<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>I want to make it clear that I\'m *not* pushing any alternate\ntheories here. I simply don\'t get it and would appreciate\nsome hints. I\'m not denying Aspect\'s experiments, or Bell\'s\nanalysis or anything. I\'m begging for help.\n\nAs the subject says, I don\'t understand the EPR paradox.\nAnd so, I\'m pretty much stumbling when it comes to Bell\'s\ntheorm and the Aspect experiment and so on.\n\nMy problem is this. Let\'s take the Aspect expt. to start.\nThe idea is, there is a pair of photons, each circularly\npolarized, and traveling roughly 180 degrees from eachother.\nThey are supposed to be in a definite total angular momentum,\nso as to agree with the fact that the single atom that\nemitted both has indeed finished up with the same angular\nmomentum as it started.\n\nThe explanation goes, if the polarizer on one side detects\na photon as vertical, then a vertical polarizer on the other\nside *must* detect the other photon as vertical.\n\nOk, it\'s that "must" that stumps me. Why "must?" As near\nas I can tell, you\'ve got a circularly polarized photon,\nit has an amplitude to be detected as vertical by a\npolarizer. The photon that goes through the polarizer\nis not in the same state as it was before. Now it is\nlinearly polarized, oriented vertical.\n\nSo what I don\'t get is, where in QM does it say the other\nphoton *must* now be in a linear vertical state? Its partner\nis not in the same state any more, so why should the two\nphotons "add up" to the same angular momentum they did?\nDon\'t we need to include in the sum the effects on the\npolarizer? Isn\'t it now the (as yet undetected) photon,\nthe working-bits of the polarizer, and the post-polarizer-\nphoton that sum to the correct total angular momentum?\n\nI\'m so clueless. Help?\ngrelbr\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>I want to make it clear that I'm *not* pushing any alternate
theories here. I simply don't get it and would appreciate
some hints. I'm not denying Aspect's experiments, or Bell's
analysis or anything. I'm begging for help.
As the subject says, I don't understand the EPR paradox.
And so, I'm pretty much stumbling when it comes to Bell's
theorm and the Aspect experiment and so on.
My problem is this. Let's take the Aspect expt. to start.
The idea is, there is a pair of photons, each circularly
polarized, and traveling roughly 180 degrees from eachother.
They are supposed to be in a definite total angular momentum,
so as to agree with the fact that the single atom that
emitted both has indeed finished up with the same angular
momentum as it started.
The explanation goes, if the polarizer on one side detects
a photon as vertical, then a vertical polarizer on the other
side *must* detect the other photon as vertical.
Ok, it's that "must" that stumps me. Why "must?" As near
as I can tell, you've got a circularly polarized photon,
it has an amplitude to be detected as vertical by a
polarizer. The photon that goes through the polarizer
is not in the same state as it was before. Now it is
linearly polarized, oriented vertical.
So what I don't get is, where in QM does it say the other
photon *must* now be in a linear vertical state? Its partner
is not in the same state any more, so why should the two
photons "add up" to the same angular momentum they did?
Don't we need to include in the sum the effects on the
polarizer? Isn't it now the (as yet undetected) photon,
the working-bits of the polarizer, and the post-polarizer-
photon that sum to the correct total angular momentum?
I'm so clueless. Help?
grelbr
Phillip Helbig---remove CLOTHES to reply
Jul2-04, 05:31 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nIn article <1a325379.0406300628.6d138b4@posting.google.com>,\ngrelbr@hotmail.com (grelbr) writes:\n\n> The idea is, there is a pair of photons, each circularly\n> polarized, and traveling roughly 180 degrees from eachother.\n> They are supposed to be in a definite total angular momentum,\n> so as to agree with the fact that the single atom that\n> emitted both has indeed finished up with the same angular\n> momentum as it started.\n>\n> The explanation goes, if the polarizer on one side detects\n> a photon as vertical, then a vertical polarizer on the other\n> side *must* detect the other photon as vertical.\n\nNot so much vertical as opposite. If photon A is up, then we know that\nphoton B must be down. How is this different from a red ball and a blue\nball, instead of two photons, so that if we measure one ball to be red\nwe know that the other must be blue? The point is, we can CHOOSE if we\nwant to measure. Instead of measuring up and down, we could measure\nleft and right. If photon A is left, then photon B is right. OK, still\nno mystery. The strange bit is that we can choose what to measure at A\nand know what will be measured at B, even though there is no way causal\ninformation can propagate from A to B in time.\n\nLet\'s do this experiment a hundred times. We measure whether A is left\nor right and whether B is up or down. There won\'t be any correlation.\nBut if we measure whether B is left or right as well, the measurements\nwill always be opposite.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <1a325379.0406300628.6d138b4@posting.google.com>,
grelbr@hotmail.com (grelbr) writes:
> The idea is, there is a pair of photons, each circularly
> polarized, and traveling roughly 180 degrees from eachother.
> They are supposed to be in a definite total angular momentum,
> so as to agree with the fact that the single atom that
> emitted both has indeed finished up with the same angular
> momentum as it started.
>
> The explanation goes, if the polarizer on one side detects
> a photon as vertical, then a vertical polarizer on the other
> side *must* detect the other photon as vertical.
Not so much vertical as opposite. If photon A is up, then we know that
photon B must be down. How is this different from a red ball and a blue
ball, instead of two photons, so that if we measure one ball to be red
we know that the other must be blue? The point is, we can CHOOSE if we
want to measure. Instead of measuring up and down, we could measure
left and right. If photon A is left, then photon B is right. OK, still
no mystery. The strange bit is that we can choose what to measure at A
and know what will be measured at B, even though there is no way causal
information can propagate from A to B in time.
Let's do this experiment a hundred times. We measure whether A is left
or right and whether B is up or down. There won't be any correlation.
But if we measure whether B is left or right as well, the measurements
will always be opposite.
Frank Hellmann
Jul2-04, 05:32 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nThat \'must\' is precisely the spooky action at a distance. you are\ndealing with entanglement, means: The two photons are NOT independent\nof each other.\nLet\'s see if I can do this from the top of my head:\n\nA circular polarization can be written as |R> = |x> + i|y> and |L> =\n|x> - i|y> in the basis in which the polarizer meassures.\nNow due to conversation of angular momentum AND parity it turns out\nyour decayed state should look like this (note that this is a\nsuperposition, and it\'s nonlocal as photons |1> and |2> can be far\napart by now):\n|1,L>|2,R> - |1,R>|2,L> = |1,x>|2,x> + i|1,x>|2,y> - i|1,y>|2,x> +\n|1,y>|2,y> - |1,x>|2,x> + i|1,x>|2,y> - i|1,y>|2,x> - |1,y>|2,y> =\n2i(|1,x>|2,y> + |1,y>|2,x>)\n\nNow if we meassure photon |1> in the x direction we are left with a\npure |2,y> state which means we \'must\' meassure the second photon\nperpendicular to the first.\n\nThe key point is that both of them are described by one wave function,\nthey are not independent photons but entangled.\n\nIt comes down to the fact that the product state of the decay is not\ntwo individual photons with definite states which together satisfy\nangular momentum conservation but a superposition.\n\n---\n\nfrank\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>That 'must' is precisely the spooky action at a distance. you are
dealing with entanglement, means: The two photons are NOT independent
of each other.
