On 09-08-2019 05:35, Bruce Kellett wrote:
On Fri, Aug 9, 2019 at 12:21 PM smitra <smi...@zonnet.nl> wrote:
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.
Yes. And in the case in which we observe which slit the particle went
through there is no interference.
Because in that case the state is either |0> or |1> or.
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.
I did make that clear. The point id that you can do that and then the
interference term integrates to zero.
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.
Nonsense. What we observe at a point x on the screen is the expectation
value of the projection operator |x><x|. The two states are orthogonal,
but that doesn't matter one iota, we always measure the modulus squared
of the inner product of the state with |x>.
Saibal
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