Re: Quantum theory of measurement

2005-10-13 Thread scerir
From: Ben Goertzel 

 The paradox is as follows:
 One does the EPR thing of creating two particles 
 with opposite spin. [...]

More or less, it is the experiment by Birgit Dopfer
(pdf on this page, unfortunately just in German)
http://www.quantum.univie.ac.at/publications/thesis/





Re: Quantum theory of measurement

2005-10-13 Thread Stephen Paul King

Hi Hal,

   After glancing over that paper it seems to me that Badagnani does not 
distinguish between classical and quantum forms of information. I strongly 
suspect that that is the reason why he thinks his idea would work.


Onward!

Stephen

- Original Message - 
From: Hal Finney [EMAIL PROTECTED]

To: everything-list@eskimo.com
Sent: Wednesday, October 12, 2005 9:59 PM
Subject: RE: Quantum theory of measurement



Now that you are experts on this, try your hand on this FTL
signalling device, http://arxiv.org/abs/quant-ph?0204108.
The author, Daniel Badagnani, is apparently a genuine physicist,
http://cabtep5.cnea.gov.ar/particulas/daniel/pag-db.html.

Hal Finney 




Re: Quantum theory of measurement

2005-10-13 Thread Saibal Mitra
Well, as you can see here:

http://cabtep5.cnea.gov.ar/particulas/daniel/curri/curreng.html

He isn't very experienced yet. I know of some experienced  professors of
have made worse mistakes :)

So, what goes wrong? Well, you don't get an interference pattern at one end
even if you don't detect the photon at the other end. To see this, just
write down the two particle state and add the phase shifts. If the detectors
on the other sides are off, then the two contributions corresponding to the
photon being detected at some position z consist of two orthogonal terms;
one term correpsonds to the other photon in pipe 1 and the other for that
photon in pipe 2

 Suppose that you add a plate to detect the photon on that side as well.
Then the probability that the photon at one end is detected at position z1
and the other is detected at z2 does contain an interference term of the
form:

Cos[delta1(z1) + delta2(z2)]

If you don't detect where photon 2 is absorbed you have to integrate over z2
and the interference term vanishes. To see an interference term you must
keep  z2 fixed. This means that you must consider only those photons for
which the entangled partners were detected at at some fixed z2. But this
means that this value must be communicated by the observer there.

- Original Message - 
From: Hal Finney [EMAIL PROTECTED]
To: everything-list@eskimo.com
Sent: Thursday, October 13, 2005 03:59 AM
Subject: RE: Quantum theory of measurement


 Now that you are experts on this, try your hand on this FTL
 signalling device, http://arxiv.org/abs/quant-ph?0204108.
 The author, Daniel Badagnani, is apparently a genuine physicist,
 http://cabtep5.cnea.gov.ar/particulas/daniel/pag-db.html.

 Hal Finney




RE: Quantum theory of measurement

2005-10-12 Thread Ben Goertzel


Hi,

Oops, I gave the wrong link

I said

 Specifically, I'll refer to the quantum eraser thought experiment 
 summarized at
 
 http://grad.physics.sunysb.edu/~amarch/

but I meant

http://www.dhushara.com/book/quantcos/qnonloc/eraser.htm

Anyway, the essential idea of the two experiments is the same.

-- Ben 



Re: Quantum theory of measurement

2005-10-12 Thread Hal Finney
Ben Goertzel writes about:
 http://grad.physics.sunysb.edu/~amarch/

 The questions I have regard the replacement of the Coincidence Counter (from
 here on: CC) in the  above experiment with a more complicated apparatus.

 What if we replace the CC with one of the following:

 1) a carefully sealed, exquisitely well insulated box with a printer inside
 it.  The printer is  hooked up so that it prints, on paper, an exact record
 of everything that comes into the CC.  Then,  to erase the printed record,
 the whole box is melted, or annihilated using nuclear explosives, or
 whatever.

The CC is not what is erased.  Rather, the so-called erasure happens
to the photons while they are flying through the apparatus.  Nothing in
the experiment proposes erasing data in the CC.  So I don't really see
what you are getting at.

 What will the outcome be in these experiments?

