On 16-08-2019 06:31, Bruce Kellett wrote:
On Wed, Aug 14, 2019 at 8:28 AM smitra <smi...@zonnet.nl> wrote:
On 13-08-2019 13:33, Bruce Kellett wrote:
Of course A(x) and B(x) refer to the same point on the screen.
That is
not a collapse, that is just what the notation means.
A(x) and B(x) considered as the representations of |A> and |B> in
the
position basis, i.e. A(x) = <x|A> and B(x) = <x|B> are still
orthogonal
states, as they represent the orthogonal states |A> and |B>:
0 = <A|B> = Integral over x of <A|x><x|B>d^3x = Integral over x of
A*(x)B(x) d^3x
I don't think this really works out. You are claiming that the
integral of the interference terms over the whole screen vanishes. If
we look at the usual derivation of the interference from two slits, we
get something like
Intensity I = 2 A^2 (sin^2(beta)/beta^2) (1 + cos(delta))
where the term involving the angle beta is the superposed diffraction
pattern from the finite width of the slits. The cos (delta) term is
the interference, but it has this form only in a small angle
approximation, and the phase difference delta is, of course, limited
by the separation of the slits. So, although the cos(delta) term may
integrate to zero over small angles, the presence of the diffraction
envelope, and the limitations of the small angle approximation, mean
that is almost certainly will not vanish when integrated over the
whole screen.
Yes, this is in the small angle approximation, if you go beyond that
then the itnegral over the screen won't vanish.
So <A|B> will not vanish in general.
No, because <A|B> is the integral over all space and this is exactly
zero. What happens is that when the small angle approximation becomes
invalid and the integral over only the screen becomes nonzero, the
integrals over surfaces parallel to the screen will have a values that
differ by a phase factor that depends on the distance in the direction
orthogonal to the screen. This then causes the integral over all space
to vanish.
Which is what I would have
thought because the paths through the separate slits are not
independent -- each particle essentially has to see both slits (go
through both slits) in order to maintain coherence. So they cannot be
orthogonal (independent).
Coherence and orthogonality have nothing to do with each other.
In practice, to see the interference pattern you need coherent
illumination over both slits. This is easy these days with lasers, but
in older books, coherence was ensured by having a preparatory single
slit followed by suitable condenser lenses. If the slits could be
treated as independent entities, this would not have been necessary.
This has nothing to do with orthogonality of the states. What matters is
that the interference pattern shouldn't get washed out due to each
wavefunction of each particle near the screen having its peaks and
fringes at different places. This can be prevented by using an
approximate monochromatic light source and making sure that the light
passes through a collimator. Without a collimator, the interference
pattern due to the light from one part of the source will be shifted
w.r.t. to the other part causing the pattern to get washed out.
Saibal
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