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I Why did Dirac want a first-order equation?

  1. Jun 19, 2016 #1
    If I understand it correctly Dirac developed his equation because he was looking for a relativistic first order wave equation for the electron, rather than a second-order one like the Klein-Gordon equation.

    Why did he wanted a first-order equation? Is it because the probability current is not positive-definite for higher than second order equations?
  2. jcsd
  3. Jun 19, 2016 #2


    Staff: Mentor

    The wikipedia article gets into why Dirac chose first order instead of second order here:
    There's also some discussion on physics stackexchange about it here having to do with the 1/2 spin of fermions which implies a first order equation:

  4. Jun 20, 2016 #3
    Ok, so to sum up in case someone googles his way to this post:

    If the equation for the wavefunction is of second order the boundary conditions must include the initial value of its first time derivative, which can be negative. The expression for the current density involves the time derivative of the wavefunction, so if the value of that derivative was negative at some point (which as we have said cold happen) then you would have a non positive definite probability current, which is nonsensical.
  5. Jun 20, 2016 #4


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    Dirac's first motivation is invalidated already by himself. There is no working interpretation of relativistic QT in terms of "wave mechanics" a la Schrödinger's non-relativistic equation. The reason is that you necessarily are lead to a many-body theory with non-conserved particle numbers and thus also the existence of antiparticles. The most convincing argument for the Dirac equation we know today is the representation theory of the Poincare group together with the assumption of local interactions and the existence of a stable ground state, leading to the CPT and spin-statstics theorem and are the basis for the very successful Standard Model of elementary particle physics.
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