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Infinite Well and Boundary Conditions

  1. Aug 8, 2013 #1
    Hi All,

    I would like to know why in the infinite well problem, after having solved the time independent SE, we are not supposed to equal to zero the x derivative of the spatial part of the wave function at -L and L (2L being the total width). We only have to make it zero at the boundary.

    Thanks a lot,

    Best wishes,

    DaTario
     
  2. jcsd
  3. Aug 8, 2013 #2

    jtbell

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    Staff: Mentor

    In this situation dψ/dx is undefined at the boundary. ψ has a perfectly sharp "kink" there, and dψ/dx is discontinuous.

    This is actually an unphysical situation. A more realistic setup would have a steep slope in V(x) at the boundary, taking it from 0 to some very large (but not infinite) value in a very short (but nonzero) distance. But that would be more difficult to solve exactly. Nevertheless, making the unphysical assumption of an infinitely-steep, infinitely-high "wall" produces solutions which are useful approximations to physically-realistic ones.
     
  4. Aug 8, 2013 #3

    jtbell

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    Staff: Mentor

    Now that I think of it, the problem with the infinite square well is not that V(x) is discontinuous, but rather that it becomes infinite outside the well. The finite square well also has discontinuous V(x), but V doesn't become infinite outside the well. In that case, both ψ and dψ/dx are continuous, and you have to use both conditions in order to find the solutions.
     
  5. Aug 8, 2013 #4
    Ok jtbell,

    but my concern here has to do with the mathematical argument used, which is not clear to me. Why for normal functions of V(x) we search for \Psi functions which are continuous and have continuous spatial derivative, but for the infinite well we only search for continuous \Psi functions ?

    In Cohen' Quantum Mechanics book they only say that the infinite discontinuous behavior of V(x) gives rise to a spatial part of \Psi with discontinuous spatial derivatives, but don't show the reason.

    Best wishes,

    DaTario
     
  6. Aug 9, 2013 #5
    Maybe you could think about the potential well as not infinite but very large, then consider how the solution behaves as the well tends towards infinity.
     
  7. Aug 9, 2013 #6
    You can show it youself.Just set up the schrodinger eqn and integrate from x-ε to x+ε taking into account the discontinuous behaviour of V.
     
  8. Aug 9, 2013 #7

    jtbell

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    Staff: Mentor

    With an infinite square well, ψ is zero outside the well. Suppose dψ/dx is continuous at the boundary, for the sake of argument. If ψ becomes positive just inside the well, it has to curve upwards smoothly, with d2ψ/dx2 > 0. If ψ becomes negative just inside the well, it has to curve downwards smoothly, with d2ψ/dx2 < 0. In either case ψ and d2ψ/dx2 have the same sign.

    Now look at the time-independent Schrödinger equation for V = 0. If E > 0, can ψ and d2ψ/dx2 have the same sign?
     
  9. Aug 9, 2013 #8
    But in this "limiting" way of thinking, at all times the two requirements will be made.
    ψ continuous and dψ /dx continuous.
     
  10. Aug 9, 2013 #9
    That is a nice starting point, at least. This seems to require that a discontinuous behavior must appear...

    Thank you,

    DaTario
     
  11. Aug 9, 2013 #10
    Yes, I agree. But the limiting approach is arguably a more rigorous way to get at the solution, without assuming special behaviour of ψ and dψ /dx at the potential well boundary.

    It is certainly more satisfying for me, since it makes clear that dψ /dx is continuous, no matter how large the potential well. Physically, attributing an infinite value to an energy is not something you should do unless there is no alternative!

    The details of the alternative can be found in: http://en.wikipedia.org/wiki/Finite_potential_well. As you can see there, the behaviour of the infinite potential well appears naturally as a limiting case of the potential well problem, as it should.
     
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