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Question about KG with negative mass^2

  1. Sep 11, 2014 #1

    ChrisVer

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    Well can someone review this?

    KG equation:
    [itex] \square \Phi + m^{2} \Phi =0, ~~ m^{2} <0 \Rightarrow m=i \mu [/itex]
    would lead to the form:
    [itex] \square \Phi = \mu^{2} \Phi [/itex].

    I'm trying to think if applying the same solution as in KG can also happen here...
    Also for on-shell particles, I seem to be getting the "same" equation as we do for normal positive masses:
    [itex] \int d^{4}k [k^{2}- \mu^{2}] \tilde{\Phi}(k) e^{ikx}=0 [/itex]
    and so [itex] k^{2} = \mu^{2} [/itex]
     
  2. jcsd
  3. Sep 11, 2014 #2

    Avodyne

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    What is your question??

    And what is the context? QFT? Classical field theory? First-quantized relativistic wave equation?
     
  4. Sep 11, 2014 #3

    ChrisVer

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    My question is that in a normal KG equation, you have solutions:

    [itex] e^{i (E t- \vec{k} \vec{x})}[/itex]
    Where [itex] E^{2} - k^{2} = m^{2} [/itex]
    and these are oscillating solutions... Now if I let [itex]m^2 <0 [/itex] then it means that [itex]E,k \in C [/itex], is that right?
    as such the solutions become exponentials...:/ however I was expecting hyperbolic solutions...

    I think I'm talking about Classical FT...
     
  5. Sep 11, 2014 #4

    Orodruin

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    ##\cosh x = (e^x + e^{-x})/2##, ##\sinh x = (e^x - e^{-x})/2## ...
     
  6. Sep 11, 2014 #5

    ChrisVer

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    and how can someone deduce from that the instability of the field? because it explodes exponentially? although I am a bit confused about x in [itex] \phi(x) [/itex] and what it actually means... eg some people say that it's unstable because if you make some displacement x--> x+dx then it won't remain in the same state...however [itex]\phi[/itex] should exist in the whole space, no?
     
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