Consider an abelian gauge field coupled with a complex field: $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+(D\varphi)^\dagger D\varphi+\mu^2 \varphi^\dagger\varphi-\lambda(\varphi^\dagger\varphi)^2.$$ In polar coordinates $\varphi=\rho e^{i\theta}$, we have $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\rho_0^2(\partial_\mu\theta-eA_\mu)^2+\cdots$$ where I have ignored the fluctuation of $\rho$ by setting $\rho=\rho_0$.

(I). The standard way to deal with it is to redefine $$B_\mu\equiv A_\mu-\frac{1}{e}\partial_\mu\theta$$ thus we have $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+e^2\rho_0^2 B^2$$ thus the massless gauge field eats a Nambu-Goldstone boson and becomes a massive gauge field which has 3 degrees of freedom.

(II). Instead of doing that, let us expand $(\partial_\mu\theta-eA_\mu)^2$ directly and impose Lorentz gauge at the same time $\partial_\mu A^\mu=0$, we have $$\mathcal{L}=-\frac{1}{4}F^2+\rho_0^2e^2 A^2+\rho_0^2 (\partial_\mu \theta)^2,$$ where the cross term vanishes because of the Lorenz gauge. And this Lorenz gauge is satisfied by the equation of motion of the gauge field $A_\mu$ naturally. It seems we have both massive gauge field and a Goldstone boson.

My questions are:

  1. How to reconcile the two approaches above?

  2. If we want to examine the dynamics of the bosons (for example, apply a boson current and see how this current propagates), will this current be a supercurrent in the presence of a dynamic gauge field? This boson current will be a supercurrent(like a current in a superconductor) in the absence of a dynamic gauge field. When we couple this boson system to a dynamic gauge field, the dynamics of phase $\theta$ of the boson field is hidden somehow in the approach (I) and it is not clear whether this boson current will be a supercurrent or not. But from approach (II), it seems this boson current will still be a supercurrent.

  • $\begingroup$ Surely you are not asking how to "reconcile" actions in different gauges? $\endgroup$ – Cosmas Zachos Jun 6 '19 at 18:18
  • $\begingroup$ @CosmasZachos In the approach (II), it seems there are 4 degrees of freedom which is inconsistent with the approach (I). $\endgroup$ – Ji Zou Jun 6 '19 at 18:28
  • $\begingroup$ Yes, all gauges but the unitarity gauge do not display the actual degrees of freedom readily. How do you transform among them? What does your text say? $\endgroup$ – Cosmas Zachos Jun 6 '19 at 18:48

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