How to derive Goldstone-Wilczek current?

Question

I am reading this Goldstone-Wilczek celebrated's paper on fractional quantum number. In particular they derived for the following Dirac Lagrangian ($\phi_1$ and $\phi_2$ are scalar fields)

$$\mathscr{L}=i\bar{\psi}\gamma^\mu\partial_\mu\psi+g\bar{\psi}(\phi_1+i\gamma_5\phi_2)\psi$$

that the expectation value of current reads ($\theta=\tan^{-1}(\phi_2/\phi_1)$)

$$\langle j^\mu\rangle=\frac{1}{2\pi}\epsilon^{\mu\nu}\partial_\nu\theta$$,

which is equation (2). It is possible to understand it via dimensional reduction or bosonization. But I would like to understand it from a field-theoretic way. In particular I have two questions:

1. How to get the Feynman diagram in Fig.3? (I am aware the similar question in here but it does not answer my question, and the Feynman diagram below is from that thread) (For clarity, curly line is current, solid line is fermion and dash-line is scalar)
2. How to calculate this Feynman diagram?

I have look on the internet and quite a few references and could not figure it out. Any helps are appreciated.

Answer 1

It is interesting to view your question above not only in light of the Goldstone-Wilczek (G-W) approach (G-W has provided a method for computing the fermion charge induced by a classical profile), but also by computing $1/2$-fermion charge found by Jackiw-Rebbi using G-W method. For simplicity, let us consider the 1+1D case, and let us consider the $Z_2$ domain wall and the $1/2$-charge found by Jackiw-Rebbi. The construction, valid for 1+1D systems, works as follows.

Consider a Lagrangian describing spinless fermions $\psi(x)$ coupled to a classical background profile $\lambda(x)$ via a term $\lambda\,\psi^{\dagger}\sigma_{3} \psi$. In the high temperature phase, the v.e.v. of $\lambda$ is zero and no mass is generated for the fermions. In the low temperature phase, the $\lambda$ acquires two degenerate vacuum values $\pm \langle \lambda \rangle$ that are related by a ${Z}_2$ symmetry. Generically we have $$ \langle \lambda \rangle\,\cos\big( \phi(x) - \theta(x) \big), $$ where we use the bosonization dictionary $ \psi^{\dagger}\sigma_{3} \psi \rightarrow \cos(\phi(x)) $ and a phase change $\Delta\theta = \pi$ captures the existence of a domain wall separating regions with opposite values of the v.e.v. of $\lambda$. From the fact that the fermion density $$ \rho(x) = \psi^{\dagger}(x)\psi(x) = \frac{1}{2\pi} \partial_{x} \phi(x) ,$$ and the current $$J^\mu=\psi^{\dagger}\gamma^\mu\psi =\frac{1}{2\pi}\epsilon^{\mu \nu} \partial_\nu \phi,$$ it follows that the induced charge $Q_{\text{dw}}$ on the kink by a domain wall is $$ Q_{\text{dw}} = \int^{x_0 + \varepsilon}_{x_0 - \varepsilon}\,dx\,\rho(x) = \int^{x_0 + \varepsilon}_{x_0 - \varepsilon}\,dx\,\frac{1}{2\pi} \partial_{x} \phi(x) = \frac{1}{2\pi} \pi = \frac{1}{2}, $$ where $x_0$ denotes the center of the domain wall.

You can try to extend to other dimensions, but then you may need to be careful and you may not be able to use the bosonization.

See more details of the derivation here in the page 13 of this paper.

Answer 2

Check this article, https://arxiv.org/pdf/1306.0998.pdf and the "Introduction to Bosonization" lecture notes from D. Sénéchal I think it will also help.