Demystifier's latest activity

But these states have zero probability of appearing, given that exactly one particle interacts with the apparatus. The Hilbert space...

Fine, but I claim that this space, in the given setup, is effectively 2-dimensional as well. That's the crucial thing. Why do you think...

Exactly!

The individual ones do. But the actual quantum state that's prepared is not one of them. It's a linear superposition of all of them...

@gentzen you have not understood my point. Let me explain it by a different, but mathematically equivalent example. Instead of the ion...

Fine, suppose that there are two atoms, one on the left and one on the right. And suppose that you observe that no atom is ionized...

I think I disagree. If there is exactly one scattered particle, then there is no state ##|{\rm neutral}_L,{\rm neutral}_R\rangle##...

So that it can be compared with my argument, so if my argument is wrong I can see why it is wrong.

I meant constructed in a sense I have constructed my operator, as a sum/integral over all possible single outcomes.

But it is not an explicit construction in a way my construction is explicit. The explicit construction should have a sum over all...

That's a lot of talk. Can you write down such an operator explicitly, in a form similar to my operator, so that we can compare them?

It doesn't change anything in my argument. I can define $$|\Psi_{\Omega}\rangle = |\Omega\rangle \otimes |{\rm apparatus}\; {\rm...

After some thinking, I think the Mott-Coleman argument is simply wrong, because the operator ##L## does not exist. To explain it in...

I still don't understand the point. The radially symmetric wave function is not supposed to be an eigenstate of ##L##, so why is he...

Can someone (e.g. @gentzen or @martinbn ) explain the Mott's argument to me in his own words? Because from the paper I haven't...