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Discontinuous wavefunction

  1. Oct 9, 2003 #1
    Quantum mechanics requires continuity of the wavefunction and its first derivative. How stringent is this requirement? If general relativity allows singularities, why not have possible discontinuity of a single wavefunction and its derivative?

    Take [psi]1(x)=cos(x) and [psi]2(x)=-cos(x). Although themselves and their derivatives respectively discontinuous, these wavefunctions define a continous probability, |[psi]1(x)|2=|[psi]2(x)|2.

    What of discontinuos quantum probability itself? Take a probability step function of finite domain, discrete at x=0. If we can say that an event occurs with probability P for -[ee]<x<0, what can we say about a probability P' for -[ee]<x<[ee]? It seems to me that this singular step makes undefinable probability P' in terms of P.
    Last edited: Oct 9, 2003
  2. jcsd
  3. Oct 10, 2003 #2


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    discontinuity of the wave function.

    First of all, it is funny you consider cos(x) as being a discontinuous function . But ok, apart from that, the continuity constraints on the wave function psi come essentially from the condition that it has to obey the Schroedinger equation, which is second order in x. In order for the second order derivative to be finite (even if discontinuous), the first order derivative and the function itself should be continuous. If you allow for infinite potentials (such as a hard wall), then the second order derivative should also become infinite, hence only the function itself should be continuous, but it is clear that this is an idealisation in the model.

  4. Oct 10, 2003 #3

    "respectively discontinuous." Just graph them together - at no value of x are cos(x) and -cos(x) in continuous agreement, or are their derivatives.
  5. Oct 10, 2003 #4


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    Ah... I suppose you're talking about a function, f(x), such that:

    f(x) = cos(x) if x<y
    f(x) = - cos(x) if x > y

    for some y.

    Sorry, the expression "respectively discontinuous" was a bit weird to me, but I guess you meant the above.

    f(x) won't be the solution to any "reasonable" Schroedinger equation. I say, "reasonable" because you could think of a highly singular potential function which doesn't represent a physical situation for which this could be heuristically true (with derivatives of Dirac impulses).
    The probability is not the only physical content of the wave function. The phase is just as important, and in the classical limit, the phase corresponds to Hamilton's principal function (in Hamilton-Jacobi theory in classical mechanics). So it is not because the probability seems to make sense that any underlying complex function makes sense as a quantum wave function.

  6. Nov 13, 2007 #5
    But this does not mean that the wave function cannot be discontinuous at the initial time t=0, right? So if at t=0 a wavefunction is discontinuous at a point, doesn't that go against the probabilistic interpretation? I mean, probability of finding a particle at that point would be undefined.
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