1. Joined
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    20 Mar '10 22:15
    Sorry for asking so many questions related to class - I know most people probably don't find these interesting. That said, this is a difficult subject to understand, and I have gotten some good answers here so far.

    My question is this: say when x < 0 we have a free particle. At x=0 the particle enters a constant potential field. When x<0 the wavefunction of the particle is f_1 and after it enters the field its wave function is f_2. My book asserts pictorially that f_1(0) = f_2(0), without giving any justification for this. Perhaps this should be obvious, but it is not to me. Could anyone explain this?
  2. Standard memberadam warlock
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    20 Mar '10 22:44
    Originally posted by amolv06
    Sorry for asking so many questions related to class - I know most people probably don't find these interesting. That said, this is a difficult subject to understand, and I have gotten some good answers here so far.

    My question is this: say when x < 0 we have a free particle. At x=0 the particle enters a constant potential field. When x<0 the wavefunction ...[text shortened]... ation for this. Perhaps this should be obvious, but it is not to me. Could anyone explain this?
    Think about the Physics of what's happening. If it isn't obvious again I can handwave you an explanation.
  3. Joined
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    23 Mar '10 06:47
    It seems to be that the probability density must be continuous, otherwise things stop making sense. There shouldn't be two separate probabilities of finding a particle at x=0. But f_1(0) does not need to equal f_2(0) to represent this, only |f_1(0)|^2=|f_2(0)|^2 -- for instance, if f_1(0) = A+iB, f_2(0) = A-iB, and the probability at x=0 is still continuous. Since f_1 is generally not equal to f_2, what is the justification for the continuity of the wavefunction itself?
  4. Joined
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    23 Mar '10 10:36
    Think about solving the Schrödinger equation. It includes a second order derivative of the wave-function. Along the lines (simplifying for convenience) of:

    df/dt = -d2f/dx2 + Vf

    V is discontinuous at 0, so d2f/dx2 will be as well. But if df/dx was discontinuous then d2f/dx2 would be a delta-function, and we wouldn't be able to get that to balance.

    Which means, I think, that not only must the wave-function be continuous at 0, but so must its first derivative.
  5. Standard memberadam warlock
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    23 Mar '10 15:54
    Originally posted by mtthw
    Which means, I think, that not only must the wave-function be continuous at 0, but so must its first derivative.
    Right on!

    For you to have a discontinuity in the first derivative of the wave function the potential needs to have rather violent discontinuities too (delta potential for instance).

    The Physical justification of this is that if the potential doesn't change very abruptly the probability current has to be continuous. That is that we can't have no sources nor sinks in our problem. If we have delta-function like potentials sources and sinks are allowed (because physically that's what a delta function is - in a handwaving kind of way.)
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