Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving Shouten identity in QFT

  1. Nov 26, 2011 #1
    Hi,

    I'm trying to prove the Shouten identity for twistors:

    [itex] \langle pq\rangle\langle rs\rangle+\langle pr\rangle\langle sq\rangle+\langle ps\rangle\langle qr\rangle=0 [/itex]

    It's easy to show that the LHS here is cyclically symmetric under [itex] q\to r\to s \to q[/itex], and also completely antisymmetric in q,r,s (just use [itex] \langle qr \rangle=-\langle rq \rangle[/itex] etc)

    But why does this imply the LHS must be zero?
     
  2. jcsd
  3. Nov 27, 2011 #2

    Bill_K

    User Avatar
    Science Advisor

    Because q, r, s each take on only two values, so if you antisymmetrize on three of them you get zero.
     
  4. Nov 29, 2011 #3
    Each twistor only has two spinorial components as far as I understand it. But then we are labeling each twistor by p,q,r,s and we have antisymmetry on q,r,s exchanges. I'm not sure I follow why this implies zero still, could you say some more please.
     
  5. Nov 29, 2011 #4

    Bill_K

    User Avatar
    Science Advisor

    A twistor is a SU(2,2) spinor and is written as a pair of dotted and undotted Weyl spinors, ZI = (λA, μA·). From the Weyl spinors we form Lorentz invariant products <λi, λj> = εAB λiA λjB and [μi, μj] = εA·B· μiA· μjB·. For short, <λi, λj> ≡ <i j> and [μi, μj] ≡ [i j].

    There are two Schouten identities,

    <i j><k l> + <i k><l j> + <i l> <j k> = 0
    [i j][k l] + [i k][l j] + [i l][j k] = 0

    They just express the fact that the λ's (and the μ's) are objects that live in a two-space, consequently any three of them must be linearly dependent, and if you try to antisymmetrize on all three of them you'll get zero.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proving Shouten identity in QFT
  1. Spin in QFT (Replies: 9)

Loading...