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Infinite square well with finite potential energy inside

  1. Mar 14, 2012 #1
    Assume that you have a one dimension box with infinite energy outside, and zero energy from 0 to L. Then my understanding of the Schrodinger equation is that the equation inside will be:
    -h^2/2m*d2/dx2ψ = ihd/dtψ

    And the energy eigenstates are given by
    ψ(x,t) = e-iwt*sin(kx)
    where k = n*π/L and w = k2h/2m and n = 1,2,3...

    Well, that's nice but what about if there is a positive potential energy U0 inside?
    -h^2/2m*d2/dx2ψ + U0ψ = ihd/dtψ

    What is the solution ψ(x,t) in explicit form? And how do you find the energy eigenstates?

    My guess at solving this would be to solve -h^2/2m (ψ)'' = (E-U)ψ as an ODE.
    I still get ψ = sin(kx), but I'm not sure whether this satisfies the boundary conditions or not. Also, I get that k2 = 2m(E-U)/h2.

    My naive assumption is that ψ(0,t) = ψ(L,t) = 0 since it can't exist pass an infinite potential wall and because then the function would be continuous, since ψ = 0 at all other points.

    If this is true, then energy eigenstates of the particle must be ψ(x) = Asin(nπ/L*x), where A is any arbitrary phase factor and n = 1,2,3...
    Thus, assuming ψ(x,t) = e-iwtψ(x)

    Thus we get h2k2/2m*ψ(x,t)+U*ψ(x,t) = hw*ψ(x,t)
    Factoring out, we get h2k2/2m + U = hw = E

    Thus w = hk2/2m + U/h = h/2m*(nπ/L)2 + U/h
    My interpretation of this is that given a certain n, the eigenstates "look" the same in complex phase space, but they are spinning around faster at a rate of U/h.

    Energy is quantized in this box to E = h/2m*(nπ/L)2 + U where n = 1,2,3...

    My main problem arises when U is negative, and chosen such that U = -h/2m*(π/L)2, thus leaving the total energy at the ground state as zero, and the wave function is "standing still" with w = 0. What's going on here? Even worse is when U is greater than E0, the angular frequency is negative, does that mean the wavefunction is spinning in the opposite direction as normal positive energy wavefunctions?
     
  2. jcsd
  3. Mar 15, 2012 #2

    Matterwave

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    Only differences in energy matter. The solutions are all the same, except all your energies are shifted up by this constant.
     
  4. Mar 15, 2012 #3
    But what is the interpretation of E = 0? Does that mean the wavevector is standing still in Hilbert Space? Is this possible or impossible? Also, do negative energy states spin in the opposite direction such that ψ(x,t) = e+i|E|t/h*ψ(x)? This would affect the time evolution of normal states (states that aren't energy eigenstates), would it not?
     
  5. Mar 16, 2012 #4

    Matterwave

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    There are no negative spin states, the ground state has some positive energy. What do you mean by spin in the other direction? There's no spin in this problem.
     
  6. Mar 16, 2012 #5
    What i meant is the intepretation of angular frequency w. the wave function is "spinning" in time through complex phase space. Im not refering to intrinsic spin. Anyway, the higher the energy, the faster the wave function rotates around in phase space. The magnitude squared stays the same at each point but wavefunction is still changing with time. Except at E = 0, where the wavefunction doesnt change over time. Anyway, do you think i solved the problem correctly?
     
  7. Mar 16, 2012 #6
    This statement does not have any physical meaning for a particle in a one dimension box, due to uncertainy principle! Or what do you mean?
     
  8. Mar 16, 2012 #7
    The particle can have a definite energy, it's position that's uncertain. When I think about a wavefunction on a one dimensional box, I imagine the x-axis as the position, which then leaves the y and z axises in my mind to represent the real and imaginary parts of the wavefunction. I like to make the y-axis the real part and the z-axis the imaginary part. At every point of x, a vector points out from the x-axis in the y-z plane to represent the value of the wavefunction at that point x. For energy eigenstates, the length of the vector at a certain point x doesn't change over time, but it's angle, or phase, does. This is what is meant by
    ψ(x,t) = e-iE/hψ(x).
    The value of |ψ(x,t)|2 = |ψ(x)|2 = constant over time, but the wavevector is still rotating in my mind around the x-axis. For example, if you state the first energy level of the box and set U = 0, then the shape of the wavefunction is ψ(x) = sin(pi/L*x), since n = 1. This is the shape a single bump (0 to pi) of a sine wave that stretches from 0 to L. With time, it rotates around the x-axis with a certain angular frequency, w = E/h. I imagine in my mind a single bump that rotates around the x-axis, keeping the length from the x-axis constant. Can you visualize that? Also, since the next energy eigenstate is n = 2, the angular frequency should be four times larger, since w = h(npi)2/2mL2.
    Thus I imagine the second energy eigenstate as being a sine wave with 2 bumps now, and rotating four times faster than the first energy eigenstate. Makes sense to me, since the number of bumps is proportional k, which is proportional to momentum, by p = hk, and since E = p2/2m, it makes since that the rotation is four times faster, since E = hw.

