On 19-08-2019 11:01, Bruno Marchal wrote:
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).
Indeed, the observer then locates him/herself in a sector of Hilbert
space corresponding the particle moving through either the left or the
right slit.
Saibal
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