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A particle in two boxes

  1. May 13, 2008 #1
    I would like to make a little poll about the following question.

    Let us consider a quantum particle in a box. At a time t1 an impenetrable barrier is inserted into the box, dividing the box in two semiboxes and the wave function of the particle in two disjoined parts. At a subsequent time t2 the semiboxes are open in order to determine in which of them there is the particle. The question is:

    Was the particle since the time t1 in the semibox where it has been found at the time t2?

    Possible answers:

    1. yes, because the barrier is impenetrable and the particle cannot jump form a semibox to the other.
    2. no, because the wave function provides the most complete description of the particle, and until t2 it does not contain the which-semibox information
    3. other...

    What do you think?
    Last edited: May 13, 2008
  2. jcsd
  3. May 13, 2008 #2
    Assuming that QM is complete then 2) is the correct answer. You can then think about what various interpretations of QM say. In the many worlds interpretation there will be a copy of you who oberves a different outcome.
    Last edited: May 13, 2008
  4. May 13, 2008 #3
    But what do you think? Another question: assume MWI. Can the observer in one of the two worlds assume that the particle was in that semibox since t1?
  5. May 13, 2008 #4
    It is also interesting to look at the opposite problem. We put the particle in one of the two boxes that are already separated. Let's define |left> as the state in which the particle is in the left box (and in the ground state) and |right> if the particle is in the left box (and in the ground state).

    Suppose we put the particle in the left box, so the initial state is |left>. Now, if the two boxes are identical then we know that the exact eigenstates of the Hamiltonian are:

    |0> = 1/sqrt(2) [|left> - |right>]

    1> = 1/sqrt(2) [|left> + |right>]

    The states |0> and |1> will have slightly different energies, so |left> and |right> won't be exact eigenstates. This means that the particle that was intially in the left box can be found in the right box some time later.

    Another thing that should be clear is that measuring the energy exactly should be nearly impossible. If you put the particle in the left box and you could exactly measure it's energy then that measurement would collapse the wavefunction into either |0> or |1>. But that means that you would now have a 50% chance of finding that the particle has moved to the right box.
  6. May 13, 2008 #5

    If I had to gamble, I would put my money on the MWI. In the MWI the two worlds are the same before the measurement is made.
  7. May 14, 2008 #6

    I do not understand why. Would you mind to explain better?
  8. May 14, 2008 #7
    Suppose the two boxes are the same. Consider the plane of symmetry defined such that the system is mirror symmetric w.r.t. a reflection in that plane. Then the Hamiltonian commutes with the parity operator, P, that reflects the wavefunction in the plane. To see this, let's define coordinates such that x = 0 is the plane of symmetry. Then the potential energy is V(x, y,z) is symmetric w.r.t. changing x to minus x

    V(x,y,z) = V(-x,y,z) (1)

    If we suppress the arguments y and z we can write:

    P V(x) psi(x) = V(-x)psi(-x) = (using (1)) = V(x) psi(-x) =

    V(x)P psi(x)

    So, P commutes with V. The operator P also commutes with the kinetic energy. The kinetic energy operator is proportional to the second derivative. We have:

    P psi''(x) = psi''(-x)

    By the chain rule:

    [psi(-x)]' = -psi'(-x)

    Differentiate both sides:

    [psi(-x)]'' = -[psi'(-x)]'

    Use chain rule again:

    -[psi'(-x)]' = psi''(-x)

    So, we see that:

    [psi(-x)]'' = psi''(-x) ----->

    [P psi (x)]'' = P psi''(x)

    Now, when two operators commute the eigenfunctions can be taken to be simultaneous egenstates of both operators. To see that in this case, suppose |psi> is an eigenstate of H with eigenvalue E, then:

    H |psi> = E |psi>

    Apply P on both sides:

    P H |psi> = E |psi>

    We have seen above that P H = H P, so we can write this as:

    H P |psi> = E P |psi>

    So, we see that if |psi> is an eigenstate with eigenvalue E, then P|psi> is also an eigenstate with the same eigenvalue E. If E is not degenerate, then this means that P|psi> must be proportional to |psi>. Since P^2 = 1, that then means that either

    P|psi> = -|psi>


    P|psi> = |psi>

    If E is degenerate, then you can conclude that P maps the eigenspace for eigenvalue E onto itself. You can then diagonalize P within that eigenspace, so you can then write down a basis for that eigenspace such that the basis vectors are simultaneous eigenstates of H and P. We say that if P|psi> = -|psi> then |psi> is odd under and if P|psi> = |psi> it is even.

    So, the eigenstates of H will be of the form I gave. They will have a small difference in energy. If you put a particle in the left box then, because that is not an exact eigenstate, the wavefunction will oscillate between left and right. So, there is a probability for the particle in one box to tunnel to the other box.
  9. May 17, 2008 #8
    I guess impenetrable barrier does not allow one particle to cross from one space to the other. Let me think about your question:

    the particle is in the box before t1 and have some wave function which generally is time dependent. At t1 the barrier appears, which physically speaking, must be taken as a real process. As such, let's suppose it very fast. If fast means really fast, then the portions of the wave function in each sub-space, at the time the barrier was inserted, will build up a wave function with two disjoint portions for the same particle. But if our understanding of what is fast, or even, if our experimental capabilities of putting fastly the barrier, does not correspond to something "quantum mechanically" fast, then what we may see is a non adiabatic transfer of amplitudes of the wave functions between different states of position and/or momentum. In this case we may not have conditions to predict what the wave function will look like just after t1.

    In both cases, It seems likely to me that there will be non zero probability of finding the particle in one semibox or in the other.

    Best regards

    Last edited: May 17, 2008
  10. May 18, 2008 #9


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    Science Advisor

    first, you must specify the wave function of the initial situation. It will be a superposition of sines and cosines, zero at the boundaries. With the barrier, the system is described by a wave function with two components; box normalized sines and cosines, one covering 0 - L, the other covering -L - 0. These two sets of eigenfunctions are orthogonal. All you need to do is to expand the original wave function on the new basis, the two sets of eigenfunctions, in order to get the probability of the particle being in one box or the other.(Technically, this problem is very similar to a spin 1/2 particle with a wave function a|+> + b|->, and asking the question: is the particle in a spin up or a spin down state?)
    Reilly Atkinson
  11. May 18, 2008 #10


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    Science Advisor

    The normalization issue is tricky. And, in fact, treating the barrier as finite, and then taking the limit to an infinite barrier, is probably a particularly good approach -- normalization is then straightforward

    Reilly Atkinson
  12. May 21, 2008 #11
    But if E is denerate, I can chose an eingensate of H which is not an eingenstate of P. For example, I argue that |left> is an eingestate of H, and therefore a particle in the left box never jumps to the right box
  13. May 21, 2008 #12
    I agree with you that we have about 50% probability to find the particle in one of the two boxes. My question is different: suppose you know at
    t2 that the particle is in the left box. Can you ''retrodict'' that it was in the same box at t1 (after the closure of the barrier)?
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