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Using Noether's theorem

\begin{equation} \partial_0 \int d^3x \left(\frac{\partial L}{\partial(\partial_0\Psi)} \delta \Psi \right) = 0 \end{equation}

we get three conserved quantites $Q_i$ from global $SU(2)$ symmetry, because the Lagrangian is invariant under infinitesimal transformations of the form $\delta \Psi = i a_i \sigma_i \Psi $. The conserved quantities that follow from the free doublet Lagrangian $L= i\bar{\Psi} \gamma_\mu \partial^\mu \Psi$ are therefore

\begin{align} Q_i&= i\bar{\Psi} \gamma_0 \sigma_i \Psi \notag \\ &= \begin{pmatrix} v_e \\ e \end{pmatrix}^\dagger \underbrace{\gamma_0 \gamma_0}_{{=1}} \sigma_i \begin{pmatrix} v_e \\ e \end{pmatrix} \end{align}

Why are the conserved quantities that follow from $i=1$ or $i=2$, never mentioned or used? For $i=1$ we have

\begin{align} Q_1&= \begin{pmatrix} v_e \\ e \end{pmatrix}^\dagger \sigma_1 \begin{pmatrix} v_e \\ e \end{pmatrix} \notag \\ &= \begin{pmatrix} v_e \\ e \end{pmatrix}^\dagger \begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix} \begin{pmatrix} v_e \\ e \end{pmatrix} \notag \\ &= v_e^\dagger e + e^\dagger v_e \end{align}

or for $i=3$ we have

\begin{align} Q_3&= \begin{pmatrix} v_e \\ e \end{pmatrix}^\dagger \sigma_3 \begin{pmatrix} v_e \\ e \end{pmatrix} \notag \\ &= \begin{pmatrix} v_e \\ e \end{pmatrix}^\dagger \begin{pmatrix} 1 & 0 \\0& -1 \end{pmatrix} \begin{pmatrix} v_e \\ e \end{pmatrix} \notag \\ &= v_e^\dagger v_e - e^\dagger e \end{align}

which is the usually used third component of weak isospin.

jak
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1 Answers1

6

The question is malformed. Noether's theorem is fine as a symmetry statement, but your gambit fails. In order for the charge you are discussing to exist, it must annihilate the vacuum of the theory, per the Fabri–Picasso theorem. Failing that, it blows up ~ does not exist: the hallmark of SSB. I gather you may have misunderstood $Q_3$ presented as a conserved quantity, which it is not: Note the toxic minus sign instead of the plus of the valid lepton number! (In practice, a propagating left-handed electron couples/transmutes to a right-handed weak-isosinglet one through the mass term involving a Higgs v.e.v. As it were, it would "absorb some $Q_3$ out of the EW vacuum"—an admittedly baroque caricature for a quantity that is ill-defined!)

In the standard model, all currents are conserved —otherwise they would not couple consistently to gauge fields; but the final step you start from, i.e. the space integral of the current zero component, may or may not exist, as per the above caveat.

In the SM, of course, the EM charge, a linear combination $Q_3+Y$, where Y is the weak hypercharge does annihilate the vacuum (so it is unbroken) and thus exists!

The independent would-be charge whose current couples to the Z , $Q_3\cos^2 \theta_W-Y \sin^2\theta_W$, by contrast, does not, just like $Q_1,Q_2$. You do not see them written down, since few cherish shadow-boxing with phantoms.

Edit: But... could you cheat? When?

A qualmful fiddler might well object that, at the very least, the current-current interaction for μ decay, etc, preserve some $Q_3$ as a fine quantum number, after all: $Q_3(\mu)=Q_3(e)+Q_3(\nu_\mu )+Q_3(\bar{\nu}_e)$, and so on. And this is not a coincidence. Could some $Q_3$ be somehow still useful as an approximate conservation law?

Indeed, the EW Lagrangian, itself, and its effective avatars, do possess the SU(2)L symmetry as indicated, and unless a Higgs coupling were involved vertices are expected to respect this symmetry, at some level. Still, any Higgs interaction is liable to SSB contamination, such as in the fermion propagation, illustrated above, which spoils the symmetry. The answer then is "with care"–caveat fiddler. Forensic excision of Higgs contamination effects would be a risky art.

Cosmas Zachos
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