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Dyson Istability

  1. Oct 4, 2011 #1
    It was argued by Dyson that http://prola.aps.org/abstract/PR/v85/i4/p631_1" [Broken], because if one considers a negative fine structure constant [itex]\alpha[/itex] then the vacuum would become unstable - that for [itex]\alpha < 0[/itex] like charge attracts and then there is no lower bound in the energy, as pair production can lead to like charge clumping together in low-energy bound states.

    I am having slight trouble understanding half of this argument. A negative fine structure constant would imply an imaginary electric charge [itex]e[/itex]. This would then imply that the Hamiltonian is no longer Hermitian. In turn, this would imply that the eigenvalues of the Hamiltonian are no longer assured to be real and one likely isn't considering the quantum mechanics of a closed system.

    Now I understand that for any good perturbative calculation one must still approximate a function analytic within some finite radius of [itex]e=0[/itex] and in this regard the physical details are irrelevant, but to invoke the physical interpretation of a world where like charge attracts seems confusing to me when one doesn't appear to have a valid Hamiltonian to discuss the dynamics. In principle, I wouldn't assume to have any clear and physical understanding of the model with imaginary charge.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 4, 2011 #2

    DrDu

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    Most perturbation series aren't analytic functions in the perturbation parameter but only asymptotic series. The problem with QED is precisely that no-one has the slightest idea how an underlying hamiltonian should look like.
     
  4. Oct 4, 2011 #3
    Isn't the QED Hamiltonian in the Coulomb gauge given in Chapters 7,8 of Weinberg?
     
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