> On 19 Aug 2019, at 06:55, smitra <[email protected]> wrote: > > On 19-08-2019 03:52, Bruce Kellett wrote: >> On Sat, Aug 17, 2019 at 10:28 AM smitra <[email protected]> wrote: >>> On 16-08-2019 09:01, Bruce Kellett wrote: >>>> On Fri, Aug 16, 2019 at 4:43 PM smitra <[email protected]> wrote: >>>> 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. >>> A quantum state is defined in the position representation by >>> assigning >>> an amplitude to all points in space. It can be the case that the >>> amplitude is zero outside of a narrow volume surrounding the the >>> screen, >>> in which case the integration can be approximated as an integral >>> over >>> the screen's surface. >>> The orthogonality can be rigorously proved as follows. If we have a >>> single particle incident on the two slits described by a time >>> dependent >>> wave function psi(x,t) = 1/sqrt(2) [A(x,t) + B(x,t)] such that at >>> A(x,0) >>> is nonzero at one slit and B(x,0) at the other slit, then A(x,0) >>> and >>> B(x,0) are obviously orthogonal. Since time evolution will preserve >>> inner products, A(x,t) and B(x,t) will remain orthogonal as a >>> function >>> of t. One can then describe the interaction with the screen as an >>> effective collapse that will happen with the largest probability >>> when >>> the peak of the wavefunction has arrived at the screen. >> The slits are orthogonal only if they are eigensatates of the position >> operator (in the x direction). That is the case only if you have a >> measurement that gives which-way information. Then there is no >> interference, as advertised. The amplitudes at the slits aren't >> orthogonal just because you say so. > > The particle will be in one of the two eigenstates if such a measurement is > made,
>From the perspective of the one which has done the measure, but actually, he >just entangle him/herself with the particle position. “The particle” is a bit >ambiguous, as “the observer” is too, in such self-entanglement. > but if we don't make that measurement then the particle ends up in a > superposition of the two eigenstates. It means that the observer is still able to get the interference. It is only his knowing which slit the particle has gone through which makes him to be unable to get the interference, but only due to self-entanglement. The measurement just makes some accessible histories inaccessible (when we don’t assume a collapse). Bruno > > 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/245556d3523e3e7692b6f29c789d95d0%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/04679737-A36A-4D80-8046-4B0B7291B82A%40ulb.ac.be.
