On Thu, Aug 15, 2019 at 7:23 AM smitra <[email protected]> wrote: > On 14-08-2019 00:44, Bruce Kellett wrote: > > > > You over-elaborate a simple schematic. My A(x) and B(x) are simply the > > amplitudes of the wave function at point x on the screen from the two > > slits. To get the intensity at x, you add the amplitudes and take the > > modulus squared -- you do not add the intensities from each slit > > separately. > > > What matters is that you can get interference of the orthogonal > components in the superposition. Whether or not you do actually get > interference is not orthogonality before measurement, but after > measurement. If the photons moving through the slits interact with > another system initially in some state |C> such that if the photon moves > through slit A the state |C> changes to |D> while it changes to |E> if > the photon moves through slit B, then the interference pattern will be > the function Re[A(x)*B(x)<C|D>], this will become zero if |C> and |D> > are orthogonal, which means that you still have a superposition of two > orthogonal terms in the MWI sector where a photon lands on some specific > spot on the screen. The system then has perfect which-way information. >
I don't think this quite works either. Look at it this way. Take |A> as the wave function for going through slit A, and |B> the corresponding wave function for going through slit B. The superposition (|A> + |B>) is the wave function that propagates to the screen. At the screen, the intensity is the square of this: (A| + |B>)(<A| + |B>) = |A|^2 + |B|^2 + <A|B> + <B|A>, and there is interference. This is because you have added together the paths through each slit, but you don't know which slit was traversed, so A and B are coherent. If you now add a detector at the slits that registers L (for Left) if A was traversed, and R (for Right) if B was traversed, the original superposition becomes (|A>|L> + |B>|R>) because the detector becomes entangled with the slit through which the photon went. If we now go to the intensity at the screen, we obviously get |A|^2 + |B|^2 + <A|B><L|R> + <B|A><R|L> The detector states recording L or R for which slit the photon traversed are orthogonal, <L|R> = <R|L> = 0, so detecting which slit the photon went through destroys the interference. If there is no detection at the slits, the amplitudes are coherent, they are not orthogonal, and there is interference. Bruce -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLRqLxhrwJk3Vo685RdXUcLeS_Z_GrnttSrouU1fit7Smw%40mail.gmail.com.
