To find a local field description of massless spin-1 particles that is Lorentz invariant, we can identify $\epsilon^\mu_{\pm}(k)$ with $\epsilon^\mu_{\pm}(k)+\alpha(k)k^\mu$. As $A^\mu$ and $\epsilon^\mu$ are related by $$ A^\mu(x) = \sum_{\lambda} \int \frac{d^3k}{(2\pi)^3 2\omega_k} \left[ \epsilon^\mu_{\lambda}(k) a_{\lambda}(k) e^{-ik\cdot x} + \epsilon^{\mu*}_{\lambda}(k) a^\dagger_{\lambda}(k) e^{ik\cdot x} \right] $$ Is it right to say the gauge transformation in position space gives $A^\mu\rightarrow A^\mu+\partial^\mu\alpha(x)$? I tried to plug $\epsilon(k)^\mu\rightarrow\epsilon^\mu(k)+\alpha(k)k^\mu$ in the expansion above, but the presence of $a(k)$ and $a^\dagger(k)$ prevented me from getting expressions like $$ \partial^\mu \alpha(x) = \int\frac{d^3k}{(2\pi)^3}(-ik^\mu)e^{-ikx} $$ So how are these two gauge transformations related? Thanks for the help!

Edit here's where the statement in the first sentence come from on my lecture note:

Consider the momentum $k^\mu = (k, 0, 0, k)$. There are two helicities, so we look for two polarization vectors $\epsilon_\pm(k)$. The transformation corresponding to rotation around the $z$-axis implies \begin{equation} \epsilon_\pm^\mu(k) e^{\pm i\theta} = R(\theta)^\mu{}_\nu \epsilon_\pm^\nu(k) \end{equation} which tells us that $e_\pm = \frac{1}{\sqrt{2}}(0, 1, \pm i, 0)$. The other little group transformation, $S(\alpha, \beta)$ acting on the polarizations tells us \begin{equation} \epsilon_\pm^\mu(k) = S(\alpha, \beta)^\mu{}_\nu \epsilon_\pm^\nu(k) \end{equation} where the left hand side is left unchanged since the single particle states do not transform under $S(\alpha, \beta)$. This equation cannot be solved for general Lorentz transformation, since it implies \begin{equation} \epsilon_\pm^\mu(k) = \epsilon_\pm^\mu(k) + \frac{(\alpha \pm i\beta)}{\sqrt{2k}} k^\mu \end{equation} We find that these polarization vectors are not Lorentz invariant -they do not remain transverse under a Lorentz transformation but pick up a piece proportional to their momentum, Therefore, we have not yet succeeded in finding a local field description of massless spin-1 particles that is Lorentz invariant. The fix is to identify $\epsilon^\mu_{\pm}(k)$ with $\epsilon^\mu_{\pm}(k)+\alpha(k)k^\mu$. This is gauge invariance (in position space this corresponds to the transformation for the fields $A^\mu\rightarrow A^\mu+\partial^\mu\alpha(x)$.

  • $\begingroup$ What's the reference? $\endgroup$
    – Errorbar
    Apr 7 at 1:10
  • $\begingroup$ Where does the statement in the first sentence come from? $\endgroup$ Apr 7 at 4:26
  • $\begingroup$ @flippiefanus Just updated :) $\endgroup$
    – IGY
    Apr 7 at 11:00

1 Answer 1


One issue is that you consider $A^\mu$ to be an operator, i.e. it includes creation and annihilation operators, while $\alpha (x)$ is just a function. This would imply the transformation $A^\mu \rightarrow A^\mu + \partial^\mu \alpha (x)$ is nonsensical.

One way would be to take a expectation value of $A^\mu$, or it might be easier to introduce operators into $\alpha (x)$. In any case you need to have some function of momentum in $\alpha (x)$. So if we write $$ \alpha (x) = \int {d^3 k \over (2 \pi )^3 } \alpha (k) e^{i kx}$$ then $$ \partial^\mu \alpha (x) = \int {d^3 k \over (2 \pi )^3 } i k^\mu \alpha (k) e^{i kx}$$.

Plugging this in we se that $A^\mu \rightarrow A^\mu + \partial^\mu \alpha (x)$ and $\epsilon ^\mu \rightarrow \epsilon^\mu + k^\mu \alpha (k)$ are equivilant.

  • 1
    $\begingroup$ Thank you! There should be one $\alpha(k)$ on the second line, right? $\endgroup$
    – IGY
    Apr 7 at 20:17
  • 1
    $\begingroup$ yeah, I fixed that. $\endgroup$ Apr 7 at 20:19

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