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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 ...


7

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. ...


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 ...


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 ...


3

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 ...


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 ...


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 ...


1

Indeed, without assuming it from first principles as in Bogoliubov formulation, the invariance property you mention holds when the interaction Lagrangian does not include derivatives of fields like in QED. In that case $${\cal H}_I = -{\cal L}_I$$ so that $$S = \sum_{n=0}^{+\infty} \frac{i^n}{n!} \int\cdots \int T \hat{\cal L}_I(x_1)\cdots \hat{\cal ...


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 ...


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 ...


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

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

The fields satisfy the wave equation. We can therefore write \begin{equation} \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} + ...


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 ...



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