How are the gauge transformations of $\epsilon(\mu)$ and $A^\mu$ related?

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 $$$$\epsilon_\pm^\mu(k) e^{\pm i\theta} = R(\theta)^\mu{}_\nu \epsilon_\pm^\nu(k)$$$$ which tells us that $$e_\pm = \frac{1}{\sqrt{2}}(0, 1, \pm i, 0)$$. The other little group transformation, $$S(\alpha, \beta)$$ acting on the polarizations tells us $$$$\epsilon_\pm^\mu(k) = S(\alpha, \beta)^\mu{}_\nu \epsilon_\pm^\nu(k)$$$$ where the left hand side is left unchanged since the single particle states do not transform under $$S(\alpha, \beta)$$. This equation cannot be solved for general Lorentz transformation, since it implies $$$$\epsilon_\pm^\mu(k) = \epsilon_\pm^\mu(k) + \frac{(\alpha \pm i\beta)}{\sqrt{2k}} k^\mu$$$$ We find that these polarization vectors are not Lorentz invariant -they do not remain transverse under a Lorentz transformation but pick up a piece proportional to their momentum, Therefore, we have not yet succeeded in finding a local field description of massless spin-1 particles that is Lorentz invariant. The fix is to identify $$\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)$$.

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

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.
• Thank you! There should be one $\alpha(k)$ on the second line, right?