# Four vectors from spinors

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!)

-
This can't possibly work since the four vector representation has spinorial indices $(1/2, 1/2)$. You can only obtain it as a product of left-handed and right-handed spinors. Taking a product of $(1/2, 0)$, $(1/2, 0)$ as you do would give you $(0,0) \oplus (1, 0)$ (i.e. you would get the usual singlet/triplet states for the product of two identical fermions). –  Marek Aug 18 '11 at 6:42
@Marek, the construction under consideration is not direct product. So, your point is correct but not for this particular problem. –  Misha Aug 18 '11 at 7:39
@Misha: hm, true that. I'll have to think about this. –  Marek Aug 18 '11 at 8:10
@Marek: If I were given one left-handed and one right-handed spinor, how could I calculate the components of the product of these two? (It seems that the above can be constructed by transforming e.g. $\xi_R$ to a left-handed spinor and then taking the direct product with $\psi_R$. But I haven't figured out the details yet.) –  David M. R. Aug 18 '11 at 17:11