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I How a particle localized in half of a box changes over time

  1. Jun 10, 2017 #1
    Screen Shot 2017-06-10 at 5.03.27 PM.png

    I think the answer is incomplete, because proving that ##\psi(x)## isn't stationary doesn't prove that the particle isn't localised, it doesn't prove that ##\psi(x)\neq0## at ##0<x<\frac{a}{2}## at some later time. ##\psi(x)## could always be changing on the left side of the box (say from a horizontal flat line to sinusoidal and back) while always be ##0## on the right side of the box. So we have to prove that this is not possible (that is, the left-side ##\psi(x)## changes while the right-side ##\psi(x)## is always ##0##). Right?
     
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  3. Jun 10, 2017 #2

    Vanadium 50

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    How would you describe what state the particle is in? How does that evolve in time?
     
  4. Jun 10, 2017 #3
    I expanded the wave function in terms of a sum of the energy eigenfunctions, each multiplied by ##e^{-i\frac{E_n}{\hbar}t}##, where ##E_n## is the energy of the ##n##th state. So each energy eigenfunction evolves at a different rate. But without a proof or a counter example, we may still argue that they evolve "conspiringly" to keep the sum always 0 on the right side of the box, as unlikely as it may seem.
     
  5. Jun 10, 2017 #4

    vanhees71

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    If you have done so (it's the perfectly right way to investigate this question), why don't you plot ##|\psi(t,x)|^2## for some times to see, what happens?
     
  6. Jun 10, 2017 #5
    But there is an infinite number of terms in the sum. I'm not familiar with any plotting software that can do that. Or do you ignore terms beyond a certain value of ##n##? But doing that would be an approximation to the actual wave function.
     
  7. Jun 10, 2017 #6
    It is safe to say that since the initial state ##\psi(x,0)## is not a stationary state, then at short times the probability amplitude must shift to the second half of the box (this might be an exercise worth trying: calculate the probability current at the box midpoint). However, to your concern, there are recurrence theorems (see link) in both classical and quantum mechanics which establish that there exists a finite time, ##T##, after which the state of the system comes arbitrarily close to the initial state.
     
  8. Jun 10, 2017 #7

    vanhees71

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    Of course, you can always only use a finite number in your numerical calculation, but you can check, how far you must go to get a satisfactory accuracy. Of course, for your initial state, you'll need a lot of wave functions, because it's a sharp edge, and Fourier-like approximations are bad for such cases.
     
  9. Jun 10, 2017 #8
    Probability current at the midpoint $$=\frac{16\hbar}{\pi ma^2}\sum_{n=2,6,10,...}^\infty\Bigg[\sum_{k=3,7,11,...}^\infty\frac{1}{k}\sin{\Big(\frac{E_k-E_n}{\hbar}t\Big)}-\sum_{l=1,5,9,...}^\infty\frac{1}{l}\sin{\Big(\frac{E_l-E_n}{\hbar}t\Big)}\Bigg],$$where ##E_n=\frac{\hbar^2\pi^2n^2}{2ma^2}##.

    Probability current at ##t=0## vanishes. But the probability current at other times is difficult to evaluate as it involves infinite number of terms, and so it is unclear whether the probability current also vanishes at all other times.
     
    Last edited: Jun 10, 2017
  10. Jun 10, 2017 #9

    Vanadium 50

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    I think we're making this harder than it needs to be.

    You have a particle in a box, and its energy eigenvalues are 1, 4, 9, 16, 25, 36, etc.

    Constrain the particle to forever be in the left-hand side of the box. Now its energy eigenvalues are 4, 16, 36, 64, etc.

    Now consider the case at hand, the particle is initially in the left hand side. If it has any admixtures of energies 1, 9, 25, 49, etc. it will not be forever in the left-hand side of the box. And if it's not forever in the left side, sometimes it must be in the right side.
     
  11. Jun 11, 2017 #10

    vanhees71

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    This is not correct. To simplify the calculation I make the interval ##x \in [0,1]## and ##\hat{H}=-\mathrm{d}_x^2##. Then the complete set of eigenfunctions are
    $$u_n(x)=\sqrt{2} \sin (n \pi x), \quad n \in \mathbb{N}_0$$
    with eigenvalues
    $$E_n=\pi^2 n^2.$$
    Then make
    $$\psi_{0n}=\int_0^1 \mathrm{d} x \Psi_0(x) u_n(x).$$
    For
    $$\Psi_0(x)=\begin{cases} \sqrt{2} & \text{for} \quad 0<x<1/2, \\
    0 & \text{for} \quad 1/2<x<1, \end{cases}$$
    which by the way is strictly speaking not allowed, because it doesn't fulfill the boundary conditions you get
    $$\psi_{0n}=\frac{4}{n \pi} \sin^2 \left (\frac{n \pi}{4} \right).$$
    Then at ##t>0##
    $$\psi(t,x)=\sum_{n=1}^{\infty} \psi_{0n} \exp \left (-\mathrm{i} n^2 \pi ^2 t \right) u_n(x).$$
     
  12. Jun 11, 2017 #11
    I think you misread what I've calculated. It's probability current ##j(x=0,t)##, not just the wave function ##\psi(x,t)##!$$j=\frac{\hbar}{2im}(\psi^*\frac{\partial\psi}{\partial x}-\frac{\partial\psi^*}{\partial x}\psi)$$But my expression could be wrong. There are two ways to get ##j(x=0,t=0)##. The first way is to use the ##\psi(x,0)## given by the question directly to find ##\frac{\partial\psi}{\partial x}## and substitute it into the definition of ##j##. This method gives us an undefined ##j(0,0)##, since ##\psi(x,0)## is not continuous at ##x=0##. The second way is to express ##\psi(x,t)## as a series of energy eigenfuctions, each term multiplied by its own time dependence, and use that series to find ##\frac{\partial\psi}{\partial x}## and hence ##j##. Then substitute ##t=0##. This method gives us ##j(0,0) = 0##. Is there a contradiction here?

