Seven lines down from the top of page 298 of P & S, it says "Single particle states containing one electron, one positron, or one transversely polarized photon are gauge-invariant, while states with timelike and longitudinal photon polarizations transform under gauge motions". Here is eqn (4.6)(78)
$\psi(x) \rightarrow e^{i\alpha(x)}\psi(x), A_{\mu} \rightarrow A_{\mu} - \frac{1}{e}\partial_{\mu}\alpha(x)$
I see that in a gauge transformation, the transformation of electrons and positrons is nothing more than a phase change and so these are manifestly gauge-invariant. However, for photons, $A_1$ and $A_2$ (the transverse photons) change in just the same way as $A_0$ and $A_3$ (the timelike and longitudinal photons). What's more, they all seem to be transformed, not gauge invariant. Probably I am looking at this in the wrong way. Can someone help me to see this in the proper light?
Answer
Just consider the gauge transformation after Fourier transforming everything. A Fourier transform turns derivatives into momenta, such that we get \begin{equation} \tilde A_\mu \rightarrow \tilde A_\mu - \frac1e k_\mu \tilde\alpha \;. \end{equation} This mean that only the component parallel to $k_\mu$ (the longitudinal one) will change, while the transversely polarized components (perpendicular to $k_\mu$) are left unchanged.
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