Conjugate Transpose for a vector

  1. Hepth

    Hepth 523
    Gold Member

    In particle physics, we commonly have the gamma matrices, whose conjugate transpose is the raised or lowered index. Does the same rule apply to ANY indexed quantity? What about to scalar/vectors like momentum.

    Is the conjugate of momentum:

    [tex]
    \left(q_\mu\right)^\dagger = q^\mu
    [/tex]

    The reason I ask is I am trying to compute:

    [tex]
    \left(\sigma_{\mu\nu} q^\nu \right)^\dagger=
    [/tex]

    I get:
    [tex]
    \left(\sigma_{\mu\nu}\right)^\dagger =-\sigma^{\nu \mu}
    [/tex]
    but, since the q is a "scalar" quantity when summed over, it doesn't change any signs under the adjoint, correct?
     
    Last edited: Sep 22, 2009
  2. jcsd
  3. CompuChip

    CompuChip 4,297
    Science Advisor
    Homework Helper

    In general, it is not true that
    [tex]
    \left(q_\mu\right)^\dagger = q^\mu
    [/tex]
    for arbitrary q.
    Operators for which this is true are called Hermitian. In quantum mechanics, all observables are Hermitian (position, momentum and energy being some of them).
     
  4. Hepth

    Hepth 523
    Gold Member

    Yeah, I think I figured that out as I finish writing the question, but I guess you can't delete posts in homework help after they've been edited? So I let it stay.

    So this means while I'm calculating a decay amplitude of a process I NEED to keep track of the indices on all my q's.

    BUT if I have an operator/hamiltonian, say, what I have above: [tex]\sigma_{\mu\nu} q^\nu[/tex] this will end up being a vector (gammas) after contraction, correct? But the conjugate process, [tex] q^\mu \sigma^{\mu \nu}[/tex] doesn't contract. I mean, sure, the sums are still implied, but it doesn't leave me with an invariant, or an operator that transforms the same way as the original, right?
     
  5. CompuChip

    CompuChip 4,297
    Science Advisor
    Homework Helper

    I don't think
    [tex]
    q^\mu \sigma^{\mu \nu}
    [/tex]
    is a valid expression. You mightmean
    [tex]
    q_\mu \sigma^{\mu \nu}
    [/tex]
    but that will transform like a vector (only one with the index upstairs).
     
  6. Hepth

    Hepth 523
    Gold Member

    thats what I think, but following my notation:

    [tex]\left(\sigma_{\mu\nu} q^\nu\right)^\dagger[/tex]
    =
    [tex]\left(q^\nu\right)^\dagger \left(\sigma_{\mu\nu}\right)^\dagger[/tex]
    =
    [tex]q^\nu \left(\sigma_{\mu\nu}\right)^\dagger[/tex]

    [tex]\left(\sigma_{\mu\nu}\right)^\dagger = -\frac{i}{2} \left[\gamma_\mu , \gamma_\nu \right]^\dagger = -\frac{i}{2}\left( \left(\gamma_\mu \gamma_\nu \right)^\dagger - \left(\gamma_\nu \gamma_\mu \right)^\dagger\right) = -\frac{i}{2}\left( \gamma_\nu^\dagger \gamma_\mu^\dagger - \gamma_\mu^\dagger \gamma_\nu^\dagger \right) [/tex]
    with [tex]\gamma_\mu^\dagger = \gamma^\mu [/tex] right?
    so
    [tex]=-\frac{i}{2}\left( \gamma^\nu \gamma^\mu - \gamma^\mu \gamma^\nu \right) =\frac{i}{2}\left( \gamma^\mu \gamma^\nu - \gamma^\nu \gamma^\mu \right) = \sigma^{\mu \nu} [/tex]

    Therefor
    [tex]\left(\sigma_{\mu\nu} q^\nu\right)^\dagger = q^\nu \sigma^{\mu \nu}
    [/tex]

    Am I doing one of these wrong?
     
