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The Schrödinger equation as the non-relativistic limit of the Dirac equation

  1. May 18, 2012 #1

    I'm reading Griffiths' introduction to elementary particles and he seems to claim that the Schrödinger equation can be seen as a non-relativistic limit of the Dirac equation. I was wondering how one could deduce this, e.g. how do we go from
    [itex]\mathcal L = \bar{\psi} \left( i \gamma^\mu \partial_\mu - m \right) \psi[/itex]
    [itex]\mathcal L = \psi^\dagger \left( i \partial_t + \frac{\nabla^2}{2m} \right) \psi[/itex]
    (and somewhere along the way the psi goes from having 4 components to having one (?))

    But maybe I misinterpreted Griffiths; he simply states that the psi of the Dirac equation non-relativistically becomes the regular quantum mechanical wavefunction, and I assumed the latter to be the psi of the Schrödinger equation.
  2. jcsd
  3. May 18, 2012 #2


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    Each spinor component satisfies the Klein-Gordan equation

    [tex] (p^\mu p_\mu - m^2 ) \psi =0,[/tex]

    where we can write

    [tex]p^\mu p_\mu = E^2 - |\vec{p}|^2.[/tex]

    In the nonrelativistic limit,

    [tex] E = m + T,~~~T\ll m,[/tex]


    [tex]E^2 \approx m^2 + 2 m T.[/tex]

    The KG equation becomes

    [tex] 2mT \psi = |\vec{p}|^2 \psi,[/tex]

    which, upon identifying

    [tex]T = i \partial_t,~~~\vec{p}= -i \nabla[/tex]

    is the Schrodinger equation for a free particle. The same argument goes through if we include a background potential.
  4. May 18, 2012 #3
    As far as I know, the Lagrangian whose Euler equation gives Schroedinger equation should be:
    \mathcal{L} = \frac{i \, \hbar}{2} \, \left( \psi^{\dagger} \, \dot{\psi} - \dot{\psi}^{\dagger} \, \psi \right) - \frac{\hbar^2}{2 m} \, \nabla \psi^{\dagger} \, \nabla \psi
    where [itex]\psi[/itex] is a two-row column spinor, and [itex]\psi^{\dagger}[/itex] is the Hermitian adjoint.

    Maybe it's the same as the one you had posted by some integration by parts, but I just wanted to point it out.
  5. May 19, 2012 #4


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    There are 2 equivalent ways, one is spelled out by fzero (going from the Dirac eqn to the KG eqn for the Dirac spinor and ignoring spin and special relativity one gets the SE directly), the other is going to Pauli's equation and then neglect spin.

    Gordon Baym's text on QM has a nice description of the Dirac eqn and its approximations.
  6. May 20, 2012 #5
    In the rest frame of the particle its 4-momentum is [itex]p^{\mu}=(m c, 0)[/itex], and the Dirac equation in momentum space is:
    \left( \hat{\gamma}^{0} - \hat{1} \right) \, \psi = 0
    Depending on the choice of the explicit form of the time-like gamma matrix, we have two linearly independent out of the 4 components of the Dirac spinor equal to zero. What are these combinations? Those are the combinations that correspond to the 2 spin projections for the antiparticle state. So, Dirac equation simply projects them out.

    Indeed the matrix:
    \hat{P} = \alpha \left( \hat{\gamma}^{0} - \hat{1} \right)
    take its square and use [itex](\hat{\gamma}^{0})^{2} = \hat{1}[/itex]
    \hat{P}^{2} = 2 \alpha^{2} \, \left( \hat{1} - \hat{\gamma}^{0} \right)
    and choose [itex]2\alpha = -1[/itex], we get a projection operator that projects the antiparticle states. Its complementary operator:
    \hat{P}' = \hat{1} - \hat{P} = \frac{\hat{1} + \hat{\gamma}^{0}}{2}

    Act with this operator on the left of the Dirac equation, use the anticommutation relations to move it to the right of the gamma matrices to obtain an equation for [itex]\hat{P}' \phi = \psi[/itex] (although this is a 4 spinor, it has only 2 independent components). Then, assume this function to be multiplied by a variable phase [itex]\psi = e^{i f(x)} \, \psi[/itex]. By choosing it appropriately, you may be able to obtain a non-relativistic approximation.
    Last edited: May 20, 2012
  7. May 21, 2012 #6
    Thank you all.
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