Noether Charge and Gauge Fields

Question

I understand the global gauge symmetry results conserved charges associated with the symmetry. My question is, why don't we have conserved charges associated with local gauge symmetry of the gauge fields $A$ itself? Why don't we have a conserved charge for the Lagrangian $\int \mathrm{Tr}\left( F\wedge\star F\right)$? Please elaborate in both abelian case and non-abelian case.

Answer 1

The abelian case:

The Lagrangian $$ \mathcal{L} = F^{\mu\nu} F_{\mu\nu} $$ is invariant off shell under local gauge symmetry $$ A_\mu(x) \mapsto A_\mu (x) - \partial_\mu \theta(x) $$ The Noether current is $$ J^\mu(x) = \frac{\delta{L}}{\delta \partial^\mu A_\nu} \delta A_\nu = F^{\mu\nu} \partial_\nu \theta(x) $$ So the Noether charge is: $$ Q = \int d^3x. J^0 = \int d^3x. F^{0i} \partial_i \theta(x) = - \int d^3x. (\partial_i F^{0i}) \theta(x) = 0 $$ where the last equality follows from equation of motion: $\partial_\mu F^{\mu\nu} = 0$.

Surprisingly! The Noether charge is 0!!!

It seems to violate the 'Noether charge theorem' which states that the Noether charge should be the quantum generator of the transformation: $$ [Q_\theta, A_\mu(x)] = -\partial_\mu \theta(x) $$ But it is reasonable, that in the proof of 'Noether charge theorem' we used the canonical commutation relation: $$ \begin{align} [Q,\phi(y)] &= \int d^3x. [\frac{\delta L}{\delta \partial_0 \phi} \mathcal{F}, \phi(y)]\\ &= [\int d^3x. \mathcal{P} \mathcal{F}, \phi(y)]\\ &= \int d^3x. [\mathcal{P}(x), \phi(y)] \mathcal{F}(x)\\ &= \int d^3x. \delta^{(3)}(x-y) \mathcal{F}(x)\\ &= \mathcal{F}(y) \end{align} $$ where the transformation is $\phi(x)\mapsto \phi(x) + \mathcal{F}(x)$ and $\mathcal{P}(x)$ is the canonical momentum. And we used canonical commutation relation $[\mathcal{P}(x), \phi(y)] = \delta^{(3)}(x-y)$. But in QED the canonical commutation relation is different: canonical momentum is $\mathcal{P}^\nu = F^{0\nu}$ and we impose(Srednicki 55.20) $$ [F^{0\nu}, A^\sigma] = (\delta^{\nu\sigma} - \frac{\partial^\nu \partial^\sigma}{\nabla^2})\delta^{(3)}(x-y) $$ At this time we can compute: $$ \begin{align} [Q,A^\sigma(y)] &= [\int d^3x. \mathcal{P}^\nu \partial_\nu\theta(x), A^\sigma(y)]\\ &= \int d^3x. [\mathcal{P}(x), A^\sigma(y)] \partial_\nu\theta(x)\\ &= \int d^3x. (\delta^{\nu\sigma} - \frac{\partial^\nu \partial^\sigma}{\partial^2})\delta^{(3)}(x-y) \partial_\nu\theta(x)\\ &= \partial^\sigma\theta(x) - \partial^\sigma\theta(x)\\ &= 0 \end{align} $$ which does not contradick to $Q = 0$ as what we previously computed.

So the reason why we don't talk about the Noether charge for local gauge symmetry is that it is always 0 and does not equal to quantum generator $[Q_\theta, A_\mu(x)] = -\partial_\mu \theta(x)$.

In fact, we can prove that the Noether charge for any local symmetry is always 0!! (See the answer by Peter Kravchuk: https://physics.stackexchange.com/q/66251. Note that the accepted answer there is incorrect...).

Consider the transformation: $$ \phi(x) \mapsto \phi(x) + \epsilon(x)\mathcal{F}(x) $$ under which we require the action is invariant when $\epsilon(x)=\epsilon$ is a constant. Then we have $$ \delta S = \int d^4x. \partial_\mu \epsilon(x) J^\mu(x) $$ It should be accentrated that the Euler-Lagrange equation is derived by requiring the stability under variation that is compactly supported!! So if $\mathcal{F}(x)$ is not, we should let $\epsilon(x)\in C_c(\mathbb{R}^4)$. This is the case that we usually use. But if instead the transformation is local, i.e. $\mathcal{F}(x)\in C_c(\mathbb{R}^4)$, as is the case for local gauge transformation, we can set $$ \epsilon(x) = \Theta(x^0) := \frac{1+sign(x^0)}{2} $$ the Heaviside function. In this situation we get on-shell: $$ 0 = \delta S = \int d^4x. \partial_\mu \Theta(x^0) J^\mu(x) = \int d^4x. \delta(x^0) J^0(x) = \int d^3x. J^0(x) = Q $$