# Why is only the third component of weak isospin used as a conserved quantity?

Using Noether's theorem

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

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.

• hint: how many generators of SU(2) can be simultaneously diagonalized? why might a diagonal generator lead to a more useful quantum number than a non-diagonal one? – innisfree Nov 28 '14 at 15:09
• $SU(2)$ has one Cartan generator. I'm not sure about your second question. Do you mean that diagonal operators can be measured at the same time and therefore only one of these three conserved quantities can be "measured" at the same time? Or: The objects in the doublet, here $v_e$ and $e$ are only eigenstates for the diagonal generator. For the other generators we can't assign a definitive number to $v_e$ and $e$, because they aren't eigenstates? – jak Nov 28 '14 at 15:17
• That is a convention. All 3 components can be equally chosen as a conserved quantity. Unfortunately, they cannot be measured simultaneously, so if you measure one of them and get the eigenvalue of that, the others won't be fixed. So it is conventional to only consider one component. – Drake Marquis Nov 28 '14 at 15:50

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 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, Fermi's effective β-decay vertex, $$G_F~ \bar{n} \gamma_\mu P_L p ~ \bar{\nu} \gamma ^\mu P_L e$$, or the current-current one for μ decay, etc, preserve some $$Q_3$$ as a fine quantum number, after all: $$Q_3(n_L)=Q_3(p_L)+Q_3(\bar{\nu}_e)+Q_3(e_L)= 1/2 -1/2 -1/2=-1/2,$$ $$Q_3(\mu)=Q_3(e)+Q_3(\nu_\mu )+Q_3(\bar{\nu}_e)$$, and so on. And this is not a coincidence. Could some $$Q_3$$ be somehow still useful as an approximate conservation law?