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Calculations with Weyl Spinor Indices in QFT

  1. Jun 30, 2011 #1
    1. The problem statement, all variables and given/known data
    The task is to show the invariance of a given Lagrangian (http://www.fysast.uu.se/~leupold/qft-2011/tasks.pdf" [Broken]), but my problem is just in one step (which i got from Peskin & Schröder, page 70) which i can not reproduce due to my lack of knowledge regarding spinors.

    The step i am talking about is 3.147 in the attached picture or written out:


    2. Relevant equations

    [itex] C \bar \psi \psi C = (-i \gamma^0 \gamma^2 \psi)^T (-i \bar \psi \gamma^0 \gamma^2)^T [/itex]
    [itex] = -\gamma^0_{ab}\gamma^2_{bc} \psi_c \bar \psi_d \gamma^0_{de}
    \gamma^2_{ea} [/itex]
    [itex] = \bar \psi_d \gamma^0_{de} \gamma^2_{ea} \gamma^0_{ab} \gamma^0_{bc} \psi_c[/itex]
    [itex] = -\bar \psi \gamma^2 \gamma^0 \gamma^0 \gamma^2 \psi [/itex]
    [itex] = \bar \psi \psi [/itex]

    3. The attempt at a solution
    Well, i browsed through Wikipedia, Google and Friends, but did not find anything.
    I know how to handle the gamma matrices (like their commutation relations or [itex](\gamma^0)^2 = 1[/itex]) .
    I have just no clue what these indices mean, why they are sorted the way they are ("ab bc cd de" instead of e.g. "ab cd ef gh") and how [itex]\psi^T [/itex] gets converted to [itex]\psi [/itex].

    Thanks!
    Regards,
    Nico
     

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  3. Jul 1, 2011 #2

    George Jones

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    It's been years since I've done any QFT, and I don't have any books here with me right now, so what I write might be completely wrong.

    First, consider some linear algebra. If [itex]A[/itex] is a square matrix, then [itex]\vec{w} = A \vec{v}[/itex] becomes

    [tex]w_i = \sum_j A_{ij} v_j[/tex]

    in component form. It looks like the matrix multiplications have been written in component form with the summation symbols omitted, which is why there are repeated indices.

    Next, is it true that for a scalar that

    [tex]C \bar{\psi} \psi C = \left( C \bar{\psi} \psi C \right)^T ?[/tex]

    Can you use this to do the calculation more simply, taking into account that fermion fields anticommute?
     
  4. Jul 5, 2011 #3
    Thanks for your answer!
    At i thought it was the standard matrix multiplication, too. But as you wrote it, it already begins with an index. So its
    [tex]w_i = \sum_j A_{ij} v_j[/tex] instead of
    [tex]w = \sum_j A_{ij} v_j[/tex], which just confuses me.

    Yeah, i think so, too. Since its a scalar transposing it shouldnt change anything :P
    But i havent come really far doing that task. I did others first, but ended up at the same spot with the same problem eventually. The same thing even appeared a few times more now.
     
  5. Jul 5, 2011 #4

    George Jones

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    If A and B are matrices, how is the product AB written using indices?
     
  6. Jul 6, 2011 #5
    ah, i think i got it.
    yeah, since (as i just mentioned in my previous post :P) the end product is a scalar, it makes sense to write it this way and it really is just a index notation for a vector-many matrices-vector product. a normal scalar product of 2 vectors would have the same features when looked at it written out in components.
    im kind of taking a break right now, but when i start again and still get stuck, i might will ask you further questions :P
    thanks!
     
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