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Dirac bilinears

  1. Aug 16, 2007 #1
    i am confused why you need to write psi-bar multiplied by psi (where psi-bar is the hermitian conjugate of psi multiplied by y_0 (dirac matrix or gamma matrix)) instead of the h.c. psi multiplied by psi in order for it to transform as a lorentz scalar.

    Note: h.c. means hermitian conjugate
     
  2. jcsd
  3. Aug 16, 2007 #2

    nrqed

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    It simply follows from the transformation of psi. If you look up how psi transforms, you will see that psi^dagger psi is not invariant but "psi bar psi" is. Just check it.
     
  4. Aug 19, 2007 #3

    dextercioby

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    You must have that [itex] \gamma_{0} [/itex] there. It follows from the Clifford algebra of the gamma matrices and how the psi transforms under infinitesimal LT-s.
     
  5. Aug 19, 2007 #4
    It is also possible to prove that [itex]\psi^{\dagger}\gamma^0 \psi[/itex] is a scalar in full Lorentz transformations. It is not much more difficult than in infinitesimal transformations. Either way, you have to merely multiply three matrices (but you have to first solve the matrices of full transformations). I've always been slightly confused about why physicists always want to do stuff in infinitesimal scale :confused:
     
  6. Aug 19, 2007 #5

    dextercioby

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    Because physicists find it easier to work with Lie algebras rather than Lie groups.
     
  7. Aug 19, 2007 #6
    But for example

    [tex]
    (1-iA\alpha) X (1+iA\alpha) = (1+B\alpha) X + O(\alpha^2) \quad\implies\quad e^{-iA\alpha} X e^{iA\alpha} = e^{B\alpha} X
    [/tex]

    isn't really correct, is it?
     
  8. Aug 19, 2007 #7

    reilly

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    See any text on the Dirac eq. You'll find that (psi adjoint)*psi is charge, the time component of a four-vector, (psi bar )*psi is a scalar. It has to do with Lorentz xforms and normalization.

    Regards,
    Reilly Atkinson
     
  9. Oct 14, 2007 #8
    kweh?

    This doesn't even make sense as a statement with Lie Algebra... Recall these are operators we're dealing with.
     
  10. Oct 18, 2007 #9

    blechman

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    This isn't really what's going on. To see how infinitesimal rotations give you finite rotations, think of a finite rotation as nothing but an infinite number of infinitesimal rotations. For example, if you want to rotate by a finite amount x, then break x up into N pieces x/N and rotate N times. For N very large, this is an infinitesimal rotation so your rotation operator is:

    [tex]
    (1+i(x/N)A)^N
    [/tex]

    where A is your rotation generator. Now imagine taking [tex] N\rightarrow\infty [/tex], and you get [tex]e^{ixA}[/tex], the usual finite rotation.

    It is very important to understand that when we are doing these infinitesimal transformations, we are NOT making any approximations - all of the information for finite transformations is there in the infinitesimal transformation! It's not that physicists are sloppy and feel content to only consider "leading order" effects. This is all mathematically well defined and justified.

    I should say that I'm being a little cavalier in the above paragraph: there could be so called "topological" complications, which is just to say the way you take the above limit could be important. But these issues are not very important in general, and they don't change anything.

    Finally, let me just say that in more advanced applications of these topics (gauge theories, QFT, etc.), it is more appropriate to formulate the physics in terms of Lie Algebras (infinitesimal) rather than Lie groups (finite). For a wonderful explanation of why this is the case, I refer you to your favorite textbook. A very nice one is H. Georgi's "Lie Algebras for Particle Physics", but there are others out there as well.
     
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