Let's see if I can do this from the top of my head:
A circular polarization can be written as |R> = |x> + i|y> and |L> =|x> - i|y> in the basis in which the polarizer meassures.
Now due to conversation of angular momentum AND parity it turns out
your decayed state should look like this (note that this is a
superposition, and it's nonlocal as photons |1> and |2> can be far
apart by now):
|1,L>|2,R> - |1,R>|2,L> = |1,x>|2,x> + i|1,x>|2,y> - i|1,y>|2,x> +|1,y>|2,y> - |1,x>|2,x> + i|1,x>|2,y> - i|1,y>|2,x> - |1,y>|2,y> =2i(|1,x>|2,y> + |1,y>|2,x>)
Now if we meassure photon |1> in the x direction we are left with a
pure |2,y> state which means we 'must' meassure the second photon
perpendicular to the first.
The key point is that both of them are described by one wave function,
they are not independent photons but entangled.
It comes down to the fact that the product state of the decay is not
two individual photons with definite states which together satisfy
angular momentum conservation but a superposition.
---
frank
grelbr
Jul4-04, 08:40 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nCerthas@gmail.com (Frank Hellmann) wrote in message news:<e2b39847.0407010725.79b58207@posting.google.com>...\n> A circular polarization can be written as |R> = |x> + i|y> and |L> =\n> |x> - i|y> in the basis in which the polarizer meassures.\n> Now due to conversation of angular momentum AND parity it turns out\n> your decayed state should look like this (note that this is a\n> superposition, and it\'s nonlocal as photons |1> and |2> can be far\n> apart by now):\n> |1,L>|2,R> - |1,R>|2,L> = |1,x>|2,x> + i|1,x>|2,y> - i|1,y>|2,x> +\n> |1,y>|2,y> - |1,x>|2,x> + i|1,x>|2,y> - i|1,y>|2,x> - |1,y>|2,y> =\n> 2i(|1,x>|2,y> + |1,y>|2,x>)\n\nIn his preprint here\n\nhttp://arxiv.org/ftp/quant-ph/papers/0402/0402001.pdf\n\nAspect uses |1,L>|2,R> + |1,R>|2,L> rather than - as you say.\nSo it should be |1,x>|2,x> + |1,y>|2,y>, meaning the photons\nhave to be parallel, not perpendicular.\n\nBut it does not change the overall thrust of your post. The state\ncan\'t be factored into a product of two states, one for each photon.\nThey are entangled.\n\nSadly, I\'m still confused, only now it\'s about different things.\nI will have to go study up on some of the books and articles that\nhave been suggested in this newsgroup. Thanks muchly.\ngrelbr\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Certhas@gmail.com (Frank Hellmann) wrote in message news:<e2b39847.0407010725.79b58207@posting.google.com>...
> A circular polarization can be written as |R> = |x> + i|y> and |L> =
> |x> - i|y> in the basis in which the polarizer meassures.
> Now due to conversation of angular momentum AND parity it turns out
> your decayed state should look like this (note that this is a
> superposition, and it's nonlocal as photons |1> and |2> can be far
> apart by now):
> |1,L>|2,R> - |1,R>|2,L> = |1,x>|2,x> + i|1,x>|2,y> - i|1,y>|2,x> +
> |1,y>|2,y> - |1,x>|2,x> + i|1,x>|2,y> - i|1,y>|2,x> - |1,y>|2,y> =
> 2i(|1,x>|2,y> + |1,y>|2,x>)
Aspect uses |1,L>|2,R> + |1,R>|2,L> rather than - as you say.
So it should be |1,x>|2,x> + |1,y>|2,y>, meaning the photons
have to be parallel, not perpendicular.
But it does not change the overall thrust of your post. The state
can't be factored into a product of two states, one for each photon.
They are entangled.
Sadly, I'm still confused, only now it's about different things.
I will have to go study up on some of the books and articles that
have been suggested in this newsgroup. Thanks muchly.
grelbr
Oz
Jul9-04, 06:02 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nFrank Hellmann <Certhas@gmail.com> writes\n\n>The key point is that both of them are described by one wave function,\n>they are not independent photons but entangled.\n\nSo it would be valid for a naive to consider \'them\' as a single\nparticle?\n\nCan I now consider what might happen when a measurement of one is\nconsidered?\n\nI asked this before, and basically got told \'no: see bell\'.\nUnfortunately nobody has come up with a brief and clear description of\nwhat bell is about.\n\nWhy can we not see the particle as a single unit, which communicated\n\'internally\' at infinite velocity? That is, along the axis joining the\n\'two halves\' space is contracted to zero distance (internally). That is\n(put horribly crudely) the particle sees itself as stationary and\ncompact but sees space travelling past at the appropriate velocity, but\ninstead of being in one direction it sees (in effect) \'external\' space\ncoming towards it "from two directions". I hope, but rather doubt, that\nthis makes sense to anybody. I can visualise it easily.\n\nThat way the conceptual problem where other observers see measurements\ntaken at different times, and even reversing which observer made the\n\'first\' measurement, simply vanish.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Frank Hellmann <Certhas@gmail.com> writes
>The key point is that both of them are described by one wave function,
>they are not independent photons but entangled.
So it would be valid for a naive to consider 'them' as a single
particle?
Can I now consider what might happen when a measurement of one is
considered?
I asked this before, and basically got told 'no: see bell'.
Unfortunately nobody has come up with a brief and clear description of
what bell is about.
Why can we not see the particle as a single unit, which communicated
'internally' at infinite velocity? That is, along the axis joining the
'two halves' space is contracted to zero distance (internally). That is
(put horribly crudely) the particle sees itself as stationary and
compact but sees space travelling past at the appropriate velocity, but
instead of being in one direction it sees (in effect) 'external' space
coming towards it "from two directions". I hope, but rather doubt, that
this makes sense to anybody. I can visualise it easily.
That way the conceptual problem where other observers see measurements
taken at different times, and even reversing which observer made the
'first' measurement, simply vanish.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<
Frank Hellmann
Jul9-04, 11:32 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nOz <oz@farmeroz.port995.com> wrote in message news:<ZU0JtmE2Sm7AFwBZ@farmeroz.port995.com>...\n> Frank Hellmann <Certhas@gmail.com> writes\n>\n> >The key point is that both of them are described by one wave function,\n> >they are not independent photons but entangled.\n>\n> So it would be valid for a naive to consider \'them\' as a single\n> particle?\n>\nIt\'s a good question which I personally can not anwser completely\nsatisfactory since I\'m just learning myself.\n\nSingle particle states are defined to be states that transform under\ncertain representations of the Lorentz Group (Gallileo Group).\n\nIn other words (using Gallileo group): This wave function we are\ntalking about here lives in the tensor product of the Hilbert space of\nfunctions on 3 Dimensional Euclidean geometry.\nH(x)H\nIf it were seperable it would live on\nH(+)H\nA rotation acts as Rv on v element H\nThen it acts as R(x)R on v element H(x)H\nThus you can tell what a single particle state is by considering it\'s\ntransformation properties under Gallileo.\n\nAt the end of the day the mathematical structure is what is important\nif you find a different way to look at it with two spaces and a single\nparticle it\'s just as well but might not be particularly usefull.\nIt will give the same predictions.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz <oz@farmeroz.port995.com> wrote in message news:<ZU0JtmE2Sm7AFwBZ@farmeroz.port995.com>...
