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Unphysical nature of ##\phi^{3}## interaction

  1. Dec 13, 2015 #1
    In page 77 of Peskin and Schroeder, it's mentioned that for a ##\phi^{3}## interaction, the energy is not positive-definite unless we add a higher even power of ##\phi##.

    Can someone please prove this statement?
     
  2. jcsd
  3. Dec 13, 2015 #2
    One reason is that ##V=\frac{1}{2}m^2\phi^2+\lambda \phi^3## is unbounded below (##V\to-\infty## as ##\phi \to-\infty##).

    On the other hand, ##V=\frac{1}{2}m^2\phi^2+\lambda\phi^3+\sigma \phi^4## is bounded below.

    Does that help in any way?
     
    Last edited: Dec 13, 2015
  4. Dec 13, 2015 #3

    Avodyne

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    Yes, that's the reason. As long as V is bounded below, we can add a constant to V that makes the minimum value of V equal zero. Then each term in the hamiltonian is positive. But if V is unbounded below, this cannot be done.
     
  5. Dec 13, 2015 #4

    stevendaryl

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    Here's a related question: What goes wrong in QM if the energy levels are not bounded below? We could reason that particles tend to radiate, and drop to lower energy levels. So if there is no lower bound to the energy levels, then the particle will just keep falling to lower and lower energy levels, and will radiate an infinite amount of energy. This would be a catastrophe, and would end up destroying the universe, I suppose. But would it be actually inconsistent with QM?
     
  6. Dec 13, 2015 #5

    ShayanJ

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    I think by QM you actually mean QFT, because AFAIK we can't have creation of photons in QM! But maybe you mean classical radiation from quantum mechanical matter? Is it reasonable?
     
  7. Dec 13, 2015 #6

    stevendaryl

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    Well, prior to actually developing QED, physicists had a simplistic model of radiation: that a particle makes a transition from energy level [itex]E_n[/itex] to [itex]E_m[/itex] and releases a photon of frequency [itex]\nu = (E_n - E_m)/h[/itex]. But I could rephrase the question: Does QM without radiation run into inconsistencies if the energy levels are not bounded from below?
     
  8. Dec 19, 2015 #7
    What immediately comes to mind is that we wouldn't have stable states like the Hydrogen ground. But I feel like there would be a more general issue with the formalism?
     
  9. Dec 19, 2015 #8

    stevendaryl

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    But nonrelativistically, excited states are stable, if you ignore radiation.
     
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