# Charge not conserved in scalar QED? [duplicate]

Since conservation of charge seems to be a well known concept, I am hoping that I am missing something and that the conclusion is incorrect. However, I have been unable to disprove this. Let me outline the situation as follows.

Overview

Consider the following Lagrangian for scalar QED and field $\varphi$: \begin{align} \mathcal{L}=\overline{D_\mu\varphi}D^\mu\varphi-U(|\varphi|^2)-\frac{1}{4}F_{\mu\nu}F^{\mu\nu},\tag{1} \end{align} where $$D_\mu=\partial_\mu-iA_\mu\varphi.\tag{2}$$

One set of Euler-Lagrange equations is: \begin{align} \frac{\delta\mathcal{L}}{\delta A_\mu}=i(\overline\varphi\partial_\mu\varphi-\varphi\partial_\mu\overline\varphi)+2A^\mu|\varphi|^2+\partial_\nu F^{\mu\nu}=0.\tag{3} \end{align}

If we consider the global gauge symmetry transformation $\varphi\to\varphi+i\epsilon\varphi$, then Noether's first theorem gives the following conservation law (for charge, see e.g. [1]) on solutions of $\delta\mathcal{S}=0$: \begin{align} \partial_\mu j^\mu:=\partial_\mu\left[i(\overline\varphi\partial_\mu\varphi-\varphi\partial_\mu\overline\varphi)+2A^\mu|\varphi|^2\right]=0.\tag{4} \end{align}

Question: does the integral of the charge density $j^0$ vanish identically? If so, does this mean that nothing is conserved?

My attempt at an answer: Yes. To see this, we integrate the continuity equation over $\mathbb{R}^3$ and apply Gauss's Theorem: \begin{align} \partial_0Q:=\partial_0\int_{\mathbb{R}^3}j^0 d^3 x=-\sum_{\mu=1}^3\int_{R^2} j^\mu \bigr\rvert_{|x_\mu|\to\infty}d^2x.\tag{5} \end{align} Here, we say that $Q=\int j^0$ is the total charge.

We assume that the fields $\varphi$ and $F$ vanish as $\max_{\mu\ge 1}(x_\mu)\to\infty$. This is a natural boundary condition for physical observables. Then the fluxes $j^\mu$ clearly vanish in this limit. Thus, our conservation law becomes: \begin{align} \partial_0\int_{\mathbb{R}^3}j^0 d^3 x=0.\tag{6} \end{align} But $$j^0=\delta\mathcal{L}/\delta A_0-\partial_\nu F^{0\nu}=-\partial_\nu F^{0\nu}\tag{7}$$ on solutions, so we can equivalently rewrite the charge integral using Gauss's Theorem as follows: \begin{align} Q=-\sum_{\nu\ge 1}\int_{\mathbb{R}^2}F^{0\nu}\bigr|_{|x_\nu|\to\infty} d^3 x.\tag{8} \end{align} Since the fields vanish at infinity, we obtain: \begin{align} Q\equiv 0.\tag{9} \end{align}

In other words, if correct, then this shows that the charge for this system is identically zero, and that nothing is conserved (it seems incorrect to call $d0/dt\equiv 0$ a conservation law rather than a tautology/identity).

There are a few other questions users asked that have some relation to this. However, I do not believe the actual charge integral's conservation was anywhere addressed. One shows that the charge density can be rewritten as a total divergence, from which the argument follows, of course. This article includes the fact that there is no conservation law for the local gauge symmetry, but does not clarify this for the global symmetry case (which should be a special case of the local results). Another gives an overview of charge conservation.

Edit:

Another question also discusses charge conservation, but it seems to be more concerned with the interpretation of the charge density. I am instead asking if the density is conserved at all.

• The fields do vanish at infinity, but the measure of $d^3x$ blows up in the same limit. So you have to be more careful. In fact, you can check that $j^0 \sim 1/r^2$ at large $r$ (i.e. near infinity) but $d^3 x \sim r^2 \sin\theta d\theta d\phi$. The power of $r^2$ cancel and give a finite result. – Prahar Sep 10 '15 at 14:22
• isites.harvard.edu/fs/docs/icb.topic624321.files/… – RoderickLee Jul 14 '16 at 3:16

your conclusions have nothing to do with scalar QED or quantum mechanics for that matter. Let's start from your $j^0 = -\partial_\mu F^{0 \mu}$, then you erroneously applied gauss law to find $Q$ vanishes at infinity. The reason is that in your surface integral $\int_{\mathbb{R}^2}dS$ need not vanish, because the field need only vanish as $1/r$ for the integral to be finite.
This is just classical electrodynamics, where $E^i= F^{0i}$ so that $\rho = j^0 = -\partial_\mu F^{0\mu} = -\nabla\cdot E$ is just Guass law and \begin{align} Q = \int_{\text{surface}} \vec{E}\cdot d\vec{A} \rightarrow \text{finite if }E\xrightarrow{r\rightarrow\infty}1/r \end{align}