My idea has now taken off and is a meme spreading across cyberspace hopefully to PANDEMIC proportions like a constructive Stuxnet & Flame among the younger physicists who are not as brainwashed as the older ones. Of course, maybe I’m wrong, but maybe I’m not. Time will tell. ;-)

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From: Quantum Computing and Quantum Information Subject: New comment on "I've posted a thought experiment for superluminal communication using delayed-choice entanglement--would like to get feedback on why it would or would not work."

Date: June 10, 2012 2:40:01 PM PDT

To: Jack Sarfatti Reply-To: LinkedIn Messages <

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Group: Quantum Computing and Quantum Information

Discussion: I've posted a thought experiment for superluminal communication using delayed-choice entanglement--would like to get feedback on why it would or would not work.

Hi Marco,

yes, that helps--I should've finished the last step :) I get:

P(|B>) = Tr{ |B> |B> + |B> =

||^2 + ||^2 + 2 Re { }

with normalization:

P(|B>) = 1/2 ( 1 + ^2 + 2 Re{ } )

so, yes if ||^2 > 0 then the Alice-controlled modulation term

2 Re { } has an effect on P(|B>).

So, according to this derivation, we would require B and B' to be Glauber states as well to give us the non-orthogonality that allows the modulation to have an effect.

Also--there is something funny with normalization, isn't there? if = 1

and = 1, then P(|B>) = 1/2( 1+1+2) = 2 ?!? which doesn't make sense but I think this is what Jack was talking about earlier that these states break the normal Born probability rule.

Do you think that Glauber states require some sort of re-normalization or are we implicitly violating some assumption used by the mathematical formalism or is there just an error in the math?

Posted by Keith Kenemer

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I've posted a thought experiment for superluminal communication using delayed-choice entanglement--would like to get feedback on why it would or would not work.

I came up with this idea a couple of years ago when I was reading about quantum computing and have been thinking about it ever since. I don't necessarily see what principle in quantum mechanics or relativity it would violate.

27 days ago

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Henning Dekant • @Marco, if you follow Jack's advice and ask Joy Christian, I fear you still won't be able to make much sense of it. Given what went down recently at Scott Aaronson's blog: http://www.scottaaronson.com/blog/?p=993

13 days ago • Like

Jack Sarfatti • I never said I agreed with Joy Christian on Bell's theorem. In fact I don't. Henning took that out of context I also suggested Roger Penrose and David Deutsch.

13 days ago

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Henning Dekant • @Jack, didn't mean to imply that you endorsed Joy's ideas about the Bell theorem. Just seems to me, that Joy shouldn't be the first go-to person if you want somebody who can get a complicated point across in the clearest possible way. On your list David Deutsch would be my first choice.

13 days ago • Like

MarcoUnfollow

Marco Barbieri • @Henning: Roger Penrose'd be my first call, but David Deutsch would be a top class choice (presents excluded). However, chances that I get to talk to them are pretty much the same that I can arrange a cricket match with Her Majesty...

13 days ago • Like

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Henning Dekant • @Marco, well I am sure there'd be no hurt in asking Joy if he wants to join you for some cricket if Her Majesty is that hard to come by :-)

13 days ago • Like

Follow Keith

Keith Kenemer • I've tried to work out Jack's idea and get something a little different but I think it may still have the normalization anomaly and a similar Alice modulation effect (let me know of any errors):

Notation:

X(x)Y indicates tensor product of matrix/vectors X and Y

XY indicates normal matrix multiplcation of matrix/vectors X and Y

X* indicates Hermitian conjugate of matrix/vector X

|X> Dirac notation for ket/state/vector X

Initial state:

|Alice,Bob> = |A>(x)|B> + |A'>(x)|B'>

Where A,A' are two qubits corresponding to Glauber(non-othogonal) states owned by Alice and B, B' are normal (orthogonal) eigenstates of another qubit owned by Bob. Note: I'm not sure you could actually create a state like this. If Bob's qubit collapses to B, Alice's state becomes A, otherwise if Bob's qubit collapses to B' then Alice's state becomes A'. I assume the normalized version would have a 1/sqrt(2) factor also. Assuming this makes sense...

Using properties:

(|X>+|Y>)* = |X>* + |Y>*

and

(|X>(x)|Y>)* = |X>* (x) |Y>* [Note: NOT the same as matrix identity (XY)* = Y* X ]

we get:

* = ( |A>(x)|B> )* + ( |A'>(x)|B'> )*

= The Alice/Bob system would have the density matrix:

rho(AliceBob) = |Alice,Bob> ( |A>(x)|B> ) ( (x)|B> ) ( + ( |A'>(x)|B'> ) ( (x)|B'> ) ( Using property:

(|W>(x)|X>) ( |Y>(x)|Z> ) = WX(x)YZ

we get:

rho(AliceBob) = ( |A> + ( |A'> To get the reduced density matrix for Bob, we need the partial trace over A, A'.

Using a property of the partial trace:

Trace(A) { |a1> |b1> rho(B,B')= rho(Bob) = |B>|B>|B'>|B'> since = 1 and =1 and =/= 0 (for Glauber states) we have:

rho(Bob) = |B>|B>|B'> which should have a normalization factor of 1/2. It looks like there may be a normalization anomaly due to the terms that are present. I think this is interesting but I'm not sure I understand the Glauber states well enough to know if this makes sense physically and if I've interpreted the idea correctly.

