Building 4-vectors out of Weyl spinors: Combining 2 independent Weyl spinors and a sigma matrix to get a 4-vector

Question

i'm struggling with this problem

In Exercise 2.3 of A Modern Introduction to Quantum Field Theory of Michele Maggiore I am asked to show that, if $\xi_R$ and $\psi_R$ are right-handed spinors, then $$ V^\mu = \xi_R^\dagger \sigma^\mu \psi_R$$ transforms as a four vector. Here, $\sigma^\mu = (1,\sigma^i)$.

Here the extract from the book

I saw also that was discussed here 4-vector from a spinor and here Four vectors from spinors but I'm still not getting how to prove that $$ (\Lambda_R \xi_R)^\dagger \sigma^\mu (\Lambda_R \psi_R) = \Lambda_\nu^\mu \xi_R^\dagger \sigma^\nu \psi_R $$ It's important for me to understand this mapping because in the definition of the Lagrangian density of Left/Right-handed Weyl spinor fields this transformation property is crucial and it has conceptual consequences in the treatment of Dirac's and Majorana fields.

I tried the following: Starting from the knowledge of the transformation property of the Right-handed Weyl spinor $$\psi_R\longrightarrow \Lambda_R \psi_R=\exp\left[{(-i\vec{\theta}+\vec{\eta})\cdot\frac{\vec{\sigma}}{2}}\right]\psi_R$$ where $$\Lambda_R=\exp\left[{(-i\vec{\theta}+\vec{\eta})\cdot\frac{\vec{\sigma}}{2}}\right]$$ Now, writing the left hand side explicitly, knowing that $\vec{\sigma}=\vec{\sigma}^\dagger$ $$ \xi_R^\dagger\Lambda_R^\dagger \sigma^\mu \Lambda_R \psi_R = \Lambda_\nu^\mu \xi_R^\dagger \sigma^\nu \psi_R $$

$$ \xi_R^\dagger\exp\left[{(+i\vec{\theta}+\vec{\eta})\cdot\frac{\vec{\sigma}^\dagger}{2}}\right] \sigma^\mu \exp\left[{(-i\vec{\theta}+\vec{\eta})\cdot\frac{\vec{\sigma}}{2}}\right] \psi_R =$$$$ \xi_R^\dagger\exp\left[{(+i\vec{\theta}+\vec{\eta})\cdot\frac{\vec{\sigma}}{2}}\right] \sigma^\mu \exp\left[{(-i\vec{\theta}+\vec{\eta})\cdot\frac{\vec{\sigma}}{2}}\right] \psi_R. $$ Expanding the exponential is the right way to solve it?

$$\xi_R^\dagger\left[I+{(+i\vec{\theta}+\vec{\eta})\cdot\frac{\vec{\sigma}}{2}}+\dots\right] \sigma^\mu \left[I+{(-i\vec{\theta}+\vec{\eta})\cdot\frac{\vec{\sigma}}{2}}+\dots\right] \psi_R=$$ $$=\xi_R^\dagger\sigma^\mu\psi_R+\xi_R^\dagger{(+i\vec{\theta}+\vec{\eta})\cdot\frac{\vec{\sigma}}{2}}\sigma^\mu\psi_R+\xi_R^\dagger\sigma^\mu{(-i\vec{\theta}+\vec{\eta})\cdot\frac{\vec{\sigma}}{2}}+\dots=$$ While writing the right hand side explicitly $$\Lambda_\nu^\mu \xi_R^\dagger \sigma^\nu \psi_R=\left(I-\frac{i}{2}\omega_{\alpha\beta}J^{\alpha\beta}\right)^\mu_\nu\xi_R^\dagger \sigma^\nu \psi_R$$ How can i fix these two expressions? Would be possible to prove this property without explicit calculation component by component but using some properties instead?

Answer 1

For a four vector, the infinitesimal Lorentz transformation can be written as $$\delta V^0= \eta^i V_i$$ $$\delta V^i= \eta^i V_0-\epsilon^{ijk} \theta_j V_k$$ We may consider the infinitesimal transformation of the right-handed Weyl spinor as well, which is $$\delta \psi_R=\frac12 (-i\theta^i+\eta^i)\sigma_i\psi_R$$ acorrding to your conventions. then $$\delta (\xi ^\dagger_R \sigma_0 \psi_R)=\frac12\xi ^\dagger_R\sigma_0 (-i\theta^i+\eta^i)\sigma_i\psi_R +\frac12 \xi ^\dagger_R (i\theta^i+\eta^i)\sigma_i\sigma_0\psi_R=\eta^i\xi ^\dagger_R \sigma_i \psi_R$$ $$\delta (\xi ^\dagger_R \sigma_j \psi_R)=\frac12\xi ^\dagger_R\sigma_j (-i\theta^i+\eta^i)\sigma_i\psi_R +\frac12 \xi ^\dagger_R (i\theta^i+\eta^i)\sigma_i\sigma_j\psi_R=\eta_j\xi ^\dagger_R\psi_R-\theta^I{\epsilon_{ij}}^k\xi ^\dagger_R\sigma_k\psi_R $$ by substituting the commutation and anti commutation relations of the Pauli matrices. And we can find that the Lorentz transformation of $\xi ^\dagger_R \sigma_j \psi_R$ is exactly the same as a four vector.