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Separate normalisation of chiral spinors?

  1. Apr 30, 2005 #1


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    When solving Dirac equation (for free massive particles) we usually impose normalisation conditions upon the eigenfunction in a single stroke.

    I am wondering, Is it possible/useful to impose separate normalisation conditions upon the left and right spinors? Should we still have a resolution of the identity, etc?
  2. jcsd
  3. Apr 30, 2005 #2


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    Yes,of course.See any book on weak interactions (the ones which used to treat neutrinos as being massless),as it deals with normalized solutions of the Weyl equations.I remember seing it in D.Bailin's book on weak interactions.

  4. May 3, 2005 #3


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    Hmm I was thinking in massive particles, not neutrinos.

    The think I am pursuing is some kind of study of non unitary chiral symmetry, this is, the one generated with [tex]i\gamma^5[/tex] instead of [tex]\gamma^5[/tex]. Do we get a conserved charge?
  5. May 5, 2005 #4


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    Let me to think aloud.

    If we dilate two spinors [tex]|R>,|L>[/tex] so that [tex]|R> \rightarrow k_R |R'>[/tex], [tex]|L> \rightarrow k_L |L'>[/tex]
    the new pair [tex]|R'>,|L'>[/tex] does not fulfil Dirac equation but instead
    [tex]D |L'> = {k_R \over k _L} m |R'>[/tex]
    [tex]D |R'> = {k_L \over k _R} m |L'>[/tex]

    Does this new pair has still some physical sense?

    Of course it is still a pair of relativistic wavefuntions of mass m. Intriguingly we have that different values of k map Dirac eq to the same bispinorial equation, because the new equation only depends on the quotient.

    The new bispinor [tex]\Psi' \equiv |L'> \oplus |R'>[/tex] still fulfils some normalisation-related properties, depending of the kind of transformations.

    Particularly if [tex]k_R, k _L[/tex] are real numbers (or diagonal self adjoint matrices) with [tex]k_R k _L=1[/tex] we have
    [tex]<\bar \Psi | \Psi> = <\bar \Psi' | \Psi'> [/tex]
    Last edited: May 5, 2005
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