Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Transferring bits using the square well

  1. Apr 9, 2010 #1
    I realize there are several practical problems with what I say below, but my question is if anything is _theoretically_ wrong with this. I'm just thinking of the problem in 1 dimension right now and ignoring the interaction potential between the particles and people involved (I don't think that's relevant to my point).

    Let's say Alice and Bob put up one side of a infinite square well at x=0. Alice remains around x=0, but Bob travels several light years away to x=a and beforehand Bob gives Alice some particles and tells her to wait for a period of time T and then measure the energy of the particles. If the energy values are in the set [tex]\{\frac{n^2\pi^2\hbar^2}{2m(a+\delta)^2}: n\in \mathbb{N}\}[/tex] (i.e. Bob put up the other end of the well at [tex]a+\delta[/tex]) then Alice is supposed to come meet Bob, but if any are not in this set (say Bob put up the well at [tex]a-\delta[/tex]) then Bob will go back to x=0 to meet Alice.

    Would that be a means for instantaneously transmitting a message, or will Alice's particles not "feel" the other end of the well until something propagates back to her, at or below the speed of light?
  2. jcsd
  3. Apr 10, 2010 #2


    User Avatar
    Science Advisor
    Gold Member

    To construct the other side of the barrier requires a time-dependent process (moving several light years also requires time-dependence obviously). As such, the wave-functions of the particles are not guaranteed to be in the eigenstates of an infinite square well the instant the other side of the barrier is put up.

    In essence, you are trying to use time-independent quantum theory (e.g. using the eigenstates which are solutions to the time-independent Schrodinger equation) to do time-dependent quantum mechanics.
  4. Apr 11, 2010 #3
    You may have already understood this, but just to clarify I'm not saying Bob travels to [tex]a[/tex] faster than light, I'm just saying that after he gets to [tex]a[/tex], and has the ability to set up the other end of the well, I'm wondering what _theoretically_ prevents him from sending the message back to Alice faster than the speed of light.

    Also, I'm assuming the other end of the well (and the potential overall) would be fixed for a period of time before and after Alice measures the energy of the particles, so from my knowledge I would think time-independent theory would work, but I don't know anything about relativistic QM, is that what prevents the particles from collapsing to one of the two sets of eigenstates? Does relativistic QM somehow "prevent the particles from knowing about the other end of the well until some extra time has passed?"
    Last edited: Apr 11, 2010
  5. Apr 19, 2010 #4
    Anybody know? Does the collapse of the wave function spread through space at a speed <= speed of light
  6. Apr 19, 2010 #5


    User Avatar
    Science Advisor
    Gold Member

    Wave-function collapse is poorly understood; however, just think of your example carefully and you should see that it's impossible. First of all, what's to guarantee that the particle will remain close to Alice so that she can actually make a measurement? What's the guarantee that the particle doesn't spread past Bob?

    You can't construct 1 side of a potential, later construct another side of the potential, and instantly expect the wave-function to settle down to an eigenstate. The wave-function will EVOLVE time-dependently!

    Here's an analogy. If I had a particle in an infinite square well, and it's already in an eigenstate, and then I expand the well very fast (not necessarily >speed of light, but just much faster than the particle can react), what will happen? Initially the particle will STILL be in the eigenstate of the ORIGINAL well. It takes a finite amount of time for the particle's wave-function to evolve out, and this evolution is described by the TDSE. It's not like the wave-function just all of a sudden turns into an eigenstate of the NEW well.
  7. Apr 19, 2010 #6
    surely Alice can keep the particles close to her, just measure their positions or apply a potential while Bob is travelling, ...

    also, i'm talking about making a measurement (measuring the energy),so I don't think the time dependent schro eqn applies, right...I mean that eqn is for the evolution of the wave function, not for the collapse of the wave function during measurement, right. I realiZe that before the measurement the tdse applies.
  8. Apr 19, 2010 #7


    User Avatar
    Science Advisor
    Gold Member

    Ok, if Alice keeps measuring the position of the particle, it collapses the particle into an eigenstate of such particle. The particle will NOT be in an energy eigenstate of the infinite square well!

    If she measures the energy of the particle, after Bob has erected the other side of the square well, she WILL NOT get energies in accordance with energy eigenvalues of the square well.

    This has been my point the entire time...there is NO reason to suspect the particle will be in energy eigenstates once Bob erects the other side of the barrier!

    Alice will just get gibberish measurements as far as she is concerned. She will NOT get measurements in the set En(a) OR En(a+da)!!! Where En(a) is the set of energy eigenvalues for a inf-square well with length a, and En(a+da) is the set of energy eigenvalues for an inf-square well with length a+da.
  9. Apr 19, 2010 #8
    Matterwave, not sure I understand your first point. I know the particles are not in an energy eigenstate of a square well when their position is measured, but not sure what that has to do with the question? What I am asking is when she measures the energy (when she wants to receive the message), what prevents the energy values from being in the aforementioned sets? I realize that something may prevent it from happening, but what? Is there some theory on how the collapse of the wave function or the measured energy depends on history?
  10. Apr 19, 2010 #9


    User Avatar
    Science Advisor
    Gold Member

    The particle is NOT in an energy eigenstate, therefore it's energy is NOT within the aforementioned sets. Why is the particle NOT in an energy eigenstate? Please refer to my previous posts.

    I have tried pretty hard to justify why the particle is NOT in an energy eigenstate...but you seem to just ignore all my points and just keep asking why...>_>
  11. Apr 19, 2010 #10
    Matterwave, sorry if I'm missing something here, but the reason I'm still asking why is because I don't see what (in the previous posts) answers the question. I know you said the time-dep Schro eqn prevents the particles from collapsing to an energy eigenstate when she measure's energy, but I don't think tdse governs the collapse process, right? According to my intro to QM textbook, when the energy is measured the wave function collapses to an eigenstate of the Hamiltonian and the measurement result is the corresponding eigenvalue. You're saying the latter is not true in this case (which I don't doubt), but I don't see what allows us to make the exception in this case. Evidentially the wave function collapse depends on history somehow, but how? Is there a set of rules about this or something? I don't see the answer in previous posts.
  12. Apr 19, 2010 #11


    User Avatar
    Science Advisor
    Gold Member

    Oh, ok I can see your point, sorry for misunderstanding...it is a little more subtle than I had at first thought.

    Let's lay down a few facts:

    Making a measurement collapses a wave-function to an eigenstate of the observable. This is true.

    The observable's eigenstates may depend on the domain you are in, this is especially obvious when talking about energy eigenstates which depends on the potential.

    The answer to your question is then, the information that the potential is there cannot be transmitted faster than the speed of light. As far as your particle is concerned, when Bob erects the other side of the barrier, until time t=a/c, the particle doesn't "feel" the effects of the other side of the barrier. If you collapsed the particle into an energy eigenstate at this point, you would get the energy eigenstate of a particle in a 1-sided box.

    Any effects of the potential will be lagged by the speed of light (at least!). This we can call the "retarded potential" (which you should learn about in E&M).

    For example, if there was a positive charge, and it moved, and you held a negative charge in your hand a distance x away, the negative charge wouldn't "feel" this movement until time t=x/c after the positive charge moved.

    Is this clearer?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook