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Time evolution and the Dirac Equation

  1. Apr 9, 2008 #1
    I have a question about the Dirac Equation. I know that if I have a given initial state in non-relativistic quantum mechanics, I can find the Fourier coefficients using that state, and then write down the wavefunction for any time. But if I have an initial state wavefunction (that is, the value of the wavefunction at t=0) in relativistic quantum mechanics, how do I find the wavefunction for any general time?

    Also, could anyone recommend books/websites that explain this in more detail? Thanks.
  2. jcsd
  3. Apr 9, 2008 #2
    Hi arunma,

    Normally you would find the time evolution by applying the "time evolution operator" to your initial wave function:

    [tex] |\Psi(t) \rangle = exp(itH) |\Psi(0) \rangle [/tex]

    So, you need to have a well-defined Hamiltonian H. However, I don't think this would work with the Hamiltonian of Dirac equation due to "zitterbewegung" problems.

    There is a discussion about the time evolution in quantum field theories (QED) in http://www.arxiv.org/abs/physics/0504062 . See, especially, chapter 9.

  4. Apr 9, 2008 #3
    Thanks for the link Eugene, I'll check it out.
  5. Apr 14, 2008 #4
    I think there is no problem (at least, no mathematical problem) in defining time evolution of the Dirac equation by way of a Hamiltonian.

    The Dirac equation can be written in the form
    \partial_0 \psi(t,x) = -i (\gamma^0 m + \gamma^0\gamma^j\partial_j) \psi(t,x)
    where psi is a 4-component spinor, and sum over j=1,2,3.

    If we take Fourier transforms in the 3 spatial dimensions, that is, define
    \hat{\psi}_t(k) = \int d^3x \; \psi(t,x) \; e^{-ikx} \qquad
    \psi(t,x) = \int \frac{d^3k}{(2\pi)^{3}} \; \hat{\psi}_t(k) \; e^{ikx}
    then the Dirac equation becomes
    \partial_0 \hat{\psi}_t(k) &= -i (\gamma^0 m + \gamma^0\gamma^j ik_j) \hat{\psi}_t(k).
    H = (\gamma^0 m + \gamma^0\gamma^j ik_j),
    and this can be solved; the solution is
    \hat{\psi}_t(k) &= \exp( -iHt) \hat{\psi}_0(k).
    where \psi_0 are the initial conditions. It's possible to evaluate the exponential in closed form, because H^2 simplifies using the result H^2 = E^2 = (k^2 + m^2).
    \exp( -iHt ) = \left( \cos( Et ) - \frac{i \sin( Et )}{E}H \right)
    Given that \hat{\psi}_t is now given explicitly for all t, this implies (by inverse fourier transform) that \psi(t,x) is constrained for all t.

    As a related comment, I thought it was quite interesting to note that evolution in Fourier space is simply pointwise multiplication by the matrix exp( -iHt ). This means that evolution in real space is convolution with the Fourier transform of exp( -iHt ), which is precisely what we mean by a propagator. I *think* that if you explicitly evaluate the Fourier transform of exp( -iHt ), you get to the retarded Green's function, but I got stuck with the math.

    Last edited: Apr 14, 2008
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