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Outcome counting, the action principle, and GR

  1. Mar 6, 2006 #1
    I hope that some of the folks (Mike, Howard, Patrick, Robin) who have engaged in the other threads [1][2][3] on outcome counting (ie Patrick's APP) see this thread.

    Starting with the Feynman path integral formulation of ordinary quantum mechanics, it is pretty straightforward to derive Hamilton's action principle, as discussed in [4][5]. Since Hamilton's principle is equivalent to Newtonian mechanics, we have (with the caveat of msg #8 by abszero in [4]):

    QM (FPI) ==> Hamilton's principle <==> Newtonian mechanics

    In a perhaps analogous manner, quantum field theory yields an action principle from which we may derive general relativity:

    QFT ==> action principle ==> GR

    The derivation of GR from QFT is more problematic than the derivation of Newtonian mechanics from standard QM, as pointed out by Physics Monkey in msg #12 in [4].

    Note that in both steps, we start out assuming quantum mechanics in one form or another. Now recall that the whole idea of outcome counting is to modify the MWI by replacing the Born rule with an alternate "probability rule," ie one where each branch is equally likely. The immediate question, discussed at length in [1][2][3], is whether it is possible to introduce some further modification so that the Born rule emerges as a valid coarse-grained approximation.

    But a second question has occurred to me. Suppose we modify QFT by replacing the Born rule with outcome counting. Can we still derive GR, along the lines of the above prescription?

    Now it is not entirely clear to me how, technically speaking, one might modify QFT by assuming outcome counting in place of the Born rule. QFT is just not "built" like that. I mean QFT is not like an automobile, where you can just open up the hood, take out the "Born-rule-erator," and replace it with an "outcome count-erator," like so many spark plugs. But there must be some sort of logically equivalent modification that would amount to the same thing. (I'm trying to get up to speed on QFT, so maybe I'll have more to say later.)

    Now I have argued in [1][2][3] that outcome counting is ontologically superior to the Born rule; indeed, that it might be considered a symmetry principle. But it's really only justified to call it a symmetry principle iff it somehow proves mathematically superior to its alternatives. As I said above, the derivation of GR from QFT is a bit problematic. So this leads to my wondering: if we (somehow) modify QFT via outcome counting, might this remove some of the difficulties inherent to the derivation of GR from QFT?

    David

    [1] my paper on the born rule
    https://www.physicsforums.com/showthread.php?t=95585

    [2] are world counts incoherent?
    https://www.physicsforums.com/showthread.php?t=101339

    [3] attempts to make the Born rule emerge from outcome counting
    https://www.physicsforums.com/showthread.php?t=101982

    [4] QM and action principles
    https://www.physicsforums.com/showthread.php?t=112257

    [5] a democracy of spacetimes?
    https://www.physicsforums.com/showthread.php?t=112556
     
  2. jcsd
  3. Mar 6, 2006 #2
    I think you have not understood fully how QFT is related to QM. This is demonstrated in your logical sentences:

    Firstly, QM was created by quantizing Hamiltonian mechanics, so it is no surprise that the FPI approach, which is analogous to Schrodinger/Heisenberg mechanics (as proved by Feynman), yields the principle of stationary action. Of course, the principle of stationary action was designed (years before QM) to yield Newtonian mechanics (in most cases).

    Secondly, QFT is the quantum mechanics of classical fields (and generalisations thereof). In fact each QFT will have as a starting point a Lagrangian. QFT as a method cannot be used to derive an action principle -- they all use an action principle in some form or another.

    --EDIT--

    It's like this:

    QM of particle gives "usual QM" (i.e. what undergraduates first study in QM courses)
    QM of classical field (and their generalisations) gives QFT

    In the classical limit, both the particle and classical fields obey an action principle. Feynman recovered the action principle in QM, so you shouldn't make the distinction between QM and QFT as you have done.

    Furthermore, GR is 2 things: a replacement of Newtonian mechanics and a theory of the dynamics of the gravitational field. It is a generalisation of Newtonian mechanics and a field theory. Whereas QFT is not a generalisation of QM, so again your logical sentences cannot be true in the way you have stated them.

    I would just like to add, that I like some of the ideas you have come up and I'm not trying to dissuade you from that, I'm just pointing out as what I see to be flaws in your reasoning. I very much hope you don't take my criticisms personally.

    Masud.
     
