Recent content by zbyszek

Take the Maxwell's equations in vacuum. Transform them to a wave equation for B. Try the solution in the form B=B_ocos(kx-\omega t). The normal particles have nonzero rest masses, so their energy is proportional to p^2 rather then to p.

Do I have a right to defend my reputation on this forum? You see, attacking somebody's work without justification is one of the lowest things a scientist can do. I am being accussed of doing just that. Is it a flame war for you, vanesch? In this world, I mean. Cheers!

Is this another joke of yours? I have hard time telling whether you are joking or lying. For those who would like to check on the moral integrity or the sense of humor (pick one) of Anonym please read post #111 under the wave packet description thread...

What kind of advice is that? Did you hear it from some one yourself? If you think you understand QM then this place is just made for you to share this blessing. Cheers!

You are barking on the wrong tree. I am not interested in your opinion anymore. Good luck!

The way out might be to step back and drop the assumption that QM applies to any single particle. QM certainly applies to an ensemble of identically prepared particles for it gives excellent predictions for probabilities (many repetitions of the same experiment). At the same time QM cannot...

You are right. If I wanted the test that would stupid. I do similar simulations on the daily basis. Without calculating Bohm's trajectories, because they do not provide any additional information. What I have is the full evolution of many-body wave functions for systems with interacting...

He means that for an interference pattern to be observed all electrons have to follow the same wave function. In particular, this means the coherence etc. In order to satisfy this requirement, the electrons have to be carefully prepared. They cannot be cast into the experimental setup...

Anonym, Let me stick to this sentence of yours. Our brains trick us into believing things that do not exist and make as blind to plain evidence residing just in front of us. I am not patronizing you. The same happens to me way too often. One way to fight that is to employ different...

Is the collapse also required in the interpretation where the wave function is not associated with any single object but only with ensembles? I might have some insides, not discussed in QM textbooks yet, as to the quantum to classical transition, and to the Schrödinger cat problem in...

So you are the man of your word. You have read it. Could you explain in more detail what experimental evidence has been rejected by Ballentine, please? Cheers!

Before I start the enlighment tell me what exactly you are questioning. Is it: 1. statistical interpretation of QM you are not comfortable with or, 2. is it my understanding of the interpretation that is not sufficient? I cannot help on 1. but will be happy to improve in case of 2. Same...

Hi Demystifier, The wave function is as much associated with a single particle as an external potential. It is like saying that periodic potential of a crystall lattice is a property of an electron in a Bloch band. Thank you for the contribution! Cheers!

Bohm already did it. The initial conditions for Bohmian trajectories are drawn from a probability density i.e. a wave function modulus squared. Thus, the wave function is associated with an ensemble rather than with a single particle. Cheers!

There is Einstein's "Reply to criticism" available on-line. The book "Quantum Theory: Concepts and Methods" by Asher Peres assumes statistical interpretation and has large parts devoted to hidden variables. I heard good things about Ballentine's textbook "Quantum Mechanics: Modern...