I Consistency of Bohmian mechanics

[A. Neumaier]

I don't know that Hamiltonian. And also you again changed the setup. Now we have an infinitely thin filter. What is this filter supposed to do?

It is a piece of metal with a hole in it. Idealized to be infinitely thin; otherwise I'd need to model the interaction of the electrons with the metal. I don't know of any discussion of an experiments with slits where the latter is done; hence the idealization. Then all the complicated stuff happens instantaneously and can be summarized by the projection. At least this is done in informal discussions when blending out partial beams (in Stern-Gerlach experiments, say).

If the filter is not moving, why should the Hamiltonian be time dependent?

You were claiming that the filters change the Hamiltonian:

if you have two slits having an effect on the particles to be measured you modify the dynamics, i.e., the Hamiltonian and thus change the probabilities compared to the situation of freely moving particles.

I was just trying to understand what you mean. Surely away from the filters the Hamiltonian is the free Hamiltonian, so the only way I could give meaning to your claim was to assume that you thought that the Hamiltonian is time-dependent with three constant pieces before, during and after the passage through the slit.

I still do not know, which experiment you really have in mind.

A very simple experiment. I have a source of electrons, two close and parallel (in this post only one) sheets of metal both with a big hole, and a screen parallel to the plates at the end. I want to know the probability that an electron emitted by the source is detected by the screen. Under the usual idealizations and the standard collapse assumption, this probability is given by the formula of Thors10 from post #65, with characteristic functions specified by the positions of the holes.

 

[A. Neumaier]

Thread reopened. Two overly argumentative posters have been thread banned and some overly argumentative posts have been deleted.

I hope that Thors10 is not banned from posting in this thread, since he is the originator of the thread, and the current discussion needs his active participation:

This is precisely what the formula of @Thors10 quoted in post #82 does. Thus it is his intended setup. The only questionable thing in his description is that he claims he can do it by repeatedly measuring a single particle twice with a single detector. So I inquired about that.

 

[PeterDonis]

Thread closed for moderation.