> Frank Hellmann <Certhas@gmail.com> writes
>
> >The key point is that both of them are described by one wave function,
> >they are not independent photons but entangled.
>
> So it would be valid for a naive to consider 'them' as a single
> particle?
>
It's a good question which I personally can not anwser completely
satisfactory since I'm just learning myself.
Single particle states are defined to be states that transform under
certain representations of the Lorentz Group (Gallileo Group).
In other words (using Gallileo group): This wave function we are
talking about here lives in the tensor product of the Hilbert space of
functions on 3 Dimensional Euclidean geometry.
H(x)H
If it were seperable it would live on
H(+)H
A rotation acts as Rv on v element H
Then it acts as R(x)R on v element H(x)H
Thus you can tell what a single particle state is by considering it's
transformation properties under Gallileo.
At the end of the day the mathematical structure is what is important
if you find a different way to look at it with two spaces and a single
particle it's just as well but might not be particularly usefull.
It will give the same predictions.
Arnold Neumaier
Jul9-04, 12:09 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nFrank Hellmann wrote:\n> Oz <oz@farmeroz.port995.com> wrote in message news:<ZU0JtmE2Sm7AFwBZ@farmeroz.port995.com>...\n>\n>>Frank Hellmann <Certhas@gmail.com> writes\n>>\n>>\n>>>The key point is that both of them are described by one wave function,\n>>>they are not independent photons but entangled.\n>>\n>>So it would be valid for a naive to consider \'them\' as a single\n>>particle?\n>>\n>\n> It\'s a good question which I personally can not anwser completely\n> satisfactory since I\'m just learning myself.\n>\n> Single particle states are defined to be states that transform under\n> certain representations of the Lorentz Group (Gallileo Group).\n\nTo qualify as a _single_ (elementary) particle they must correspond to\nirreducible representations.\n\nThis means that a state of two entangled photons will not be a single\nparticle state. Entanglement is specific to reducible representations,\nsince it is a statement about states in a tensor product of representations.\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Frank Hellmann wrote:
> Oz <oz@farmeroz.port995.com> wrote in message news:<ZU0JtmE2Sm7AFwBZ@farmeroz.port995.com>...
>
>>Frank Hellmann <Certhas@gmail.com> writes
>>
>>
>>>The key point is that both of them are described by one wave function,
>>>they are not independent photons but entangled.
>>
>>So it would be valid for a naive to consider 'them' as a single
>>particle?
>>
>
> It's a good question which I personally can not anwser completely
> satisfactory since I'm just learning myself.
>
> Single particle states are defined to be states that transform under
> certain representations of the Lorentz Group (Gallileo Group).
To qualify as a _single_ (elementary) particle they must correspond to
irreducible representations.
This means that a state of two entangled photons will not be a single
particle state. Entanglement is specific to reducible representations,
since it is a statement about states in a tensor product of representations.
Arnold Neumaier
Tom Trotter
Jul9-04, 01:03 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\ngrelbr@hotmail.com (grelbr) wrote in message news:<1a325379.0406300628.6d138b4@posting.google.com>...\n> I want to make it clear that I\'m *not* pushing any alternate\n> theories here. I simply don\'t get it and would appreciate\n> some hints. I\'m not denying Aspect\'s experiments, or Bell\'s\n> analysis or anything. I\'m begging for help.\n>\n> As the subject says, I don\'t understand the EPR paradox.\n> And so, I\'m pretty much stumbling when it comes to Bell\'s\n> theorm and the Aspect experiment and so on.\n>\n> My problem is this. Let\'s take the Aspect expt. to start.\n> The idea is, there is a pair of photons, each circularly\n> polarized, and traveling roughly 180 degrees from eachother.\n> They are supposed to be in a definite total angular momentum,\n> so as to agree with the fact that the single atom that\n> emitted both has indeed finished up with the same angular\n> momentum as it started.\n>\n> The explanation goes, if the polarizer on one side detects\n> a photon as vertical, then a vertical polarizer on the other\n> side *must* detect the other photon as vertical.\n>\n> Ok, it\'s that "must" that stumps me. Why "must?" As near\n> as I can tell, you\'ve got a circularly polarized photon,\n> it has an amplitude to be detected as vertical by a\n> polarizer. The photon that goes through the polarizer\n> is not in the same state as it was before. Now it is\n> linearly polarized, oriented vertical.\n\nThe light (opposite moving photons) emitted by a single\natom\'s electron decay process (via the laser excitation\nof atomic calcium cascades in Aspect) is polarized\nidentically. The photons are "correlated in polarization"\nvia the emission process.\n\nIf the polarizers are oriented at the same angle, then\nif the polarizer on one side transmits enough light (ie.,\na photon\'s worth) during a given coincidence window to\nregister a detection, then the polarizer on the other\nside will (in the ideal) also transmit enough light\nduring that coincidence window to register a detection.\n\n>\n> So what I don\'t get is, where in QM does it say the other\n> photon *must* now be in a linear vertical state? Its partner\n> is not in the same state any more, so why should the two\n> photons "add up" to the same angular momentum they did?\n> Don\'t we need to include in the sum the effects on the\n> polarizer? Isn\'t it now the (as yet undetected) photon,\n> the working-bits of the polarizer, and the post-polarizer-\n> photon that sum to the correct total angular momentum?\n\nIt\'s the spins of the paired photons between emission and\npolarization that sum to conserve the angular momentum of\nthe atom that emitted them, and that\'s responsible for their\nentanglement. The light transmitted by the polarizers is no\nlonger entangled.\n\nIt\'s the "sameness" of polarization via emission that\'s\nbeing measured (filtered through the separated polarizers).\n\nThis is why LHV formulations (such as Bell\'s) where Lambda\nis the "angle" of polarization of the photons following\nemission and prior to polarization/detection don\'t work --\nthat is, they produce inequalities that will be experimentally\nviolated.\n\nThe angle of polarization of paired photons via emission is\nirrelevant wrt determining *coincidental* detection. It\'s only\nthe sameness of polarization of paired photons via emission\nthat matters in the combined context.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>grelbr@hotmail.com (grelbr) wrote in message news:<1a325379.0406300628.6d138b4@posting.google.com>...
> I want to make it clear that I'm *not* pushing any alternate
> theories here. I simply don't get it and would appreciate
> some hints. I'm not denying Aspect's experiments, or Bell's
> analysis or anything. I'm begging for help.
>
> As the subject says, I don't understand the EPR paradox.
> And so, I'm pretty much stumbling when it comes to Bell's
> theorm and the Aspect experiment and so on.
>
> My problem is this. Let's take the Aspect expt. to start.
> The idea is, there is a pair of photons, each circularly
> polarized, and traveling roughly 180 degrees from eachother.
> They are supposed to be in a definite total angular momentum,
> so as to agree with the fact that the single atom that
> emitted both has indeed finished up with the same angular
> momentum as it started.
>
> The explanation goes, if the polarizer on one side detects
> a photon as vertical, then a vertical polarizer on the other
> side *must* detect the other photon as vertical.
