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Foundations of Relativistic QM

  1. Aug 8, 2010 #1
    I'd liked to know whether the postulates of standard QM are still valid in Relativistic QM
    By postulates I mean what is ussually stated in texbooks as follows
    1)Physical states are determined by a vector in sate space E
    2)A measurable physical quantity A is described by an observable A acting on E
    3)The possible results in a measurement are the eigenvalues of the corresponding observable
    4)The postulate that states the provabilities of the differents possible results
    5)Reduction of the wave packet or wave function colapse
    6)Schrodinger equation

    [tex]i\hbar\frac{d|\psi(t)>}{dt}=H(t)|\psi(t)>[/tex]
     
  2. jcsd
  3. Aug 8, 2010 #2
  4. Aug 9, 2010 #3

    Demystifier

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    You meant except for 6, right?
     
  5. Aug 9, 2010 #4

    Fredrik

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    They're all valid in relativistic QM, including 6, but I prefer to replace 6 with the requirement that there's a representation of (the covering group of) the symmetry group of spacetime on the Hilbert space mentioned in 1-5. From this point of view, relativistic QM and non-relativistic QM are just the quantum theories associated with two different groups, or equivalently, with two different spacetimes. The Schrödinger equation shows up in both theories because the group of translations in time is a subgroup of both groups.

    I could also nitpick some of the details of 1-5, but you don't seem to be assuming that your statement of the axioms is perfect, so I guess I don't need to tell you that it isn't.
     
  6. Aug 9, 2010 #5

    vanhees71

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    I'd leave out 5). An assumption about a state collapse neither necessary nor well defined in standard quantum mechanics.
     
  7. Aug 9, 2010 #6
    Thank you guys, you clear it up for me.
     
  8. Aug 9, 2010 #7
    Are you sure about that? How do you describe how a particular outcome of an experiment is found? Decoherence turns quantum probabilities into classical probabilities. But that doesn't mean you get a specific result out. It's well known you can't assume the result was there all along because of the EPR paradox.
     
  9. Aug 9, 2010 #8

    vanhees71

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    Behind my statement is the socalled "Minimal Statistical Interpretation" (MSI) of quantum theory which has been proposed by Ballentine in

    Ballentine, Leslie E.: The Statistical Interpretation of Quantum Mechanics, Rev. Mod. Phys. 42, volume 42, APS, 358–381, 1970

    and is also used in his very nice introductory textbook on quantum mechanics:

    Ballentine, Leslie E.: Quantum Mechanics, World Scientific, 1998

    This interpretation just takes the Born rule of quantum theory seriously and does not introduce "metaphysical meaning" to probabilities like some other interpretations of quantum theory, particularly the Copenhagen Interpretation (CI) which has the well-known problems as discussed by Einstein, Podolsky, and Rosen in their famous paper on quantum theory. In my opinion this (very well-justified) critique is not an argument against quantum theory itself but to the (imho unnecessary) idea of a "collapse of the quantum state" or "quantum-state reduction" by measurement.

    I think (roughly) the most convincing principles of quantum are the following

    (1) The state of a system is described by a ray in Hilbert space (or equivalently by a Statistical Operator of the form [tex]\hat{R}=|\psi \rangle \langle \psi |[/tex], where [tex]|\psi \rangle[/tex] is normalized Hilbert-space vector).

    (2) The observables of the system are described by (essentially) self-adjoint operator on that Hilbert space. The possible outcomes of a measurement of an observable [tex]A[/tex], represented by such an operator [tex]\hat{A}[/tex] are values in the spectrum of the operator. In the following I use the usual language of physicists, and call also the generalized eigenvectors of operators for values in the continous part of the spectrum simply eigenvectors, and also spectral values simply eigenvalues. In the following, I label the eigenvectors of [tex]\hat{A}[/tex] for the eigenvalue [tex]a[/tex] with [tex]|a,\alpha \rangle[/tex], where [tex]\alpha[/tex] is any set of (continuous and/or discrete real) parameters necessary to label the different eigenvectors to the same eigenvalue [tex]a[/tex]. It is assumed in the following that these states are complete in the sense that

    [tex]\int \text{d} a \int \text{d} \alpha |a,\alpha \rangle \langle a,\alpha |=1.[/tex]

    (3) If a System is in the state, represented by the normalized state vector [tex]|\psi \rangla [/tex], then the probability (density) to measure the value [tex]a[/tex] of the observable [tex]A[/tex] is given by

    [tex]P(a)=\int \text{d} \alpha |\langle a,\alpha|\psi \rangle|^2.[/tex]

    This is Born's rule.

