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Time Evolution of Quantum systems.

  1. Jul 14, 2005 #1
    In non-relativistic quantum mechanics time evolution is given by the usual [tex]e^{\frac{-i\hat{H}t}{\hbar}}[/tex] (for non time dependent hamiltonians). How does one time evolve a quantum system in the context of relativity, where time and space have been placed on equal footing? We clearly cannot use the above expression since it is a result of the schrodinger equation which is not relativistically invariant.....help?

    Phil :!!)
     
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  3. Jul 14, 2005 #2

    dextercioby

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    Before I or someone else provide a serious attempt to clarify things, at least I need to be sure that you'd understand everything that may possibly be written.

    "Time evolution of quantum systems" is an awfully imprecise statement. By virtue of the first two postulates of quantum mechanics, a quantum system is described through physical states and physical observables. So the time evolution is ascribed either to quantum physical states (Schrödinger picture), or to quantum (usually Hamiltonian) observables (Heisenberg picture), or to both states & observables (interaction/Dirac-Tomonaga-Schwinger picture).

    So get a serious QM book and make sure postulates + formulations, descriptions & representations are crystal clear. :wink:

    Daniel.
     
  4. Jul 14, 2005 #3
    I assure you that I understand what is going on with respect to non-relativistic quantum mechanics, and am also familiar with non-relatvistic quantum field theory. However, I do not know where to begin for a relativistic setting. The dirac/klein gordon equations govern the evolution of systems in quantum mechanics; but as far as i can tell the dirac equation is used for fermi systems whereas the klein gordon equation is used for bosons. This poses some confusion to me since particle statistics are built into the commutator relations in our quantum field theory, so why do we have two different equations which govern time evolution?

    /boggled

    Phil
     
  5. Jul 14, 2005 #4

    dextercioby

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    Unfortunately things are not that simple. The KG equation is the field equation for the simplest possible classical field theory. No electric charge, no constraints, spin 0 boson. The full classical analysis of this simple model (free real scalar field) should be clear to you before you jump into quantization. The classical free Dirac field (actually, both models don't have a physical significance at classical level) is the most complicated one may have: two sorts of particles, II-nd class constraints, fermion 1/2 spin.

    Assuming the classical analysis of the KG field is clear, which picture of the 3 in QM would you choose?

    Daniel.
     
  6. Jul 17, 2005 #5

    reilly

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    In relativistic field theory time and space are described by c#parameters. And, no matter what "picture" (Schrodinger, Heisenberg, Interaction) you use, the name of the game is dynamics, evolution in time, or whatever you choose to call it.

    The answer to your question lies with the development of covariant perturbation theory. And, we are talking the basis of field theory, which can be found in any number of references -- there is no quick way to get to covariant dynamics. Weinberg's, Gross's texts on field theory, Schweber's QED and the Men Who Made It, Feynman's original papers, .....

    Yes, done properly, the appropriate exponential of the interaction Hamiltonian is covariant, and forms the basis for developing the S-matrix.
    Regards,
    Reilly Atkinson
     
  7. Jul 18, 2005 #6

    vanesch

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    In the limit of T-> infinity (the S-matrix). It's harder to talk about *finite* time evolution in QFT, although there may be techniques now that I don't know of.

    cheers,
    Patrick.
     
  8. Jul 18, 2005 #7
    I see nobody offers you a simple and easy answer.

    Of course, there is only a evolution of quantum systems. That is the Schödinger like picture.

    The evolution (for fields) is the same in the context of relativistic quantum mechanics. See Weinberg The quantum theory of fields volume 1, p49.

    There is not relativistic quantum mechanics. In fact, it cannot exist.
     
  9. Jul 18, 2005 #8
    All of these replies seem to indicate to me that relativistic systems are evolved in the usual way (ie, shrodinger, hiesenberg or whatever picture). This has allowed me to rephrase my question: Does it not matter that the schrodinger equation is not relativistically invariant?

    Phil :!!)
     
  10. Jul 18, 2005 #9
    Errm, what? As a (simple) example, the Dirac equation is the result of making Schrodinger's Equation Lorentz invarient.
     
  11. Jul 18, 2005 #10
    This was also my understanding of the matter, however the Dirac equation is only for spin-1/2 paticles, correct? Is there a more general equation which is Lorentz invariant?

    Phil
     
  12. Jul 18, 2005 #11

    dextercioby

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    Nope. For a free particle, the Lorentz covariant form of the Schrödinger equation is the Klein-Gordon equation. For heaven's sake, don't you know that Schrödinger initially obtained the relativistic equation and just then considered the nonrelativistic case? Read the first chapter in Weinberg's book, first volume and the references therein.

    Daniel.
     
  13. Jul 18, 2005 #12

    dextercioby

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    Yes. Massive, parity invariant , electrically charged 1/2 spin particles.

    Nope, there's no general case. The equations of motion are simply the Lagrange field equation for each classical field corresponding to all finite dimensional nonunitary reps of the restricted Lorentz group.

    Daniel.
     
  14. Jul 19, 2005 #13
    So to study a relativistic sytem what can be done? Is the best I can do to start with a relativistic lagrangian and derive the propagator via the feymann path integral method using this relativistic lagrangian? Seems like the best idea I can think of?

    Phil :!!)
     
  15. Jul 19, 2005 #14

    George Jones

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    Another way of looking at the field equations for non-interacting fields is that they naturally characterize the infinite-dimensional unitary irreducible representations of the (universal cover of the) Poincare group.

    Regards,
    George
     
  16. Jul 23, 2005 #15
    It is the nonrelativistic Schrodinger equation that is not Lorentz invariant.
     
  17. Jul 23, 2005 #16
    No. It is not so simple. Dirac equation is incosistent and the developing of relativistic quantum mechanics is imposible. This is the reason of that Dirac equation was abandoned in relativistic quantum field theory, where the basic dinamical equation is a Schrodinger like one.

    [tex]
    i \hbar \frac{\partial \Psi}{ \partial t} = H \Psi
    [/tex]

    Note: Dirac equation is not the result of making Schrodinger's Equation Lorentz invarient: that is the well-known relativistic Schrodinger equation.

    Dirac equation was derived from the searching of a first order (in time) representation of Klein-Gordon equation more spin corrections for hidrogen atom.
     
    Last edited: Jul 23, 2005
  18. Jul 23, 2005 #17
    Use the canonical formalism. Obtain the corresponding hamiltonian from the lagrangian and use it directly in the S-matrix.

    The path integral can offer wrong answers.
     
  19. Jul 23, 2005 #18
    The Lorentzian version of the Schrödinger equation is the relativistic Schrödinger equation, which is not the Klein-Gordon equation. In fact, the relativistic Schrodinger equation has no problems with negative probabilities for example and is very used in semirelativistic quantum chemistry .
     
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