Four vectors from spinors

Question

In Exercise 2.3 of A modern introduction to Quantum Field Theory by 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, $\sigma^\mu = (1,\sigma^i)$.

I have shown that it does so for boosts along the x-axis by explicitly transforming the two spinors and showing that the components of $V^\mu$ transform correctly. It seems easy to do the same thing for boosts along other directions and rotations around the separate axes.

However, I would like to show this for a general Lorentz-transformation, i.e. I would like to show $$ (\Lambda_R \xi_R)^\dagger \sigma^\mu (\Lambda_R \psi_R) = \Lambda_\nu^\mu \xi_R^\dagger \sigma^\nu \psi_R \text{ }\text{ }\text{ }\text{ }\text{ }\text{ (2)} $$ Trying to do this explicitly seems even less elegant than what I have done so far. I have tried commuting $\sigma^\mu$ to the right of $\Lambda_R$ in (2), but that gives a rather complicated expression.

Is there a slightly more elegant way to see why (2) is true without computing all the components explicitly? (I am not looking for a solution, but would rather have a hint!)

Answer 1

You're essentially done, because any proper, orthochronous Lorentz transformation is the product of a rotation and a boost. If you are not familiar with this fact, try looking up the proof of polar decomposition of matrices and you will see that it generalizes easily. If you get stuck, add a comment below and I'll update this answer with more details.

Answer 2

Did you try to use the fact that the product of two spinors transforms in the same way as a vector? It is fairly general rule (in fact, is it a definition), but to use it for transformation law of Pauli matrices you need carefully work with the indices.