  7. turin

    turin 2,326
    Homework Helper

    Is that independent of the basis that you choose for the gammas? (I would be surprised if it is, but also happy to know about this.) Is there a reason that you have chosen to avoid such a construction as:
    [tex]
    A\gamma^{\mu\dagger}A=\gamma^{\mu}
    [/tex]
    (where, for me [itex]A[/itex] usually means [itex]\gamma^{0}[/itex])?
     
  8. turin

    turin 2,326
    Homework Helper

    It depends on your notation.1 At least I agree that this is an archaic notation (since the unification of GR and QM is in fashion). Also, the OP obviously wants to make a distinction between raised and lowered indices, but I'm not clear what this distinction should really be in the OP's context.



    1 See, for example, J. J. Sakurai, Advanced Quantum Mechanics, Addison-Wesley (1967) [ISBN 0-201-06710-2].
     
  9. Hepth

    Hepth 523
    Gold Member

    If you're in the dirac basis, http://en.wikipedia.org/wiki/Gamma_matrices,
    and you're metric is {1,-1,-1,-1} diagonal (for calculating this cross section), then isn't it true that
    [tex]
    \gamma^0^\dagger = \gamma_0 = \gamma^0
    [/tex]
    [tex]
    \gamma^1^\dagger = -\gamma^1 = \gamma_1
    [/tex]
    [tex]
    \gamma^2^\dagger = -\gamma^2 = \gamma_2
    [/tex]
    [tex]
    \gamma^3^\dagger = -\gamma^3 = \gamma_3
    [/tex]


    Ahh, so you're saying, I have to choose a basis first. Well, can't I just choose the dirac basis and stick with it? Or should/ MUST I leave it completely general. I mean, I'm using dirac spinors for my wavefunctions.
     
  10. Hepth

    Hepth 523
    Gold Member

    Using obviously [tex]\gamma_\mu = g_{\mu \nu} \gamma^\nu [/tex]
     
  11. turin

    turin 2,326
    Homework Helper

    I'm not saying that you have to, but I always do. What I'm really suggesting is that you can just use
    [tex]
    \gamma^{\mu\dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0}
    [/tex]
    and not worry about the raising and lowering with the Hermitian conjugate (for Lorentz tensor indices anyway). However, this does provide some kind of uniformity to the peculiar raising and lowering of spinor and flavor indices that one encounters in SUSY. I just never realized that I should also be raising and lowering a Lorentz tensor index under Hermitian conjugation.
     
  12. turin

    turin 2,326
    Homework Helper

    So, you're saying that
    [tex]\gamma^{\mu\dagger} = g_{\mu \nu} \gamma^\nu [/tex]
    so that the Hermitian conjugate acts on the gamma matrices in the same way that the metric tensor does. Then, I think that this is just a special property of the gamma matrices (and even perhaps a special subset of representations), and you should not extend this to general Lorentz-vector-valued objects. For instance, if [itex]q^\mu[/itex] is a Hermitian Lorentz-vector-operator, then your rule would suggest that
    [tex]q^\mu = g_{\mu \nu} q^\nu [/tex]
    which would only be true in certain frames (i.e. the rest frame of q using your metric signature).
     
  13. Hepth

    Hepth 523
    Gold Member

    No, i wasn't trying to generalize, just talking about gamma matrices and actually doing the matrix math you see, in the dirac basis at least, that it is true, for a {1,-1,-1,-1} metric.

    But, like was said earlier, i just ended up using the [tex]\gamma_0 \gamma_\mu^\dagger \gamma_0 = \gamma_\mu [/tex]

    Now I get to simplify 256 combinations of gamma matrices....
     
  14. turin

    turin 2,326
    Homework Helper

    There are computer algebra systems for that. Are you familiar with REDUCE?
     
  15. Hepth

    Hepth 523
    Gold Member

    No, but I heard about FeynCalc for mathematica which I was informed should do it.
     
  16. turin

    turin 2,326
    Homework Helper

    That works if you're willing to sell your soul ... No, seriously, that should work feyn. REDUCE is simply the free alternative that came to mind.
     
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