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Weyl Spinors

  1. Sep 13, 2010 #1
    Hey guys,

    I have a question about said spinors.

    In supersymmetry introductions one finds (e.g. for two left-handed spinors [tex] \eta [/tex], [tex] \nu [/tex]) that [tex]\eta\nu=\nu\eta[/tex] due to their Grassmannian character and the antisymmetry of the spinor product.

    If I look, however, at modern field theoretical methods that rely on the fact that SO(3,1) is doubly covered by SL(2,C) to express 4-vectors in terms of Weyl spinors, I find for the spinor product of two left-handed Weyl spinors (in a slightly different notation) [tex]\langle \nu |[/tex], [tex]\langle \eta |[/tex] is given by
    [tex]\langle \nu |\eta\rangle=-\langle \eta |\nu\rangle[/tex].

    I don't see where the minus-sign enters, since in my view only the notation is different, whereas the definition of the spinor product is the same in the two cases...

    Does someone know where the difference comes from?
    Cheers,
    earth2
     
  2. jcsd
  3. Sep 13, 2010 #2
    I had a great many arguments about those minus signs. We finally did track it down. Suppose we're working with spinors A and B whose components are complex (commuting) numbers. The definition of the inner product is AB = epsilon_mn A^m B^n = -epsilon_nm A^n B^m = -epsilon_nm B^m A^n = -BA.

    Now, suppose that instead they're either grassmann numbers or fermionic fields. In that case, the second to last step requires an anticommutation, which kills off the minus from the previous line, leaving you with = BA.

    Thus, there are different kinds of spinors depending on your goal. If the spinor is to describe a classical fermion, use grassmann numbers. If it's to describe a quantum fermion, use anticommuting fields. If you're just decomposing a null vector in terms of spinors (for example, p^mu -> sigma_mu p^mu = lambda_a lambda_adot), then use complex numbers.
     
  4. Sep 13, 2010 #3
    Thanks for the reply! :)

    I still don't get one subtle thing about the "goal":
    Why don't I use grassmann numbers when i decompose a nullvector? I mean, all I do is to use the isomorphism between SO(3,1) and SL(2,C), i.e. i express all my four vectors via Weyl spinors...But suddenly they don't have grassmannian entries anymore...
    I am confused...

    Edit: Btw, what is the precise definition of a Weyl spinor? Perhaps that sheds some light on the problem...?
     
  5. Sep 14, 2010 #4
    All I mean is something that transforms under the 2 rep of SO(1,3)/the fundamental of SL(2,C).

    Suppose I made the fourvector p^mu = (E, 0, 0, E). In order to write it in terms of Weyl spinors, i'd form sigma dot p = ((2E, 0), (0, 0)) and then factor it into a column vector times a row vector. In this case, I'd say lambda = ((sqrt(2E)), (0)) and lambdabar = (sqrt(2E), 0). I can do that, but if I want p to transform like a fourvector under lorentz transformations, I have to insist that lambda transforms like a left-handed spinor and lambdabar like a right-handed one. (Maybe I have it backwards; I forget.) Doing that reproduces the correct transformation properties on p.

    What's clear from this example is that the components of lambda and lambdabar MUST be commuting numbers; if we had used grassmann numbers we wouldn't have complex-valued numbers at the end of the day, but rather more grassmann numbers.

    If, instead, you were to start the day by saying I want lambda to represent the state of a fermion, then you find that, via the spin-statistics connection, that spinor components must anticommute with other spinor components. The two methods of doing that are to make it grassmann-valued in the classical case, or anticommuting operators in the quantum case.
     
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