Jarwulf said:
Do wavefunctions have to have every conceivable possibility? Say for instance you have a chair. Does the wavefunction of the chair necessarily have a possibility where the chair breaks apart spontaneously? Or a set of worlds where the chair breaks apart if MWI is true? Or can the wavefunction simply consist of possibilities where the chair does not splinter apart?
Does the wavefunction of a being have to have a possibility where the being changes their mind about something or can all possibilities of the wavefunction simply be ones where the being's mind stays the same?
When we solve for the equation of motion using Newtonian mechanics, what we first do is account for all the forces acting on the system, i.e. we do F=ma. So say you have an object falling to the ground, you then have
F_g = ma
where F_g is the force due to gravity. But if you want to include more realistic situation, you put in other facts, such as frictional force due to air friction F_f, and maybe the object itself has its own propulsion F_p. Then you write
F_g + F_f + F_p = ma
and then you solve (if you can) for the equation of motion.
The same thing occurs for the wavefunction. You first start with the Hamiltonian/Schrodinger equation. You need to know all of the potential landscape that the system has. This may or may not be trivial. In one of the simplest case, say for an infinite square well potential (which every student in intro QM classes should know), you write down the kinetic term and then the potential representing that square well. That's the whole system! So the wavefunction that you solve
describes the system fully based on what you have given as the starting point, i.e. what you wrote for the kinetic and potential term.
But here's where it can get complicated, especially when you start adding complexity to the system.
1. You don't know what the exact Hamiltonian is, and so you have to make either an estimate or an approximation. This is true when you are dealing with a gazillion particles, as in condensed matter physics. It is impossible to write an exact Hamiltonian for a many-body system. So in such a case, you make some clever approximation for the potential, such as using the mean-field approximation. You say that, even though a particle in the system sees all the potential from other particles, we can simply make the approximation that, on average, it sees a constant "mean field" from all of the particles.
So your Hamiltonian will consist of the kinetic term, and a mean-field approximation of the potential term. Therefore, your wavefunction can only be as good as what you have done in the beginning. It cannot predict or describe something beyond that. In many situations, the mean-field approximation is perfectly valid and can account for a large number of phenomena. But in other situations, this approximation breaks down. It is not because the wavefunction is inadequate, it is rather our starting point and our knowledge of the system is inadequate.
2. You know the exact Hamiltonian, but you cannot get a full, exact wavefunction. In many instances, you can write the exact wavefunction, but solving the differential equation is often a major problem. One also encounters this in classical Newtonian mechanics (try to find exact, closed solution for the 3-body or more problem). This is where you either do numerical solutions, or in other cases, you make an approximate solution as a simplification, or even only consider special cases that gives you nice, analytical solutions. So obviously, it is not inconceivable that the solution could miss something when such simplifications are applied.
So in principle, the wavefunction should be able to describe ALL of the observables as described in the Hamiltonian. It depends on how well you can construct a Hamiltonian that accurately and fully describe the system you are looking at, and how well you can arrive as the wavefunction solution.
Zz.