# 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!)

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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

## 2 Answers

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.

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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.

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