My problem is understanding the transformation behaviour of a Dirac spinor (in the Weyl basis) under parity transformations. The standard textbook answer is
ΨP=γ0Ψ=(0110)(χLξR)=(ξRχL),
which I'm trying to understand using the transformation behaviour of the Weyl spinors χL and ξR. I would understand the above transformation operator if for some reason χ→ξ under parity transformations, but I don't know if and how this can be justified. Is there any interpretation of χ and ξ that justifies such a behaviour?
Some background:
A Dirac spinor in the Weyl basis is commonly defined as
Ψ=(χLξR), where the indices L and R indicate that the two Weyl spinors χL and ξR, transform according to the (12,0) and (0,12) representation of the Lorentz group respectively. A spinor of the form
Ψ=(χLχR), is a special case, called Majorana spinor (which describes particles that are their own anti-particles), but in general χ≠ξ.
We can easily derive how Weyl spinors behave under Parity transformations. If we act with a parity transformation on a left handed spinor χL:
χL→χPL
we can derive that χPL transforms under boosts like a right-handed spinor
χL→χ′L=e→θ2→σχL
χPL→(χPL)′=(e→θ2→σχL)P=e−→θ2→σχPL, because we must have under parity transformation →σ→−→σ. We can conclude χPL=χR Therefore, a Dirac spinor behaves under parity transformations
Ψ=(χLξR)→ΨP=(χRξL), which is wrong. In the textbooks the parity transformation of a Dirac spinor is given by
ΨP=γ0Ψ=(0110)(χLξR)=(ξRχL).
This is only equivalent to the transformation described above of χ=ξ, which in my understand is only true for Majorana spinors, or if for some reason under parity transformations χ→ξ. I think the latter is true, but I don't know why this should be the case. Maybe this can be understood as soon as one has an interpretation for those two spinors χ and ξ...
Update:
A similar problem appears for charge conjugation: Considering Weyl spinors, one can easily show that iσ2χ⋆L transforms like a right-handed spinor, i.e. iσ2χ⋆L=χR. Again, this can't be fully correct because this would mean that a Dirac spinor transforms under charge conjugation as
Ψ=(χLξR)→Ψc=(χRξL),
which is wrong (and would mean that a parity transformation is the same as charge conjugation). Nevertheless, we could argue, that in order to get the same kind of object, i.e. again a Dirac spinor, we must have
Ψ=(χLξR)→Ψc=(ξLχR),
because only then Ψc transforms like Ψ. In other words: We write the right-handed component always below the left-handed component, because only then the spinor transforms like the Dirac spinor we started with.
This is in fact, the standard textbook charge conjugation, which can be written as
Ψc=iγ2Ψ⋆=i(0σ2−σ20)Ψ⋆=i(0σ2−σ20)(χLξR)⋆=(−iσ2ξ⋆Riσ2χL)=(ξLχR).
In the last line I used that, iσ2χ⋆L transforms like a right-handed spinor, i.e. iσ2χ⋆L=χR. The textbook charge conjugation possible hints us towards an interpretation, like χ and ξ have opposite charge (as written for example here), because this transformation is basically given by χ→ξ.
This post imported from StackExchange Physics at 2014-11-17 09:07 (UTC), posted by SE-user JakobH