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Why does Klein-Gordon's equation not obey quantum mechanics?

  1. Sep 7, 2012 #1
    Hello Everyone,

    I was reading in Srednicki's QFT book, Chapter 1 and he was explaining why the Klein-Gordon equation doesn't obey quantum mechanics. He said the fact that the time derivative is second order means it disobey's Shrodinger's equation which is first order in the time derivative. He continues to say that this might not seem important but it has significant consequences. One consequence is that the norm:

    [tex] \langle \psi, t | \psi, t \rangle = \int d^{3}x \langle \psi, t | \bar x \rangle \langle \bar x | \psi, t \rangle = \int d^{3}x \psi^{*}(x)\psi(x) [/tex]

    is not in general time-independent (This is equation 1.23 in Srednicki's for reference). Thus the probability is not conserved, and hence we conclude that quantum mechanics is not obeyed (because quantum mechanics tells us that probability is conserved).

    I have three small questions here please. The first is how he got rid of the time dependence in the second equality above. My second question is how is the last equality time dependent? And finally my thrid question is why time dependence (in the norm of a state) implies probability is not conserved. What if probability is conserved at everytime, t, but the probability distribution is time dependent?

    Thank you in adavance,
  2. jcsd
  3. Sep 7, 2012 #2
    In the equation you've copied from Srednicki, x (unbolded) is a general spacetime point, not just a spatial position [Srednicki uses bold-x to denote a pure spatial position, as in the middle expression]. So the right hand side still contains the time dependence.

    If we want to interpret |psi(x, t)|^2 as a probability density, it must be that the probability of finding the particle *somewhere*, at any specific time t, is 1. So the integral of |psi(x, t)|^2, integrated over all space, must be 1, independent of t. If this isn't true then we can't interpret |psi(x, t)|^2 as a probability density.
  4. Sep 7, 2012 #3
    Thank you very much for the wake-up call on the 4-vector notation!

    Regarding the time-dependence, I understand the point made now, thank you. However, how would having an equation of motion which is first order in time solve this problem of the time dependence in the probability? Or, why does a second order derivative in time cause this problem.

    Thank you again very much!
  5. Sep 8, 2012 #4


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    iman, To say that the Klein-Gordon equation "does not obey quantum mechanics" is a bit misleading. What Srednicki meant to say was that the Schrodinger wavefunction ψ has certain properties, and the Klein-Gordon field φ cannot be interpreted as a single-particle probability amplitude since it does not share them.

    For a probability interpretation, one would expect to find a continuity equation ∂ρ/∂t + ∇·J = 0, where ρ, J are the probability density and current respectively. For the Klein-Gordon equation the best we can do is

    ρ = (iħ/2mc2)(φ* ∂φ/∂t - φ ∂φ*/∂t)
    J = (-iħ/2m)(φ* φ - φ φ*)

    And in fact using Eφ = iħ∂φ/∂t we get ρ = (E/mc2)φ*φ, which in the nonrelativistic limit reduces to the Schrodinger probability density φ*φ. However, due to the second-order nature of the Klein-Gordon equation, φ and ∂φ/∂t can be prescribed arbitrarily, meaning that ρ can just as well be negative. This rules out its interpretation as a probability density.

    Accounts that follow the historical order of things make it sound as if the Klein-Gordon equation itself was to blame. Rather it was the mistaken assumption prevalent in the 1920's that a relativistic single-particle quantum theory was possible. In 1934, Pauli reinterpreted the K-G equation as a field equation and quantized it.
    Last edited: Sep 8, 2012
  6. Sep 10, 2012 #5
    Bill_K, thank you very much for your explanation!
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