It won't make any difference, because the CC is not used in the way you
imagine.  It doesn't have to produce a record and it doesn't have to erase
any records.

Let me tell you what really happens in the experiment above.  It is
actually not so mystical as people try to make it sound.

We start off with the s photon going through a 2 slit experiment and
getting interference.  That is standard.

Now we put two different polarization rotations in front of the two slits
and interference goes away.  The web page author professes amazement,
but it is not really that surprising.  After all, interference between two
photons would typically be affected by messing with their polarizations.
It is not all that surprising that putting polarizers into the paths
could mess up the interference.

But now comes the impressive part.  He puts a polarizer in front of the
other photon, the p photon, and suddenly the interference comes back!
Surely something amazing and non-local has happened now, right?

Not really.  This new polarizer will eliminate some of the p photons.
They won't get through.  The result is that we will throw out some
of the measurements of s photons, because if the p photon got eaten
by its polarizer, the CC doesn't trigger as there is no coincidence.
(This is the real reason for the CC in this experiment.)

So now we are discarding some of the s photon measurements, and keeping
some.  It turns out that the ones we keep do show an interference pattern.
If we had added back in the ones we discarded, it would blur out the
interference fringes and there would be no pattern.

The point is, there is no change to the s photon when we put the polarizer
over by p.  Its results do not visibly change from non-interference
to interference, as the web page might imply.  (If that did happen,
we'd have the basis for a faster than light communicator.)  No, all
that is happening is that we are choosing to throw out half the data,
and the half we keep does show interference.

The only point of the CC, then, is to tell us which half of the data
to throw out of the s photon measurements.  Destroying the CC and all
of the other crazy things you suggest have nothing to do with the
experiment.  The CC is not what is erased and it does not create a
permanent record.  It is only there to tell us whether a p photon got
through its polarizer or not, so that we know whether to throw away
the s photon measurement.

Hal Finney



RE: Quantum theory of measurement

2005-10-12 Thread Ben Goertzel

Hal,

  What will the outcome be in these experiments?

 It won't make any difference, because the CC is not used in the way you
 imagine.  It doesn't have to produce a record and it doesn't have to erase
 any records.

OK, mea culpa, maybe I misunderstood the apparatus and it was not the CC
that records
things, but still the records
could be kept somewhere, and one can ask what would happen if the records
were
kept somewhere else (e.g. in a macroscopic medium).  No?

 The point is, there is no change to the s photon when we put the polarizer
 over by p.  Its results do not visibly change from non-interference
 to interference, as the web page might imply.  (If that did happen,
 we'd have the basis for a faster than light communicator.)  No, all
 that is happening is that we are choosing to throw out half the data,
 and the half we keep does show interference.

Yes but we are choosing which half to throw out in a very peculiar way --
i.e. we are throwing it out by un-happening it after it happened,
by destroying some records that were only gathered after the events
recorded in the data already happened...

Ben




RE: Quantum theory of measurement

2005-10-12 Thread Ben Goertzel


What if instead of throwing out the information you shoot it into a black
hole?

Then presumably the information is really gone so the result should be as if
the information were quantum erased??

Unless there are white holes of course!! ;-)


 Yes but we are choosing which half to throw out in a very peculiar way --
 i.e. we are throwing it out by un-happening it after it happened,
 by destroying some records that were only gathered after the events
 recorded in the data already happened...

 Ben




RE: Quantum theory of measurement

2005-10-12 Thread Hal Finney
Ben Goertzel writes:
 Hal,
  It won't make any difference, because the CC is not used in the way you
  imagine.  It doesn't have to produce a record and it doesn't have to erase
  any records.

 OK, mea culpa, maybe I misunderstood the apparatus and it was not the CC
 that records
 things, but still the records
 could be kept somewhere, and one can ask what would happen if the records
 were
 kept somewhere else (e.g. in a macroscopic medium).  No?

I don't think this makes sense, at least I can't understand it.


  The point is, there is no change to the s photon when we put the polarizer
  over by p.  Its results do not visibly change from non-interference
  to interference, as the web page might imply.  (If that did happen,
  we'd have the basis for a faster than light communicator.)  No, all
  that is happening is that we are choosing to throw out half the data,
  and the half we keep does show interference.

 Yes but we are choosing which half to throw out in a very peculiar way --
 i.e. we are throwing it out by un-happening it after it happened,
 by destroying some records that were only gathered after the events
 recorded in the data already happened...

You have to try to stop thinking of this in mystical terms.  IMO people
present a rather prosaic phenomenon in a misleading and confusing way,
and this is giving you an incorrect idea.  Nothing is un-happening.
No records are destroyed after they were gathered.

Forget that anybody told you this was a quantum eraser and think
about what really happens.  When all the polarizers are in place, half
of the p photons get eaten and half get through.  This gives us a way
to split up the s measurements into two halves.  It turns out that each
half independently shows interference, but that the two interference
patterns are the opposite of each other.  When you combine the two halves
back together, the peaks of one half fill in the valleys of the other,
and the data set as a whole shows no interference.

Look at it concretely as it might happen in the lab.  We record a bunch
of s measurements and also record whether we get a coincidence with a p
photon getting through, in the CC.  Maybe we write a little check mark
next to the s measurements where there was a p photon coincidence.

We go through afterwards to analyze the data.  If we just plot all
the s measurements we see a smooth curve, no interference.  Now we
go through and cross off the ones where there was no p coincidence.
We cross off s measurement number 1, then numbers 3 and 4, then 5, 7,
10 through 12, and so on.  When we plot the remaining measurements,
now we see an interference pattern.

In other words, the coincidence with the p photon identifies a subset
of the s measurements which shows interference.  The total collection
of s measurements still shows no interference.

There is no real erasing going on.  Whoever coined the term quantum
eraser was a master of public relations, but unfortunately he confused
millions of lay people into getting the wrong idea about the physics.

Hal Finney



Re: Quantum theory of measurement

2005-10-12 Thread Saibal Mitra
Hal gives the correct explanation of what's going on. In general,  all you
have to do to analyze the problem is to consider all contributions to a
particular state and add up the amplitudes. The absolute value squared of
the amplitude gives the probability, which may or may not contain an
interference term.

A simple minded formal description can be given as follows:

If you pass a photon through two slits then close to the screen its state
would be of the form:

Integral over z of [1 + Exp(i delta(z))] |s,z

Here z denotes the postion on the screen, s is just a label for the photon
and Exp(i delta(z)) is the phase shift between the two paths which gives
rise to the interference term. Delta(z) will be zero exactly inbetween the
two slits and will be nonzero elsewhere. The probability of having the
photon at z is obtained (up to normalization) by taking the absolute value
squared of the prefactor of |s,z, which is 2 + 2 cos(delta(z)).

Let's do the same for the two entagled photons. The entangled photon pair
can be denoted as:

 |p_x,s_y + |p_y,s_x

here x and y denote the polarization states.

If you pass s through the two slits then the state becomes:

 [1 + Exp(i delta(z))] |p_x,s_y,z +  [1 + Exp(i delta(z))] ||p_y,s_x,z

This has to be integrated over z, but let's focus only at the contribution
at some fixed position z. The prefactors of both state vectors |p_x,s_y,z
and |p_y,s_x,z are of the same form as in the single photon case and thus
you get an interference term Cos(delta(z) as above. If you put the quarter
wave plate in then instead of

 [1 + Exp(i delta(z))] |p_x,s_y,z you get:

 |p_x,s_r,z +  Exp(i delta(z))  |p_x,s_l,z

And the complete state vector becomes:

|p_x,s_r,z +  Exp(i delta(z))  |p_x,s_l,z+

|p_y,s_l,z +  Exp(i delta(z))  |p_y,s_r,z


All the four state vectors are orthogonal. The probability that you detect
the photon s at z is just the sum of the absolute value squared of the four
terms, which is constant and doesn't contain an interference term.

