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Infinite vs Finite potential well

  1. Oct 27, 2008 #1

    I have learned about the infinite quantum well last week and today the finite quantum well was introduced. There is however one thing I don't completely understand, or which I'm not completely happy with.

    In the case of the finite well, the shrodinger equation is solved for both inside and outside the well, yielding the usual sine/cosine inside and now a decreasing exponent outside the well, meaning the particle can leak through.

    In the case of the infinite well, instead of 'trying to solve' the shrodinger equation, it is simply stated that [itex]\psi = 0[/itex] outside the well.

    Is this all there is to it, or can you determine this by actually trying to solve the shrodinger equation (although I have no idea how that would work) using an infinite potential?
    If you can't calculate it using the shrodinger equation, is the fact that the potential is infinite enough to make it absolutely certain that the particle cannot be there? Classically, the particle cannot be outside a finite well either (assuming it does not have enough energy) but by solving the shrodinger equation we can show that it can, so when I hear this quick 'psi = 0' argument without further elaboration, something does not 'agree' inside my head...
    It seems that either people simply assume that because the potential is infinite, the particle can't be there, or that there is something else going on that was not explained to me yet (i hope the latter...)

    I just feel that, in this 'all new world' of quantum physics where strange effects are so common, you shouldn't just say that the particle can't be outside the infinite well without further explaining...
  2. jcsd
  3. Oct 27, 2008 #2


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    The infinite well is a limiting case of the finite well. You seem to be fine with the solution to the finite well, so start there.

    Assume the potential barrier is at some finite potential V. Thus, the wave function for the areas outside of the well are decreasing exponentials. Now, what happens to the wave function in the regions of potential=V as you let V approach infinity. i.e. What is:

    [tex]\lim_{V\rightarrow\infty}\Psi_{outside well}[/tex]

    You should be able to see that the Psi=0 assumption your instructor made is justified by realizing that the result of this limit will describe the wave function outside of an infinite well.
  4. Oct 27, 2008 #3
    Alright that makes sense, the book even makes that argument I notice now... So I was merely thinking the wrong way around; thinking that the infinite square well was the basis and the finite square well is something entirely different where a new strange effect (the 'bleeding') occured.

  5. Oct 28, 2008 #4
    you know i wrote a whole giant response and then i found out about attenuation at the walls of a finite well am just as confused as OP. not really but what in the hell is signal attenuation when applied to matter waves.
  6. Oct 28, 2008 #5

    George Jones

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    Outside the well (for E < V) is a classically forbidden region. [itex]\left| \psi \right|^2 > 0[/itex] means that quantum mechanically there is a non-zero probability of finding the particle in the classically forbidden region, i.e., a non-zero probability that quantum tunneling occurs. The probability density decays exponentially as one moves further into the classically forbidden region.
  7. Oct 28, 2008 #6
    That's another 'issue' that occured to me today. Maybe it should be in its own thread but I feel it is related so I'm going to just post it here.

    We haven't 'learned' about quantum tunneling yet but I think I know roughly what it means.

    If you have a finite well, where the potential is some value -V between two points and zero everywhere else, then if my understanding of tunneling is correct, it is not possible for the particle to tunnel, because there is nowhere it can tunnel to. If you have a potential wall for example of finite width then the particle can tunnel through the potential barrier even if it hasn't got enough energy simply because the wavefunction is not zero at the otherside of the barrier. But in the case of a single finite well there is no 'otherside', so the particle cannot tunnel, correct?

    I also believe that in the case of a potential barrier, you cannot find the particle IN the barrier, only outside it (whether in the classically allowed region or the classically forbidden region). If the particle is in the classically forbidden region it is has tunneled through, but it never was in the barrier itself, correct?

    So how can we say that there is a non-zero chance of finding the particle outside the finite well, if it cannot tunnel there because there is nowhere to tunnel to?

