With weak isospin doublets $ \bar{\Psi}^{L}=\Psi^{\dagger L} \gamma^0$ no more true

Question

I'm considering the doublet: $$\Psi_1^L \equiv \begin{pmatrix} \psi_{v_l}^L\\ \psi_{l}^L \end{pmatrix}$$ I know that under $SU(2)$ transformation: $$\Psi^{'L} = e^{\frac{i}{2}\vec{a} \cdot \vec{\sigma}}\Psi^{L}$$ I want to show that $$ \bar{\Psi}^{'L} = \bar{\Psi}^{L}e^{-\frac{i}{2}\vec{a} \cdot \vec{\sigma}}$$ Attempt:

$$ \bar{\Psi}^{L}=\Psi^{\dagger L} \otimes \gamma^0 = \Psi^{L} e^{-\frac{i}{2}\vec{a} \cdot \vec{\sigma}}\otimes \gamma^0 = \Psi^{L} e^{-\frac{i}{2}\vec{a} \cdot \vec{\sigma}}\otimes 1_2 \otimes \sigma_3 $$ $$\sigma_i \otimes \sigma_3 \neq \sigma_3 \otimes \sigma_i $$

Answer 1

The direct product $$ e^{-\frac{1}{2}\vec{\alpha} \cdot \vec{\sigma}} \otimes \gamma_0 $$ means that the isospin Pauli 2x2 matrices are orthogonal to the Dirac 4x4 matrices.

To reinforce the point, you may rewrite the isospin Pauli matrices $\sigma_i$ as $$ \sigma_i \equiv \sigma_i \otimes I_{4} $$ and the Dirac matrices $\gamma_\mu$ as $$ \gamma_\mu \equiv I_{2} \otimes \gamma_\mu $$

Therefore the commutative relationship between the two sets of matrices is manifest.

To help you visualize $SU(2)$ doublet further, you can regard the doublet as (note that if we work in the chiral basis, the 3rd and 4th Dirac components of left-handed fermions are zero) $$ \Psi_L \equiv \begin{pmatrix} \psi_{\nu_l}\\ \psi_{e_l} \end{pmatrix} \equiv \begin{pmatrix} \psi_{\nu_l1}, \psi_{\nu_l2},\psi_{\nu_l3}, \psi_{\nu_l4}\\ \psi_{e_l1},\psi_{e_l2},\psi_{e_l3},\psi_{e_l4} \end{pmatrix} $$ with Pauli matrices multiplying from left of $\Psi_L$ (i.e. $\sigma\Psi_L$) and Dirac matrices multiplying from the top of $\Psi_L$ (i.e. $(\gamma\Psi_L^T)^T$). You may interpret the multiplication rules as Pauli matrices $\sigma_i$ mixing between columns of $\Psi_L$ and Dirac matrices $\gamma_\mu$ mixing between rows of $\Psi_L$.

Of course, in text books, $\Psi_L$ is typically manipulated via the tensor notation $$ \Psi_{L\alpha\beta} $$ with indices $\alpha$ and $\beta$ contracting with isospin and Dirac matrix indices, respectively. Then you can forget about the matrix related mumbo jumbo and peacefully spectate the forthcoming Tuesday speech and Wednesday finale.

Answer 2

Start with the R isosinglet $\psi_R$ to fix your thinking. No group action to worry about. Its antiparticle is Left handed, of course, as the projector slipped off with a sign flip by the $\gamma_0$, $$ \overline{(\psi_R)}= (\overline{\psi})_L, $$ which prevents its kinetic term $\overline{\psi}_L \partial \!\!/ ~\psi_R ~ $from vanishing.

Now move to your L isodoublet (with R antiparticles), $$\Psi_L \equiv \begin{pmatrix} \psi_{\nu_l}\\ \psi_{l} \end{pmatrix}_L$$ so $$\Psi'_L = e^{\frac{i}{2}\vec{a} \cdot \vec{\sigma}}\Psi_{L},$$ whose hermitian conjugate is $${\Psi'}_L^\dagger = \Psi_{L}^\dagger e^{-\frac{i}{2}\vec{a} \cdot \vec{\sigma}}.$$ Multiply on the right with $\gamma_0$, which, of course, commutes with all isospin matrices, as all Dirac matrices do, $${\Psi'}_L^\dagger \gamma_0 = \Psi_{L}^\dagger \gamma_0 e^{-\frac{i}{2}\vec{a} \cdot \vec{\sigma}} \implies \\ \overline{ {\Psi'}}~ _R = \overline{\Psi}~_{R} e^{-\frac{i}{2}\vec{a} \cdot \vec{\sigma}} . $$

Can you now see how $$ \overline{\Psi}_R \cdot \partial \!\!/ ~\Psi_L ~ $$ is non-vanishing and su(2) invariant?

• Edit in response to comment: C is a 4x4 matrix only acting on Dirac spinor space and leaving the isospin tensor factor completely alone, $\gamma^\mu C= -C\gamma^{\mu T}$. But you are right that the conjugation operator which maps the spinor to its adjoint also takes the adjoint of the iso tensor factor. (If it helped you, you might think of each of the 4 Dirac spinor components as being proportional to the same iso-spinor (two-vector).)