>
> Ok, it's that "must" that stumps me. Why "must?" As near
> as I can tell, you've got a circularly polarized photon,
> it has an amplitude to be detected as vertical by a
> polarizer. The photon that goes through the polarizer
> is not in the same state as it was before. Now it is
> linearly polarized, oriented vertical.
The light (opposite moving photons) emitted by a single
atom's electron decay process (via the laser excitation
of atomic calcium cascades in Aspect) is polarized
identically. The photons are "correlated in polarization"
via the emission process.
If the polarizers are oriented at the same angle, then
if the polarizer on one side transmits enough light (ie.,
a photon's worth) during a given coincidence window to
register a detection, then the polarizer on the other
side will (in the ideal) also transmit enough light
during that coincidence window to register a detection.
>
> So what I don't get is, where in QM does it say the other
> photon *must* now be in a linear vertical state? Its partner
> is not in the same state any more, so why should the two
> photons "add up" to the same angular momentum they did?
> Don't we need to include in the sum the effects on the
> polarizer? Isn't it now the (as yet undetected) photon,
> the working-bits of the polarizer, and the post-polarizer-
> photon that sum to the correct total angular momentum?
It's the spins of the paired photons between emission and
polarization that sum to conserve the angular momentum of
the atom that emitted them, and that's responsible for their
entanglement. The light transmitted by the polarizers is no
longer entangled.
It's the "sameness" of polarization via emission that's
being measured (filtered through the separated polarizers).
This is why LHV formulations (such as Bell's) where \Lambda
is the "angle" of polarization of the photons following
emission and prior to polarization/detection don't work --
that is, they produce inequalities that will be experimentally
violated.
The angle of polarization of paired photons via emission is
irrelevant wrt determining *coincidental* detection. It's only
the sameness of polarization of paired photons via emission
that matters in the combined context.
scerir
Jul9-04, 03:53 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n"Oz":\n> So it would be valid for a naive to consider \'them\'\n> as a single particle? Can I now consider what might\n> happen when a measurement of one is considered?\n\nThis (the bi-photon issue) is what Sir Karl Raimund\nPopper asked [\'Zur Kritik der Ungenauigkeitsrelationen\',\nin \'Die Naturwissenschaften\', 22, 807-808,(1934)]\nshortly before EPR paper. In this paper Popper\ninvented an interesting gedanken experiment, involving\ntwo entangled particles (or, say, a bi-particle).\n\nEinstein did not think (1934) that Popper\'s gedanken\nexperiment could be carried out, since to predict\nposition and momentum of the A-entangled-particle,\nboth time and energy of the B-entangled-particle have\nto be measured simultaneously, which appeared to be\nimpossible (a sort of Bohrian line of reasoning, contra\nEPR!).\n\nAccording to Nathan Rosen (the \'R\' in EPR) it might\nhave been possible that Popper\'s gedanken experiment\ninfluenced Einstein and EPR. According to Max Jammer\n(and Popper himself) Popper\'s gedanken experiment is\nan \'overdetermined\' EPR argument.\n\nKim and Shih, at last, performed the experiment\nhttp://www.arxiv.org/abs/quant-ph/9905039 ,\nand Asher Peres wrote something else ...\nhttp://www.arxiv.org/abs/quant-ph/9910078\n\nRegards,\ns.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Oz":
> So it would be valid for a naive to consider 'them'
> as a single particle? Can I now consider what might
> happen when a measurement of one is considered?
This (the bi-photon issue) is what Sir Karl Raimund
Popper asked ['Zur Kritik der Ungenauigkeitsrelationen',
in 'Die Naturwissenschaften', 22, 807-808,(1934)]
shortly before EPR paper. In this paper Popper
invented an interesting gedanken experiment, involving
two entangled particles (or, say, a bi-particle).
Einstein did not think (1934) that Popper's gedanken
experiment could be carried out, since to predict
position and momentum of the A-entangled-particle,
both time and energy of the B-entangled-particle have
to be measured simultaneously, which appeared to be
impossible (a sort of Bohrian line of reasoning, contra
EPR!).
According to Nathan Rosen (the 'R' in EPR) it might
have been possible that Popper's gedanken experiment
influenced Einstein and EPR. According to Max Jammer
(and Popper himself) Popper's gedanken experiment is
an 'overdetermined' EPR argument.
Kim and Shih, at last, performed the experiment
http://www.arxiv.org/abs/http://www.arxiv.org/abs/quant-ph/9905039 ,
and Asher Peres wrote something else ...
http://www.arxiv.org/abs/http://www.arxiv.org/abs/quant-ph/9910078
Regards,
s.
Oz
Jul11-04, 03:57 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nTom Trotter <tom129@juno.com> writes\n\n>It\'s the "sameness" of polarization via emission that\'s\n>being measured (filtered through the separated polarizers).\n>\n>This is why LHV formulations (such as Bell\'s) where Lambda\n>is the "angle" of polarization of the photons following\n>emission and prior to polarization/detection don\'t work --\n>that is, they produce inequalities that will be experimentally\n>violated.\n\nOK. You are getting near to explaining bells inequalities, which nobody\nhas done here (simply) before.\n\nAre you saying that bell assumed two particles leaving with a set angle\nlambda. That is one at lambda+pi/2 and one lambda-pi/2?\n\n>From this he derived the appropriate statistics,\nwhich turned out *not* to agree with experiment?\n\n>The angle of polarization of paired photons via emission is\n>irrelevant wrt determining *coincidental* detection. It\'s only\n>the sameness of polarization of paired photons via emission\n>that matters in the combined context.\n\nWhat experiment showed was that if one was up, the other must be down.\n\nJust out of interest I imagine this has also been done for entangled\nelectrons, to give the same result? Nuclear decay, perhaps?\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<\nozacoohdb@despammed.com still functions.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Tom Trotter <tom129@juno.com> writes
>It's the "sameness" of polarization via emission that's
>being measured (filtered through the separated polarizers).
>
>This is why LHV formulations (such as Bell's) where \Lambda
>is the "angle" of polarization of the photons following
>emission and prior to polarization/detection don't work --
>that is, they produce inequalities that will be experimentally
>violated.
OK. You are getting near to explaining bells inequalities, which nobody
has done here (simply) before.
Are you saying that bell assumed two particles leaving with a set angle
\lambda. That is one at \lambda+\pi/2 and one \lambda-\pi/2?
>From this he derived the appropriate statistics,
which turned out *not* to agree with experiment?
>The angle of polarization of paired photons via emission is
>irrelevant wrt determining *coincidental* detection. It's only
>the sameness of polarization of paired photons via emission
>that matters in the combined context.
What experiment showed was that if one was up, the other must be down.
Just out of interest I imagine this has also been done for entangled
electrons, to give the same result? Nuclear decay, perhaps?
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<
ozacoohdb@despammed.com still functions.
