# Measuring conserved charges of gauge bosons with conserved charge operators

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?

• If you ignored the local symmetry, there would still be a global symmetry of the group you wrote, and the Ws would still transform under that group, and so contribute to the Noether current: you skipped the term contributed by the Ws in the Noether charge. Can you compute it? – Cosmas Zachos Jan 17 at 0:14
• The Noether charge would be $\int \psi^{\dagger}T^{a}\psi d^{3}x$ wouldn't it? The $\frac{\partial L}{\partial(\partial\phi_{\mu})}$ will not have any factors of the W field. Isn't it the case that the global symmetry doesn't apply because the gauge is fixed before quantisation? This I believe is also the reason that massless bosons have but two (transverse) polarization sates. In the case of QED this comes from choosing the Coulomb gauge as I'm sure you know. This is also discussed here: physics.stackexchange.com/questions/471575/… – Stijn Boshoven Jan 17 at 11:16
• Focus on the classical theory first, and the adjoint transformation of the Ws... – Cosmas Zachos Jan 17 at 11:24
• Then you do not agree with the accepted answer to the question I referred to in my comment? It already seemed questionable to me. However if there are indeed conserved currents associated with global symmetries in QFT then why are they not discussed in QED where this lead to an infinite number of conserved currents $j^\mu=-fe(\bar\psi\gamma^\mu\psi)+\partial_\nu f(\partial^\mu A^\nu-\partial^\nu A^\mu)$ for every $f$ is a real scalar field? – Stijn Boshoven Jan 17 at 11:50
• Pretty sure that would be $j^{\mu}_{i}=-2g\epsilon^{ijk}W^{\mu}_{j}W^{\nu}_{k}\partial_{\nu}\Lambda$ right? – Stijn Boshoven Jan 17 at 13:05

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 your first transformation comes from $\sigma\cdot W_{\mu}\rightarrow e^{-ig\sigma\cdot \theta}\sigma\cdot W_{\mu}e^{ig\sigma\cdot \theta}+\frac{i}{g}\partial_{\mu} \left(e^{-ig\sigma\cdot \theta}\right)e^{ig\sigma\cdot \theta}$? And (I'm not asking you to prove this but just so I what I'm working towards) this theory can be quantised with W1, W2 and W3 creation and annihilation operators s.t. we can calculate: $Q^{i}|W^{+} \rangle = \frac{1}{\sqrt{2}} Q^{i} \left(|W^{1} \rangle-i|W^{2} \rangle\right) = |W^{+} \rangle$ showing $|W^{+} \rangle$ has $I_{3}=1$? – Stijn Boshoven Jan 20 at 12:19