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Subject: Re: Henry (Stapp) where is the error in my "proof" about entanglement signaling?



On Jun 27, 2011, at 6:06 PM, Henry P. Stapp wrote:

On Mon, 27 Jun 2011, JACK SARFATTI wrote:

On Jun 27, 2011, at 12:49 PM, Henry P. Stapp wrote:



On Sat, 25 Jun 2011, JACK SARFATTI wrote:

On Jun 25, 2011, HENRY STAPP wrote:


To accommodate the Bem results, or otherwise allow for violations of the
no-FTL-signally theorem one must violate the basic rules of contemporary quantum mechanics, not merely set up clever experiments.


I agree with Henry on this. Indeed my Journal of Cosmology paper (April 2011 Vol 14 - Penrose-Hameroff ed) agrees with this. So if Henry wants to say that

<A1|A2> =/= 0 violates OQT so be it.

However, it appears that we can do that in the lab with modern techniques.

HPS:

I am not saying that <A1|A2> =/= 0 violates OQT!
Just the opposite!


Well in that case what do you disagree with in my description of the double two slit experiment?


HPS: If you happen to have a standard double slit experiment then in this case you happen to have a condition <A1|A2>=0,


But that is not the gedankenexperiment. The one I am modeling with the algebra is David Kaiser's Fig 9.1 So, the issue is what specific algebra in my model do you think is in error?
I think I have followed your general proof in this particular instance. However, in order to get no FTL signal I need <A1|A2> = 0 in the case that the Heisenberg microscope is switched off on the transmitter side. In OCT this condition is satisfied I think. However, with new techniques of weak and partial measurements and perhaps with entangled laser beams this may not be so. We also have John Cramer's variation on the Dopfur experiment which is functionally equivalent to mine. Now both Cramer and I may be in error here of course. The issue is where is the error in my algebra?

HPS but that fact does does not enter into my proof of the No-FTL-Signalling theorem, which goes through just as well, and unaltered, for the case <A1|A2> =/= 0. In order to make the predictions in R to depend on which choice of experiment performed in L you must deviate from the statistical rules of OQT, no matter whether <A1|A2> is zero or not. My proof that the orthodox rules lead to the No-FTL-signalling property does not fail, or need to be changed in any way, if <A1|A2> =/= 0. The proof holds equally well in this case: it does not involve any condition <A1|A2>=0 in the first place, even though that condition does happen to be satisfied in the simpest case of a true double slit experiment.

Your statements create the impression that you thought that by going to the case <A1|A2>=/=0 you could evade my proof of NO FTL signalling.
For reasons given, that is not the case.

JS: I suspect I am in error, but I can't find it in my algebra for Fig 9.1. For same reason I suspect that Cramer's experiment will not work, though he makes a plausible argument.

Do you disagree with my statement that terms like


<A2|A1><B1'|x'><x'|B2'>  (7)

suffice for a local interference pattern on the receiver side provided that

<A2|A1> =/= 0 ?

In case the readers do not have easy access to the original argument I repeat it here

On Jun 25, 2011, at 10:02 AM, JACK SARFATTI wrote:

Let y be the support of the test particles that scatter the electron in the Heisenberg microscope on the transmitter side. Do not confuse y with x the position of the massive electron part of an entangled pair of electrons. Orthogonality of the transmitter slit states

<A2|A1> = 0  (1)


is only true for a reliable strong measurement of which way the transmitter electron A goes OR if no strong which-way measurement is even attempted. That is when the multiple integral with respect to y and x of

<A2|x,y><y,x|A1> = 0  (2)

This is not the case in an unreliable partial and/or weak which-way measurement where the effective measurement operator is not Hermitian.

e.g. in some weak measurements the effective measurement operator has an imaginary part.


The EPR pair interferometers have the entangled state for slits (1,2) & (1',2') at positions on screens x & x' for photons A & B in the same individual pair

<x,x'|A,B> = <x|A1><x'|B2'> + e^i@(y)<x|A2><x'|B1'>     (3)

where @(y) is a stochastic phase whose degree of randomness is controlled by tuning the efficiency of the slit detector on the transmitter side.

Now the no-signal theorem in this case is trivial as shown by Henry Stapp.

Form the Born probability. Ignore @ for now. It's not important for this part of the argument.

P(x,x') ~ |<x,x'|A,B> |^2  (4)

What is important are the cross terms, e.g.

<A2|x><x|A1><B1'|x'><x'|B2'>  (5)

Integrate over the transmitter positions x and also the scatterers y if the Heisenberg microscope is switched on at the transmitter and assume completeness of the eigenstates for the transmitter electron of the entangled pair |x>

i.e. integral of the projection operator

|x><x| = 1   (6)

Therefore the net result of this partial trace is

<A2|A1><B1'|x'><x'|B2'>  (7)

The usual assumption is that the A slit bras and kets are orthogonal i.e.

<A2|A1> = 0  (8)

In that case there is never a LOCAL interference pattern on EITHER side under any conditions.




I am pointing out that neither the condition <A1|A2>=0,
nor any analog of such a condition, enters into my
general proof of the No-FTL signalling theorem.

OK, but what is wrong with my above model of the above experiment in David Kaiser's Fig 9.1?