Oz
Jul13-04, 03:37 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nFrank Hellmann <Certhas@gmail.com> writes\n\n>>Oz\n>> So it would be valid for a naive to consider \'them\' as a single\n>> particle?\n>>\n>It\'s a good question which I personally can not anwser completely\n>satisfactory since I\'m just learning myself.\n>\n>Single particle states are defined to be states that transform under\n>certain representations of the Lorentz Group (Gallileo Group).\n\nI would imagine this not to be so of a \'widely separated\' entangled\npair. This means that in this context my concept of \'single\' is not that\nused in QM. That\'s fine.\n\n>In other words (using Gallileo group): This wave function we are\n>talking about here lives in the tensor product of the Hilbert space of\n>functions on 3 Dimensional Euclidean geometry.\n>H(x)H\n>If it were seperable it would live on\n>H(+)H\n\nHang on though. Haven\'t you just implied its \'not separable\' since it\ndoesn\'t live on H(+)H? That is, its neither \'single\' nor \'double\'?\n\nOr, put another way the pair is neither (using the normal definitions) a\nsingle nor a double particle? So, one can say its not a double particle,\neven if its not a single particle? Does this make me half right?\n\n>A rotation acts as Rv on v element H\n>Then it acts as R(x)R on v element H(x)H\n>Thus you can tell what a single particle state is by considering it\'s\n>transformation properties under Gallileo.\n\nThank you for your very clear illustrations. I\'m very much clearer about\nthe whole thing (which doesn\'t mean I really understand it of course).\n\nNow I imagine that I could pick a special basis of rotations R, such\nthat on some axes it looks like one/two single particles but in others\nit does not. I am doing this visually, so plenty of scope for error. I\nguess that rotations with an axis along the centreline of the \'two\nparticles\' look like single particles, whilst the others do not. That\'s\nprobably very unclear. Perhaps I should try and ape your kets in the\nfull and certain knowledge that I will have all the key details wrong,\njust try to look at what I am trying to say rather than what I am\nactually saying!\n\n(RxR HxH) ->\n(R_1xR_1 H_1xH_1) + (R_2xR_2 H_2xH_2) + (R_3 H_3) +(R_3 H_3)\n\nThat is on axes perpendicular to the axis joining the \'two particle\npaths\' you see *both* an up and a down rotated, but on the other axes\nyou see a single particle rotated. That is the particle is in some sense\nboth single and double.\n\nOf course it is totally beyond my capability to show this is so.\n\n>At the end of the day the mathematical structure is what is important\n>if you find a different way to look at it with two spaces and a single\n>particle it\'s just as well but might not be particularly usefull.\n\nConceptually it makes a big difference. One can imagine a particle\ncommunicating (or appearing to) \'internally\' at ftl, using a variety of\nplausible arguments. I am uncomfortable about two separate particles\ndoing so.\n\n>It will give the same predictions.\n\nI sincerely hope so, else it would be wrong!\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com<<\nozacoohdb@despammed.com still functions.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Frank Hellmann <Certhas@gmail.com> writes
>>Oz
>> So it would be valid for a naive to consider 'them' as a single
>> particle?
>>
>It's a good question which I personally can not anwser completely
>satisfactory since I'm just learning myself.
>
>Single particle states are defined to be states that transform under
>certain representations of the Lorentz Group (Gallileo Group).
I would imagine this not to be so of a 'widely separated' entangled
pair. This means that in this context my concept of 'single' is not that
used in QM. That's fine.
>In other words (using Gallileo group): This wave function we are
>talking about here lives in the tensor product of the Hilbert space of
>functions on 3 Dimensional Euclidean geometry.
>H(x)H
>If it were seperable it would live on
>H(+)H
Hang on though. Haven't you just implied its 'not separable' since it
doesn't live on H(+)H? That is, its neither 'single' nor 'double'?
Or, put another way the pair is neither (using the normal definitions) a
single nor a double particle? So, one can say its not a double particle,
even if its not a single particle? Does this make me half right?
>A rotation acts as Rv on v element H
>Then it acts as R(x)R on v element H(x)H
>Thus you can tell what a single particle state is by considering it's
>transformation properties under Gallileo.
Thank you for your very clear illustrations. I'm very much clearer about
the whole thing (which doesn't mean I really understand it of course).
Now I imagine that I could pick a special basis of rotations R, such
that on some axes it looks like one/two single particles but in others
it does not. I am doing this visually, so plenty of scope for error. I
guess that rotations with an axis along the centreline of the 'two
particles' look like single particles, whilst the others do not. That's
probably very unclear. Perhaps I should try and ape your kets in the
full and certain knowledge that I will have all the key details wrong,
just try to look at what I am trying to say rather than what I am
actually saying!
That is on axes perpendicular to the axis joining the 'two particle
paths' you see *both* an up and a down rotated, but on the other axes
you see a single particle rotated. That is the particle is in some sense
both single and double.
Of course it is totally beyond my capability to show this is so.
>At the end of the day the mathematical structure is what is important
>if you find a different way to look at it with two spaces and a single
>particle it's just as well but might not be particularly usefull.
Conceptually it makes a big difference. One can imagine a particle
communicating (or appearing to) 'internally' at ftl, using a variety of
plausible arguments. I am uncomfortable about two separate particles
doing so.
>It will give the same predictions.
I sincerely hope so, else it would be wrong!
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com<<
ozacoohdb@despammed.com still functions.
Tom Trotter
Jul15-04, 04:57 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nOz <oz@farmeroz.port995.com> wrote in message news:<8Kpg65ICw77AFwUX@farmeroz.port995.com>...\n> Tom Trotter <tom129@juno.com> writes\n>\n> >It\'s the "sameness" of polarization via emission that\'s\n> >being measured (filtered through the separated polarizers).\n> >\n> >This is why LHV formulations (such as Bell\'s) where Lambda\n> >is the "angle" of polarization of the photons following\n> >emission and prior to polarization/detection don\'t work --\n> >that is, they produce inequalities that will be experimentally\n> >violated.\n>\n> OK. You are getting near to explaining bells inequalities, which nobody\n> has done here (simply) before.\n\nThere\'s really nothing to explain wrt Bell inequalities.\nThey\'re just arithmetic relationships wrt quantities\nof groups of things.\n\n>\n> Are you saying that bell assumed two particles leaving with a set angle\n> lambda. That is one at lambda+pi/2 and one lambda-pi/2?\n\nIn terms of light and photons, Bell\'s lambda is the property\nof the light coming from the emitter, and incident on the\npolarizers, a (at A) and b (at B), that, if it were known,\nwould allow more accurate predictions of individual results.\n\nSo, lambda effectively refers to the *polarization* of the\noppositely directed beams of light (in, say, the\nAspect experiment) via emission.\n\n>\n> From this he derived the appropriate statistics,\n> which turned out *not* to agree with experiment?\n\nBell\'s theorem is an arithmetic relationship which\nmust be satisfied if the relationship between lambda\nand a and lambda and b is relevant to the determination\nof coincidental detection. Experiment shows that\nit isn\'t. (But this can be deduced without\nreferring to experiments.) It\'s the relationship\nbetween the emitted photons (that is, it\'s their\ncombined orientation, not their individual orientations)\nwrt the polarizers that matters in determining\ncoincidental detection. This *relationship* is\na global or nonlocal parameter pertaining to\npaired photons. It doesn\'t vary. The relationship\nis that paired photons are polarized identically.\n\nIn other words, the correlations in the combined\ncontext don\'t depend on the same thing that\nmore accurate predictions of results of individual\nmeasurements would depend on.\n\nThe things that are happening to produce individual\nresults are still happening in the combined context.\nThey just aren\'t relevent when talking about the\ncombined context.\n\n>\n> >The angle of polarization of paired photons via emission is\n> >irrelevant wrt determining *coincidental* detection. It\'s only\n> >the sameness of polarization of paired photons via emission\n> >that matters in the combined context.\n>\n> What experiment showed was that if one was up, the other must be down.\n\nExperiment has shown that paired photons\n(ie., photons emitted from the same atom)\nare polarized identically -- which is\nwhat the emission model pertaining to the\nway Aspect produced photon pairs says.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz <oz@farmeroz.port995.com> wrote in message news:<8Kpg65ICw77AFwUX@farmeroz.port995.com>...