    (4) There exists an operator [tex]\hat{H}[/tex], the Hamilton operator of the system, such that the time derivative of the observable, [tex]A[/tex], i.e., [tex]\dot{A}[/tex], is represented by the operator

    [tex]\mathring{\hat{A}}=\frac{1}{\text{i} \hbar} [\hat{A},\hat{H}]+(\partial_t \hat{A})_{\text{expl}}.[/tex]

    This latter postulate, we won't need in the following, and thus, I'll not comment on it.

    At the heart of the present discussion is of course Born's rule (postulate 3). As I said above, the MSI just takes this postulate seriously. Within quantum theory we cannot know the system's state better than is described by a pure state as described by postulate 1. The consequence of postulate 3 then is that quantum theory allows us to make probabilistic statements about the system and only probabilistic statements. This immediately implies that the preparation of a system in a certain pure state does not make is possible to predict the outcome of the measurement of an observable with certainty but only with a certain probability. In practice this tells us that we have to prepare a lot of systems, which are independent of each other (i.e., they must not interact in any way with each other) in this pure state and then measure the observable for each of these systems. Then, by statistical analysis one can check whether the predicted probabilities for finding a certain value for the observable, predicted by quantum theory is correct.

    Experience shows that this indeed works very well so far: Quantum theory has undergone the hardest tests of its predictions of any physical theory so far, and it's thus the most successful and comprehensive theory we have in the whole history of physics! There is no need for "collapse of the wave function/reduction of the state" anywhere!
     
  10. Aug 9, 2010 #9
    actually, 6) isn't really valid unless you know how to reinterpret the form of [tex]\Psi[/tex] in the relativistic case, it's not a simple function on configuration space as in the non-relativistic case.

    Pedantically speaking you might say the most general form of the Schrodinger eqn applies in relativistic QM, but more usefully you would talk about Klein-Gordon, Dirac eqns etc

    A discussion can be found in this thread:
    https://www.physicsforums.com/showthread.php?t=224334
     
  11. Aug 9, 2010 #10

    jcsd

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    Hi vanhees71,

    Not a leading question: how would you derive from the above that, for example, the state space of a two particle quantum system is the tensor product of the state spaces of the two particles? Is it implicit in the other postulates that only such a state space can fufil the necessary requirements?
     
  12. Aug 10, 2010 #11

    Fredrik

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    That can't be derived from those axioms alone. You either take your statement to be a new axiom, or you introduce several new simple axioms and derive your statement from them. This article (thanks Meopemuk) does the latter.
     
  13. Aug 10, 2010 #12

    dextercioby

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    Hi, normally compound quantum systems require a special postulate with 2 points, one for different particles, the other for identical ones.

    Actually, the quest for axiomatization can be undertaken in several ways. The one provided by <vanhees71> is largely due to Dirac and von Neumann. I particularly prefer the one relying on the works of Wigner, Weyl and Bargmann: using group theory. Actually, using group theory one can easily pass from Galilei relativity to special relativity. Identical particles' states can also be obtained by studying the Hilbert space projective representations of the permutation group, task first completed by Isiah Schur in mathematics and applied to physics by Hermann Weyl.
     
  14. Aug 10, 2010 #13
    Hi, thanks for the post, it was interesting. I'm still not sure however! Does the interpretation you describe have anything to say about changes in information? Any time you gain information, probabilities get updated. That is how I see wave function collapse, I don't think it's all that mysterious. It's not clear from what you said what happens when information is exchanged.
     
  15. Aug 11, 2010 #14

    jcsd

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    Thanks, I suspected as much as I've seen it listed as an axiom so many times. Can see how the other axioms could motivate it, wasn't sure if it could be derived from the others.
     
  16. Aug 11, 2010 #15

    jcsd

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    Any links to that approach? My motto is the more mathematical the axioms (of a physical theeory) the better!
     
  17. Aug 11, 2010 #16

    dextercioby

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    I haven't seen this formulation properly developed in a book, but I would suggest you reading the second chapter of B. Thaller's <Dirac Equation>. It provides valuable insight into group theory applied to relativistic QM. Also the second chapter of Weinberg's treatise is rich in information about group theory.

    I would add for you an article I hold dear (attached).
     
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