    Last edited: Mar 6, 2006
  4. Mar 6, 2006 #3

    Hurkyl

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    Also, I don't think the probability postulate matters. According to what I've read about algebraic quantum field theory (AQFT), a state is p is defined to be the function:

    [itex]\rho(T) = [/itex] the expectation of the operator T

    So the actual factoring into bras and kets, and methods of counting probabilities in various Hilbert spaces, are physically irrelevant. And since this framework is logically equivalent to the usual one, the answer to your question is:

    If:

    Usual QFT --/--> GR

    then

    QFT with new counting --/--> GR
     
  5. Mar 6, 2006 #4
    Although I'm not sure that Hurkyl's notation came through properly on my computer, I think I agree with his point. QFT describes (in one representation) the time evolution of a state vector. On it one can define 'expectation values' of operators, but the relation of those to actual expectation values obtained by averaging repeated measurements is entirely outside the scope of QFT. The problems in making that connection between a formal 'expectation value' and an operational one are usually not even considered by QFT practioners. Thus the nature of the explanation for the Born operational probabilities seems unlikely to directly affect any reconciliation between QFT and GR.
     
  6. Mar 6, 2006 #5

    Physics Monkey

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    Hi straycat,

    I explained in the other thread the extent to which the path integral, quantum field theory, and so forth can be thought of as justifying the classical general theory of relativity. You will find that this justification is essentially a statement about the amplitudes in the theory; in other words, what one is really saying is that the amplitude is dominated by the classical configuration. I can be more precise about this if pressed.

    At this level of justification the connection between amplitudes and experiment hasn't yet entered. In the usual formulation, one takes the square of the amplitude as the probability following Born. By adding various kinematical doodads to this primitive object, you arrive at the scattering cross section which is measured experimentally. I would hazard to say that from a practical point of view in the classical limit you don't really have to choose to between the Born rule and outcome counting because they give the same answer.
     
  7. Mar 6, 2006 #6
    Hey Masud,

    I've been doing a little internet reading on QFT, eg:
    http://universe-review.ca/R15-12-QFT.htm

    As I read through the section on Field equation, it is looking to me right now like you start out with your Lagrangian (density), and you proceed rather immediately with the application of the action principle; which makes me wonder whether the action principle in QFT could be considered an axiom, rather than a consequence of some deeper axioms ... ?


    I understand that Feynman recovered the path action principle in the QM of particles, but I'm still confused where exactly the action principle appears in QFT, ie whether it is assumed or recovered (as you say) from something else ...

    Don't worry, I wasn't offended in any way :smile: . I freely admit that my understanding of QFT is rudimentary and jumbled, as I am sure I have illustrated with my postings :yuck: But I hope to be able to fix that enough to make my ideas more precise. (I have Zee, as well as Schroder and Peskin, on the way from amazon, maybe they will help ...)

    David
     
  8. Mar 6, 2006 #7
    Yup, I have a question here. I like making analogies, so I'll try to make one here.

    In the QM of particles, each possible individual path has an associated amplitude.

    So in the QM of fields, does each possible field configuration have an associated amplitude? I know (or think I know) that each field configuration has an associated action S, so I suppose the amplitude would simply be [itex]e^{-i S / /hbar}[/itex]. Is this a defined entity in QFT, and if so, do they interfere the way they do in the QM of particles, ie do you sum them together to make a kernel K (to use terminology from Feynman and Hibbs)?

    david
     
  9. Mar 6, 2006 #8
    Hey Mike and Hurkyl,

    Does "A --/--> B" mean "A does not imply B" ?

    From one of the companion threads to this discussion, I've been swinging back and forth in my thinking / understanding of whether QFT ==> GR. Gettin' me dizzy ...

    ds
     
  10. Mar 7, 2006 #9
    The field, not any specific configuration, is described by the action. So just as with regular QM we whack the action in the transition amplitude and we get FPI for QFT. We don't need to worry about whether we're in QFT or QM in this instance. Particle excitations occur at either end points, which we can look at as particles propogating (or the transition form one state or field configuration to another) between them, as in QM.
     
  11. Mar 7, 2006 #10

    Hurkyl

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    That's how I meant it.
     
  12. Mar 12, 2006 #11
    I got Zee and Schroeder/Peskin in the mail yesterday :biggrin: ... Zee has several pages (the beginning of cap I.3) explaining how we go from the lagrangian to the lagrangian density -- the most explicit that I have run across. I still need some time to go through the mathematical details (busy at work as usual), but I think it will help me answer some of the questions I've been asking. More when I can.

    David
     
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