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Why doesn't QM make particles zig-zag as they travel?

  1. Jul 18, 2007 #1
    Pardon my ignorance, I'm pretty sure this has an obvious answer, but it's something I've been wondering about.

    In QM it's said that particles don't exist in a definite state between measurements, but rather exist as a wave function, with a probability of being found in a certain place when they're measured. Neutrinos are very weakly interacting, and travel through the partial vaccuum of space. Since they're not being "measured" when they're not interacting with other particles, they should exist in their wave-function state during those times. How come, then, don't they zig-zag as the fly accross the universe? My intuition tells me that between interactions (measurements), their position and momentum should be determined by a probability function, and there should be a possibility of them changing direction during those times. Obviously, this doesn't seem to happen (from what I've read), so I'm guessing my intuition is wrong. Could anyone shed some light on this for me? Thanks.
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  3. Jul 18, 2007 #2
    I think you are mixing up Newtonian physics with quantum mechanics.

    In quantum mechanics, there is no particle travelling around in some direction. All you can speak of is a wave, which assign a complex number to the 3D real space. the wave can move and evolve according to the Schrödinger's equation. The wave function probably would not zig-zag around. (unless perhaps for a very weird potential). You can think of a strongly peaked wave function as a representation of a particle.

    anyway, when you take a measurement of momentum, it could be in the opposite directions (depending on the wave function), the probability of getting a momentum measurement opposite to the general direction in which the wave travels is probably pretty low though. but if you calculate <p> (the mean value of momentum for an ensemble of identical system) , you will get a classically consistent answer and <p> would not zig-zag around.
    Last edited: Jul 18, 2007
  4. Jul 18, 2007 #3
    To speak of a "zig-zag" path - or any other path, for that matter - means that you're thinking of the particle as having definite positions between observations, which is inherently non-quantum mechanical. On the other hand, you can think of the uncertainty in the particle's position, which grows the longer it propagates, as the effect of all the possible paths it could follow. This is the basis of the path integral formulation of QM. In that picture, the probability of finding a particle at a given position is given by the sum (i.e. the integral) of all possible paths leading to that point, including your zig-zag paths. It turns out that when you do this, you get the same answer as from the Schroedinger Eq. So, yeah, you have to consider the contribution of zig-zag paths, but the more outlandish they are, the less they contribute to the end result.
  5. Jul 18, 2007 #4
    So a particle's wave function describes the entire possible path the particle could take, not just the segments between measurements? I think I'm still far from clear on this, but you can only lead a horse to water, and I'm a pretty dumb horse :)
  6. Jul 18, 2007 #5


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    I think that the problem is that you are thinking about paths between measurements. You have to give up that notion. To talk about the path of a particle one must observe its position at different times. Therefore to even talk about a path there must be measurements. In between measurements, the concept of path is not even defined. You can't associate a path to a wavefunction because once a measurement is made, the wavefunction changes. So no measurement -> no concept of path.
  7. Jul 18, 2007 #6
    The trace of a charged particle in a bubble chamber detector can be considered a sequence of measurements of particle's positions and is usually a smooth curve. Why is that not a zig zag?
  8. Jul 19, 2007 #7


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    In Schiff's QM book, there's a 2nd-order perturbation calculation of
    this sort of thing. I.e: he works out the probability of two particles in
    a cloud chamber interacting with an incoming charged test particle.
    It turns out that the probability is negligible unless the two cloud
    chamber particles lie on a line that corresponds to the test particle's
    momentum direction.

    So QM doesn't rule out zig-zag completely. But the probability of it
    happening and being observed is vanishing small.
  9. Jul 19, 2007 #8


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    There is an interpretation of QM that says that a sort of "zig zag" particle motion between experiments is exactly what happens:
  10. Jul 19, 2007 #9
    Both the Bohm interpretation and Feynman's Path Integral formulation are ways of interpreting what happens between observations, but it's critical to remember that observations are the only things we can verify - not what happens between them. The predictions of the Schroedinger Eq. have been well verified, so any alternate interpretation must reproduce the same predictions, or it will likely be wrong. The alternate interpretation might help you to think about what's going on, or it might just be more comfortable, but in the end, the predicted observations are the same.

