# Only transverse photons are gauge-invariant (Peskin page 298)

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?

• Welcome to Physics SE! You can add math inside Dollar brackets just like in LaTeX: $A_0$ It's called MathJax here. Others could understand your question if you included the equation. – Stefan Bischof Sep 14 '14 at 12:01
• Thank you Stefan. I have made the edits you suggested. I appreciate your help. If you have other suggestions, I will be glad to hear them. – Lifetime Beginner Sep 14 '14 at 12:20
• I'm not sure why you call components of $A_\mu$ "photons". The photons are states in the Hilbert space obtained by Gupta-Bleuler or BRST quantization, and not components of the gauge field. – ACuriousMind Sep 14 '14 at 14:16
• Thanks ACuriousMind, for taking a look at this. I looked up Gupta-Bleuler in P & S in the index, but it isn't there. There is an entry for BRST though. It is on page 517 much later than the page I'm working on (298). In the section on BRST, it refers directly back to this section though. It says that since QED is an abelian theory, BRST is not necessary. BRST is needed for non-abelian theories. – Lifetime Beginner Sep 15 '14 at 14:00
• As for $A_{\mu}$ not being a photon, you are correct. The actual electrons and photons are states. Like this $|1> = \psi(x)|0>$ for fermions, and $|1> = A_{\mu}|0>$ for photons. Then under a gauge transformation we get $e^{i\alpha(x)}\psi(x)|0> = e^{i\alpha(x)}|1>$ which is just a phase change. For photons we get $A_{\mu}|0> - \frac{1}{e}\partial_{\mu}\alpha(x)|0> = |1> - \frac{1}{e}\partial_{\mu}\alpha(x)|0>$ – Lifetime Beginner Sep 15 '14 at 14:07

Just consider the gauge transformation after Fourier transforming everything. A Fourier transform turns derivatives into momenta, such that we get $$\tilde A_\mu \rightarrow \tilde A_\mu - \frac1e k_\mu \tilde\alpha \;.$$ 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.