    In other words, is it fine to have one method gives us undefined as the answer while another method gives us ##0##?

    One of the steps in the second method that I have used may not be correct. The step in which I converted a product of series into a double series: ##(\Sigma A)(\Sigma B)=\Sigma\big(\Sigma(AB)\big)##.
     
    Last edited: Jun 11, 2017
  13. Jun 11, 2017 #12
    I think this method is brilliant but it doesn't work for this question unfortunately, because the initial wave function given is not ##0## at both ##x=-\frac{a}{2}## and ##x=0##, but all wave functions satisfying the following two properties:
    (1) being constraint on the left of the box and
    (2) expressible as a series/sum of energy eigenfunctions of eigenvalues 4, 16, 36, 64, etc.,
    must be ##0## at both ##x=-\frac{a}{2}## and ##x=0##.

    Let's denote the property that the wave function is ##0## at both ##x=-\frac{a}{2}## and ##x=0## as property P. What you have established is that if a particle is constraint to the left with property P, then its energy eigenvalues are only 4, 16, 36, 64, etc. Therefore, if the energy eigenvaues are not only 4, 16, 36, 64, etc., then the particle is not constraint to the left with property P. But unfortunately, the particle in the question is not constraint with property P. Hence, the question has yet to be answered.
     
    Last edited: Jun 11, 2017
  14. Jun 11, 2017 #13

    vanhees71

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    As I said, just do the calculation. It's really easy. Attached you find a Mathematica Notebook (although the initial wave function is not a very good choice, because (a) it's not in the Hilbert space under investigation and thus at best an idealization of a more realistic example and (b) needs a lot of eigenfunctions to be approximated, because it's not converging very well, which is due to the jump of the wave function, which is always trouble for (generalized) Fourier decompositions).

    Since it seems still not to be possible to upload a Mathemica Notebook, I put it here:

    http://th.physik.uni-frankfurt.de/~hees/pf-faq/misc/wave-packet-box.nb
     

    Attached Files:

  15. Jun 11, 2017 #14

    Vanadium 50

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    I think it does.

    A necessary condition for a particle constrained to be in the region -a/2 < x < 0 is the energy eigenvalue condition I described above. It may or may not be a sufficient condition (in fact, it's not, but I haven't demonstrated it) but it is necessary. Do you agree with that?

    Now all I have to do is demonstrate that this necessary condition is unsatisfied. The easiest way to show that is to show that there is non-zero overlap between the given wavefunction and the E=1 eigenvector. You can take the integral (it's just a cosine) or you can see it by inspection. Do you agree with that?
     
  16. Jun 12, 2017 #15
    It is necessary only for a particle constrained in a special way, that is, with ##\psi(x)=0## at the boundaries. But it doesn't work for this question because ##\psi(x)\neq0## at the boundaries.

    But on the other hand, it is unphysical for ##\psi(x)\neq0## at the boundaries as all physically allowed ##\psi(x)##'s should be continuous and hence vanish at the boundaries. So you method works for all continuous ##\psi(x)##'s but not for discontinuous ones that do not vanish at the boundaries. So strictly speaking, your method does not answer the question.
     
  17. Jun 12, 2017 #16

    dextercioby

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    Can you write down the Hamiltonian operator for the quantum system for which the proposed wavefunction would be mathematically acceptable?
     
  18. Jun 12, 2017 #17

    Vanadium 50

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    The problem describes a wavefunction which is discontinuous at the boundaries, and complaining about it is a bit like complaining that stretchless ropes and frictionless planes don't exist. This, like so many physics exercises, is an idealization.

    Thus far, you have two answers. One is to adopt the more physical solution I posted. The other is to grind it out the way Vanhees suggested: expand the given wavefunction in energy eigenvectors, evolve them in time, and show that the wavefunction is non-zero on the right side. For most people, one answer is enough.
     
  19. Jun 12, 2017 #18
    I am working on a WebGL based finite difference time dependent Schrödinger solver, and I thought it would be interesting to see how it handles this. I put up an example that allows you to see this wave function evolve over time. It wasn't exactly what I expected. You can also change the initial conditions to see how other wave functions behave in the same potential.
     
  20. Jun 13, 2017 #19

    Vanadium 50

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    Your app says "For example, Ψ=Ψ1, is extremely low energy so the change over time is very slow." But Ψ1 is a stationary state. It shouldn't evolve at all.
     
  21. Jun 13, 2017 #20
    This is a great example of why I see simulations like this as valuable teaching tools. I am sure you are already aware of this, but I'll be pedantic here for pedagogical purposes. When we say that ##\Psi_1## is an energy eigenstate, we mean that ##i\hbar \frac{\partial\Psi}{\partial t}= E\Psi##. While this is a strong constraint on the time dependence, it also shows that there is some time dependence.

    Specifically, the time dependence takes the form ##\exp \left (-\mathrm{i} \frac {E t} {\hbar} \right)##. This has a couple of interesting consequences.
    • The real and imaginary parts of the wave function rotate into one another at an energy dependent rate.
    • The values for any measurable quantity do not evolve over time.
    This second point is what makes it a stationary state.

    In the simulation, this is exactly what we see. The green line, ##\Psi^*\Psi##, remains stationary, while the red and purple lines, the real and imaginary parts of the wave function, rotate into one another.

    BTW, this simulation is intended for educational use and carries an extremely permissive license.
     
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