> On 9 Aug 2019, at 04:21, smitra <[email protected]> wrote: > > On 09-08-2019 04:07, Bruce Kellett wrote: >> From: BRUNO MARCHAL <[email protected]> >>>> On 8 Aug 2019, at 13:59, Bruce Kellett <[email protected]> >>>> wrote: >>>> On Thu, Aug 8, 2019 at 8:51 PM Bruno Marchal <[email protected]> >>>> wrote: >>>> If the superposition are not relevant, then I don’t have any >>>> minimal physical realist account of the two slit experience, or >>>> even the stability of the atoms. >>>> Don't be obtuse, Bruno. Of course there is a superposition of the >>>> paths in the two slit experiment. But these are not orthogonal >>>> basis vectors. That is why there is interference. >>> But each path are orthogonal. See the video of Susskind, where he >>> use 1 and 0 to describe the boxes where we can find by which hole >>> the particles has gone through. Then, without looking at which hole >>> the particle has gone through, we can get the interference of the >>> wave which is obliged to be taken as spread on both holes, and that >>> represent the superposition of the two orthogonal state described >>> here as 0 and 1. >> I seldom watch long videos of lectures. But if Susskind is saying that >> the paths taken by the particle through the two slits are orthogonal >> then he is flatly wrong. Writing the paths as 1 and 0 does not make >> them orthogonal. And if they were orthogonal they could not interact, >> and you would not get interference. Two states |0> and |1> are >> orthogonal if their overlap vanishes: <0|1> = 0. Interference comes >> from the overlap, so if this vanishes, there is no interference. >> Either Susskind is terminally confused, or you have misrepresented >> him. > > > We can measure which slit the particle moved through, therefore the two > states correspond to different eigenstates with different eigenvalues of the > observable for this, and they are therefore orthogonal.
Exactly. > The interference pattern is apparent in a wavefunction psi(x) = 1/sqrt(2) > [|<x|0> + <x|1>], on a screen we can measure |psi(x)|^2, and this contains > the term I(x) = Re[<0|x><x|1>]. Integrated over all space, this term will > vanish as it's the real part of the inner product between |0> and |1>. But as > a function of x, the term [<0|x><x|1> will in general not be zero. Exactly. I don’t understand why Bruce says that going through slit 1 and going through slit 2 are not orthogonal. At least that explains why he see so much confusion in the posts of other people. That confirms the rule that if people insult you, or use dismissive tone, or say negative things like “you are confused”, it means that they are the one confused. (Apology for that meta-remark) 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/95467b1b4472900733873c8fd529d7e0%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/0F84264B-D054-4182-B714-8797A48FA8B1%40ulb.ac.be.
