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Questions on the Dirac equation

  1. Apr 7, 2014 #1

    Matterwave

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    I must admit that I have never had a great familiarity with the Dirac equation. No matter how many times I study it, I get bogged down in the algebra and never seem to get a good understanding of it. So here's a few questions in my mind at the moment. I am referring here to the Dirac equation as a single particle wave equation (not as a field equation).

    1) We have that there are 4 internal states of a spin 1/2 particle, 2 of which correspond to particle with two possible spin orientations, and 2 of which correspond to its antiparticle with 2 possible spin orientations. Does that mean we can have a particle which is a linear combination of particle+antiparticle states? This in my mind would be analogous to a particle which is a linear combination of spin up and spin down for example. If not, then why not? But if so, what is the physical meaning of an particle in a superposition of, for example, electron+positron states? By charge conservation, it seems this should be forbidden. But then again when we first learned QM, we found immediately that strict Energy conservation appears to be broken for linear combinations of different energy states, but that the expectation value of energy is conserved. Is there some analogous "the expectation value of charge is conserved" statement here?

    2) Can we transform a particle into its antiparticle via a Lorentz transformation? In my mind, this would be analogous to transforming a spin up along z particle to a spin up along x particle by rotating my x axis into the z axis. If not, then why not?

    3) Since the Lorentz boosts cannot be represented by unitary group of transformations, to construct a probability current and density which remains normalized under a Lorentz boost, we cannot use ##u^\dagger \gamma^\mu u## but instead we use ##\overline{u}\gamma^\mu u## where ##\bar{u}=u^\dagger\gamma^0##. Is this basically modifying the metric on our space from the usual metric to a modified metric where I have to multiply by a gamma matrix in the middle of all inner products? This would seem to, in turn, change our correspondences of kets to bras. The physical state which is dual to ##u## is now ##\overline{u}## and not ##u^\dagger##?
     
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  3. Apr 7, 2014 #2

    Avodyne

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    If you want to understand the Dirac equation as a single-particle wave equation, then there is no antiparticle; instead, there are negative energy states of the particle. Superposing states of different energy presents no problem, as usual in quantum mechanics.

    If you want to understand the negative-energy states as "really" being antiparticles, then you must abandon the interpretation of the Dirac equation as a single-particle wave equation.
     
  4. Apr 7, 2014 #3

    Matterwave

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    So what about question 2 and 3 then? Can we give a single particle a negative energy simply by making a Lorentz transformation?
     
  5. Apr 8, 2014 #4

    Avodyne

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    No. An orthochronous Lorentz transformation (that is, one that does not involve time reversal) does not change the sign of the energy.
     
  6. Apr 9, 2014 #5

    Matterwave

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    Thanks for the help. I have another question about the Dirac equation.

    Deriving the Dirac equation, one finds that there are equivalent representations of the spinors and transformation matrices corresponding to the Lorentz group. The most commonly used ones are the chiral representation, and the standard representation, which are related by a similarity transformation with a unitary matrix.

    I don't know very much about representation theory, but there's at least one other representation of the Lorentz group using 4x4 matrices that I know of, namely, the regular Lorentz transformations on 4-vectors in space-time. Why is this representation not used?
     
  7. Apr 9, 2014 #6

    WannabeNewton

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    These are not representations of the Lorentz group per say. What you're referring to are representations of the Clifford algebra for the Dirac (gamma) matrices ##\{\gamma^{\mu}, \gamma^{\nu}\} = 2g^{\mu\nu}## from which we can define a representation of the Lorentz algebra by ##S^{\mu\nu} = \frac{1}{4}[\gamma^{\mu},\gamma^{\nu}]## which can be easily checked. Now a representation of the Lorentz algebra has to act on some field. The field ##\psi^{\alpha}(x)## that this representation acts on is the Dirac field (here ##\alpha## is a spinor index) i.e. ##\psi^{\alpha} \rightarrow S^{\alpha}_{\beta}\psi^{\beta}(\Lambda^{-1}x)##; here ##\Lambda## is a 4-vector representation of the Lorentz group.

    The Dirac fields do not transform under ##\Lambda## they transform under ##S## as defined above; in other words these representations act on different fields (the latter on Dirac fields and the former on vector fields). ##S## and ##\Lambda## are fundamentally different representations. For example ##S## takes ##\psi^{\alpha} \rightarrow -\psi^{\alpha}## under a full rotation whereas ##\Lambda## as usual takes ##A^{\mu} \rightarrow A^{\mu}## under a full rotation (here ##\mu## is a Lorentz index).

    As an aside, the 4-vector representation (1/2,1/2) can be built from the spinorial representations (1/2,0) and (0,1/2) of the Lorentz algebra as can the 4-tensor representations and so the spinorial representation is the most fundamental so it's not that it isn't used at all (the photon field transforms under the 4-vector representation) but rather that it applies to vector fields, not spinor fields.
     
  8. Apr 10, 2014 #7

    Matterwave

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    Hmmm, thanks for your answer. I believe you mean that the field transforms as ##\psi^\alpha\rightarrow \exp(i\omega_{\mu\nu}S^{\mu\nu})^\alpha_\beta\psi^\beta(\Lambda^{-1}x)## right? So it seems that representations of the Lorentz algebra using the gamma matrices does not, in fact, exponentiate into the Lorentz group itself...but the simply connected group with the same Lorentz algebra right? Which group is that? I know, for example, for SO(3) that the simply connected group with the same Lie algebra is SU(2), but I don't know what's the case for O(1,3)...su(2)xsu(2)? (That would be my guess)
     
  9. Apr 10, 2014 #8

    WannabeNewton

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    Well that's for an arbitrary rotation angle; when I said "full rotation" above I meant a rotation by ##2\pi##.

    ##SL(2,C)##
     
  10. Apr 10, 2014 #9

    Matterwave

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    Ok thanks! :D
     
  11. Apr 10, 2014 #10

    dextercioby

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    It can't be [itex] \mbox{SU}(2)\times\mbox{SU}(2) [/itex], because this one is compact, while the restricted Lorentz group ([itex] \mbox{SO}_{\uparrow}(1,3)[/itex]) isn't.
     
    Last edited: Apr 10, 2014
  12. Apr 10, 2014 #11

    Bill_K

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    The covering group of the restricted Lorentz group SO(3,1) is SL(2,C).
     
  13. Apr 11, 2014 #12

    ChrisVer

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    A basis for sl(2,C) is the [itex]\sigma^{\mu}= (1, \vec{\sigma})[/itex] so a general element of sl(2,C)can be written as:
    [itex] A= x_{0} \sigma_{0} + \vec{x} \vec{\sigma} \in SL(2,C)[/itex]
    In matrix form:

    [itex]A= [/itex]
    [itex]x_{0}+x_{3} , x_{1}-ix_{2} [/itex]
    [itex]x_{1}+ix_{2}, x_{0}-x_{3} [/itex]

    From which you can take from detA=1:
    [itex] (x_{0}+x_{3}) (x_{0}-x_{3} ) - (x_{1}-ix_{2})(x_{1}+ix_{2}) = 1 [/itex]
    or that:
    [itex] x_{0}^{2}-x_{3}^{2} - x_{1}^{2}- x_{2}^{2} = 1 [/itex]
    and here you have that strange "sphere" thing, which gives you the [itex]SO(3,1)[/itex] due to the different signs of the metric...
     
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