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Conservation of momentum and uncertainty.

  1. Sep 24, 2012 #1
    In situations where the momentum of particles is only known to some degree of certainty, what can you say about the total momentum of the system?

    Is it only conserved to an extent that it always falls in a certain range?
     
  2. jcsd
  3. Sep 24, 2012 #2
    In quantum mechanics, conservation laws still hold true, but you can say that they become "statistical" in some sense. So although it is not true that the system will have the same momentum every time you measure it, the expectation value (in simple terms, the average) of the momentum will still be conserved.
     
  4. Sep 25, 2012 #3
    Momentum, energy, and angular momentum are all exactly conserved in QM, just as in classical mechanics. The fact that systems can be in superpositions of different momenta, energy, or angular momentum doesn't change this.
     
  5. Sep 25, 2012 #4
    But you have to admit, don't you, that the meaning of conservation law is different in classical mechanics and quantum mechanics? After all, classically we say that the system has the same value of momentum at all times, whereas in QM we say that the system may not even have a well-defined value of momentum at all times.
     
  6. Sep 25, 2012 #5

    Ken G

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    Yes, it seems to me that saying an energy expectation value is exactly conserved is not quite the same thing as saying the energy itself is exactly conserved. Any completely closed system that has a definite energy, such that we have something exact to conserve in the first place, should be in a "stationary state"-- meaning that it doesn't do anything at all until it interacts with something external. Thus in formal quantum mechanics, the concept of a closed system that has anything happening in it is having a little fight with the concept of a conserved exact energy for that closed system, though it's not clear if that is really what is meant by energy being "exactly conserved." If it is only the expectation value that is exactly conserved, then that's an awful lot like being statistically conserved, because no finite number of trials used to ascertain the exact conservation will obtain exact conservation. Perhaps the issue is semantic, since the same could be said for classical mechanics experiments-- they will only verify exact energy conservation in a kind of statistical limit. Perhaps we can say that the theory of quantum mechanics is a theory of exact energy conservation, but in practice, quantum systems, like classical systems, can only be demonstrated to conserve energy in a statistical sense.
     
    Last edited: Sep 25, 2012
  7. Sep 25, 2012 #6
    Why can classical mechanics only verify energy conservation in a statistical limit?

    If you measure the energies of all the particles in some closed system at some time [itex]t_0[/itex] and then let them interact with each other for some period of time [itex]t[/itex] then measure all of their energies again at time [itex]t_0 + t[/itex], the total energy will be exactly the same as before. (Let's pretend we can measure the particle's energies without disturbing them, which you can in the classical way of thinking).

    That doesn't work in QM though, because you have no way of knowing exactly what the particles are doing at [itex]t_0 + t[/itex] even if you knew everything about them at [itex]t_0[/itex].
     
  8. Sep 25, 2012 #7

    Ken G

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    Because we have measurement uncertainty, which certainly never approaches anywhere near as small as quantum mechanical uncertainties.
    It's an issue of what counts in the theory-- what we are pretending to be true, or what is actually a demonstrable success of a theory. It is an unproven assumption of standard interpretations of the theory that it is completely deterministic (and not at all a necessary assumption to use the theory effectively, as we no longer make the assumption that classical physics is an exact theory). My point is that this is a rather ironic interpretation of classical physics, because in any real effort to verify that the theory is working, you will never be able to treat it as a theory that acts on exact values of observables-- you would always falsify the theory if you really interpreted it that way! Instead, the theory is interpreted as mapping from an uncertain domain into an uncertain domain, and although there is no specified smallest uncertainty as there is in QM, the uncertainty encountered is always way larger than in QM.
    It has never been known if classical physics has that property either, it is merely the simplest way to cast the theory. I can easily create a version that makes all the same testable predictions as the usual interpretation of those laws, without invoking that assumption. It was always a kind of lazy and untested assumption that never played any important role in classical physics, and now that we know it isn't true, we still use classical physics in exactly the same way we did before.
     
    Last edited: Sep 25, 2012
  9. Sep 25, 2012 #8
    What I'm talking about are measurements made within the framework of that theory and their idealizations: In classical mechanics, there's no fundamental reason why you can't know the position of a particle exactly, whereas in quantum mechanics position isn't even well defined, which is the source of the uncertainty, not some flaw in your instrument or technique.
     
  10. Sep 25, 2012 #9

    Ken G

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    But why do we say that it is possible to know a position exactly in classical physics? Is that ever a necessary component of any aspect of classical physics? It certainly isn't true that we can do it in practice, so why on Earth would we have a need to say it is true "in principle"? Just what kind of principle is that anyway?

    If one says that it is a postulate of classical physics that the position and momentum can be specified exactly in principle, I ask, when is that postulate ever invoked? I say that is no kind of postulate of classical physics-- the only necessary postulate is that whatever is the inherent and unspecified uncertainty in classical physics, it is smaller than the observational limitations in every real situation we would use to verify the predictions of the theory. Which is quite true! My version of classical physics walks, talks, and quacks exactly like regular classical physics in every way, yet includes inherent uncertainty that is simply never encountered in any experiment because it is always lost in the very real noise that all experimenters cope with routinely. That seems like a much more realistic version of any theory, if you ask me-- and explodes the myth that classical physics necessarily invokes exact determinism.

    The irony is that quantum mechanics gets dinged for being clear about the uncertainties it encounters, while classical physics is given a pass for being completely vague about those uncertainties! We mistake vagueness about uncertainty for the absence of uncertainty, but that of course has never been true in any real usage of classical physics. Hence I say the uncertainty principle is a feature of quantum mechanics, not a bug that needs fixing. It doesn't mean that quantum mechanics has to be the last theory, but it does mean that we should be happy if any "superior" theory be equally clear about its inherent uncertainties as quantum mechanics is!
     
    Last edited: Sep 25, 2012
  11. Sep 25, 2012 #10
    The measurement of momentum is more and more exactly known the more particles you are measuring at one time because the more particles you measure the less you know about the exact position of each one. There are bizzillions of atoms in a baseball. You can be certain of its total momentum because you know absolutely nothing about the exact position of ANY of those particles.
     
  12. Sep 25, 2012 #11

    Ken G

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    Good point, but note that the total uncertainty in any possible measurement of the location of the center of mass of a baseball is way larger than the uncertainties encountered in the HUP. I'm told the LISA satellite would be the first effort to ever get macroscropic locations with quantum mechanical precision, but that's still something of a pipe dream at this point.
     
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