Physics Asked by user332153 on May 2, 2021
My question is this:
What do we get if instead of having a quantum state as a vector in a Hilbert space, we choose to relax this vector condition to a module over a ring instead of a vector space over a field?
I received this answer and I try to study it. Can you explain better this idea or provide me a better solution?
You need more ingredients than that mathematical structure:
- You need a way to draw probabilities. This probably means you want a
specifically, so you have some analogue of an inner product.- You also want some analogue of the Schrodinger equation. This requires two things
· Assuming you still want continuous time, this would involve
wanting some degree of continuity in your output space, so you need
topology (and, in particular, completeness). So now we have a complete
topological module as well.
· You presumably want some analogue of the
factor of i in the SE (from which you specify Hamiltonians as linear
maps of some kind, presumably with some requirements on their
properties). It’s unclear to me what property this general mapping
should have, but it probably requires some further constraints on your
space.For a complex Hilbert space, all of these come automatically I
suppose completeness might technically be a little stronger than what
you need, but you’ll want something like it.Reason being, you presumably want $lim_t→t_0f(t)=f(t_0)$ for your time
evolution. For that to be true, you first need the LHS to converge to
something (and then the right thing, but that comes later).[The reason I say “might be a little stronger” is that, depending on
your constraints, you might not need every trajectory, but at least
along some. E.g., in complex-space, you technically only need it along
norm-preserving trajectories, since time-evolution is
norm-preserving.]In fact, if you want time-evolution to be smooth, you probably
actually want to be even more constraining and have a real manifold
and not just a complete topological space (because real manifolds are
the spaces on which you naturally define differential equations in the
way we usually think of them).But again: requiring a real manifold along with everything else seems
to be getting dangerously close to just having a complex vector space
again. I don’t know offhand of another model that would have such
properties.[Perhaps you could get around some of it with algebraic geometry,
where you can also have derivative operators, since you can define
them as formal operators on polynomials without requiring those
polynomials be over the reals?
What do you think about it?
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP