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Klein gordon equation

  1. May 15, 2008 #1
    I am being really thick here

    I have this wave equation, the massless klien gordon equation

    [tex]\partial_{\mu}\partial^{\mu}\phi(x)=0[/tex]

    where the summation over [tex]\mu[/tex] is over 0,1,2,3

    the general solution is a superposition of plane waves yes? i.e

    [tex]\phi(x)=\int d^4 p \overline{\phi}(p)exp(i p_{\mu}x^{\mu})[/tex]

    where [tex]\overline{\phi}[/tex] is the weighting function.

    When you susbsitute this back into the klein gordon equation you get down two factors of p, i.e

    [tex]p_{\mu}p^{\mu}[/tex] which equals zero. (mass shell constraint), thus satisfying the equation of motion.

    My question is, is [tex]\overline{\phi}(p)[/tex] arbitrary? I don't really understand why this is so, let alone believe it.

    Hope peeps understand the question.
     
  2. jcsd
  3. May 15, 2008 #2

    neu

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    The superpostion equation you wrote is simply the fourier transform of [tex] \phi (x)[/tex].

    [tex] \phi (p)[/tex] are not arbitrary as [tex] \phi (p) = FT^{-1}[\phi (x)][/tex]

    Have I misunderstood your question?
     
  4. May 15, 2008 #3
    yeah I understand that.

    I'm being really thick here.

    I want the general solution [tex]\phi(x)[/tex] to the equation, ie as a superposition of plane wave solutions, but in fourier space.

    Ultimately I want to know what [tex]\overline{\phi}(p)[/tex] is.
     
  5. May 15, 2008 #4
    What I mean is, if [tex]\overline{\phi}(p)[/tex] is one weighting function whose fourier transform solves the klein gordon equation and [tex]\overline{\psi}(p)[/tex] is another different weighting function is the fourier transform of [tex]\overline{\psi}[/tex] also a solution.
     
  6. May 15, 2008 #5
    Well, any [tex]\phi(p)[/tex] will make [tex]\phi(x)[/tex] a solution. These are merely coefficients in your Fourier expansion. However, once you write the Hamiltonian in terms of [tex]\phi(p)[/tex] and [tex]\pi(p)[/tex], you will find that it simplifies greatly (decoupled harmonic oscillators, one for each p), and quantization is the next step.
     
  7. May 15, 2008 #6
    great as long as you're sure about that. That's what I hoped.

    I have guessed some expression between [tex]\phi(p)[\tex] and a whole load of other stuff and I want to test my conjecture on the computer. So, presumably I can just invent some suitable function for [tex]\phi(p)[\tex] stick it into a c-program and check it works.
     
  8. May 15, 2008 #7

    nrqed

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    As others said, phi(p) is arbitrary. The way to see this is the following: plane waves (with the condition on p^2) are solutions of the equation and the equation is linear, therefore arbitrary linear combinations of plane waves will satisfy the equation. Therefore phi(p) is arbitrary.
     
  9. May 15, 2008 #8

    strangerep

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    Adding my $0.02 to what others have already said, the [itex]\overline{\phi}(p)[/itex] is not
    entirely "arbitrary", because you have imposed the mass-shell constraint [itex]p_{\mu}p^{\mu}=0[/itex].
    Think of that as a "constraint hypersurface" in 4D momentum space.

    I.e., [itex]\overline{\phi}(p)[/itex] is undefined for values of p which are not on
    the constraint hypersurface.
     
    Last edited: May 15, 2008
  10. May 15, 2008 #9

    nrqed

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    True but this is not a condition on the functional form [itex]\overline{\phi}(p)[/itex], it's a restriction on the argument p. I mean, as long as p is a valid p, any function phi(p) is valid, right?

    A different consideration arises if we consider localized wavepackets phi(x). Then there must be a condition on phi(p).
     
  11. May 15, 2008 #10

    strangerep

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    Yes and yes.
     
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