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  How does the Lorentz group act on a 4-vector in the spinor-helicity formalism pα˙α?

+ 2 like - 0 dislike
2704 views

Given a 4-vector pμ the Lorentz group acts on it in the vector representation: pμ(JV[Λ])μνpνΛμνpν.

However, I can always represent a 4-vector pμ using left and right handed spinor indices, writing pα˙ασμα˙αpμ.
So the question is: in what representation does the Lorentz group act on pα˙α?


There are a lot of questions about this and related topics around physics.se, with a lot of excellent answers, so let me clear up more specifically what I am asking for.

I already know that the answer to this question is that the transformation law is pα˙α(ApA)α˙α

with ASL(2,C) (how is mentioned for example in this answer by Andrew McAddams). I also understand that so(1,3)sl(2,C),
(which is explained for example here by Edward Hughes, here by joshphysics, here by Qmechanic).

So what is missing? Not much really. Two things:

  1. How do I obtain (3) and what is the specific form of A, i.e. its relation with the vector representation Λμν? Defining the following (˜p)pμ,ΛΛμν,

    σσα˙α,ˆppα˙α,
    we can rewrite (1) and (2) in matrix form as ˆpσ˜pσΛ˜p=(σΛσ1)ˆp,
    however, this disagrees with (3) which I know to be right, so what is wrong with my reasoning?

  2. Why does the transformation law (3) has a form AU1AU,

    while the usual vector transformation (1) has a form VΛV? I suspect this comes from a similar reason to that explained here by Prahar, but I would appreciate a confirmation about this.

This post imported from StackExchange Physics at 2015-01-13 11:51 (UTC), posted by SE-user glance
asked Jan 12, 2015 in Theoretical Physics by glance (65 points) [ no revision ]
Related: physics.stackexchange.com/q/28505/2451 and links therein.

This post imported from StackExchange Physics at 2015-01-13 11:51 (UTC), posted by SE-user Qmechanic
@Qmechanic I edited the question to better specify my problem

This post imported from StackExchange Physics at 2015-01-13 11:51 (UTC), posted by SE-user glance
The question (v2) seems closely related to physics.stackexchange.com/q/153736/2451

This post imported from StackExchange Physics at 2015-01-13 11:51 (UTC), posted by SE-user Qmechanic
@Qmechanic I already saw it and I agree. Unfortunately there are no answers there.

This post imported from StackExchange Physics at 2015-01-13 11:51 (UTC), posted by SE-user glance

1 Answer

+ 1 like - 0 dislike

Your equation (3) comes from the following steps. First, a dotted index transforms in the complex conjugate representation of an undotted index. For a tensor product, each index transforms according to its own representation. Thus pa˙aAabˉA˙a˙bpb˙b=Aabpb˙bA˙b˙a

where on the left side of the equal sign we have elementwise complex conjugation. Putting the conjugate matrix on the right we have to take a transpose to get order of indices right.

In reasoning about (4) and (5) you are neglecting the transformation of σμa˙a. The correct description of the relation pa˙a=σμa˙apμ is that the 4-vector representation is equivalent to the (12,0)(0,12) representation, by means of the linear transformation σμa˙a:V(12,0)(0,12)

meaning that σμa˙a belongs to the space (12,0)(0,12)V, on which the (double cover of the) Lorentz group acts. In fact, it acts like σμa˙aAa˙bσνb˙bA˙b˙a(Λ1)μν
so that Aμa˙apμ indeed has the correct transformation law.

This post imported from StackExchange Physics at 2015-01-13 11:51 (UTC), posted by SE-user Robin Ekman
answered Jan 12, 2015 by Robin Ekman (215 points) [ no revision ]
Thank you very much, that definitely sorted it out. Just one thing: could you also provide some reference where I can find more on the subject (particulary where I can find an exposition of the transformation rules of σμα˙α that you quoted)?

This post imported from StackExchange Physics at 2015-01-13 11:51 (UTC), posted by SE-user glance
That σμa˙a transforms in that manner is really implicit in the indices it has, so I don't know if it's written out anywhere. My best guess is Penrose & Rindler, but I haven't checked.

This post imported from StackExchange Physics at 2015-01-13 11:51 (UTC), posted by SE-user Robin Ekman

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