Applying the FTC to an arbitrary solution of the Schrodinger equation

[AxiomOfChoice]

Suppose you've got a function \psi(t) that satisfies i\dot \psi = H \psi for some self-adjoint Hamiltonian H. I'd like to apply the fundamental theorem of calculus to this guy and write something like
<br /> \psi(t) - \psi(0) = \int_0^t \psi&#039;(s)ds.<br />
Can I do this, given only the very bare conditions I've placed on \psi? Or are there some other things I'd need to assume about \psi to make it kosher?

 

[lugita15]

I'm not sure about whether the time-derivative of the wave function must be continuous with respect to time; I suspect it does, only because for each energy in the spectrum the associated time dependence is an exponential function, so we're dealing with a linear combination of exponential functions if the energy spectrum is discrete or at worst an integral of an exponential function with respect to dE if the energy spectrum is continuous. So let me talk about space is instead. In order for the fundamental theorem of calculus to fail to apply with respect to position, the set of discontinuities of the derivative must have positive measure. And the derivative of the wavefunction can only be discontinuous at a point if the Hamiltonian is infinite. So to construct a situation where a wavefunction fails to obey the FTC, we could contruct something absurd like a potential that is an uncountable sum of delta functions, so that it is infinite on the Fat Cantor set (so that the wave function becomes something like Volterra's function.

 

[lugita15]

So to construct a situation where a wavefunction fails to obey the FTC, we could contruct something absurd like a potential that is an uncountable sum of delta functions, so that it is infinite on the Fat Cantor set (so that the wave function becomes something like Volterra's function.

I'm a bit less sanguine about this answer than I was a few months ago. Does anyone know whether the notion of an "uncountable sum of delta functions" can be made precise?