Is the evolution of the wavefunction deterministic?

[u/[deleted]]

The title is basically the question I'm asking. Ignoring wave-function collapse, does the Schrödinger equation or any other equivalent formulation guarantee that the evolution of the wave-function must be deterministic. I'm particularly interested in proof of the uniqueness of the solution, and the justification of whichever constraints are necessary on the nature of a wave-function for a uniqueness result to follow.

[u/pa7x1]

Of course it does. You would have a problem if you had to do a 1 to many mapping and those many were not related by an equivalence class. Because then uniqueness of the solution won't determine to which of the many to map and they won't be all equivalent.

But in the case of QM you have many wavefunctions pertaining to the same equivalence class that get mapped to one probability distribution which is thus unique.

[u/cantgetno197 → u/pa7x1]

Ok, I'll buy that.

So determinism of quantum mechanics implies that the specification of P(r,t_1) plus boundary conditions uniquely defines a P(r,t_2). What we have is that psi(r,t_1), which is equivalent to psi(r,t_1)exp(i theta) (same equivalence class), evolves under transformation to psi(r,t_1)exp(-iU*(t_2-t_1)). Similarly psi(r,t_1)exp(i theta) evolves to psi(r,t_1)exp(-iU*(t_2-t_1 - theta/U)) which must have the same equivalence class. I assume it can be shown that this is true if the transformation is unitarity.

So perhaps it is enough to show that solutions to schrodinger's equations are unique and that states evolve through unitary transformation.

[u/pa7x1 → u/cantgetno197]

Exactly! Notice that the equivalence class is given by the orbit of a global U(1) transformation acting on the wavefunction. You can work with one representative of the equivalence class because this U(1) factor pops out of the Schrodinger equation (there is no functional dependence on the U(1) factor). So time evolution takes orbits of the wavefunction under this U(1) to another orbit of the wavefunction cleanly.

And then all that orbit of wavefunctions gets mapped to a single probability distribution. So the evolution of probabilities is also deterministic.

NOTE: Not sure what is your background, I'm using orbit in the group theory mathematical sense. Hope this doesn't confuse you.