On Fri, Aug 9, 2019 at 12:21 PM 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.
Yes. And in the case in which we observe which slit the particle went through there is no interference. > 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, > We do not integrate over all space since we observe the interaction with a screen. It is really quite simple. If a state is a sum of two components, |psi> = (|A> + |B>), then we measure <psi|psi> = (<A| + <B|)(|A> + |B>) = <A|A> + <B|B> + 2<A|B>. If <A|B> does not vanish (the components are not orthogonal), then there is interference. For orthogonal components <A|B> = 0, and there is no interference. Introducing separate states for the two slits does not aid comprehension here. Bruce > 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. > > 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/CAFxXSLTCt9B2rjPs2pc8rNPwCOkEB%3Dp-wJTtqFsKe6r8328R1Q%40mail.gmail.com.
