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$$\newcommand{\slashed}[1]{#1\!\!\!/}$$ Since $\Gamma_\mu$ has a Lorentz index $\mu$, it must involve $\gamma_\mu$, $p_\mu, p'_\mu$ (or equivalently, the linear combinations $p_\mu\pm p'_\mu$) such …

Edit Circularly polarized photons have $$\textbf{S}\cdot\hat{\textbf{p}}=\pm \hbar\tag{1}$$ and it also satisfies $$\boldsymbol{\epsilon}\cdot\hat{\textbf{p}}=0\tag{2}$$ where $\textbf{S}$ is the spin, …

However, in addition to intrinsic parity coming from spin, orbital parity can arise for single-particle states if it carries an orbital angular momentum such as photons. … See the questions below What is the orbital angular momentum (OAM) of individual photons? Does orbital angular momentum has no meaning for single photons? …

Photons have zero chemical potential and their number is not conserved. … I mean is there a theoretical problem for photons, phonons and magnons to Bose condense? …

This is what happens for a Proca field. The corresponding "electric field" will not remain divergenceless i.e., $\boldsymbol{\nabla}\cdot\textbf{E}\neq 0$ but the corresonding "magnetic field" will re …

In the radiation gauge, the 3-vector potential has the most general Fourier mode expansion given by $$\textbf{A}(\textbf{x})=\int\frac{d^3\textbf{p}}{(2\pi)^3\sqrt{2E_{\textbf{p}}}}\sum\limits_{r=1}^{ …

The Fourier mode expansion of the free electromagnetic field in radiation gauge is given by $$\textbf{A}(x)=\int\frac{d^3p}{(2\pi)^3\sqrt{2\omega_\textbf{p}}}\sum\limits_{\lambda=1,2}[\boldsymbol{\eps …

(2) From the representation theory of Poincare group, it is known that for photons $\textbf{S}\cdot\hat{\textbf{p}}$ has eigenvalues $h=\pm 1$ where $\textbf{S}$ denotes the spin operator. …

It can be shown that the total angular momentum of a free electromagnetic field is given by (for example, in the book A Modern Introduction to quantum field theory by Maggiore, page 98, Eq. (4.82)) $$ …

On the other hand, the state of polarization of single photons are specified in a different manner. …

How is the the state of polarization of a beam of light measured? How is the state of polarization of a single photon measured?

Is it not possible to derive BE distribution for photons in a blackbody cavity without assuming the quantum nature of light. …

My suspiction is that even if the rotation generators for photons are $J_i$, they cannot satisfy SU(2) algebra because if it were then photons must have transformed like representation of SU(2). …

What is the underlying group and group representation which describes the transformation of photons under rotation. …

We know photons have spin s=1. … But we know that photons also carry spin angular momentum. Why we do not include the spin of the photon and write the conservation equation as $$\vec J_i=\vec J_f+\vec L+\vec S ?$$ …