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Why should wave function be continuous?

  1. Nov 28, 2006 #1
    I think it needn't be continuous even if the probability should.
    Wave function can be complex while probability is its absolute value.:bugeye:
  2. jcsd
  3. Nov 28, 2006 #2


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    Sure, and if a complex function has a singularity, what is its absolute value there?:rolleyes:
  4. Nov 28, 2006 #3
    I'm sorry that I can't catch you whole meaning exactly.

    Let me construct an example as follow (f(x) is the wave function over x):
    When 0<x<1/2, f(x)=e^(i*x),
    when 1/2<x<1, f(x)= -2x+2,
    and otherwise, f(x)=0.
    there is only one singularity at x=1/2 with the absolute value is 1.

    By the way,I don't know how to paste a picture or a formula quickly in this forum.How could I get information about that?
  5. Nov 28, 2006 #4


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    There are at least two problems here: there's no Hamiltonian for one. But more serious, in the function above, obviously the limit from the right and the limit from the left at x=1/2 don't agree. the only way this could happen physically is with some kind of weird barrier at x=1/2. The basic Schrodinger Eq. only allows discontinuous solutions for disjoint regions; that is continuity within a region, but not region-to-region. This, of course, is a well known aspect of differential equations.

    Physically it will take a large perturbation to shift the absolute value of by a substantial amount over a very short region of space. So, it makes sense to me to extrapolate, and suggest that an amost infinite potential change is required to make an almost discontinuous change in a wave function. That's the intuitive reason why discontinuous wave functions are troublesome, and really refer to differences between two or more disjoint regions of space.
    Reilly Atkinson
  6. Nov 28, 2006 #5
    One of the more important reasons that the wave function needs to be continuous is that

    [tex] \hat{p} \equiv - i \hslash \nabla[/tex]

    so what happens to the expectation value of momentum if you have a discontinuous wave function?

    Also, since the hamiltonian is a partial differential equation that is linear, there are a lot of theorems, one of which pretty much requires that the solutions be continuous and twice-differentiable.
  7. Nov 28, 2006 #6
    Thank you,reilly and StatMechGuy.
    You reply help me greatly.

    But when we deal with delta potential,why we consider the wave function continuous at x=0 where the potential is infinite and the differential of the wave function is discontinuous?
    Is it just for simplity in analyse and calculate.
  8. Nov 28, 2006 #7
    Since the wave function should be a solution of the Schrödinger equation, it must be differentiable, thus also be continuous.
  9. Nov 28, 2006 #8
    You can go back to the Schrodinger equation and then integrate immediately around the delta function. If you assume the wave function is continuous, you get for sufficiently crazy potentials that the derivative of the wave function is discontinuous. That said, you never actually see a delta function potential in nature.
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