> Tom Trotter <tom129@juno.com> writes
>
> >It's the "sameness" of polarization via emission that's
> >being measured (filtered through the separated polarizers).
> >
> >This is why LHV formulations (such as Bell's) where \Lambda
> >is the "angle" of polarization of the photons following
> >emission and prior to polarization/detection don't work --
> >that is, they produce inequalities that will be experimentally
> >violated.
>
> OK. You are getting near to explaining bells inequalities, which nobody
> has done here (simply) before.
There's really nothing to explain wrt Bell inequalities.
They're just arithmetic relationships wrt quantities
of groups of things.
>
> Are you saying that bell assumed two particles leaving with a set angle
> \lambda. That is one at \lambda+\pi/2 and one \lambda-\pi/2?
In terms of light and photons, Bell's \lambda is the property
of the light coming from the emitter, and incident on the
polarizers, a (at A) and b (at B), that, if it were known,
would allow more accurate predictions of individual results.
So, \lambda effectively refers to the *polarization* of the
oppositely directed beams of light (in, say, the
Aspect experiment) via emission.
>
> From this he derived the appropriate statistics,
> which turned out *not* to agree with experiment?
Bell's theorem is an arithmetic relationship which
must be satisfied if the relationship between \lambda
and a and \lambda and b is relevant to the determination
of coincidental detection. Experiment shows that
it isn't. (But this can be deduced without
referring to experiments.) It's the relationship
between the emitted photons (that is, it's their
combined orientation, not their individual orientations)
wrt the polarizers that matters in determining
coincidental detection. This *relationship* is
a global or nonlocal parameter pertaining to
paired photons. It doesn't vary. The relationship
is that paired photons are polarized identically.
In other words, the correlations in the combined
context don't depend on the same thing that
more accurate predictions of results of individual
measurements would depend on.
The things that are happening to produce individual
results are still happening in the combined context.
They just aren't relevent when talking about the
combined context.
>
> >The angle of polarization of paired photons via emission is
> >irrelevant wrt determining *coincidental* detection. It's only
> >the sameness of polarization of paired photons via emission
> >that matters in the combined context.
>
> What experiment showed was that if one was up, the other must be down.
Experiment has shown that paired photons
(ie., photons emitted from the same atom)
are polarized identically -- which is
what the emission model pertaining to the
way Aspect produced photon pairs says.
grelbr
Jul15-04, 04:57 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nOz <oz@farmeroz.port995.com> wrote in message news:<8Kpg65ICw77AFwUX@farmeroz.port995.com>...\n[snips]\n> What experiment showed was that if one was up, the other must be down.\n\nWell, actually they showed quite a bit more than that. They showed\nthat the angular dependance, at any angle, of the rate of coincidence\nis just what QM predicts. At least within the experimental error\nbars.\n\nIn other words, if you have a polarizer on the left that is vertical,\nthen a polarizer on the right that is at some angle theta to the\nvertical, then the coincident detection rate is a function of theta.\nThe measured behaviour of this function is as predicted by QM.\nThe preprint by Aspect I mentioned up thread shows this function.\n\nFor photons, you don\'t really get up or down, but vertical\nor horizontal or circular or like. That is, if a photon is\nlinearly polarized, then it can be detected to be up-and-down\npolarized, or left-and-right, or some angle in between.\n\n> Just out of interest I imagine this has also been done for entangled\n> electrons, to give the same result? Nuclear decay, perhaps?\n\nPeople have done some experiments with other decays. I saw one\nin the arxiv about production of quarks and subsequent decay\nof a resonance. As expected, it followed QM and violated\nBell\'s inequality for certain ranges. I could not really\nfollow the paper in any detail though, other than the conclusion\nthat they did observe a violation of Bell\'s inequality.\n\nThis type of experiment presents more challenges for charged particles.\nDetection equipment is more cumbersome (Stern Gerlach type equipment),\nyou need to have vacuum to allow the electrons (or whatever) to move,\ndecays producing electron pairs are higher energy, and so on. With\nphotons, you can much more easily produce the test photons, you can\nlet the photons move through nearly any transparent material (air,\nglass, etc.), poloarizers and detectors are easy to obtain, easy to\nconfigure, easy to test. ("Easy" in the relative sense.) And you\ndon\'t have to worry about energies that involve ionizing radiation,\nso it\'s easier to convince your institution let you do the experiment.\ngrelbr\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz <oz@farmeroz.port995.com> wrote in message news:<8Kpg65ICw77AFwUX@farmeroz.port995.com>...
[snips]
> What experiment showed was that if one was up, the other must be down.
Well, actually they showed quite a bit more than that. They showed
that the angular dependance, at any angle, of the rate of coincidence
is just what QM predicts. At least within the experimental error
bars.
In other words, if you have a polarizer on the left that is vertical,
then a polarizer on the right that is at some angle \theta to the
vertical, then the coincident detection rate is a function of \theta.
The measured behaviour of this function is as predicted by QM.
The preprint by Aspect I mentioned up thread shows this function.
For photons, you don't really get up or down, but vertical
or horizontal or circular or like. That is, if a photon is
linearly polarized, then it can be detected to be up-and-down
polarized, or left-and-right, or some angle in between.
> Just out of interest I imagine this has also been done for entangled
> electrons, to give the same result? Nuclear decay, perhaps?
People have done some experiments with other decays. I saw one
in the arxiv about production of quarks and subsequent decay
of a resonance. As expected, it followed QM and violated
Bell's inequality for certain ranges. I could not really
follow the paper in any detail though, other than the conclusion
that they did observe a violation of Bell's inequality.
This type of experiment presents more challenges for charged particles.
Detection equipment is more cumbersome (Stern Gerlach type equipment),
you need to have vacuum to allow the electrons (or whatever) to move,
decays producing electron pairs are higher energy, and so on. With
photons, you can much more easily produce the test photons, you can
let the photons move through nearly any transparent material (air,
glass, etc.), poloarizers and detectors are easy to obtain, easy to
configure, easy to test. ("Easy" in the relative sense.) And you
don't have to worry about energies that involve ionizing radiation,
so it's easier to convince your institution let you do the experiment.