Now let's pass photon p through the polarizer (45 degrees w.r.t. x). This
amounts to measuring the polarization state of photon p in the basis |p_x +
p_y and |p_x - p_y. If you don't discard one of these two states and keep
them both then, as Hal rightly points out, nothing changes. If you
substitute:

|p_x = |a + |b;

|p_y = |a - |b

in the state vector above you get:



|a,s_r,z +  Exp(i delta(z))  |a,s_l,z+

|a,s_l,z +  Exp(i delta(z))  |a,s_r,z+


|b,s_r,z +  Exp(i delta(z))  |b,s_l,z+

- |b,s_l,z -  Exp(i delta(z))  |b,s_r,z = (collecting the prefactors of
like state vectors)

=

[1 + Exp(i delta(z))]|a,s_r,z + [1 + Exp(i delta(z))]|a,s_l,z +

[1 - Exp(i delta(z))]|b,s_r,z - [1 -  Exp(i delta(z))] |b,s_l,z

The absolute value squared of the terms with the p photons in the |a state
contains the term 2 Cos (delta(z)), but for the b terms this is - 2
Cos(delta(z). If you don't observe the polarization of the p photon, the
interference terms would thus cancel. This is obvious since all we have done
is to write down the same state in a different basis. But if you do observe
the polarization of the p photon, then can measure the probability of
detecting s at position z and p in polarization state |a then you have to
add up the absolute value squared of |a,s_r,z and |a,s_l,z. Then you do
get the Cos(delta(z) interference term.




- Original Message - 
From: Hal Finney [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; everything-list@eskimo.com
Sent: Wednesday, October 12, 2005 08:16 PM
Subject: Re: Quantum theory of measurement


 Ben Goertzel writes about:
  http://grad.physics.sunysb.edu/~amarch/
 
  The questions I have regard the replacement of the Coincidence Counter
(from
  here on: CC) in the  above experiment with a more complicated apparatus.
 
  What if we replace the CC with one of the following:
 
  1) a carefully sealed, exquisitely well insulated box with a printer
inside
  it.  The printer is  hooked up so that it prints, on paper, an exact
record
  of everything that comes into the CC.  Then,  to erase the printed
record,
  the whole box is melted, or annihilated using nuclear explosives, or
  whatever.

 The CC is not what is erased.  Rather, the so-called erasure happens
 to the photons while they are flying through the apparatus.  Nothing in
 the experiment proposes erasing data in the CC.  So I don't really see
 what you are getting at.

  What will the outcome be in these experiments?

 It won't make any difference, because the CC is not used in the way you
 imagine.  It doesn't have to produce a record and it doesn't have to erase
 any records.

 Let me tell you what really happens in the experiment above.  It is
 actually not so mystical as people try to make it sound.

 We start off with the s photon going through a 2 slit experiment and
 getting interference.  That is standard.

 Now we put two different polarization rotations in front of the two slits
 and interference goes away.  The web page author professes

RE: Quantum theory of measurement

2005-10-12 Thread Jesse Mazer

Hal Finney wrote:



Ben Goertzel writes:
 Hal,
  It won't make any difference, because the CC is not used in the way 
you
  imagine.  It doesn't have to produce a record and it doesn't have to 
erase

  any records.

 OK, mea culpa, maybe I misunderstood the apparatus and it was not the CC
 that records
 things, but still the records
 could be kept somewhere, and one can ask what would happen if the 
records

 were
 kept somewhere else (e.g. in a macroscopic medium).  No?

I don't think this makes sense, at least I can't understand it.


I think Ben's question here does make sense. See below...




  The point is, there is no change to the s photon when we put the 
polarizer

  over by p.  Its results do not visibly change from non-interference
  to interference, as the web page might imply.  (If that did happen,
  we'd have the basis for a faster than light communicator.)  No, all
  that is happening is that we are choosing to throw out half the data,
  and the half we keep does show interference.

 Yes but we are choosing which half to throw out in a very peculiar way 
--

 i.e. we are throwing it out by un-happening it after it happened,
 by destroying some records that were only gathered after the events
 recorded in the data already happened...

You have to try to stop thinking of this in mystical terms.  IMO people
present a rather prosaic phenomenon in a misleading and confusing way,
and this is giving you an incorrect idea.  Nothing is un-happening.
No records are destroyed after they were gathered.


