From: sarfatti@well.sf.ca.us (Jack Sarfatti)
Subject: parallel slit kets OK after all?
Organization: Whole Earth 'Lectronic Link
Date: Sat, 21 Nov 1992 21:35:46 GMT


Sarfatti Lectures in Super Physics (Lecture 2)

#7 Quantum connection communication is required if the Princeton PEAR
reports of "precognitive remote viewing" are a real phenomenon. Quantum
connection communication may also be required to explain how the universe
comes into being and what our destiny and ultimate purpose may be.  Quantum
connection communication may be the source of meaning to existence.

Whoops! It appears that one can get a sensible interpretation of the 2 slit
experiment with parallel slit kets after all!

OK Must the slit kets must be orthogonal? Is there is no sensible way that
they can be parallel? Let's re-examine the 2-slit experiment and the Mach-
Zehnder interferometer using standard approach of orthogonal kets and then
the non-standard approach of parallel slit kets.  The latter is sufficient
to get a quantum connection signal easily. The former, makes it more
difficult to get the quantum connection signal but still not impossible,
especially if we monitor detectable hidden orders in the receiver
fluctuations that may depend on parameters controlled by settings of
transmitter devices.


|         |               |
|           |1>           |
  |i>     |            |x>|
|           |2>           |
|         |               |
          slits           screen

Fig.7a   2-slit experiment

In Schrodinger picture where kets evolve in time (time evolution notation
suppressed for simplicity):

|i> = |1><1|i> + |2><2|i>

1 = |1><1| + |2><2|

<1|2> = 0

<x|i> = <x|1><1|i> + <x|2><2|i>

The probability density at the screen is

p(x) = |<x|i>|^2 = <i|x><x|i>

= <i|1><1|x><x|1><1|i> + <i|2><2|x><x|2><2|i>

+ <i|1><1|x><x|2><2|i> + <i|2><2|x><x|1><2|1>

But Integral{|x><x|dx} = 1

Therefore,

Integral{p(x)dx} = <i|[|1><1|+|2><2|]|i>+<i|1><1|2><2|i>+<i|2><2|1><2|1>

= <i|i> = 1

Note that <1|2> = 0 (i.e. orthogonality of the slit kets is necessary to
get a vanishing integral of the interference terms over the entire screen.

On the other hand, suppose that the slit kets were not orthogonal but
parallel, so that

|1> = e^i@|2>

<1| = <2|e^-i@

assume phase @ is a c-number not an operator (if we do second-quantization
it becomes an operator which might change things?)

Therefore,

1 = |1><1| + |2><2| = 2|1><1|

<1|2> = e^-i@

Therefore,

Integral{p(x)dx} = 2<i|1><1|i>+<i|1>e^-i@<2|i>+<i|2>e^i@<2|1> = 1

Let us simplify to the case of equal illumination so that

<i|1> = <i|2> = 1/sqrt2

with zero phases. Therefore, the condition is simply @ = pi/2 or

|1> = i|2>

In general, the relative phase between the parallel slit kets will be
dependent on the initial conditions of the input state (i.e., the two
complex variables <i|1> and <i|2>).

note that for our simple special case

p(x) = |<1|x>|^2[1 + cos@]

which is not at all unreasonable!

Thus, it appears, that assuming parallel slit kets for the 2-slit
experiment gives a sensible answer. Note that @ = @(x) so that we require

Integral{|<1|x>|^2[1 + cos@(x)]dx} = 1

more specifically since the fringes rearrange a fixed total energy

Integral{|<1|x>|^2dx} = 1

Integral{|<1|x>|^2 cos@(x)dx} = 0

to be continued