On 15-08-2019 06:25, Bruce Kellett wrote:
On Thu, Aug 15, 2019 at 7:23 AM smitra <smi...@zonnet.nl> wrote:
On 14-08-2019 00:44, Bruce Kellett wrote:
You over-elaborate a simple schematic. My A(x) and B(x) are
simply the
amplitudes of the wave function at point x on the screen from the
two
slits. To get the intensity at x, you add the amplitudes and take
the
modulus squared -- you do not add the intensities from each slit
separately.
What matters is that you can get interference of the orthogonal
components in the superposition. Whether or not you do actually get
interference is not orthogonality before measurement, but after
measurement. If the photons moving through the slits interact with
another system initially in some state |C> such that if the photon
moves
through slit A the state |C> changes to |D> while it changes to |E>
if
the photon moves through slit B, then the interference pattern will
be
the function Re[A(x)*B(x)<C|D>], this will become zero if |C> and
|D>
are orthogonal, which means that you still have a superposition of
two
orthogonal terms in the MWI sector where a photon lands on some
specific
spot on the screen. The system then has perfect which-way
information.
I don't think this quite works either. Look at it this way. Take |A>
as the wave function for going through slit A, and |B> the
corresponding wave function for going through slit B. The
superposition (|A> + |B>) is the wave function that propagates to the
screen. At the screen, the intensity is the square of this:
(A| + |B>)(<A| + |B>) = |A|^2 + |B|^2 + <A|B> + <B|A>,
and there is interference. This is because you have added together the
paths through each slit, but you don't know which slit was traversed,
so A and B are coherent.
If you now add a detector at the slits that registers L (for Left) if
A was traversed, and R (for Right) if B was traversed, the original
superposition becomes
(|A>|L> + |B>|R>)
because the detector becomes entangled with the slit through which the
photon went. If we now go to the intensity at the screen, we obviously
get
|A|^2 + |B|^2 + <A|B><L|R> + <B|A><R|L>
The detector states recording L or R for which slit the photon
traversed are orthogonal, <L|R> = <R|L> = 0, so detecting which slit
the photon went through destroys the interference. If there is no
detection at the slits, the amplitudes are coherent, they are not
orthogonal, and there is interference.
<A|B> = 0 due to orthogonality (and this is preserved as the screen is
approached as the state evolves in a unitary way), what we measure at
the screen is <A|x><x|B> + <B|x><x|A> as a function of the position x on
the screen. If we add up all the counts on the screens from all
positions then the interference effect averages out to zero due to
orthogonality, but that doesn't matter. We can still see fringes and
peaks on the screen.
Saibal
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