Although it may be true that no records are destroyed after they're 
gathered, what is true is that an *opportunity* to find out retroactively 
which path the signal photon took is eliminated when you choose to combine 
the paths of the idler photons instead of measuring them. For reference, 
look at the diagram of the setup in fig. 1 of this paper:


http://xxx.lanl.gov/PS_cache/quant-ph/pdf/9903/9903047.pdf

In this figure, pairs of entangled photons are emitted by one of two atoms 
at different positions, A and B. The signal photons move to the right on the 
diagram, and are detected at D0--you can think of the two atoms as 
corresponding to the two slits in the double-slit experiment, while D0 
corresponds to the screen. Meanwhile, the idler photons move to the left on 
the diagram. If the idler is detected at D3, then you know that it came from 
atom A, and thus that the signal photon came from there also; so when you 
look at the subset of trials where the idler was detected at D3, you will 
not see any interference in the distribution of positions where the signal 
photon was detected at D0, just as you see no interference on the screen in 
the double-slit experiment when you measure which slit the particle went 
through. Likewise, if the idler is detected at D4, then you know both it and 
the signal photon came from atom B, and you won't see any interference in 
the signal photon's distribution. But if the idler is detected at either D1 
or D2, then this is equally consistent with a path where it came from atom A 
and was reflected by the beam-splitter BSA or a path where it came from atom 
B and was reflected from beam-splitter BSB, thus you have no information 
about which atom the signal photon came from and will get interference in 
the signal photon's distribution, just like in the double-slit experiment 
when you don't measure which slit the particle came through. Note that if 
you removed the beam-splitters BSA and BSB you could guarantee that the 
idler would be detected at D3 or D4 and thus that the path of the signal 
photon would be known; likewise, if you replaced the beam-splitters BSA and 
BSB with mirrors, then you could guarantee that the idler would be detected 
at D1 or D2 and thus that the path of the signal photon would be unknown. By 
making the distances large enough you could even choose whether to make sure 
the idlers go to D3D4 or to go to D1D2 *after* you have already observed 
the position that the signal photon was detected, so in this sense you have 
the choice whether or not to retroactively erase your opportunity to know 
which atom the signal photon came from, after the signal photon's position 
has already been detected.


This confused me for a while since it seemed like this would imply your 
later choice determines whether or not you observe interference in the 
signal photons earlier, until I got into a discussion about it online and 
someone showed me the trick.  In the same paper, look at the graphs in 
Fig. 3 and Fig. 4, Fig. 3 showing the interference pattern in the signal 
photons in the subset of cases where the idler was detected at D1, and Fig. 
4 showing the interference pattern in the signal photons in the subset of 
cases where the idler was detected at D2 (the two cases where the idler's 
'which-path' information is lost). They do both show interference, but if 
you line the graphs up you see that the peaks of one interference pattern 
line 

Re: Quantum theory of measurement

2005-10-12 Thread scerir
From: Ben Goertzel 
 [...] but still the records
 could be kept somewhere, 
 and one can ask what would happen 
 if the records were kept somewhere else [...]

Not sure, but the quote below - about the information
'in principle' - might be helpful.

The superposition of amplitudes is only valid if there
is no way to know, even in principle, which path the particle
took. It is important to realize that this does not imply
that an observer actually takes note of what happens.
It is sufficient to destroy the interference pattern,
if the path information is accessible in principle from
the experiment or even if it is dispersed in the environment
and beyond any technical possibility to be recovered, but
in principle 'still out there'.
- Anton Zeilinger, (Rev.Mod.Phys.,1999, p.S-288)






RE: Quantum theory of measurement

2005-10-12 Thread Ben Goertzel

Thanks very much Jesse!

You answered the question I *would have* asked had I rememberd my quantum
physics better ;-)

I think your answer is related to a paradox a friend mentioned to me.

The paradox is as follows:

One does the EPR thing of creating two particles with opposite spin. Send
one far away to Alpha Centauri and send the other through a
Stern-Gerlach magnet and let the spin up and spin down outputs
interfere to form a double slit. If the far away particle is measured
up vs. down, our local particle must definitely go through the up hole
or the down hole and we get no interference pattern. If he measures
the far away particle sideways we get a superposition of states and we
get interference. Thus by rotating his measurement he should be able
to communicate to us faster than the speed of light. We should see our
pattern blinking between interference and not. What's wrong with that
argument?

