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Paradox seemingly violating relativity

  1. Sep 29, 2008 #1

    i thought of a paradox that i'm not sure i can resolve for myself.

    You have a 1-dim potential well of length L. You measure a particle to be within [tex]x=L/2\pm \sigma[/tex] with equal probability. Assume sigma is very tiny. After a short time [tex]\Delta t[/tex] the wavefunction has evolved to be non-zero even at [tex]x=L/3[/tex], say (with tiny probability). But if a measurement of the position of the particle actually yielded [tex]x=L/3\pm \sigma'[/tex], as unlikely it may be, wouldn't relativity be violated, in that the particle would have had to travel with velocity [tex] \frac{L/6}{\Delta t} [/tex], which for short enough [tex] \Delta t [/tex] might in fact be larger than the velocity of light?

    My guess would be, that this is making too much of non-relativistic quantum mechanics and that it would all work out if instead of speaking about velocity v one talked about the momentum p, which might become as large as necessary to satisfy the uncertainty relation, without making the velocity larger than c (in the same way as the momentum of a constantly accelerated particle in relativity can become as large as necessary without the velocity increasing very much).

    Is this the resolution? If yes, is it possible to make a more convincing argument for that?
    If no, what is the resolution?


  2. jcsd
  3. Sep 29, 2008 #2
    I would assume you are working at the microscopic scale where you must follow QM- HUP rules not Relativity. That would give you the freedom to consider “backwards time” in applying relativity, as in Feynman Diagrams, as long as any paradoxes are resolved or cancelled out before any result can reach the macro level.
  4. Sep 29, 2008 #3

    Vanadium 50

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    Why do you assume that the time you call [tex]\Delta t[/tex] can be made arbitrarily small?
  5. Sep 30, 2008 #4
    Having measure the position the first time, the subsequent evolution of the wave is is not the original.
  6. Sep 30, 2008 #5
    Having once measured the position, the oringinal evolution of the wave is replaced by another that is dependent upon the results of the measurement.
  7. Sep 30, 2008 #6

    Hans de Vries

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    You're guess is right. I presume you are talking about Schrödinger's equations which
    indeed violates special relativity. It's a good enough approximation in many cases,
    but insufficient in other ones. For instance: Gold would has a silver metallic color
    according to Schrödingers equation. The relativistic Dirac equation predicts the right color.

    Regards, Hans
  8. Sep 30, 2008 #7


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    You are entirely right. This is simply because you use a non-relativistic hamiltonian here ; or in other words, the Green function of the Schroedinger equation is not limited within the lightcone. In quantum field theory, this is not so: the Green functions remain within the lightcone.

    However, it is not the Schrodinger equation itself ( hbar/i d/dt psi = H psi) which is the cullprit, but rather the form of H, which is derived from non-relativistic mechanics.

    The classical diffusion equation has the same problem.
  9. Sep 30, 2008 #8
    @RandallB I think I understand what you mean, if you say that certain processes, seemingly violating known laws, may arise, which then don't contribute in the end to what is effectively observed (Path Integral Formalism). But I don't think that the act of finding a particle at a place where it can not arrive by traveling at a velocity smaller c is one of those processes.

    @Vanadium 50 if you have a wavefunction which has a rectangular shape (in the psi-x plane) in this potential well, at t=0, then generally it will have a non-zero value within this well for any t>0 (of course there might be nodes, but only a finite amount of them, so this does not disturb the general line of the argument).

    @Phrak I agree that a measurement changes the initial wavefunction. But what I imagined is that at t=0 you happen to know that the particle is located in the interval [tex] x=L/2\pm \sigma [/tex] and using the Schrödinger equation you know how it will evolve. So you can find out at any time afterwards how the wave function looks (no intermediate measurement allowed, of course).

    @Hans de Vries Wow, I hadn't heard of the discrepancy you mentioned about the shine of metal. Very impressive.

    @vanesh I see how the classical diffusion equation has the same problem.
    I guess I will have to study QFT soon :-)

    Thanks a lot everyone!

  10. Oct 3, 2008 #9
    "You measure a particle to be within x = L/2 +/- sigma with equal probability."

    How does one do this?
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