4
$\begingroup$

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 $A^\mu\rightarrow A^\mu+\partial^\mu\alpha(x)$.

$\endgroup$
3
  • $\begingroup$ 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

1
$\begingroup$

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.

$\endgroup$
2
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.