Along the lines of your solution to my other, related puzzle, I'll try to
analogize a solution to this puzzle.

I guess the idea must be: there is no change to what a particular particle
does
when you observe its faraway coupled pair in a certain way.  Its individual
results do not visibly
change from non-interference to interference.  (If that did happen,
you'd have the basis for a faster than light communicator, as you say.)

Instead, when you observe some of the particles sideways and some
vertically,
you must be creating correlational information that exists only
statistically
as a correlation between that's happening in Alpha Centauri and what's
happening locally.

So, maybe there is some weird cancellation here, like in the case you
described in your email.
Perhaps, if one restricts attention to the cases where
the faraway particle is measured right then interference is seen, and if
one restricts
attention to the cases where the faraway particle is measured left then
interference
is seen; but if one looks across all cases where the faraway particle is
measured
sideways, then the peaks and troughs of the different cases might cancel out
and
you'd get no interference?

-- Ben

 I think Ben's question here does make sense. See below...

 
 
The point is, there is no change to the s photon when we put the
 polarizer
over by p.  Its results do not visibly change from non-interference
to interference, as the web page might imply.  (If that did happen,
we'd have the basis for a faster than light communicator.)  No, all
that is happening is that we are choosing to throw out half
 the data,
and the half we keep does show interference.
  
   Yes but we are choosing which half to throw out in a very
 peculiar way
 --
   i.e. we are throwing it out by un-happening it after it happened,
   by destroying some records that were only gathered after the events
   recorded in the data already happened...
 
 You have to try to stop thinking of this in mystical terms.  IMO people
 present a rather prosaic phenomenon in a misleading and confusing way,
 and this is giving you an incorrect idea.  Nothing is un-happening.
 No records are destroyed after they were gathered.

 Although it may be true that no records are destroyed after they're
 gathered, what is true is that an *opportunity* to find out retroactively
 which path the signal photon took is eliminated when you choose
 to combine
 the paths of the idler photons instead of measuring them. For
 reference,
 look at the diagram of the setup in fig. 1 of this paper:

 http://xxx.lanl.gov/PS_cache/quant-ph/pdf/9903/9903047.pdf

 In this figure, pairs of entangled photons are emitted by one of
 two atoms
 at different positions, A and B. The signal photons move to the
 right on the
 diagram, and are detected at D0--you can think of the two atoms as
 corresponding to the two slits in the double-slit experiment, while D0
 corresponds to the screen. Meanwhile, the idler photons move to
 the left on
 the diagram. If the idler is detected at D3, then you know that
 it came from
 atom A, and thus that the signal photon came from there also; so when you
 look at the subset of trials where the idler was detected at D3, you will
 not see any interference in the distribution of positions where
 the signal
 photon was detected at D0, just as you see no interference on the
 screen in
 the double-slit experiment when you measure which slit the particle went
 through. Likewise, if the idler is detected at D4, then you know
 both it and
 the signal photon came from atom B, and you won't see any interference in
 the signal photon's distribution. But if the idler is detected at
 either D1
 or D2, then this is equally consistent with a path where it came
 from atom A
 and was reflected by the beam-splitter BSA or a path where it
 came from atom
 B and was reflected from beam-splitter BSB, thus you have no information
 about which atom the signal photon came from and will get interference in
 the signal photon's distribution, just like in the double-slit experiment

RE: Quantum theory of measurement

2005-10-12 Thread Jesse Mazer

Ben Goertzel wrote:




Thanks very much Jesse!

You answered the question I *would have* asked had I rememberd my quantum
physics better ;-)

I think your answer is related to a paradox a friend mentioned to me.

The paradox is as follows:

One does the EPR thing of creating two particles with opposite spin. Send
one far away to Alpha Centauri and send the other through a
Stern-Gerlach magnet and let the spin up and spin down outputs
interfere to form a double slit. If the far away particle is measured
up vs. down, our local particle must definitely go through the up hole
or the down hole and we get no interference pattern. If he measures
the far away particle sideways we get a superposition of states and we
get interference. Thus by rotating his measurement he should be able
to communicate to us faster than the speed of light. We should see our
pattern blinking between interference and not. What's wrong with that
argument?

