How many worlds in the many-worlds interpretation of quantum mechanics?

[iamnotbatman]

Very easily.

A wave function can be written in a variety of bases. Do you know what a basis is? When I say a wave function is continuous, I am of course referring to it being continuous in some basis. For the particle in a box, the wave function in position space is continuous, and in energy space is discrete. The energy eigenvalues are discrete. The position eigenvalues are continuous.

It really seems to me that you might not have a full grasp of quantum mechanics.

[Pleonast]

@iamnotbatman

A wave function can be written in a variety of bases. When I say a wave function is continuous, I am of course referring to it being continuous in some basis. For the particle in a box, the wave function in position space is continuous, and in energy space is discrete. The energy eigenvalues are discrete. The position eigenvalues are continuous.

Thank you for clarifying, but the representation of the wave function is independent of the eigenvalues.

In any case your tone and continued jibes indicate that you are not interested in a respectful conversation.

[Asympotically_fat]

@Pleonast

How do you reconcile this:

With this:

?

This is actually incredibly simple: the eigenvectors of some operators don’t span the state space.

However the postion operator does provide an eigenbasis.

[Half_Man_Half_Wit]

@Pleonast

In any particular experiment, not all values are potentially observable.

You should really provide a cite, or at least an argument, for this statement. What makes it so that out of a continuous energy spectrum, only a discrete set of values is observable in an experiment? Why is some value E observable, but E + e for arbitrarily small e not? You may not be able to distinguish them in a measurement, but this a) has nothing to do with quantum mechanics, b) does not change the fact that the set of possible values is continuous, not discrete.

@Pleonast

Quantum measurements are always discrete. It’s impossible to construct a single device to make a single quantum measurement that could potentially return any value from a continuous set.

How do you tell, for a single measurement, whether it came from a continuous or discrete set? Could you create, in a purely classical world, an instrument that could return a value from a continuous set?

That means that someone using a many worlds interpretation of quantum mechanics does not need to worry about a continuum of worlds. Any observation that “splits” the world will always produce a countable number of worlds.

No. If the eigenvalues of the original system are continuous, then the eigenvalues of the system+measurement device after measurement will be continuous, and the eigenvalues of the system+measurement device+experimenter after looking at the measurement device will be continuous and so on, until eventually you end up with a universal wavefunction with continuous eigenvalues.

[iamnotbatman]

@Pleonast

Thank you for clarifying, but the representation of the wave function is independent of the eigenvalues.

In any case your tone and continued jibes indicate that you are not interested in a respectful conversation.

What do you expect when your retorts are as incoherent/irrelevant as:

“The representation of the wave function is independent of the eigenvalues.”

So? The point is that the wave function can have both continuous and discontinuous eigenvalues. Your continuing to harp on the fact that the energy eigenvalues are discrete is totally irrelevant.