    Where did my understanding fail?
  8. Oct 29, 2008 #7


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    Maybe it is easy to illustrate with a real life example, I'll take that one of alpha decay:


    Alpha Decay - The strong force, despite its strength, has a very short range; it can't even reach from one end of a fair-sized atomic nucleus to the other. If a proton is at the edge of a big nucleus, it can feel the pulling strong force only from the particles in the neighborhood, but there is an electromagnetic force, which tends to push it out, all the way from the other side of the nucleus. There is a sensitive balance between these two competing forces. The nucleus needs not to acquire extra energy to escape; the quantum mechanical effect called tunneling allows a certain probability of escape through a potential wall. Alpha decay, in which just a small chunk breaks off from the main nucleus, is a rather mild case of fission; in more dramatic examples, the nucleus can break more or less in half. The broken-off chunk most often is packed into a helium nucleus (alpha particle) because it is in a more stable form. Figure shows the effect of tunneling through the Coulomb barrier; the nucleus has a small probability of escape to the outside depending on the height and width of the wall.

    So think of this as an alpha particle inside the nucleus, it will feel a repulsive force from charges lying on larger radius that itself. BUT there is a small change for it to overcome the repulsive potential and escpace.

    You must understand that examples given in introductory QM is just for training to solve the shrödinger equation and get familiar with the concepts, and most of all, get rid of "all" intuitive working in modern physics - the formalism is the important stuff.

    You can solve the shrödinger equation for a finite well, then increase the strength of well (lim-> infinity) and look what happens.
  9. Oct 29, 2008 #8

    George Jones

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    Why do you say this? Because the potential is zero outside the well?
  10. Oct 29, 2008 #9
    No, because of something I heard my teacher say earlier. I don't know if I interpreted him incorrectly or if he was simply incorrect (your 'tone' suggest that I am incorrect, right? :p ).

    My teacher was joking about quantum tunneling and told us that if he would walk into the wall of the classroom there was a non-zero chance of him tunneling through and appearing at the other side. Then some smart-*** said 'what if you tunnel half-way through the wall and then become stuck?' to which he replied that he was never inside the wall at all, he merely 'jumped' from inside the classroom to outside, never actually being inside the wall so there was no chance of him getting stuck.

    So what is wrong here, besides the fact that quantum tunneling probably can't be described so easily by a "real-world" example of warping through a wall...? Is the particle actually moving through the potential barrier, or is it suddenly outside the barrier?

    I'm becoming to think this is much more complicated than my teacher made it look because it doesn't make sense to me to say that the particle was moving through the barrier while you can't even know that for sure (because the particle's location can't ever be pinned down to a single point)...

    On another note, is it actually possible for you to walk through a wall by quantum tunneling? I would say yes, if every single particle in your body tunnels through the wall at the exact same time. Obviously the chance of this happening is next to nothing but it's still a non-zero chance right? So if I would try it right now I could get extremely lucky and tunnel through, right? Or is there something else that makes the chance of all these particles tunneling at the same time zero after all?
  11. Oct 29, 2008 #10
    no because there's nowhere to tunnel to.

    you know you have surprisingly good understanding of the concepts for them having been presented to you so unrigorously. no there is nothing that makes the probability zero. but extremely small and nonzero are essentially the same thing.

    my question is if you could measure something about the particle in the barrier would we observe the particle there occasionally? has someone done these experiments?
  12. Oct 29, 2008 #11
    No, it isn't. If the probability if me tunneling through my wall right here and now are zero then I can try and I know with 100% certainty that I will fail.
    If the probability is not zero but simply extremely small, then if I try it here and now, there is a possibility that it works and that I tunnel through. Even if the probability is extremely small, there's still a chance that the first time you try it it works!
    (I'm not saying I'm going to try it, I know the chance is extremely tiny but it what you say is true than it could happen!)

    That's actually precisely what I was trying to say in my last post but didn't find the correct wording, thanks :)
    But since we can't say the position of the particle with 100% certainty, how can we ever be sure that it even is in the barrier, even if the probability of it being there (psi^2) is very large?
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