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Weyl spinor notation co/contravariant and un/dotted

  1. Aug 21, 2012 #1
    Hello,
    sorry for my english..
    I have a problem with weyl's spinors notation.
    I'm confused, becouse i read more books (like Landau, Srednicki and Peskin) and it's seems to me that all of them use different and incompatible notations..

    If i define

    [itex]M=\exp\left(-\frac{1}{2}(i\theta+\beta)\sigma\right)[/itex]

    as a generic lorentz transformation in left spinor rappresentation

    if [itex] \psi_\alpha [/itex] represent left covariant spinor that transform with M

    [itex] \psi^\alpha [/itex] represent left contravariant spinor that transform with M^(-1) right?

    so how do i represent covariant and contravariant right spinor in dotted notation?
    and how do they transform in connection with M matrix?

    if i transform covariant left spinor with [itex]\epsilon^{\alpha\beta}[/itex] I obtain a contravariant left spinor or not?

    the inner product involves dotted-dotted spinors (covariant and contravariant) or dotted-undotted spinors?

    thank you :)
     
  2. jcsd
  3. Aug 21, 2012 #2
    I would recommend using Wess and Bagger for notational purposes. It is short and conforms to the notations of many authors. Also if your looking for a free resource S. Martins "SUSY Primer" is also good for at least having the notation.

    The general rules for the left transformation are:

    [itex]\psi_{\alpha}[/itex] transfoms with M

    [itex]\psi^{\alpha}[/itex] transforms with M^-1

    For the right people usually put a bar on the spinor and the transform rules are:

    [itex]\bar{\psi}_{\dot{\alpha}}[/itex] transforms with [itex] M^{*} [/itex]


    [itex]\bar{\psi}^{\dot{\alpha}}[/itex] transforms with [itex] M^{* -1} [/itex]

    What is important to remember is that [itex]\psi_{\alpha}[/itex] and [itex]\bar{\psi}^{\dot{\alpha}}[/itex] are what we normally think of left and right handed spinors, respectively.

    Finally the epsilon symbol raises and lowers indeces and the inner product involves only like indeces. In fact, it makes no sense to have dotted-undotted since these objects live in different representations
     
  4. Aug 21, 2012 #3
    I forgot to mention that my convention conforms to Wess and Bagger and may disagree with other conventions
     
  5. Aug 21, 2012 #4
    with bar as hermitian conjugate right?
     
  6. Aug 21, 2012 #5
    No, the bar is there to tell you that it is right handed. The reason is because people prefer to remove indices from their notation. In doing so there must be a way to tell the difference between spinors without looking at whether or not there are dots on the indices.
     
  7. Aug 21, 2012 #6
    Ohh! Thank you! :smile:
    I hate notation problems!
    Finaly i've understood it!
     
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