Measuring conserved charges of gauge bosons with conserved charge operators

Question

In QED we don't care about this because the photons carry no charge and so it is no problem that the conserved charge operator:

$$Q=\int\psi^{\dagger}\psi \ d^{3}x$$

That ends up consisting of number operators measuring the positrons and electrons in the state are blind to the number of photons in any state. You could say that this is because the conserved charge contained no factors of $A$, the photon field, and so could not contain any number operators for photons.

I cannot see how this is resolved in any theory where the bosons do carry nonzero conserved charge.

Consider the $SU(2)_{L}\times SU(1)_{y}$ theory and forget about the Higgs mechanism for a moment:

The Noether current associated with the symmetry:

$$\psi_{L}\rightarrow e^{-igT^{a}}\psi_{L}$$

Where $T^{a}$ with $a=1,2,3$ are the Pauli matrices.

Leads to conservation of each of the three components of weak isospin. These conserved currents, however, will not involve any factors of the W boson fields. Therefore, upon quantising the theory, the associated quantum field theoretical weak isospin operator can never give anything except zero when we let this operator act on a state containing any of the three W bosons, correct?

Is this just a shortcoming of this operator and do we instead take this conservation law, see how the bosons change the doublets and then infer from these interactions what the weak isospin of the W boson in question is?

Answer 1

You threw five and a half pots on the fire, intricately entangled, which are addressed by over a dozen questions and answers on this site, but I will focus on a very narrow one, which I suspect might be at the bottom of your question. That is, I am going to avoid quantization, gauge invariance, matter fields, and symmetry breaking, and focus on the globally symmetric under SU(2) classical Yang-Mills action. So the intriguing Noether's second theorem doesn't enter, and you just calculate the standard ("first") Noether current, which integrates to a charge, transforming classical fields under Poisson bracketing.

Consider, then, the Yang-Mills theory as a theory of vector fields transforming under the adjoint of global SU(2), in a world where nobody had noticed it is also gauge (locally) invariant: $$ W^i_\mu \mapsto W^i_\mu + \epsilon^{ijk} \theta^j W_\mu^k +O(\theta^2). $$ The action ${\cal L}=-(\partial_\mu W^i_\nu - \partial_\nu W^i_\mu +g \epsilon^{ijk} W^j_\mu W^k_\nu )^2/4$ is varied with respect to the gradients of all fields and multiplied/saturated with the variation of all such fields, by constant infinitesimal angle $\theta^j$, $$ \theta^i j^i_\mu= {\delta {\cal L} \over \delta \partial_\mu W^{i }_\nu} \delta W^{i}_\nu = -F^i_{\mu\nu}\epsilon^{ijk}W^{k~~\nu} \theta^j \\ \leadsto j^i_\mu = \epsilon^{ijk}F^j_{\mu\nu} W^{k~~\nu} , $$ on-shell conserved, of course. (It is covariant under the relic global symmetry, but not under the ignored local symmetry.)

The corresponding isospin charge then is $Q^i=\int\!\! d^3 y ~~j^i_0(y)$, so that it infinitesimally rotates the vector fields adjointly, as expected, $$ \{Q^i, W_\mu^n(x) \}\propto \epsilon ^{imk} W_\mu^k $$ classically, by use of $\{ \partial_0 W_\mu^i (y), W^m(x) \}\propto \delta^{im}\delta^3(x-y)$.

• So indeed, the global Noether charge classically isorotates the vector fields.