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Klein-Gordon equation

  1. Dec 20, 2011 #1

    Matterwave

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    Hello, I have a question. In Ryder chapter 2, he develops the KG equation and says something along the lines of "the density, in order to be relativistic, must transform like the time component of a 4 vector" and he immediately gives:
    [tex]\rho=\frac{i\hbar}{2m}\left(\phi^*\frac{\partial \phi}{\partial t}-\phi\frac{\partial \phi^*}{\partial t}\right)[/tex]

    Where did this come from? Seems like he just pulled it out of thin air...o.o
     
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  3. Dec 20, 2011 #2

    vanhees71

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    Well, that's anyway not a very good explanation. First of all, in relativistic quantum theory, a one-particle description doesn't make sense (except for free particles), and thus one should immediately use quantized fields rather than classical fields.

    Then, the idea is to get relavistically covariant field equations (for the field operators). Thus the most convenient starting point is a Poincare invariant action with a Lagrange density that depends only on the fields and the first space-time derivatives. For the free field there should be no higher powers of field operators and their derivatives than 2nd. Thus, the most simple possibility is

    [tex]S[\phi]=\int \mathrm{d}^4 x [(\partial_{\mu} \phi^\dagger)(\partial^{\mu} \phi) - m^2 \phi^{\dagger} \phi ].[/tex]

    Hamilton's principle leads to the Klein-Gordan equation,

    [tex](\Box+m^2) \phi=(\Box+m^2) \phi^{\dagger}=0.[/tex]

    Now, if you want to interpret the field operator to describe, e.g., the electromagnetics of charged scalar bosons, you need to couple this to the electromagnetic field in a gauge invariant way since the electromagnetic field is described by a massless vector field, and if you do not describe it as a gauge field already the representation theory of the Poincare group tells you said all hell breaks loose. So better describe it as a gauge field.

    The standard description is to substitute

    [tex]\partial_{\mu} \rightarrow D_{\mu}=\partial_{\mu}+\mathrm{i} q A_{\mu}.[/tex]

    Then the variation of the corresponding action and setting [tex]A_{\mu}=0[/tex] leads to the conserved electromagnetic current for a free Klein-Gordon field,

    [tex]j_{\mu} = \mathrm{i} q (\phi^{\dagger} \partial_{\mu} \phi - (\partial \mu \phi^{\dagger}) \phi).[/tex]

    Imho that's the most physically motivated explanation for using this expression as the four-current density of a Klein-Gordon field, and its time component is (the operator representing) the charge density of the corresponding Bose particles.
     
  4. Dec 20, 2011 #3
    One way of seeing p as the time component:
    If I recall right...
    Your problem can be solved by writting the continuity equation:
    dp/dt = div[J]
    where p is the density and J is the current.
    or in other words the 4-D equation:
    D[itex]_{μ}[/itex] J[itex]^{μ}[/itex]=0

    where

    J[itex]^{μ}[/itex]= (p, Jx,Jy,Jz)

    As you can see straightforward from this, p is indeadly transformed as the time component of a 4vector.

    Now why it has that form, I don't have Ryder to check straightforward, but generally by the form of Klein Gordan equation you can see that the Current is the divergence of the quantity:
    [itex]\nabla[/itex][J]=[itex]\frac{ih}{2m}[/itex] (φ* [itex]\frac{d^{2}φ}{dt^{2}}[/itex] - φ [itex]\frac{d^{2}φ*}{dt^{2}}[/itex])

    and from there you see that the time derivative of your given p is truly giving you the divergence of current, and comes as the solution of the continuity equation
     
  5. Dec 20, 2011 #4

    Matterwave

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    @Vanhees: I am aware that the one particle description is inadequate (Ryder says this later on in the chapter after giving arguments such as the density is not positive definite and so cannot be interpreted as a probability density), I was just wondering where he got that particular expression all of a sudden.

    @Morgoth: I think I got it. If I posit the continuity equation and the (3) current, then I can get the density. Is there perhaps some way to get the density without positing the current? It seems you need one to get the other.
     
  6. Dec 20, 2011 #5
    i don't think there is, since the DJ=0 shows a conservation and that conservation means that "particles" are not created or lost, as you already know from the continuity equation... It is rather fundamental ...
     
  7. Dec 20, 2011 #6

    Matterwave

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    I just looked at Peskin and Schroeder and they showed how to derive this "conserved current" using Noether's theorem on the KG Lagrangian. I should have realized that this could be done earlier. =]
     
  8. Dec 20, 2011 #7
    well in fact I don't know but I personally have not sufficient knowledge over QFT, so i would not fall into Lagrangians for KG (for now).
    Of course conserved quantities in QM hide some kind of Symmetry of Lagrangian and Noether's theorem can be used...
     
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