    So ... think about zig-zag paths between observations if that works for you. They're definitely allowed in QM, but there will be infinite numbers of them, as there will be infinite numbers of smoother, straighter paths. When you work out the probabilities of all of those paths and predict what we'll actually see (expectation values), the most likely outcome will be a classical type path, i.e. smooth and straight.

    There is a very tiny probability of a sequence of observations that zig and zag, but there's also a tiny probability that you'll tunnel through your chair in the next ten seconds and end up on the floor - but you don't worry about that, do you? :wink:
  11. Jul 19, 2007 #10


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    I don't think that the path integral formulation says what happens between observations. If one would try to interpret it in this way, one would say that the particle passes all different paths at once, which would be logically meaningless.
  12. Jul 19, 2007 #11
    Momentum conservation does not allow it.
  13. Jul 19, 2007 #12


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    QM is full of weird things when questions are asked on times BETWEEN measurements. So, perhaps momentum is not conserved between measurements? Indeed, when the wave function is such that the momentum is uncertain, Bohmian interpretation says that the momentum is certain but not conserved.
  14. Jul 20, 2007 #13
    Well, no, of course it doesn't say that the particle actually follows all those paths. I just meant to point out the weirdness that occurs whenever we try to explain the observable results of quantum theory by using the natural language of classical physics, e.g. well-defined paths. Maybe I should just have said that it's not only correct to consider the possibility of a zig-zag path, but you must consider it - along with all other possible paths - to predict the actual outcome. Or maybe that's not any better ... it's hard to speak sensibly about these things.:uhh:
  15. Jul 21, 2007 #14
    Yeah, but when momentum is uncertain the wave function is constrained somehow (for example when the particle travels through a narrow slit). In this situation there is a momentum transfer between the particle and the slit, therefore the particle's momentum is not conserved. This is true even in classical physics. Doesn't BM predict straight trajectories for a particle departing from a point source?
  16. Jul 21, 2007 #15
    No, because you are not observing nor measuring the particle's position. What you observe is the emitted EM radiation generated by the interaction of the particle with the medium through which it passes.
    Why would it be ?

    Point is that one cannot speak about a particle's position, let alone its trajectory, unless you measure the position. But once you measure its position, all info on previous measurements is gone. So you certainly cannot build up the trajectory. In short : wavefunction collapse !

  17. Jul 21, 2007 #16
    Why is that not considered a measurement of position? Don't we measure objects' positions constantly based on their interactions with photons? It seems to me that at least the classical concept of position is being imprinted on the emitted EM radiation. You could follow the emitted EM radiation back in time to the point of emission to determine the position of the particle that generated it. Of course I'm not sure if the the concept of position as piece of quantum information is different from what I've described.
  18. Jul 21, 2007 #17
    Because the electron does NOT interact with photons here but with the atoms that constitute the medium through which it passes. These atoms can get excited and emit EM radiation during the de-excitation process. You observe that radiation arising because of the atomic de-excitation. Interaction with phonons (ie crystal lattice vibrations) is also possible.

    My point is that you observe radiation that is the CONSEQUENCE of the electron-medium interaction.

  19. Jul 22, 2007 #18
    Do you mean that those atoms moves, so if a pulse of EM radiation emitted from one of them in a specific point, can actually corrispond to a particle that has passed 1 metre away?
    I don't think it would be quite useful, in physics!
    I really cannot understand why you don't consider that interaction as a measure of position. You consider "measure of position" only in the case of interaction with photons? So, how would you perform a photon's measure of position, considered there isn't any photon-photon scattering (at least at common energies)? Maybe with a photographic film? Isn't the photon or particle's position measured through the interaction with atoms in that case?
  20. Jul 22, 2007 #19
    Well, you tell me, what are we actually observing ? An electron that makes a certain trajectory or EM radiation that is emitted ? I guess you agree with me that an electron is not the same as EM radiation, no ?

    I don't quite understand what your problem is here. We KNOW there is an electron through the observed EM radiation, we know how it behaves and we can check that behaviour with respect to theoretical models. Theory and experiment give the same results for the same initial parameters and THAT IS IT !

  21. Jul 23, 2007 #20


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    Not necessarily. Only when the wave function is an eigenstate of the radial-momentum operator.
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