Mixing of $\gamma$ and $Z^{0}$ boson: Isospin?

Question

In our lecture, we had for the Lagrange density of the $\gamma$ and the $Z^{0}$ boson: $$\mathcal L = -\sum_{\psi = e_{L}, e_{R}, \nu_{L}, \nu_{R}}\bar{\psi}\gamma^{\mu}\left( gT^{3}W_{\mu}^{3} + g'\frac{Y}{2}B_{\mu} \right)\psi.$$ We then introduced the mixing: $$\begin{pmatrix} W_{\mu}^{3} \\ B_{\mu} \end{pmatrix} = \begin{pmatrix} \cos\theta_{W} & \sin\theta_{W} \\ -\sin\theta_{W} & \cos\theta_{W}\end{pmatrix}\begin{pmatrix} Z_{\mu} \\ A_{\mu}\end{pmatrix}\qquad (1.1)$$

And then we wrote: $$ \mathcal L = -\left(\sum_{\psi = e_{L}, e_{R}, \nu_{L}, \nu_{R}}\bar{\psi}\gamma^{\mu}\left( g\sin\theta_{W}I_3 + g'\cos\theta_{W}\frac{Y}{2}\right)A_{\mu}\psi + \bar{\psi}\gamma^{\mu}\left( g\cos\theta_{W}I_3 - g'\sin\theta_{W}\frac{Y}{2} \right)Z_{\mu}\psi \right) \qquad (1.2)$$

Unfortunately, I am not sure how the weak isospin $I_3$ enters in $(1.2)$ if we start from $(1.1)$, as $T^{3} = \frac{1}{2}\sigma^{3}$ is a Pauli-matrix..

Answer 1

$I_3$ denotes the eigenvalue of the $T_3$ matrix corresponding to the fermionic field in the sum. In the very first equation you write the sum is not over electrons and neutrinos. Instead you have the left doublets on which $T_3$ acts as the Pauli matrix, \begin{equation} l_k= \begin{pmatrix}\nu_{Lk}\\e_{Lk}\end{pmatrix} \end{equation} (where $k$ denotes the generation) and you have singlets on which $T_3$ acts as zero (!) \begin{equation} \bar{e}_{Rk}, \bar{\nu}_{Rk} \end{equation}

When the electroweak symmetry is broken it becomes useful to consider the components of the doublets individually. Because they are written in the eigenbasis of $T_3$ you can simply replace it with the corresponding eigenvalue $I_3$ that equals $+\frac{1}{2}$ for $\nu_{Lk}$, $-\frac{1}{2}$ for $e_{Lk}$ and zero for the right leptons.

Answer 2

I would like to add an explanation to @OON's statement that $I_{3}$ is an eigenvalue to $T^{3}$: If we consider at least an $SU(2)_{L}$-doublet of the form

$$ T^{3}\psi = T^{3}\begin{pmatrix} \nu_{L} \\ e_{L} \end{pmatrix}= \frac{1}{2}\sigma^{3}\begin{pmatrix} \nu_{L} \\ e_{L} \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \nu_{L} \\ e_{L} \end{pmatrix} = \frac{1}{2}\begin{pmatrix}\nu_{L} \\ -e_{L}\end{pmatrix} = I_3\begin{pmatrix} \nu_{L} \\ e_{L} \end{pmatrix},$$ since we know that the upper element of the $SU(2)_{L}$-doublet has a weak isospin of $I_{3} = +\frac{1}{2}$, and the lower elements $I_{3} = -\frac{1}{2}$.