$\mathcal{L}_{I}^{C C} $ global $U(1)$ invariance

Question

It is well known that the SM lagrangian has the $U(1)$ symmetry. Considering for the moment the global symmetry $U(1)$, I am having some trouble seeing it in the following terms:

$$\mathcal{L}_{I}^{C C}=-\frac{g}{\sqrt{2}} \sum_{\ell=e, \mu, \tau}\left(W_{\mu}^{+} \bar{\nu}_{\ell L} \gamma^{\mu} \ell_{L}+W_{\mu}^{-} \bar{\ell}_{L} \gamma^{\mu} \nu_{\ell L}\right)$$

after the transformation I should obtain $\mathcal{L}^{'}=\mathcal{L}$:

$$\mathcal{L}_{I}^{C C}=-\frac{g}{\sqrt{2}} \sum_{\ell=e, \mu, \tau}\left(W_{\mu}^{+} \bar{\nu}_{\ell L} e^{i \alpha_l}\gamma^{\mu} \ell_{L}e^{-i \beta_l}+W_{\mu}^{-} \bar{\ell}_{L} e^{i \beta_l}\gamma^{\mu} \nu_{\ell L}e^{-i \alpha_l}\right).$$

This should implies that $\alpha_l=\beta_l$ which is not true, because we know that lepton are charged and neutrinos no.

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In case of interaction lagrangian of QED as far as I know, you can see that it will conserve the electric charge doing: $$\mathcal{L}_{I}^{',QED}= - q_e \bar{\psi}e^{iq_e}{A\!\!\!/}e^{-iq_e}\psi=- q_e \bar{\psi}{A\!\!\!/}\psi =\mathcal{L}_{I}^{QED}$$

Honestly I have never transformed a gauge boson under a global symmetry, however, the only way in which the previous Lagrangian is invariant, according to a transformation that is connected to the electric charge, is : $$\mathcal{L}_{I}^{',C C}=-\frac{g}{\sqrt{2}} \sum_{\ell=e, \mu, \tau}\left(W_{\mu}^{+} e^{-iq_W}\bar{\nu}_{\ell L} e^{iq_{\nu}}\gamma^{\mu} \ell_{L}e^{-iq_{\ell}}+W_{\mu}^{-}e^{iq_W} \bar{\ell}_{L} e^{iq_{l}}\gamma^{\mu} \nu_{\ell L}e^{-iq_{\nu}}\right)= \mathcal{L}_{I}^{C C}$$ with $q_w=e,q_{\ell}=-e,q_{\nu}=0$

maybe is correct in this way, but I'm not sure...

Answer 1

The cornerstone symmetry associated with the electric charge in the SM is, as your text details, I'm sure, $$ Q=T_3+ \frac{Y_w}{2}, $$ where $Y_W$ is the weak hypercharge, and $T_3$ is the z-component of the weak isospin (the SU(2)).

The fields in your currents have eigenvalues $(t_3, y_w)$ under these, $$ W^+: \qquad (1, 0)\qquad \leadsto q= 1\\ e: \qquad (-1/2, -1)\qquad \leadsto q= -1\\ \nu: \qquad (1/2, -1)\qquad \leadsto q= 0, $$ and reversed for the conjugates.

Plugging these into your expression shows all Q, Y_W, T₃ balance to 0s.

Can you repeat this check for each and every term of the SM? Your text did it in a hyper efficient way, which you might have missed...

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A propos of your comment: I'm a bit confused as to what your "doubt" could possibly be... As demonstrated, the charge of $W^+$ is, indeed, 1, and that's why we label it that way. You added the charges of the term you have written correctly.

If, instead, mysteriously, you were concerned about the local charge Q invariance of your action, not of any relevance to the term you already wrote, you skipped the crucial gauge coupling of your SM-QED action, which I'm sure your text writes down and discusses in painful detail, $$ -ie[ \partial_\mu A_\nu (W_\mu^+ W_\nu^- -W_\nu^+ W_\mu^- ) \\ + A_\nu(-W_\mu^+\partial_\nu W^-_\mu +W_\mu^-\partial_\nu W^+_\mu +W_\mu^+\partial_\mu W^-_\nu - W_\mu^-\partial_\mu W^+_\mu )]. $$ The first term (upstairs) is separately gauge invariant, while the second term is the EM (Q) current coupling involved in the gauge invariance of the W kinetic terms!