In his book, Srednicki introduces the notion of twistor in chapter 50. It is described as a simply commuting spinor, as opposed to anti-commuting. How do we know that this object is simply commuting? Since it carries a spinor index, I expected the opposite.
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Presumably OP is worried that the spin-statistics theorem states that spin $1/2$ fields should be described by Grassmann-odd (as opposed to Grassmann-even) fields. Recall that in the Dirac field, the Grassmann-odd nature sits in the creation and annihilation operators.
On the other hand, the Dirac spinors $u_s({\bf p})$ and $v_s({\bf p})$, which Ref. 1 discusses, are Grassmann-even. In particular the commuting spinor $\phi_a$ that Ref. 1 mentions around eq. (50.8) is a Grassmann-even left-handed Weyl spinor.
References:
- M. Srednicki, QFT, 2007; Chapter 50. A prepublication draft PDF file is available here.
answered Oct 18, 2019 at 22:36