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Klien Gordon Equation solution under Lorentz transformation

  1. Jul 4, 2011 #1
    Hi,

    I am using Griener's Relativistic Quantum Mechanics and I have a question. Using the Klien Gordon Equation [itex](p^{\mu}p_{\mu}-m_{0}c^{2})\psi=0[/itex], he says that the transformation law for the wavefunction i.e [itex]\psi(x)[/itex] transforming to [itex]\psi'(x')[/itex] must have the form [itex] \psi'(x')=\lambda \psi(x)[/itex] with [itex]|\lambda|=1[/itex]. I don't understand why this is the case. Can anyone help me see why this must be so?

    Thank you
     
  2. jcsd
  3. Jul 4, 2011 #2

    dextercioby

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    It doesn't follow really. I'm thinking that any massive spin n representation of the Poincare group must obey the mass sheet condition: p^2 - m^2 = 0.

    Think about the Dirac field for a second: by squaring the Dirac equation you get the Klein-Gordon equation which is satisfied by the field, but under restricted Poincare transformation, the Dirac field transforms like a (1/2,0) directsum (0,1/2) spinor, not like a (pseudo)scalar.
     
  4. Jul 4, 2011 #3
    I am a doing this as a bit of holiday reading so I'm actually completely unfamiliar with what you're saying. So basically, how do we know how the wavefunction transforms when we go from one frame to another? What tells us that it must be the same up to a minus sign?

    Thank you
     
  5. Jul 4, 2011 #4

    dextercioby

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    Actually, we don't. The scalar field's existence is a consequence of a quantum theory invariant under Poincare trasformations. Its behavior under such transformations is again a result, not an assumption. The minus sign comes from considering the full Poincare group, or a subgroup containing the spatial reflection.
     
  6. Jul 5, 2011 #5
    Thank you dextercioby
     
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