On 13-08-2019 02:49, Bruce Kellett wrote:
On Tue, Aug 13, 2019 at 10:43 AM smitra <[email protected]> wrote:
On 13-08-2019 01:41, Bruce Kellett wrote:
On Tue, Aug 13, 2019 at 9:09 AM smitra <[email protected]> wrote:
On 12-08-2019 08:29, Bruce Kellett wrote:
Look at this another way. It is just an illustration of
complementarity. Measuring which slit the photon went through
is
a
position measurement at the slits. Measuring the interference
pattern
at the screen is equivalent to a momentum measurement at the
slits.
Such measurement operators do not commute -- the measurements
are
complementary and cannot be performed simultaneously.
It doesn't matter for orthogonality of the states whether or not
they
are measured.
Of course it does. The slits are not orthogonal states unless
they are
measured position eigenstates. If they are not measured, they are
individually superpositions of many position eigenstates
(including
eigenstates that overlap both slits), so the slits themselves are
no
longer orthogonal. Orthogonal states cannot interfere, that is
why a
position measurement at the slits makes the interference pattern
on
the screen disappear.
The fact remains, that orthogonal states cannot interfere:
(<A| + <B|)(|A> + |B>) = <A|A> + <B|B> + 2 <A|B>
and the interference term <A|B> vanishes if |A> and |B> are
orthogonal. You can't get away from this basic fact about quantum
mechanics.
<A|B> is zero in the two slit experiment, if you integrate the
interference term over the screen you get zero.
Thing is that the interference we can observe at some position x on
the
screen is Re[<A|x><x|B>], which for general x is nonzero despite
the
fact that <A|B> = 0.
So you agree that if the overlap vanishes you do not get interference.
You go to some lengths to try and avoid this fact. Saying that the
integral over the screen vanishes is beside the point.
<A|B> is the integral over all space of <A|x><x|B> and this is zero, but
the interference we observe at some point x on the screen is
Re[<A|x><x|B>], and that's in general nonzero even for orthogonal |A>
and |B>.
Saibal
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