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I have a perhaps stupid question about Noether's theorem.

In Abelian gauge theory, say

$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\Psi}(iD\!\!\!\!/-m)\Psi, \tag{1.0} $$

where $D_{\mu}=\partial_{\mu}+iA_{\mu}$ is the covariant derivative, the equations of motion are

$$\partial_{\mu}F^{\mu\nu}=J^{\nu} \tag{1.1}$$

$$(iD\!\!\!\!/-m)\Psi=0 \tag{1.2} $$

where

$$J^{\mu}=\bar{\Psi}\gamma^{\mu}\Psi \tag{1.3} $$

is a conserved current,

$$\partial_{\mu}J^{\mu}=0 \tag{1.4} $$

following (1.1).

In fact, the spinor current (1.3) is also the Noether current associated with a global $U(1)$ symmetry. One can check that the current (1.3) is indeed gauge invariant.

In nonabelian case, we consider the spinor-Yang-Mills theory

$$\mathcal{L}=-\frac{1}{4}\mathrm{Tr}(F_{\mu\nu}F^{\mu\nu})+\bar{\Psi}(iD\!\!\!\!/-m)\Psi. \tag{2.0} $$

One finds the equations of motion

$$D_{\mu}F^{\mu\nu}=J^{\nu} \tag{2.1} $$

$$(iD\!\!\!\!/-m)\Psi=0 \tag{2.2} $$

where

$$J^{\mu}=\sum_{a=1}^{N}\left(\bar{\Psi}\gamma^{\mu}T_{a}\Psi\right)T_{a} \tag{2.3} $$

is a Lie-algebra-valued spinor current, and $\left\{T_{a}\right\}_{a=1,\cdots,N}$ is the set of generators of the gauge group.

In this case, the current $J^{\mu}$ is "covariantly conserved",

$$D_{\mu}J^{\mu}=0 \tag{2.4} $$

because

$$D\star J=DD\star F=F\wedge\star F-\star F\wedge F=F^{a}\wedge\star F^{b}[T_{a},T_{b}]\equiv 0$$

where the last equality holds because $F^{a}\wedge\star F^{b}$ is an inner-product $\langle F^{a},F^{b}\rangle$, which is symmetric.

However, there should be no constant of motion (i.e. locally conserved charge) associated with the spinor current (2.3) because of the covariant derivative.

On the other hand, there still can be a global part of the gauge symmetry of (2.0). Considering a "global gauge transformation" acting on (2.0), by Noether's theorem of the global symmetry, one finds a locally conserved current

$$J^{\mu}=\sum_{a=1}^{N}\left(\bar{\Psi}\gamma^{\mu}T_{a}\Psi\right)T_{a} \tag{*} $$

which is identical to the spinor current (2.3).

Why would Noether's theorem fail in such a case?

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