On Fri, Aug 16, 2019 at 4:43 PM smitra <[email protected]> wrote: > On 16-08-2019 06:31, Bruce Kellett wrote: > > On Wed, Aug 14, 2019 at 8:28 AM smitra <[email protected]> wrote: > > > >> On 13-08-2019 13:33, Bruce Kellett wrote: > >>> > >>> Of course A(x) and B(x) refer to the same point on the screen. > >> That is > >>> not a collapse, that is just what the notation means. > >> > >> A(x) and B(x) considered as the representations of |A> and |B> in > >> the > >> position basis, i.e. A(x) = <x|A> and B(x) = <x|B> are still > >> orthogonal > >> states, as they represent the orthogonal states |A> and |B>: > >> > >> 0 = <A|B> = Integral over x of <A|x><x|B>d^3x = Integral over x of > >> > >> A*(x)B(x) d^3x > > > > I don't think this really works out. You are claiming that the > > integral of the interference terms over the whole screen vanishes. If > > we look at the usual derivation of the interference from two slits, we > > get something like > > > > Intensity I = 2 A^2 (sin^2(beta)/beta^2) (1 + cos(delta)) > > > > where the term involving the angle beta is the superposed diffraction > > pattern from the finite width of the slits. The cos (delta) term is > > the interference, but it has this form only in a small angle > > approximation, and the phase difference delta is, of course, limited > > by the separation of the slits. So, although the cos(delta) term may > > integrate to zero over small angles, the presence of the diffraction > > envelope, and the limitations of the small angle approximation, mean > > that is almost certainly will not vanish when integrated over the > > whole screen. > > Yes, this is in the small angle approximation, if you go beyond that > then the itnegral over the screen won't vanish. > > > > > So <A|B> will not vanish in general. > No, because <A|B> is the integral over all space and this is exactly > zero. What happens is that when the small angle approximation becomes > invalid and the integral over only the screen becomes nonzero, the > integrals over surfaces parallel to the screen will have a values that > differ by a phase factor that depends on the distance in the direction > orthogonal to the screen. This then causes the integral over all space > to vanish. >
I think you need to prove that. In my understanding, A(x) = <x|A> is to be interpreted as the amplitude for a wave through slit A to get to the screen at x. There is nothing 3-dimensional about this. The 'x' is just the distance from the centre of the screen in the plane of the screen. Nothing else is relevant. You do not have to integrate over all space because you use a complete set of states in the x-direction: \int |x><x| dx = 1. Bruce > > Which is what I would have > > thought because the paths through the separate slits are not > > independent -- each particle essentially has to see both slits (go > > through both slits) in order to maintain coherence. So they cannot be > > orthogonal (independent). > > Coherence and orthogonality have nothing to do with each other. > > > > > In practice, to see the interference pattern you need coherent > > illumination over both slits. This is easy these days with lasers, but > > in older books, coherence was ensured by having a preparatory single > > slit followed by suitable condenser lenses. If the slits could be > > treated as independent entities, this would not have been necessary. > > This has nothing to do with orthogonality of the states. What matters is > that the interference pattern shouldn't get washed out due to each > wavefunction of each particle near the screen having its peaks and > fringes at different places. This can be prevented by using an > approximate monochromatic light source and making sure that the light > passes through a collimator. Without a collimator, the interference > pattern due to the light from one part of the source will be shifted > w.r.t. to the other part causing the pattern to get washed out. > > Saibal > > -- > 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/a321127d45d9f99a187bf1fe519580f1%40zonnet.nl > . > -- 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/CAFxXSLRcdXsp_85q2Bv1uPk12NTxTU6XH7thdLQX7BGzC0jU_Q%40mail.gmail.com.
