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What is the role of proper time in quantum theory

  1. Feb 9, 2009 #1
    What role does proper time play in quantum theory? Does proper time have any meaning in QM?

    By quantum theory I should perhaps say "relativistic quantum mechanics" since I don't know enought QFT to ask a proper question.

    If there were a proper time [tex]\tau[/tex] parameter or dynamical variable, then a rest-mass operator [tex]i\hbar\frac{\partial}{\partial \tau}[/tex] would be its conjugate in the same way that energy is "conjugate" to time in non-relativistic QM. A specific rest mass then, such as we see assumed in the Klein-Gordon equation, would represent a eigentstate of this rest-mass operator.

    But there is no degree of freedom associated with rest mass, right? We never see superpositions of states of different rest mass, right?

    Of course, proper time is not completely analogous to time-as-parameter in non-relativistic QM. In non-relativistic QM, time is a global parametrization of the whole system, whereas proper time ,as usually understood, is specific to each particle.

    So what is the consequence of this to our understanding of proper time within quantum theory?
     
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  3. Feb 9, 2009 #2

    Demystifier

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  4. Feb 9, 2009 #3

    Fredrik

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    As I'm sure you know, proper time is a property of a curve in Minkowski space, but the motion of a particle in quantum mechanics, can't represented by such a curve. So the concept seems to be of little use. However, we can't get rid of the SR postulate that says that what a clock measures is the proper time of the curve that represents its motion. Even in QM, we have to think of the clocks we use to test the predictions of the theory as if they behave classically (and to a good enough approximation, they do).
     
  5. Feb 9, 2009 #4

    Demystifier

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    You may equally say that x(t) does not have a meaning in QM. But it does, e.g. as an operator in the Heisenberg picture. In a similar sense, you can introduce x(tau) and t(tau) as a relativistic analog of the Heisenberg picture.
     
  6. Feb 10, 2009 #5
    When I read these papers I think I understand them ok. Yet I don't see how they show any better understanding of the question posed above. Can you elaborate?

    Also, "unparticle" is a new concept to me. But isn't one conclusion that follows from the paper that, given the mathematical basis for QFT together with the usual interpretation, there is nothing that prevents us from considering varying mass and non-integer particles? Which is something I've noticed before. Within the theory, are specific mass and integer particles extra postulates? (Is that the same thing as saying they require a superselection rule?) If the operator [tex](a^{\dagger}(k))^{\frac{2}{3}}[/tex] makes sense mathematically (and I am not sure that it does), what rules out a state containing two-thirds of a particle?
     
  7. Feb 11, 2009 #6

    Demystifier

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    Have you been reading some of the papers in Ref. [5]? I think they are quite close to your idea.

    The idea of the paper above is that only integer number of particles makes sense. In particular, [tex](a^{\dagger}(k))^{\frac{2}{3}}[/tex] does not make sense.
     
  8. Feb 11, 2009 #7
    Thanks, D.
     
  9. Feb 12, 2009 #8

    atyy

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    Last edited by a moderator: Feb 12, 2009
  10. Feb 12, 2009 #9

    atyy

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    Perhaps see section 4 "Probing Masses through Neutrino Mixing" of Haxton and Holstein's http://arxiv.org/abs/hep-ph/9905257
     
  11. Feb 12, 2009 #10
    The relativistic version of QM, the Dirac equation is invariant under Lorentz transformation (in contrast to the Schrödinger equation which is not). I think the use of Hamiltonian (energy-momentum picture, rather than forces, position and momentum) in an assumed fixed frame S, makes it not necessary to deal with proper time tau=gamma(u)*t.

    But when it comes to derive forces, in terms of expectation values of operators, I think it would be wise to do this with proper time, .i.e., What are the eqm for the expectation values <x> you get from the relativistic version of Ehrenfest theorem? You should obtain the version with proper time!
     
  12. Feb 12, 2009 #11
    Very interesting. Especially the neutrino mixing stuff. Thank you!
     
  13. Feb 13, 2009 #12
    I would say none. At least in the most standard case that all textbooks follow, where you have quantum fields and multiparticle fock states, no single particles with clocks.

    As Atvy pointed out, Srednicki says there is a second option for a relativistic quantum theory, on which he unfortunately gives not much detail and says it is much more difficult then the field approach.
     
  14. Feb 13, 2009 #13

    jtbell

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    In addition to neutrino mixing, there's the mixing of [itex]K^0[/itex] mesons, which has been studied experimentally since the 1950s.
     
  15. Feb 14, 2009 #14
    Yes, no 5th dimension as time only go forward (not in my mathematics). NOT NEED 5th dimension, time NOT this - all wrong - no good. Time NOT space - no.
     
  16. Feb 16, 2009 #15

    Demystifier

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    Time no space - no good. Time is space - good. Einstein good, Newton no good. :biggrin:
     
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