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Klein-Gordon Equation & Continuity Equation

  1. Nov 20, 2012 #1
    Hello,

    My question is on the Klein-Gordon equation and it's relation to the continuity equation, so for a Klein-Gordon equation & continuity equation of the following form, I have attained the following probability density and probability current relations (although not normalised correctly? If that's the correct term):

    [tex]-\frac{\partial^2 }{\partial t^2}\psi +\bigtriangledown ^2\psi-m^2\psi=0 \: ,\: \frac{\partial }{\partial t}\rho +\mathbf{\bigtriangledown } \cdot \mathbf{j}=0[/tex]

    [tex]\rho=i(\psi^{*}\dot{\psi}-\psi\dot{\psi^{*}})\,,\; \mathbf{j} =i(\psi\bigtriangledown \psi^{*}-\psi^{*}\bigtriangledown\psi)[/tex]

    We then impose conservation of probability by making the time derivative of the probability equal to zero, and the following manipulation of this derivative is shown below:

    [tex]\frac{\partial P}{\partial t}=\int_{V}\frac{\partial \rho}{\partial t}\, d^3x=-\int_{V}\bigtriangledown\cdot \mathbf{j}\, d^3x=-\oint_{S}\mathbf{j}\cdot d\mathbf{s}=0[/tex]

    This above equation makes use of the divergence theorem in the last pair. This is true if j=0, but does j in general need to be equal to zero? I thought 'j' would vary around the surface and so the integration of all 'j' at each part on the surface would equal zero but not necessarily all 'j' =0.

    To be honest I'm quite unsure at what 'j' represents in terms of probability!

    Can anyone explain this?

    Thanks,
    SK
     
  2. jcsd
  3. Nov 20, 2012 #2
    Also I want to show that the probability density is not always positive but can be negative, so I tried substituting a plane wave solution of the form:

    [tex]\psi\sim e^{-iEt+i\mathbf{p\cdot x}}[/tex]

    into my probability density equation in the post above, but this gave me a positive answer - though if I use the negative energy plane wave solutions I will attain negative probability density (negative energy plane wave solutions attained by switching the signs in the exponent above), is this the correct way of showing this?

    The -ve energies correspond to the -ve probabilities?
     
  4. Nov 20, 2012 #3

    Demystifier

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    Sekonda, if you take a SUPERPOSITION of two plane waves with different positive energies, you will obtain that rho is not positive at some points. This implies that rho is NOT a probability density.

    For some details, see also
    http://arxiv.org/abs/0804.4564
     
  5. Nov 20, 2012 #4
    there is a problem with single particle representation of klein-gordon eqn.you can see that ρ depends on both ψ and ∂ψ/∂t,and klein-gordon eqn itself is a second order differential eqn which requires both ψ and ∂ψ/∂t to be specified at say t=0.so they can take arbitrary value and hence can make ρ negative.it is here identified as charge density rather than probability density.
     
  6. Nov 20, 2012 #5
    Thanks guys,

    I think that link is a bit too complex for me but I think I understand how to go about the superposition of waves, I'll try it. Though is my negative energy plane-wave not also a simple and fair enough way of showing that the probability density can be negative - but this only occurs for negative energy solutions.

    Cheers,
    SK
     
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