- #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]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.
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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