- #1

jf117

- 17

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For the benefit of those of you who haven't followed previous arguments on this forum, I will synthetically describe the experimental setting. A quantum particle (photon ,electron, it really doesn't matter) is flying with constant momentum p=hbar*k and energy E=P^2/(2m) along the z direction towards a plate of equation z=0, containing two narrow perforations at (0,-d/2,0) and (0,d/2,0). As Marcella explains, the plate+perforations constitute a state preparation system which forces the particle wave function to be the sum of deltas of Dirac. The extreme localization of the particle in the initial state, expecially for x and z, but also for y, implies that its momentum is free to assume nearly all values of the momentum space, with higher probability on certain planes of equation py=const.

We can, therefore, find with equal probability the particle's momentum as, for instance, (a,py0,c) and (a',py0,c'), where p0y is a y-component with high probability value, and a,c,a',c' are arbitrary real numbers. What is now interesting is that in the two cases just exemplified the energy takes different values, E=(a^2+pyo^2+c^2)/(2m) and E'=(a'^2+py0^2+c'^2)/(2m), and also different from the initial value E=p^2/(2m). There is here a clear violation of the conservation principles, both for momentum and energy.

In his paper Marcella by-pass this point without any explanation, by imposing the energy conservation principle (in fact the momentum magnitude is maintained unaltered, while the diffraction angle is the only responsible for the variation of the y component), and thus violating the uncertainty principle.

I have been trying to search through various literature to better understand the whole matter, but I have had no success so far. I hope I'll get help by some of you.

Many thanks,

james

(P.S. Don't worry too much about E=P^2/(2m) for the photon case, where m=0. There you can consider E=hw, etc. This difference does not change the problem, because E=pc for a photon)