Clarifying the Meaning of "Random" in Quantum Physics

[stevendaryl]

A single solution of the classical Maxwell equations may after some time (e.g., after passing a beam splitter) have significant energy in many of your cells simultaneously, even if it is localized at time $t=0$ in a single one. This is contrary to your assumed particle behavior. You must assume that upon a split the particle remains in a single cell though with a certain probability only, while the wave remains a single object divided into two beams.

Sorry, I still don't see that. In terms of my cellular model (which is basically Bell's "local beable") the classical record within a cell would include both the amplitude and direction of the E and B fields inside the cell at every moment in time, the presence or absence of particles, and for each particle, the position at each moment of time. I think that it would be true (classically) that the record for the next time interval would depend only on the records for neighboring cells in this time interval. The classical version would only give particle-like properties to electrons and would only give wavelike properties to the electromagnetic field, but both types of phenomena can be modeled by a cellular model, it seems to me.

 

[stevendaryl]

Where did you get all that? Do you understand how a simple plane wave doesn't fulfill locality as defined in the model you linked(and that you claimed made no assumptions)?

I'm with zonde. I don't know what you mean by saying that a plane wave is nonlocal.

The sense of nonlocal that is relevant for Bell's inequality is this: If you confine your attention to a small region of space, then what happens to that region at time t+\delta t depends only on facts about what's going on at points less than c \delta t away from that region, at time t. I don't see how that is violated by plane waves.

 

[TrickyDicky]

I'm with zonde. I don't know what you mean by saying that a plane wave is nonlocal.

The sense of nonlocal that is relevant for Bell's inequality is this: If you confine your attention to a small region of space, then what happens to that region at time t+\delta t depends only on facts about what's going on at points less than c \delta t away from that region, at time t. I don't see how that is violated by plane waves.

There is no finite region of a plane wave where you can apply that sense of locality. That's how it is violated.

 

[A. Neumaier]

Bell's theorem applied to the EPR experiment shows that it can't be described this way

I hadn't noticed that you were talking about a myriad of tiny cubes with arbitrary neighboring interaction; indeed, in this case you can approximate a field.

But then your just cited claim is no longer justified; Bell's assumptions are very different. You'd have to give a new proof.

 

[stevendaryl]

There is no finite region of a plane wave where you can apply that sense of locality. That's how it is violated.

I don't understand what you mean. Maxwell's equations are local, regardless of whether your solution is a plane wave or not.

You have a region, say the set of points (x,y,z) such that x^2 + y^2 + z^2 < R^2
You're trying to predict the behavior of the electromagnetic field in that region at time t+\delta t.
Then it is sufficient to know the values of the electromagnetic field in the region x^2 + y^2 + z^2 < (R+c \delta t)^2 at time t

[edit:] What I said is not true if you have charges within the region. So for simplicity, let's assume that we also know that there are no charges in the region x^2 + y^2 + z^2 < (R+c \delta t)^2 at time t.

I don't understand the sense in which plane waves make any difference.

 

[TrickyDicky]

I don't understand what you mean. Maxwell's equations are local, regardless of whether your solution is a plane wave or not.

You have a region, say the set of points (x,y,z) such that x^2 + y^2 + z^2 < R^2
You're trying to predict the behavior of the electromagnetic field in that region at time t+\delta t.
Then it is sufficient to know the values of the electromagnetic field in the region x^2 + y^2 + z^2 < (R+c \delta t)^2 at time t

[edit:] What I said is not true if you have charges within the region. So for simplicity, let's assume that we also know that there are no charges in the region x^2 + y^2 + z^2 < (R+c \delta t)^2 at time t.

I don't understand the sense in which plane waves make any difference.

That's propagation of signals by any field. The definition of locality usually (in
Einstein definition in Wikipedia or the
one linked by zonde and use in Bell's
theorem) talks about objects, and for a
plane wave in vacuum the only way to
define an object includes infinite
extension, for al its properties as a wave.

 

[TrickyDicky]

I don't understand what you mean. Maxwell's equations are local, regardless of whether your solution is a plane wave or not.

They are local in the operational sense of finite signal propagation speed. But that sense is respected by quantum violations of Bell's inequalities.
Mathematically it is clear how monochromatic plane waves violate the (special relativistic) Minkowskian sense of locality(Einstein locality). Such solutions of Maxwell's equation in vacuum are singular in Minkowski space hyperplanes.