2.3.1 Isospin and SU(2), Redux

The tale we told of isospin in Section 2.1 only concerned the strong force, which binds nucleons together into nuclei. We learned about an approximation in which nucleons live in the fundamental rep ${\mathbb{C}}^2$ of the isospin symmetry group ${\rm SU}(2)$, and that they interact by exchanging pions, which live in the complexified adjoint rep of this group, namely $\sl (2,{\mathbb{C}})$.

But this tale is mere prelude to the modern story, where weak isospin, defined in Section 2.2.2, is the star of the show. This story is not about the strong force, but rather the weak force. This story parallels the old one, but it involves left-handed fermions instead of nucleons. The left-handed fermions, with $I_3 = \pm \frac{1}{2}$, are paired up into fundamental representations of ${\rm SU}(2)$, the weak isospin symmetry group. There is one spanned by left-handed leptons:

\begin{displaymath}\nu_L, e^-_L \in {\mathbb{C}}^2, \end{displaymath}

and one spanned by each color of left-handed quarks:

\begin{displaymath}u^r_L, d^r_L \in {\mathbb{C}}^2, \quad u^g_L, d^g_L \in {\mathbb{C}}^2, \quad
u^b_L, d^b_L \in {\mathbb{C}}^2. \end{displaymath}

The antiparticles of the left-handed fermions, the right-handed antifermions, span the dual representation ${\mathbb{C}}^{2*}$.

Because these particles are paired up in the same ${\rm SU}(2)$ representation, physicists often write them as doublets:

\begin{displaymath}\left( \! \begin{array}{c} \nu_L \\ e^-_L \end{array} \! \rig...
...ad \left( \! \begin{array}{c} u_L \\ d_L \end{array} \! \right)\end{displaymath}

with the particle of higher $I_3$ written on top. Note that we have suppressed color on the quarks. This is conventional, and is done because ${\rm SU}(2)$ acts the same way on all colors.

The particles in these doublets then interact via the exchange of $W$ bosons, which are the weak isospin analogues of the pions. Like the pions, there are three $W$ bosons:

\begin{displaymath}W^+ = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \rig...
... = \left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right). \end{displaymath}

They span the complexified adjoint rep of ${\rm SU}(2)$, $\sl (2,{\mathbb{C}})$, and they act on each of the doublets like the pions act on the nucleons, via the action of $\sl (2,{\mathbb{C}})$ on ${\mathbb{C}}^2$. For example,

Again, Feynman diagrams are the physicists' way of drawing intertwining operators. Since all the ${\mathbb{C}}^2$'s are acted on by the same ${\rm SU}(2)$, they can interact with each other via $W$ boson exchange. For example, quarks and leptons can interact via $W$'s:

This is in sharp contrast to the old isospin theory. In the new theory, it is processes like these that are responsible for the decay of the neutron:

The fact that only left-handed particles are combined into doublets reflects the fact that only they take part in weak interactions. Every right-handed fermion, on the other hand, is trivial under ${\rm SU}(2)$. Each one spans the trivial rep, ${\mathbb{C}}$. An example is the right-handed electron

\begin{displaymath}e^-_R \in {\mathbb{C}}.\end{displaymath}

Physicists call these particles singlets, meaning they are trivial under ${\rm SU}(2)$. This is just the representation theoretic way of saying the right-handed electron, $e^-_R$, does not participate in weak interactions.

In summary, left-handed fermions are grouped into doublets (nontrivial representations of ${\rm SU}(2)$ on ${\mathbb{C}}^2$), while right-handed fermions are singlets (trivial representations on ${\mathbb{C}}$). So, the left-handed ones interact via the exchange of $W$ bosons, while the right-handed ones do not.

The First Generation of Fermions -- ${\rm SU}(2)$ Representations
Name Symbol $I_3$ ${\rm SU}(2)$ rep
Left-handed leptons $\left( \! \begin{array}{c} \nu_L \\ e^-_L \end{array} \! \right)$ $\pm \frac{1}{2}$ ${\mathbb{C}}^2$
Left-handed quarks $\left( \! \begin{array}{c} u_L \\ d_L \end{array} \! \right)$ $\pm \frac{1}{2}$ ${\mathbb{C}}^2$
Right-handed neutrino $\nu_R$ $0$ ${\mathbb{C}}$
Right-handed electron $e^-_R$ $0$ ${\mathbb{C}}$
Right-handed up quark $u_R$ $0$ ${\mathbb{C}}$
Right-handed down quark $d_R$ $0$ ${\mathbb{C}}$