- #1

Loren Booda

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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]

What of discontinuos quantum

Take [psi]

_{1}(x)=cos(x) and [psi]_{2}(x)=-cos(x). Although themselves and their derivatives respectively discontinuous, these wavefunctions define a continuous 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.
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