# Bosonization and gauge symmetry

The bosonization map relates the fermionic current $$\bar{\psi}\gamma\psi$$ to the bosonic current $$\partial\phi$$, and also the components of $$\psi$$ to $$e^{i\sqrt{\pi}\left(\phi\pm\bar\phi\right)}$$. Here I'm using $$\bar{\phi}$$ to denote the dual field $$\partial_\mu \phi = \epsilon_{\mu\nu}\partial^\nu\bar\phi$$.

Now I was under the impression that global $$U(1)$$ vector symmetry $$\psi \rightarrow e^{i\theta}\psi$$ is associated with shift symmetry $$\phi\rightarrow \phi+\frac{1}{\sqrt{\pi}}\theta$$, and similarly for axial symmetry and shifts in $$\bar{\phi}$$.

But we can extend this global symmetry to a gauge symmetry $$\theta(x)$$. Now the current $$\bar{\psi}\gamma\psi$$ is gauge invariant, but $$\partial\phi\rightarrow \partial\phi+\frac{1}{\sqrt{\pi}}\partial\theta$$ is not!

Does gauge symmetry make sense in bosonization?

I'm particularly confused since I've seen use of bosonization of a fermion coupled to a gauge field to explain things like the chiral anomaly and lack of dependence on the vacuum angle. (see e.g. Witten Nucl Phys B149, 285)

For instance suppose we have a gauge invariant Lagrangian like $$\bar{\psi}i(\displaystyle{\not}\partial+i\displaystyle{\not}A)\psi -\frac{1}{4}F^2$$ and after naive bosonization we get something like (forgive me for ignoring the constant factors) $$\frac{1}{2}(\partial{\phi})^2+\partial_\mu\phi A^\mu -\frac{1}{4}F^2$$ but this is missing the $$\frac{1}{2}A^2$$ term that would be needed for gauge invariance!

• I'm guessing that the current used in the bosonization map in the presence of a gauge field becomes $\partial \phi - \frac{1}{\sqrt{\pi}}A$, but it would be nice to see this justified somehow, and understand how other quantities are affected – octonion Sep 16 at 17:22
• @ACuriousMind, Thanks again for your answer. You might be interested that what I was claiming about $\phi$ being like a Stueckelberg field and also the form of the current above is correct. It is implicitly in a paper by Kazuo Fujikawa et al (arXiv:hep-th/0305008). My $\phi$ is his $\alpha$ and my $\bar\phi$ is his $\xi$ – octonion Sep 18 at 19:36
• @ACuriousMind, Strictly speaking that means your answer was not correct, by the way, but I found it useful as a starting point. The gauge symmetry is there for $\phi$ but not $\bar\phi$ (we were never gauging chiral symmetry in the first place) – octonion Sep 18 at 19:45

The resolution to how this theory is still gauge-invariant lies in rewriting the Lagrangian and assuming nice boundary conditions. The action has a term $$\int \partial_\mu\phi A^\mu = \int \mathrm{d}\bar{\phi}\wedge A$$, which becomes $$\int \bar{\phi}\wedge\mathrm{d}A = \int \bar{\phi} \wedge F$$ upon integration by parts and vanishing of the boundary term. So the equivalent Lagrangian is $$\frac{1}{2}(\partial\bar{\phi})^2 + \frac{1}{2}\bar{\phi} \epsilon^{\mu\nu}F_{\mu\nu} - \frac{1}{4}F^2$$ which is manifestly gauge-invariant under gauge transformations of $$A$$ without any need for $$\phi$$ to transform at all.
• I don't see it. What does $F$ mean, the contraction with a Levi-Civita? Did you switch to the dual field $\bar{\phi}$? – octonion Sep 16 at 17:29
• @octonion Yes, since we're in two dimensions, $\partial_\mu A^\mu$ is the value of the dual of the field strength, which I lazily also called $F$ but more properly could be written ${\star}F$. – ACuriousMind Sep 16 at 17:31
• I think you need a Levi-Civita symbol to appear for this to work. This would be so if I bosonized using the dual field $\bar{\phi}$ instead of $\phi$. In fact this is what Witten did in that paper I cited. – octonion Sep 16 at 17:44
• I'm a little skeptical that $\phi$ does not have gauge symmetry. I think $\phi$ is like a Stueckelberg field to make the vector field massive. If I'm not mistaken, a massive vector field in 2D is the same as a massive boson, and this is exactly what 2D QED (the Schwinger model) reduces to. – octonion Sep 16 at 17:51