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How to prove conservation of electric charge using Noether's theorem according to classical (non-quantum) mechanics? I know the proof based on using Klein–Gordon field, but that derivation use quantum mechanics particularly.

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up vote 6 down vote accepted

By the word classical we will mean $\hbar=0$, and we will use the conventions of Ref. 1.

The Lagrangian density for Maxwell theory with various matter content is

$$\tag{1} {\cal L} ~=~{\cal L}_{\rm Maxwell} + {\cal L}_{\rm matter} , $$

$$\tag{2} {\cal L}_{\rm Maxwell}~=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu},$$

$$\tag{3} {\cal L}_{\rm matter}~=~{\cal L}_{\rm matter}^{\rm QED}+{\cal L}_{\rm matter}^{\rm scalar QED} + \ldots, $$

$$\tag{4} {\cal L}_{\rm matter}^{\rm QED} ~:=~ \overline{\Psi}( i\gamma^{\mu} D_{\mu}-m)\Psi , $$

$$\tag{5} {\cal L}_{\rm matter}^{\rm scalar QED}~:=~ -(D_{\mu}\phi)^{\dagger} D^{\mu}\phi -m^2\phi^{\dagger}\phi -\frac{\lambda}{4} (\phi^{\dagger}\phi)^2, $$

with covariant derivative

$$ \tag{6} D_{\mu}~=~d_{\mu}-ieA_{\mu}. $$

(Here we are too lazy to denote various matter masses $m$ and charges $e$ differently.) The matter equations of motion (eom) are

$$ \tag{7}( i\gamma^{\mu} D_{\mu}-m)\Psi ~\approx~0, \qquad D_{\mu}D^{\mu}\phi~\approx~m^2\phi+\frac{\lambda}{2} \phi^{\dagger}\phi^2, \qquad \ldots.$$

(The $\approx$ symbol means equality modulo eom, i.e. an on-shell equality.)

The infinitesimal global off-shell gauge transformation is

$$ \delta A_{\mu} ~=~0, \qquad \delta\Psi~=~-i\epsilon \Psi, \qquad \delta\overline{\Psi}~=~i\epsilon \overline{\Psi}, $$ $$ \tag{8} \delta\phi~=~-i\epsilon \phi,\qquad \delta\phi^{\dagger}~=~i\epsilon \phi^{\dagger}, \qquad \ldots, \qquad\delta {\cal L} ~=~0, $$

where the infinitesimal parameter $\epsilon$ does not depend on $x$.

The Noether current is the electric $4$-current$^1$

$$ \tag{9} j^{\mu}~=~e\overline{\Psi}\gamma^{\mu}\Psi - ie\{\phi^{\dagger} D^{\mu}\phi-(D^{\mu}\phi)^{\dagger}\phi\}+\ldots. $$

Noether's Theorem is a theorem about classical field theory. It yields an on-shell conservation law

$$ \tag{10} d_{\mu}j^{\mu}~\approx~0.$$

Hence the electric charge

$$\tag{11} Q~=~\int\! d^3x~ j^0$$

is conserved on-shell.


  1. M. Srednicki, QFT.


$^1$ Interestingly, the electric $4$-current $j^{\mu}$ depends on the gauge potential $A_{\mu}$ in case of scalar QED matter.

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I'm wondering if PhysiXxx would not prefer to call a "classical" proof a proof with real fields... then without gauge symmetry. I believe in "classical" Maxwell equations the conservation of charge $dQ / dt = 0$ is a postulate, leading to $\partial_{\mu} j^{\mu} = 0$ through integration over a finite volume. – FraSchelle Mar 24 '13 at 17:26
It is true that [Maxwell eqs. $d_{\mu}F^{\mu\nu}\stackrel{A_{\lambda}}{\approx}-j^{\nu}$] $\Rightarrow$ [Continuity eq. $d_{\mu}j^{\mu}\stackrel{A_{\lambda}}{\approx}0$] $\Rightarrow$ [Electric charge conservation], where $j^{\mu}:=\frac{\delta S_{\rm matter}}{\delta A_{\mu}}$. In fact, here $j^{\mu}$ could be an unspecified background source, that knows nothing about the matter theory. However, OP asked specifically to use Noether's (first) theorem in the proof. According to Noether's (first) theorem, we have [global gauge symmetry of the action]$\Rightarrow$ [Electric charge conservation]. – Qmechanic Apr 24 '13 at 17:19
Interestingly, we can analyze the question of electric charge conservation in a matter theory in a fixed (but arbitrary, possibly off-shell) gauge background $A_{\lambda}$. The $\approx$ symbol in the answer only requires the matter sector being on-shell. – Qmechanic Mar 8 '15 at 21:37
Minkowski sign convention (-,+,+,+). – Qmechanic Sep 11 '15 at 9:39

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