    The energy of a particle to me is related in how fast in time it's wavefunction is rotating around in phase space, the y-z plane in this case. A wavefunction standing still actually doesn't seem so bad anymore, it's all relative to the other rotations anyway.

    My new question is, why doesn't the phase seem to matter in an absolute sense? Why is it that only relative phase matters? Doesn't nature care at all about which way we start the rotation of the wavefunction? The answer to me appears to be not, just like there is no such thing as absolute velocity, it's all relative.
     
  9. Mar 16, 2012 #8

    Matterwave

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    You can't have E=0 in a infinite square well with V=0. The lowest eigenstate has E=hf
     
  10. Mar 16, 2012 #9
    Right, exactly, you can't have E = 0 when with V = 0. But what happens when we set V exactly to the negative of the first energy eigenstate when V = 0? Thus we set V = -h2(pi/L)2/2m, for a given L and m. What then, shouldn't the potential energy perfectly cancel out the kinetic energy from setting k = npi/L to n = 1? What's up with that? Wouldn't you agree the energy level there would be E = 0?

    You said all the energy levels shift up with a positive potential energy, so shouldn't they shift down with a negative potential energy? I don't think this violates the uncertainty principle by the way, the infinite square box always creates energy eigenstates with perfectly defined momentum, and only finitely undefined position. So actually in a sense the infinite square box somewhat always violates the uncertainty principle, but that's alright because it's an impossible situation. It's just a useful model for understanding the eigenstates of free particles as you set L → ∞.

    I just find that the E = 0 situation to be interesting in the case of a finite L, but in the case of L → ∞, the V would also have to become infinitely negative. So E = 0 should be impossible with a truly free particle, right? Also, you're not much of a texter are you? Only one or two sentences, why don't you write more? :)

    Also, I mean h = hbar everytime, I'm not ignorant, I just don't know how do make that symbol appear. Do you know how to do that? Please answer me that at least.
     
  11. Mar 16, 2012 #10
    Also, I think made a huge mistake in my analysis, as you set L → ∞, V → 0. Which means it's totally possible to get E = 0 for a free particle as long has you extend a uniform negative potential energy across the entire x dimension. In fact there would even be negative energy states. Every state above V would be possible, I'm so silly sometimes haha.
     
  12. Mar 16, 2012 #11

    Matterwave

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    Ok, it still doesn't matter though because all of the energy eigenstates are stationary states. None of the probabilities change in time for any of them, so the case with E=0 is the same as any other energy eigenstate. As is apparent, absolute energy should not matter ever.

    [itex]\hbar[/itex]

    Just quote to see how that symbol is made.
     
  13. Mar 16, 2012 #12
    It's there any shortcuts you use to writing [itex]\hbar[/itex]? It's seems kinda cumbersome to have to keep having to copy and paste that over and over. Also, [itex]\hbar[/itex] rocks, it's way better than h, wouldn't you agree?

    Also, there's two reasons I'm worrying about this amplitude rotation business. Sure, the amplitude squared doesn't change over time, but the amplitude itself certainly changes with time, right? Except when E = 0, that's the only time when even the amplitude doesn't change over time, in addition to the square of it also not changing.

    The first reason I care is because of how to model Fourier compositions of multiple energy states. The energy basis of ψ is extremely useful because the Schrodinger equation shows how each individual energy eigenstate amplitude changes over time. Thus any ψ can be described as a combination of many to infinitely many of these simple eigenstates. The way I imagine ψ change over time is by imagining each of its energy eigenstate components as rotationing at different rates, and the difference in rates leads to a different value of ψ(x) at different times t. Thus the usefulness of refering to ψ(x,t) is born. When you have an energy eigenstate of E = 0 though, that component is just chilling there, not rotating at all while all the other parts are.
    Since Schrodinger chose the ψ(x,t) = eikx-iwt convention when writing his equation, the positive energy states are rotating clockwise in my imagined y-z plane, and the negative ones are rotating counterclockwise.

    The second reason, which is more important and kind of related to first is, how do you model a potential energy V in a infinite square box that's changing with time?
    For example, say V(x,t) = V0sin(ωt), for some ω. Let's also say that V is uniform throughout the box. What happens to an energy eigenstate then? The energy of the box is changing, so the energy of the particle should also be changing. Also, the energy eigenstates are changing in how fast they rotate.

    The most important question is how do you model any given ψ going through that perturbation?
     
  14. Mar 16, 2012 #13

    Matterwave

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    If you like this rotating analogy, just note that one can always go into a reference frame where one of the vectors is not rotating, and the other vectors do rotate. There's no need for EVERY vector to rotate. Having one not rotating is fine. What you are doing by shifting the energies up and down is basically moving into these co-rotating frames, if you will.

    As I said, shifting energies up and down cannot do anything physical to the system. Notice that taking fourier series, one always gets sums and differences of energies in the interference terms, and it is only these sums and differences that matter. The non-interference terms will be squared and any e^ikx terms will cancel out.

    If you have a time-dependent potential, then you no longer have energy eigenstates. In that case, the problem becomes more difficult, and will, in general, not have easily obtainable analytic solutions. One usually resorts to time dependent perturbation theory for these problem.
     
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