grelbr
Oz
Jul15-04, 03:22 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\ngrelbr <grelbr@hotmail.com> writes\n>\n>Oz <oz@farmeroz.port995.com> wrote in message news:<8Kpg65ICw77AFwUX@farmeroz.po\n>rt995.com>...\n>[snips]\n>> What experiment showed was that if one was up, the other must be down.\n>\n>Well, actually they showed quite a bit more than that. They showed\n>that the angular dependance, at any angle, of the rate of coincidence\n>is just what QM predicts. At least within the experimental error\n>bars.\n>\n>In other words, if you have a polarizer on the left that is vertical,\n>then a polarizer on the right that is at some angle theta to the\n>vertical, then the coincident detection rate is a function of theta.\n>The measured behaviour of this function is as predicted by QM.\n>The preprint by Aspect I mentioned up thread shows this function.\n\nHmm. I am intrigued. How would one work this out? Lets think.\nI haven\'t done this sort of thing for years, no, decades.\n\nIf particle one arrived at the vertical polariser (A) and passed through\nthen we could say the probability of it getting through for an incident\nangle of L(ambda) would be sin(L) (=l) (from a very *very* old memory).\nAt polariser B, set at some angle theta (T) to A, it would be hmm, let\nme think, sin(T+pi/2+L) (=t)?\nHmm, yes, that feels about right. So the probability of getting two\ndetections would be sin(L)sin(T+L+pi/2), and similarly for one at A and\nnone at B, and vice versa (and no detections).\nHmmm. I\'m not at all sure that sums to unity, better check.\nlt + l(1-t) + t(1-l) + (1-l)(1-t) = 1\n\nOh, it does, so perhaps I\'m right.\nMore likely I have exposed a terrible and embarrassing ignorance.\n\nCan we use this to give a simple example of bells inequality. For\nexample perhaps the above is NOT the correct way to model it. My\nassumption of an \'incoming angle of L\', may well be suspect. I\'m not\nvery happy about it myself. It doesn\'t feel quite right because its\neffectively assuming two (separate) particles leaving in antiphase.\n\nI would ideally want to put the particles through some more polarisers\nbecause I think I could separate out a \'double-particle\' from two single\nones that way. For two single particles, sent through a series of\npolarisers, I ought to be able to get their polarisation the same, but I\ncould *never* do that for an entangled pair that would always end up in\nantiphase.\n\n>For photons, you don\'t really get up or down, but vertical\n>or horizontal or circular or like.\n\nOf course you are right.\n\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com<<\nozacoohdb@despammed.com still functions.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>grelbr <grelbr@hotmail.com> writes
>
>Oz <oz@farmeroz.port995.com> wrote in message news:<8Kpg65ICw77AFwUX@farmeroz.po
>rt995.com>...
>[snips]
>> What experiment showed was that if one was up, the other must be down.
>
>Well, actually they showed quite a bit more than that. They showed
>that the angular dependance, at any angle, of the rate of coincidence
>is just what QM predicts. At least within the experimental error
>bars.
>
>In other words, if you have a polarizer on the left that is vertical,
>then a polarizer on the right that is at some angle \theta to the
>vertical, then the coincident detection rate is a function of \theta.
>The measured behaviour of this function is as predicted by QM.
>The preprint by Aspect I mentioned up thread shows this function.
Hmm. I am intrigued. How would one work this out? Lets think.
I haven't done this sort of thing for years, no, decades.
If particle one arrived at the vertical polariser (A) and passed through
then we could say the probability of it getting through for an incident
angle of L(ambda) would be sin(L) (=l) (from a very *very* old memory).
At polariser B, set at some angle \theta (T) to A, it would be hmm, let
me think, sin(T+\pi/2+L) (=t)?
Hmm, yes, that feels about right. So the probability of getting two
detections would be sin(L)sin(T+L+\pi/2), and similarly for one at A and
none at B, and vice versa (and no detections).
Hmmm. I'm not at all sure that sums to unity, better check.
lt + l(1-t) + t(1-l) + (1-l)(1-t) = 1
Oh, it does, so perhaps I'm right.
More likely I have exposed a terrible and embarrassing ignorance.
Can we use this to give a simple example of bells inequality. For
example perhaps the above is NOT the correct way to model it. My
assumption of an 'incoming angle of L', may well be suspect. I'm not
very happy about it myself. It doesn't feel quite right because its
effectively assuming two (separate) particles leaving in antiphase.
I would ideally want to put the particles through some more polarisers
because I think I could separate out a 'double-particle' from two single
ones that way. For two single particles, sent through a series of
polarisers, I ought to be able to get their polarisation the same, but I
could *never* do that for an entangled pair that would always end up in
antiphase.
>For photons, you don't really get up or down, but vertical
>or horizontal or circular or like.
Of course you are right.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com<<
ozacoohdb@despammed.com still functions.
Tom Trotter
Jul15-04, 03:23 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\ngrelbr@hotmail.com (grelbr) wrote in message news:<1a325379.0407140959.5d3df304@posting.google.com>...\n> Oz <oz@farmeroz.port995.com> wrote in message news:<8Kpg65ICw77AFwUX@farmeroz.port995.com>...\n> [snips]\n> > What experiment showed was that if one was up, the other must be down.\n>\n> Well, actually they showed quite a bit more than that. They showed\n> that the angular dependance, at any angle, of the rate of coincidence\n> is just what QM predicts. At least within the experimental error\n> bars.\n>\n> In other words, if you have a polarizer on the left that is vertical,\n> then a polarizer on the right that is at some angle theta to the\n> vertical, then the coincident detection rate is a function of theta.\n\nI\'m not disagreeing with anything you\'ve said,\n(it seems like you understand EPR ok),\nbut there might be a clearer, way\nto phrase/illustrate some of the pertinent\nconsiderations.\n\nTheta is the angular difference between the\norientation of the polarizers, a and b, as you\nlook through the transmission line from one\npolarizer to the other.\n\nNow consider one pair of photons, correlated\nin linear polarization via emission. (Because,\ncircular polarization can be expressed\nin terms of linear polarization).\n\nAs you look through the transmission line,\nwith a line of sight from one polarizer to the\nother, and with the polarizers in alignment\n(identically oriented), then this alignment\ncan be represented by a single line (ie., the\nline representing the orientation of a is\ncongruent with the line representing the\norientation of b when the polarizers are\naligned) bisecting the circle that represents\nthe transmission line.\n\nNow let another bisecting line represent\nthe identical plane of polarization of\na pair of opposite moving photons.\n\nIt\'s easy to see that as you rotate the\nline representing the plane of polarization\nof the light incident on the polarizers,\nthen the probability of detection (pertaining\nto the amplitude/intensity of the light\ntransmitted by the polarizers) will change\naccordingly, and this probability will\nbe the same at both ends when the polarizers\nare in alignment. So, when the polarizers\nare aligned, the probability of coincidental\ndetection is (in the ideal) 1.\n\nNow have the line representing the polarization\nof the light incident on the polarizers remain\nstationary, and rotate one of the polarizer\nsetting lines. As Theta increases, then\nthe rate of coincidental detection decreases,\nand vice versa, as a circular function of\nTheta -- and this relationship holds\nno matter what the polarization of the\nlight is, so long as photons emitted by the\nsame atom are correlated in polarization\nvia emission.\n\n> The measured behaviour of this function is as predicted by QM.\n> The preprint by Aspect I mentioned up thread shows this function.\n\nIt\'s also perhaps helpful to note that this is\nrelated to classical optics via polariscopes and Malus\' Law.\n\n>\n> For photons, you don\'t really get up or down, but vertical\n> or horizontal or circular or like. That is, if a photon is\n> linearly polarized, then it can be detected to be up-and-down\n> polarized, or left-and-right, or some angle in between.\n\nNote also that the photons that are detected (transmitted\nby the polarizers) are not the same photons that are coming\nfrom the emitter.\n\nWhat\'s being measured in the combined context isn\'t the\npolarization of individual photons at a or b, but rather\nthe relationship of paired photons wrt Theta. This\nis what makes the context a nonlocal one -- in the sense\nthat nonlocal does *not* connote instantaneous\ntransmission of anything.\n\nThis is sort of the archetypal quantum mechanical\nscenario -- ie., entanglement -- correlating the behavior\nof two entities that have a common source, or have\ninteracted, or are separate parts of a larger entity.\n\n>\n> > Just out of interest I imagine this has also been done for entangled\n> > electrons, to give the same result? Nuclear decay, perhaps?\n>\n> People have done some experiments with other decays. I saw one\n> in the arxiv about production of quarks and subsequent decay\n> of a resonance. As expected, it followed QM and violated\n> Bell\'s inequality for certain ranges.\n\n[... snip]\n\nThe violations of Bell inequalities are superfluous, since\nthe inequalities are based on formulations that include\na parameter, lambda, that\'s irrelevant wrt the combined\ncontext.\n\nSo, I\'m not sure -- do we understand EPR? Do we understand\nBell?\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>grelbr@hotmail.com (grelbr) wrote in message news:<1a325379.0407140959.5d3df304@posting.google.com>...
