On 09-08-2019 04:07, Bruce Kellett wrote:
From: BRUNO MARCHAL <marc...@ulb.ac.be>
On 8 Aug 2019, at 13:59, Bruce Kellett <bhkellet...@gmail.com>
wrote:
On Thu, Aug 8, 2019 at 8:51 PM Bruno Marchal <marc...@ulb.ac.be>
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. 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.
Saibal
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