# Can you gauge a $U(1)_L$ symmetry?

I was recently calculating the one loop correction for the propagator of a gauge boson,

$$\hspace{5cm}$$ I assumed arbitrary left and right couplings, $$g _L$$ and $$g _R$$. I found that the one loop correction was, $$\begin{equation} = \frac{ - 4i }{ ( 4\pi ) ^{ d / 2}} \int dx \frac{ \Gamma ( 2 - d / 2 ) }{ \Delta ^{ 2 - d /2 }} \left[ ( g _L ^2 + g _R ^2 ) x ( 1 - x ) \left( g _{ \mu \nu } - \frac{ p _\mu p _\nu }{ p ^2 } \right) + g _{ \mu \nu } ( g _L ^2 - g _R ^2 ) m ^2 \right] \end{equation}$$ Now we also know that for a $$U(1)$$ invariant theory we should have the Ward identity and hence this one loop correction would be of the form, $$\begin{equation} \Pi _{ \mu \nu } = \Pi ( p ^2 ) \left( g _{ \mu \nu } - \frac{ p _\mu p _\nu }{ p ^2 } \right) \end{equation}$$ So it appears that in order for this propagator to arise from a $$U(1)$$ invariant theory, we must have $$g _L = g _R$$ (unless I made a mistake). I found this very strange. Is it not possible to gauge a $$U(1) _L$$ symmetry and if so why?

As a working example I invented the theory below which seems like it could be gauged: $$\begin{equation} {\cal L} = i \psi ^\dagger \bar{\sigma} ^\mu \partial _\mu \psi + i \chi \sigma ^\mu \partial _\mu \chi + \phi ^0 \chi \chi + \phi ^{+ + } \psi \psi + \phi \mbox{ terms} \end{equation}$$ where $$\psi$$ is a left chiral spinor and $$\chi$$ is a right chiral spinor. The particles transform under the symmetry as, $$\begin{equation} \psi \rightarrow e ^{ i \alpha } \quad \chi \rightarrow \chi \quad \phi ^0 \rightarrow \phi ^0 \quad \phi ^{ + + } \rightarrow e ^{ - 2i \alpha } \phi ^{ + + } \end{equation}$$ This seems like a perfectly fine theory, whose symmetry can be gauged but if my conclusion above is correct for some reason I shouldn't be able to gauge this $$U(1 ) _L$$ symmetry. Did I make a mistake or is there some deep reason this can't be done?

• A quick answer: Read about the axial anomaly. It is better to think of the vector and axial combinations: $g_{V,A} = (g_L \pm g_R)/2$. – suresh Aug 26 '14 at 0:14
• @suresh: Thanks, that's a neat answer! I knew about the basics behind the axial anomaly but I didn't think of making the connection. – JeffDror Aug 26 '14 at 10:22
• I guessed as much and knew that my brief comment should suffice :-) I strongly recommend reading Jackiw's Les Houches lectures which are reprinted in a book titled "Current Algebras, Anomalies" amazon.com/Current-Algebra-And-Anomalies-Treiman/dp/9971966972 – suresh Aug 27 '14 at 0:10
• JeffDror, I think it's not good to have this question unanswered when you've evidently found the answer. Would you care to post the answer as an answer here? – ACuriousMind May 8 '15 at 19:59

The issue here is actually the validity of the model. In order for a theory to have massive fermions, the mass term needs to be invariant (we assume gauge symmetry is a good symmetry here, at least up to loop corrections). Gauge symmetry forbids Majorana mass terms so the only masses you can write down are Dirac masses. However, Dirac fermions must have $g _L = g _R$ to preserve gauge symmetry. If $g _L = g _R$ the problematic term above drops out and we are left with a gauge invariant correction to the propagator. Conversely if the fermion is massless but $g_L\neq g _R$ the problematic term still drops out, keeping the propagator gauge invariant.
There is one subtlety to the above. Here I computed the $2$-point function. It turns out the lowest order potentially problematic diagram with chiral gauge theories are the $3$-point functions, or the "triangle diagrams". These generically lead to violations of gauge invariance for chiral theories (where $g_L\neq g_R$), unless the gauge charges are aptly chosen.