma^0\gamma^0=1$ and ...

0

Well, they could develop a vev, depending of the interactions you consider. Take e.g. this term in the lagrangian $L=\frac{1}{\Lambda^2}[\bar{\psi}\psi- v^3]^2$. Of course this would imply that you also break Lorentz, which you may want to avoid.

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Explanation for the electromagnetism aspect: If the vacuum carried charge under some generator, that would mean that the generator would not annihilate the vacuum. That would mean that even if such a generator corresponds to a symmetry of the theory, the vacuum however is not symmetric under that operation. Then the gauge boson corresponding to this ...

0

I think it is a general fact about grassmannian field, and this has nothing to do with Lorentz invariance or other symmetries (you can invent a lot of QFTs without this kind of symmetry, but the VEV of a fermionic operator will be always zero (in the absence of sources)). In a functional integral formulation, the VEV of a grassmannian field $\psi$ is ...

0

Well, as regards the second part of the first question, "and QCD color neutral", the if the vacuum state of any Standard Model field had a colour charge surely it would be in opposition to the Confinement property of QCD that we observe. That is, we never see colour charge in an experiment, we only ever see colour singlet (white) field states. I'm not sure ...

1

Well, the Logic is reversed compared to the one you are implicitly using. It is the direction of the Higgs'vev that's defining what we call electric charge. In practice SU(2)xU(1) is broken to a certain U(1) that we can always choose to point in a certain direction, and accordingly assign the electric charges afterwards. Since the Higgs can't carry color it ...

3

Why can't fermions have a non-zero vacuum expectation value (VEV)? Lorentz invariance. If anything other than a Lorentz scalar has a non-zero VEV, Lorentz invariance would be spontaneously broken. For example, suppose we have a Lorentz invariant term in a Lagrangian for a vector $$\mathcal{L} \supset m^2 A_\mu A^\mu.$$ Now suppose the vector obtains a ...

2

I think, you are mixing quantum electrodynamics with classical (relativistic) electrodynamics. The equation of motion you stated above is from the classical electrodynamics where particles are considered point charges. In quantum electrodynamics particles are just field excitation, so you cannot ask for their four-velocity but rather for the time evolution ...

3

It is confusing and frankly disgusting to talk about transformations of points in spacetime. This results from abusing that in Minkowski space the tangent space can be identified with spacetime itself; the result is that one confuses coordinates and points in spacetimes. This is a terrible habit and should be avoided. The proper statement is that under a ...

0

Antiquarks can be distinguished from quarks, so you only need to antisymmetrize two at a time. That's no problem, and even if you had 3 quarks it wouldn't be. Furthermore, you only need the total state to be antisymmetric. You could have antisymmetry in space, symmetry in spin and symmetry in color, and the whole thing would be antisymmetric. (Like how you ...

2

The bounds of the integral have no dependence on any of the variables, and hence we may move the differential operator into the integrand, $$\frac{\delta}{\delta \eta (z)} \int \mathrm{d}^4 y \, S_F (z-y) \eta(y) = \int \mathrm{d}^4 y \, S_F (z-y) \delta^{(4)}(z-y)$$ Evaluating the integral using the standard delta distribution identity, we obtain your ...

1

This is an interesting question, and although I don't know a rigorous answer, we can discuss some typical cases. Usually, the inverse exists, but the cases where this inverse does not exist are not necessarily pathological (sound models can have the problem that the inverse does not exist). For standard field theories (say, $\phi^4$, O(N) models, classical ...

0

Unfortunately I cannot comment to ask you for more details. What exactly do you mean 'equilibrium point at -V'? Is this the potential, $V(\phi^* \phi)$, or the VEV, $v$ ? Is it the fact that we put $$\mu^2 < 0$$ where $$V(\phi^* \phi) = \mu^2 (\phi^* \phi) + \frac{\lambda}{4}(\phi^* \phi )^2$$ that is bothering you? The Vacuum Expectation Value ...

1

Well firstly, you're trying to put two concepts together that do not usually go. Lorentz invariance is a special property of relativistic quantum mechanics. You do not have Lorentz invariance in non-relativistic Quantum Mechanics. As regards your first expression, I do not think that this is correct. For one, it would give an infinite amplitude on-shell, ...

0

I think this is where Srednicki is being sloppy, a better way to do this is to take $$\langle f|i\rangle = \langle 0|T ~a_{1'}(+\infty)a_{2'}(+\infty)a_{1}^\dagger(-\infty)a_{2}^\dagger(-\infty)~|0\rangle$$ As the definition of the transition amplitude. Remember Srednicki does not try very hard to motivate his definition of the initial and final states ...

3

Seeing as $$\langle k|k_1k_2\rangle = \langle 0| a(\mathbf{k}) a^{\dagger}(\mathbf{k_1}) a^{\dagger}(\mathbf{k_2}) |0\rangle$$ and $$a(\mathbf{k})a^{\dagger}(\mathbf{k_1}) = a^{\dagger}(\mathbf{k_1})a(\mathbf{k}) + f(\omega)\delta(\mathbf{k}-\mathbf{k_1})$$ you'll get one term that vanishes because you cannot destroy the vacuum and one term that simply ...

2

I'll answer this question by example. Some standard gauge choices are the $R_\xi$ gauge and axial gauge with propagators $$\Delta^\xi_{\mu\nu} (k) = - \frac{i }{ p^2 - i \varepsilon} \left[ g_{\mu\nu} - \left( 1 - \xi \right) \frac{ k_\mu k_\nu }{ k^2 } \right]$$  \Delta^{\text{axial}}_{\mu\nu} (k) = - \frac{i }{ p^2 - i \varepsilon} \left[ g_{\mu\nu} - ...

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If you are interested in physical applications you could also include: Bratteli-Robinson: Operator algebras and quantum statistical mechanics It is a two-volume quite complete book, mathematically minded, discussing lots of applications of operator algebras theory to several physical systems, especially arising from statistical mechanics. Haag: Local ...

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I don't think that Sasha made a mistake. I'll use the dotted/undotted notation which may clarify the possible SL(2,C) invariants. Let $\xi^{A}$, $\theta^{A}$, $\eta^{\dot{A}}$ and $\phi^{\dot{A}}$ be Weyl spinors. The Levi-Civita tensors $\epsilon_{AB}$ and $\epsilon_{\dot{A}\dot{B}}$ transform trivially under SL(2,C) so they can be used to lower indices. ...

0

let's say the trace is the expectation value. the action will be invariant so by calculating the expectation value of the action one would expect a minima on the path taken by a particle. This would be independent of time, the same physics will describe the dynamics tomorrow. hence, periodicity in the temporal direction is a way of saying that if something ...

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i will have to disagree with some of the answers posted in this question. First, this involves a matter of interpretation of the quantum formalism (and a prevailing "interpretation", the Copenhagen one) Although this interpretation (which i find unsatisfactory and non-physical) may seem prevailing (and indeed it might be), is not because it offers a better ...

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