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Commutation relations in relativistic quantum theory

  1. Feb 25, 2008 #1
    Given the Hamiltonian [itex]H = \vec{\alpha} \cdot \vec{p} c + \beta mc^2[/itex],

    How should one interpret the commutator [itex][\vec{x}, H][/itex] which is supposedly related to the velocity of the Dirac particle? [itex]\vec{x}[/itex] is a 3-vector whereas H is a vector so how do we commute them. Is some sort of tensor product in order? ie

    [itex][\vec{x}, H] = x_i H_{jk} - H_{jk}x_i[/itex]?
     
  2. jcsd
  3. Feb 7, 2009 #2
    The Hamiltonian is not a vector but an operator on a Hilbert space which we can write as a matrix.

    Commutators like [tex][\vec{x}, H] = ...[/tex] are shortcuts for [tex][x_i, H] = ..., \quad i \in {1,2,3}[/tex].

    Note that [tex]x_i[/tex] is also an operator on a Hilbert space. Thus a commutator [tex][A,B] = C[/tex] is also a shortcut for [tex][A,B] \Psi = C \Psi[/tex] so the commutator tells you if there's a difference between [tex]AB \Psi[/tex] and [tex]BA \Psi[/tex] and how big this difference is.
     
  4. Feb 7, 2009 #3
    I've never seen any of this before. Interesting. I guess x=x*I, where x is an orbital operator, and I is the 4x4 identity. The beta emm cee ^2 should therefore commute with x=x*I so that does not contribute to the velocity dx/dt. The momentum that the alpha is multiplying does not commute with x, but is given by [x,p]=ih. So in the end i got (if I didn't mess up):

    [tex]\frac{dx^i}{dt}=c\alpha^i [/tex]

    So that the velocity is the 4-4 matrix alpha.

    This is really weird. Maybe the canonical/Heisenberg formulation has been stretched too far?

    The Hamiltonian almost looks like a Legendre transformation with the Lagrangian equal to mc^2.
     
  5. Feb 7, 2009 #4
    [itex]H_D = \vec{\alpha} \cdot \vec{p} c + \beta mc^2[/itex] is the Dirac Hamiltonian. The original Dirac equation looks like the Schrödinger equation but it contains the Dirac Hamiltonian instead of the Schrödinger Hamiltonian:

    [itex]H_D \Psi(\vec{x},t) = i \hbar \partial_t \Psi(\vec{x},t)[/itex]

    Using a special choice for the [itex]\alpha^i[/itex] and [itex]\beta[/itex] matrices, the Dirac [itex]\gamma[/itex] matrices, you can write it as

    [itex](i \gamma^{\mu} \partial_{\mu} - m) \Psi(\vec{x},t) = 0[/itex]
     
  6. Feb 7, 2009 #5
    Things kinda get confusing sometimes, and I think it has to do with the fact that we are interpreting the Dirac equation as an equation for a particle, instead of an equation for a field. We're treating the Hamiltonian as 4x4. In fact, if you sandwich this 4x4 Hamiltonian between two spinors [tex]\psi^{\dagger} H \psi[/tex], then you get the 1x1 Hamiltonian density for a field. Of course to get the Hamiltonian for a field, you just integrate the Hamiltonian density, so the Hamiltonian is 1x1 for a field, and 4x4 for a particle. Or at least that's what I think is going on.
     
  7. Feb 8, 2009 #6

    strangerep

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    Science Advisor

    This is an old "puzzle". You might want to read up on the Foldy-Wouthuysen transformation
    (but not from Wiki - it's F-W page is not very good). Try a good RQM textbook like that of
    Paul Strange. Also take a look at the Newton-Wigner position operator in parallel.
     
  8. Feb 8, 2009 #7

    samalkhaiat

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    Science Advisor

    Indeed you are right! Since the eigenvalues of [itex]\alpha_{i}[/itex] are [itex]\pm 1[/itex], we are faced with a paradox. That is the absolute value of the velocity of a relativistic electron always equals the velocity of light! Also, since the [itex]\alpha_{i}[/itex] do not commute with each other, the components of the velocity would not be simultaneously measurable! Clearly this nonsense (in cotradiction with the Ehrenfest theorem) would not give the classical result for the mean values.

    To have a "consistent" relativistic QM, i.e., relativistic one-electron theory, we must necessarily define all physical quantities by EVEN operators. That is

    [tex]
    \hat{O} \rightarrow [\hat{O}] = \frac{1}{2} (\hat{O} + \Omega \hat{O} \Omega )
    [/tex]

    where the Hermitian and unitary sign operator is defined ( in the momentum representation) by

    [tex]
    \Omega = \frac{\alpha_{i}p_{i} + m \beta}{E_{p}}
    [/tex]

    Note that [itex][H_{D}] = H_{D}[/itex] and [itex][\hat{p}_{i}] = \hat{p}_{i}[/itex], but

    [tex]
    [\alpha_{i}] = \hat{p}_{i} \frac{\Omega}{E_{p}}
    [/tex]

    Thus

    [tex]
    [\frac{d \hat{x}_{i}}{dt}] = \frac{\hat{p}_{i}}{E_{p}} \Omega
    [/tex]

    and the velocity is [itex]p_{i}/E_{p}[/itex] for positive and [itex]-p_{i}/E_{p}[/itex] for negative free particle solutions. Notice that the classical picture corresponds only to the positive solutions. One could carry on to bring about the connection with Zitterbewegung motion arround the classical trajectory by solving the EOM in the Hiesenberg representation.

    regards

    sam
     
  9. Feb 8, 2009 #8
    Thanks. Awhile ago I was reading a relativistic QM book by Bjorken and Drell, but I quit after the first couple of chapters because I wanted to jump to QFT (and also the Bjorken and Drell book seemed really old). So I think I missed this chapter on Foldy-Wouthuysen and Zitterbewegung. I'll be sure to read it when I have some spare time though as it seems interesting.
     
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