Electrodynamics

$\mathcal{L}=-\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu };F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$

We impose the Lorentz gauge: ${\partial }_{\mu }{A}^{\mu }=0$.

Then, ${\partial }_{\mu }{F}^{\mu \nu }={\partial }_{\mu }({\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu })={\partial }^2{A}^{\mu }=0$.

Let $A_{\mu }={ϵ}_{\mu }(p)e^{ipx}$. Then, the gauge condition becomes $ϵ.p=0$ and the e.o.m becomes $p^2=0$.

There is a residual gauge symmetry that preserves the gauge condition: $A_{\mu }\to A_{\mu }+{\partial }_{\mu }\omega$, where ${\partial }^2\omega =0$.

Now, consider a plane wave solution of the form $\omega =\alpha (p)e^{ipx}$ with $p^2=0$. This corresponds to $ϵ\to ϵ+i\alpha p$, which is a gauge transformation on the polarisation vector. So, $ϵ$ and $i\alpha p$ are physically equivalent.

Since $ϵ$ satisfies $ϵ.p=0$ and an additional gauge invariance, it has two physical degrees of freedom; ${ϵ}_{\pm }$.

Let us now choose a frame in which $p^{\mu }=(E,0,0,E)$. Then,

$ϵ.p=0⟹{ϵ}^{\mu }=(0,b,c,0)+a(1,0,0,1)$.

We can choose a residual gauge transformation $\alpha =-\frac{a}{iE}$. Then,

${ϵ}^{\mu }\to {ϵ}^{\mu }-\frac{a}{E}{p}^{\mu }=(0,b,c,0)$.

Then, we can write ${ϵ}_{\pm }^{\mu }=\frac{1}{\sqrt{2}}(0,1,\pm i,0)$.

Thus, we have the following properties:

• ${ϵ}_{+}=({ϵ}_{-}{)}^{\ast }$
• ${ϵ}_{+}.{ϵ}_{-}=-1$
• ${ϵ}_{+}.{ϵ}_{+}={ϵ}_{-}.{ϵ}_{-}=0$. Since they are null vectors, they can be written in a bi-spinor form.

\epsilon_{+}^{\dot{\alpha}\alpha}=\epsilon_{+}^\mu(\bar{\sigma}_{\mu})^{\dot{\alpha}\alpha}=-\sqrt{ 2 } \frac{\tilde{\lambda}^\dot{\alpha}\mu^\alpha}{\langle\lambda \mu\rangle} and \epsilon_{-}^{\dot{\alpha}\alpha}=\epsilon_{-}^\mu(\bar{\sigma}_{\mu})^{\dot{\alpha}\alpha}=\sqrt{ 2 } \frac{\lambda^\alpha \tilde{\mu}^\dot{\alpha}}{[\tilde{\lambda}\tilde{\mu}]},

where p^{\dot{\alpha}\alpha}=\lambda^\alpha \tilde{\lambda}^\dot{\alpha} is the momentum and q^{\dot{\alpha}\alpha}=\mu^\alpha \tilde{\mu}^\dot{\alpha} is the reference momentum encoding the residual gauge symmetry.

We can also check if the above definition satisfies the properties mentioned before.

helicity $h{ϵ}_{\pm }=(\pm 1){ϵ}_{\pm }$.

The $q$ dependence of ${ϵ}_{\pm }$ is a reflection of the gauge freedom. Consider an arbitrary variation $\mu \to \mu +\delta \mu$. Without loss of generality, we can expand $\delta \mu$ in a basis of $\mu ,\lambda$ as

$\mu \to \mu +a\mu +b\lambda =(1+a)\mu +b\lambda$ for arbitrary $a\text{ and }b$.

Then, \epsilon_{+}^{\dot{\alpha}\alpha}\to-\sqrt{ 2 } \frac{\tilde{\lambda}^{\dot{\alpha}}((1+a)\mu^\alpha+b\lambda^\alpha)}{\langle\lambda,(1+a)\mu+b\lambda\rangle}=-\sqrt{ 2 } \frac{(1+a)\tilde{\lambda}^\dot{\alpha}\mu^\alpha+b\lambda^\alpha \tilde{\lambda}^\dot{\alpha}}{(1+a)\langle\lambda\mu\rangle}

=-\sqrt{ 2 } \frac{\tilde{\lambda}^\dot{\alpha}\lambda^\alpha}{\langle\lambda \mu\rangle}- \frac{\sqrt{ 2 }b}{(1+a)} \frac{\lambda^\alpha \tilde{\lambda}^\dot{\alpha}}{\langle\lambda \mu\rangle}=\epsilon_{+}^{\dot{\alpha}\alpha}- \frac{\sqrt{ 2 }b}{(1+a)\langle\lambda \mu\rangle} p^{\dot{\alpha}\alpha}

where the second term acts as a gauge transformation term.