Along the lines of your solution to my other, related puzzle, I'll try to
analogize a solution to this puzzle.

I guess the idea must be: there is no change to what a particular 
particle

does
when you observe its faraway coupled pair in a certain way.  Its individual
results do not visibly
change from non-interference to interference.  (If that did happen,
you'd have the basis for a faster than light communicator, as you say.)

Instead, when you observe some of the particles sideways and some
vertically,
you must be creating correlational information that exists only
statistically
as a correlation between that's happening in Alpha Centauri and what's
happening locally.

So, maybe there is some weird cancellation here, like in the case you
described in your email.
Perhaps, if one restricts attention to the cases where
the faraway particle is measured right then interference is seen, and if
one restricts
attention to the cases where the faraway particle is measured left then
interference
is seen; but if one looks across all cases where the faraway particle is
measured
sideways, then the peaks and troughs of the different cases might cancel 
out

and
you'd get no interference?

-- Ben


Yeah, I'd agree with your guess about what would happen here. Certainly you 
won't see interference vs. non-interference on the screen depending on what 
measurement was made on the other particle, since  it's been proven that 
quantum effects cannot be used to transmit information faster than light; 
so, I think you would see no interference in the total pattern on the screen 
no matter what, but if you looked at specific subsets of cases once you 
found out the result of the measurement on the second particle, you'd 
probably be able to see interference patterns in those subsets where the 
measurement of the second particle did not allow you to determine which slit 
the first one went through.


Jesse




RE: Quantum theory of measurement

2005-10-12 Thread Jesse Mazer

Ben Goertzel wrote:

My own understanding is that whether Fred, a pigeon or a printer is 
involved

in the experiment  should be basically irrelevant.  That is, I don't think
registration in consciousness (whatever  that means) is the important
thing, but rather registration in the sense of statistical correlation
with some macroscopic system effectively obeying classical probability
theory.  However, I realize  that not everyone agrees with me on this; my
reading of Penrose, for instance, is that he would  predict a different
outcome for 2 versus 1, because he believes that Fred's brain (via
unspecified  quantum gravity related effects) does something special to
collapse the wave function, which the  printer does not.


By the way, I also wanted to point out that this actually isn't Penrose's 
position, although given his unusual views about the relationship between 
consciousness and QM it's a pretty common misunderstanding. It's been a 
while since I read his book, but my memory is that Penrose advocates a 
gravitationally-induced collapse view where the wavefunction of an 
entangled system spontaneously collapses once it exceeds a certain threshold 
in mass (due to interactions with the outside world which result in more and 
more particles becoming entangled with the original system), which he 
speculates is likely to be around the Planck mass, about 5*10^-8 kg (close 
to the mass of a flea). He also speculates that the theory of quantum 
gravity which deals with this may be noncomputable, and that the human brain 
may somehow take advantage of such noncomputable effects to solve problems 
that could not be solved by a computational system.


The consciousness collapses the wavefunction view has been advocated by a 
few serious physicists like Eugene Wigner, and of course by lots of new-agey 
types.


Jesse




Re: Quantum theory of measurement

2005-10-12 Thread Stephen Paul King

Hi Ben,

   Could it be that black holes do not destroy information but merely 
randomize it, ala decoherence? For a hint see:


http://www.arxiv.org/abs/quant-ph/0410172


Onward!

Stephen

- Original Message - 
From: Ben Goertzel [EMAIL PROTECTED]

To: everything-list@eskimo.com
Sent: Wednesday, October 12, 2005 2:52 PM
Subject: RE: Quantum theory of measurement





What if instead of throwing out the information you shoot it into a 
black

hole?

Then presumably the information is really gone so the result should be as 
if

the information were quantum erased??

Unless there are white holes of course!! ;-)



Yes but we are choosing which half to throw out in a very peculiar way --
i.e. we are throwing it out by un-happening it after it happened,
by destroying some records that were only gathered after the events
recorded in the data already happened...

Ben