It seems to me that in that particular model you must demand

<A2|A1> = 0

The proof that I gave earlier of the No-FTL-signalling theorem does not bring in any condition <A1|A2> = 0,
and it applies perfectly well, without comment, to the
case at hand where <A1|A2>=/= 0.

It applies perfectly well to the case where the projection
operator representing the answer "yes" to the experimenter's
probing question is P = |X><X|/<X|X> with
|X> = a|A1> + b|A2>, with a and b any complex numbers not
both zero.

I do not know where the condidtion <A1|A2> =0 that
Sarfatti mentions came from. It certainly does not
enter into my general argument, and emphatically not in
the case that Sarfatti is considering, in which
<A1|A2> =/= 0. The given proof of the theorem
goes through, provided only that the statistical
rules of OQT are maintained.

The theorem states that if the statistical rules
of OQT are valid, then the probabilities of the various
possible outcomes of an experiment performed in a spacetime
region R cannot depend upon which experiment is randomly chosen
and performed in a spacetime region L if R lies outside the
forward lightcone of region L, even though the OUTCOMES of
the experiments in the two regions may be intricately correlated,
due to interactions that occurred in the past.


On Jun 27, 2011, at 12:49 PM, Henry P. Stapp wrote:
On Sat, 25 Jun 2011, JACK SARFATTI wrote:
See Fig 9.1 in David Kaiser's book How the Hippies Saved Physics to follow the equations below.
On Jun 25, 2011, HENRY STAPP wrote:
To accommodate the Bem results, or otherwise allow for violations of the
no-FTL-signally theorem one must violate the basic rules of contemporary quantum mechanics,
not merely set up clever experiments.

I agree with Henry on this. Indeed my Journal of Cosmology paper (April 2011 Vol 14 - Penrose-Hameroff ed)
agrees with this. So if Henry wants to say that
=/= 0 violates OQT so be it.
However, it appears that we can do that in the lab with modern techniques.
HPS:
I am not saying that =/= 0 violates OQT!
Just the opposite!

Well in that case what do you disagree with in my description of the double two slit experiment?
Do you disagree with my statement that terms like
  (7)
suffice for a local interference pattern on the receiver side provided that
=/= 0 ?
In case the readers do not have easy access to the original argument I repeat it here
On Jun 25, 2011, at 10:02 AM, JACK SARFATTI wrote:
Let y be the support of the test particles that scatter the electron in the Heisenberg microscope on the transmitter side. Do not confuse y with x the position of the massive electron part of an entangled pair of electrons. Orthogonality of the transmitter slit states
= 0  (1)
is only true for a reliable strong measurement of which way the transmitter electron A goes OR if no strong which-way measurement is even attempted. That is when the multiple integral with respect to y and x of
= 0  (2)
This is not the case in an unreliable partial and/or weak which-way measurement where the effective measurement operator is not Hermitian.
e.g. in some weak measurements the effective measurement operator has an imaginary part.
The EPR pair interferometers have the entangled state for slits (1,2) & (1',2') at positions on screens x & x' for electrons A & B in the same individual pair
= + e^i@(y)     (3)
where @(y) is a stochastic phase whose degree of randomness is controlled by tuning the efficiency of the slit detector on the transmitter side.
Now the no-signal theorem in this case is trivial as shown by Henry Stapp.
Form the Born probability. Ignore @ for now. It's not important for this part of the argument.
P(x,x') ~ | |^2  (4)
What is important are the cross terms, e.g.
  (5)
Integrate over the transmitter positions x and also the scatterers y if the Heisenberg microscope is switched on at the transmitter and assume completeness of the eigenstates for the transmitter electron of the entangled pair |x>
i.e. integral of the projection operator
|x> Therefore the net result of this partial trace is
  (7)
The usual assumption is that the A slit bras and kets are orthogonal i.e.
= 0  (8)
In that case there is never a LOCAL interference pattern on EITHER side under any conditions.
HPS:
I am pointing out that neither the condition =0,
nor any analog of such a condition, enters into my
general proof of the No-FTL signalling theorem.

OK, but what is wrong with my above model of the above experiment in David Kaiser's Fig 9.1?
It seems to me that in that particular model you must demand
= 0
HPS:
The proof that I gave earlier of the No-FTL-signalling theorem does not bring in any condition = 0,
and it applies perfectly well, without comment, to the
case at hand where =/= 0.
It applies perfectly well to the case where the projection
operator representing the answer "yes" to the experimenter's
probing question is P = |X> with
|X> = a|A1> + b|A2>, with a and b any complex numbers not
both zero.
I do not know where the condition =0 that
Sarfatti mentions came from. It certainly does not
enter into my general argument, and emphatically not in
the case that Sarfatti is considering, in which
=/= 0. The given proof of the theorem
goes through, provided only that the statistical
rules of OQT are maintained.
The theorem states that if the statistical rules
of OQT are valid, then the probabilities of the various
possible outcomes of an experiment performed in a spacetime
region R cannot depend upon which experiment is randomly chosen
and performed in a spacetime region L if R lies outside the
forward lightcone of region L, even though the OUTCOMES of
the experiments in the two regions may be intricately correlated,
due to interactions that occurred in the past.

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