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Dual spinor and gamma matrices

  1. Aug 14, 2017 #1
    Here it is a simple problem which is giving me an headache,


    Recall from class that in order to build an invariant out of spinors we had to introduce a somewhat
    unexpected form for the dual spinor, i.e. ߰ψ = ψ⋅γ0
    Then showing that ߰ is invariant depends on the result that (ei/4⋅σμν⋅ωμν) ⋅γ0 = γ0⋅e-i/4⋅σμν⋅ωμν
    Prove this by expanding out the exponential for the first three terms and using the (anti)commutation relations of the gamma matrices.




    2. Relevant equations
    The Metric signature is (-+++)

    You can read the problem (number 2) by clicking the link below.

    https://inside.mines.edu/~aflourno/Particle/HW4.pdf

    3. The attempt at a solution

    My issue is that I can't see why there is a - sign in the exponential after transpose/conjugating and moving the gamma matrix.

    Moving the gamma matrix γ0 to the left cancels the negative sign from conjugating the σ0j in the exponential hence the + sign should not change.
    Which step am I missing?

    You can read the solution by clicking the link below.

    https://inside.mines.edu/~aflourno/Particle/HW4solutions.pdf


    Thank you for reading and any replies.
     
  2. jcsd
  3. Aug 14, 2017 #2

    jambaugh

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    Gold Member

    The hint would seem to explain it... the [itex]\gamma^0[/itex] will commute with the rotation generators and anti-commute with the boost generators. But the conjugate transpose of (i times) the rotation generators are their negatives (they times i must be anti-hermitian to generate unitary rotations). Contrawise the boost generators (times i) are hermitian (under conjugate transpose) and thus will not change sign under conjugate transpose. Either one or the other but not both sign changes will occur in all terms.

    [EDIT, added:] This makes all the generators (times i) anti-pseudo-Hermitian in the pseudo-Hilbert space wherein we represent relativistic spinnors. The [itex]\gamma^0[/itex] matrix is effectively the indefinite part of the pHilbert space's inner product expressed by the Dirac bar adjoint vs the usual conjugate transpose. The way it is expressed here is still horribly basis dependent to some extent. The [itex]\gamma^0[/itex] matrix as it is used here is really a very different object than the [itex]\gamma^0[/itex] matrix used to represent the time inversion transformation in pin(3,1). I believe the object that is in this representation equal to [itex]\gamma^0[/itex] is referred to as [itex]\beta[/itex] and will differ from [itex]\gamma^0[/itex] in other representations.

    It is similar to how we use a matrix to represent a rank 2,0 tensor [itex]g^{\mu\nu}[/itex] as well as a rank 1,1 tensor (operator) [itex]R^\mu_\nu[/itex]. You might even have two which have the same matrix form in a given basis (time inversion operator vs space-time metric in the (-,+++) signature). But they are distinctly different types of objects with different behaviors under change of basis transformations.
     
    Last edited: Aug 14, 2017
  4. Aug 15, 2017 #3
    .


    The anticommutation relations were clear to me but I have just realised I was not taking into account the i before before the generators thus the -sign after the conjugate/transposition.
    A silly mistake.


    Thank you
     
  5. Aug 15, 2017 #4

    jambaugh

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    Gold Member

    Well its the silly mistakes that are easiest to correct.
     
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