4 days ago • Like

MarcoUnfollow

Marco Barbieri • Hi Keith,

this is Bob's reduced state. It does make sense, as for A=A', you find a pure superposition of |B> and |B'>, if =0, the state is a incoherent mixture of |B> and |B'>.

In order to get the probability of measuring |B>, you need to calculate Tr(rho|B>, which is 1+||^2+Re(). If =0, this probability is simply 1/2. If you allow for ||^2>0, you do get a modulation, but the measurement will be imperfect: you'll have either some errors or some inconclusive results.

Hope this might help.

4 days ago • Like

Follow Keith

Keith Kenemer • Hi Marco,

yes, that helps--I should've finished the last step :) I get:

P(|B>) = Tr{ |B> |B> + |B> =

||^2 + ||^2 + 2 Re { }

with normalization:

P(|B>) = 1/2 ( 1 + ^2 + 2 Re{ } )

so, yes if ||^2 > 0 then the Alice-controlled modulation term

2 Re { } has an effect on P(|B>).

So, according to this derivation, we would require B and B' to be Glauber states as well to give us the non-orthogonality that allows the modulation to have an effect.

Also--there is something funny with normalization, isn't there? if = 1

and = 1, then P(|B>) = 1/2( 1+1+2) = 2 ?!? which doesn't make sense but I think this is what Jack was talking about earlier that these states break the normal Born probability rule.

Do you think that Glauber states require some sort of re-normalization or are we implicitly violating some assumption used by the mathematical formalism or is there just an error in the math?

26 minutes ago • Unlike

1

Jack Sarfatti • Yes, Keith gets my point. Marco the signal is sent by Alice the sender changing the overlap what changes the response seen by Bob. Now there are several ways of doing that. I'm sure experimentalists will come up with actual designs once they see the theoretical idea - the new possibility. First one has to know that entanglement signaling is a theoretical possibility within known physics. Right now everyone is brainwashed it is not. See, for example, David Kaiser's article in June 2012 Scientific American on the right way to get it wrong. On violation of Born probability see Antony Valentini's many papers on it that are online.

http://arxiv.org/abs/quant-ph/0203049http://tinyurl.com/739a4sh

The normalization anomaly simply means that the Born probability interpretation breaks down. There is also a lot of noise about something called the PBR theorem that seems to say that in some way the Born probability interpretation is inconsistent even in orthodox quantum theory, but I have not studied it yet and do not understand it.

Matthew Pusey - Wikipedia, the free encyclopedia

en.wikipedia.org/wiki/Matthew_PuseyThis preliminary result has been referred to as Pusey's theorem or thePBR theorem, and has been cited by theoretical physicist Antony Valentini as "the most ...

Quantum Times Article on the PBR Theorem | Matt Leifer

mattleifer.info/.../quantum-times-article-on-the-pbr-theorem...

by Matthew Leifer - in 1,016 Google+ circles - More by Matthew Leifer

Feb 26, 2012 – I recently wrote an article (pdf) for The Quantum Times (Newsletter of the APS Topical Group on Quantum Information) about the PBR theorem.

PBR Theorem | Taking up Spacetime

takingupspacetime.wordpress.com/2011/11/21/pbr-theorem/Nov 21, 2011 – A few days ago, there appeared an interesting result by Pusey, Barrett, and Rudolph concerning the foundations of quantum mechanics. A nice ...

Quantum Measurement (4): PBR Theorem (again) | Taking up ...

takingupspacetime.wordpress.com/.../quantum-measurement-4-pbr-t...Nov 22, 2011 – Quantum Measurement (4): PBR Theorem (again). Sorry to be repetitive, but I was just preparing this entry as JohnManchak uploaded the ...

Alternative Experimental Protocol for a PBR-Like Result

arxiv.org › quant-phby DJ Miller - 2012 - Cited by 1 - Related articles

Feb 29, 2012 – Abstract: Pusey, Barrett and Rudolph (PBR) have recently proven an important new theorem in the foundations of quantum mechanics.

[DOC]

PBRCOM

www-physics.lbl.gov/~stapp/PBRCOM.docFile Format: Microsoft Word - Quick View

Assessing the significance of the theorem of Pusey. Barrett ... The logical structure of the PBR argument, as explained by M. Leifer [2], rests upon a formula of the ...

About PBR theorem - Google+

https://plus.google.com/104569184257973656413Feb 1, 2012 – About PBR theorem - discussions aroundarXiv:1201.6554v1.

Newsletter of the Topical Group on Quantum Information American ...

www.aps.org/units/gqi/newsletters/upload/vol6num4.pdfby N Argaman

Apr 16, 2012 – explore the whole space of possible interpretations, and to rule out possibilities via rigorous theorems, such as the. PBR theorem, rather than ...

Quantum mechanics: Get real : Nature Physics : Nature Publishing ...

www.nature.com/nphys/journal/vaop/ncurrent/full/nphys2325.htmlby S Aaronson - 2012

May 6, 2012 – At one extreme, Antony Valentini told Nature News that he thought the word ' seismic' seemed apt, and called the PBR theorem the most ...

1. - Why I Am Not a Psi-ontologist - PIRSA - Perimeter ...

pirsa.org/12050021May 8, 2012 – This talk will address the question of whether the PBR theorem ... I will argue that the PBR theorem provides additional clues for "what has to ...