> Oz <oz@farmeroz.port995.com> wrote in message news:<8Kpg65ICw77AFwUX@farmeroz.port995.com>...
> [snips]
> > What experiment showed was that if one was up, the other must be down.
>
> Well, actually they showed quite a bit more than that. They showed
> that the angular dependance, at any angle, of the rate of coincidence
> is just what QM predicts. At least within the experimental error
> bars.
>
> In other words, if you have a polarizer on the left that is vertical,
> then a polarizer on the right that is at some angle \theta to the
> vertical, then the coincident detection rate is a function of \theta.
I'm not disagreeing with anything you've said,
(it seems like you understand EPR ok),
but there might be a clearer, way
to phrase/illustrate some of the pertinent
considerations.
\Theta is the angular difference between the
orientation of the polarizers, a and b, as you
look through the transmission line from one
polarizer to the other.
Now consider one pair of photons, correlated
in linear polarization via emission. (Because,
circular polarization can be expressed
in terms of linear polarization).
As you look through the transmission line,
with a line of sight from one polarizer to the
other, and with the polarizers in alignment
(identically oriented), then this alignment
can be represented by a single line (ie., the
line representing the orientation of a is
congruent with the line representing the
orientation of b when the polarizers are
aligned) bisecting the circle that represents
the transmission line.
Now let another bisecting line represent
the identical plane of polarization of
a pair of opposite moving photons.
It's easy to see that as you rotate the
line representing the plane of polarization
of the light incident on the polarizers,
then the probability of detection (pertaining
to the amplitude/intensity of the light
transmitted by the polarizers) will change
accordingly, and this probability will
be the same at both ends when the polarizers
are in alignment. So, when the polarizers
are aligned, the probability of coincidental
detection is (in the ideal) 1.
Now have the line representing the polarization
of the light incident on the polarizers remain
stationary, and rotate one of the polarizer
setting lines. As \Theta increases, then
the rate of coincidental detection decreases,
and vice versa, as a circular function of
\Theta -- and this relationship holds
no matter what the polarization of the
light is, so long as photons emitted by the
same atom are correlated in polarization
via emission.
> The measured behaviour of this function is as predicted by QM.
> The preprint by Aspect I mentioned up thread shows this function.
It's also perhaps helpful to note that this is
related to classical optics via polariscopes and Malus' Law.
>
> For photons, you don't really get up or down, but vertical
> or horizontal or circular or like. That is, if a photon is
> linearly polarized, then it can be detected to be up-and-down
> polarized, or left-and-right, or some angle in between.
Note also that the photons that are detected (transmitted
by the polarizers) are not the same photons that are coming
from the emitter.
What's being measured in the combined context isn't the
polarization of individual photons at a or b, but rather
the relationship of paired photons wrt \Theta. This
is what makes the context a nonlocal one -- in the sense
that nonlocal does *not* connote instantaneous
transmission of anything.
This is sort of the archetypal quantum mechanical
scenario -- ie., entanglement -- correlating the behavior
of two entities that have a common source, or have
interacted, or are separate parts of a larger entity.
>
> > Just out of interest I imagine this has also been done for entangled
> > electrons, to give the same result? Nuclear decay, perhaps?
>
> People have done some experiments with other decays. I saw one
> in the arxiv about production of quarks and subsequent decay
> of a resonance. As expected, it followed QM and violated
> Bell's inequality for certain ranges.
[... snip]
The violations of Bell inequalities are superfluous, since
the inequalities are based on formulations that include
a parameter, \lambda, that's irrelevant wrt the combined
context.
So, I'm not sure -- do we understand EPR? Do we understand
Bell?
Frank Hellmann
Jul15-04, 03:23 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n> >In other words (using Gallileo group): This wave function we are\n> >talking about here lives in the tensor product of the Hilbert space of\n> >functions on 3 Dimensional Euclidean geometry.\n> >H(x)H\n> >If it were seperable it would live on\n> >H(+)H\n>\n> Hang on though. Haven\'t you just implied its \'not separable\' since it\n> doesn\'t live on H(+)H? That is, its neither \'single\' nor \'double\'?\n>\n\nYes it\'s double. Two particles live on the tensor product space, and\ntransform under the (reducible) tensor product of the irreducible\nrepresentation of the relevant group.\nTensor products is how you group degrees of freedom together in QM.\n\n> Or, put another way the pair is neither (using the normal definitions) a\n> single nor a double particle? So, one can say its not a double particle,\n> even if its not a single particle? Does this make me half right?\n>\n\nAs mentioned before you are somewhat on the right track since what is\nimportant here is that you need to consider them as one wavefunction,\nhowever this one wave function does properly represent two particles\nsince it transforms under R(x)R.\n\n(snip)\n> (RxR HxH) ->\n> (R_1xR_1 H_1xH_1) + (R_2xR_2 H_2xH_2) + (R_3 H_3) +(R_3 H_3)\n>\n\nHmmm....\n\nI can kind of see what you mean but it\'s not how tensor products work.\nFirst of all let\'s keep things simple, take the tensor product of two\nthree dimensional Hilbert spaces.\nThe basis of this 9 dimensional tensor space is:\n(x1,x2)(x1,y2)(x1,z2)(y1,x2)(y1,y2)(y1,z2)(z1,x2)(z1,y2)(z1,z2)\nThe basis of the H(+)H space is\n(x1)(x2)(y1)(y2)(z1)(z2)\n\nNow keep in mind that this is not your physical 3D space but a 3\ndimensional Hilbert space (we can\'t use 2D since 2+2=2*2, think a Spin\n1 particle with the states 1,0,-1).\n\nA General rotation that leaves the z axis invariant will leave only\n(z1,z2) invariant in the tensor space, but both (z1) and (z2) in the\nsum space.\nI\'m somewhat out of my experience here, it would take me a while to\nwritedown the actual formulae. And I don\'t quite see what you were\ndoing there except that you seemed to have a slightly skewed\ngeometrical intuition of the tensor product.\n\nThe problem is that we are really talking about infinite dimensional\nHilbert spaces of functions on low/three dimensional vector spaces.\nSomehow it is the case that the tensor product of two such hilbert\nspaces looks like a Hilbert space on a six dimensional vector space. I\ncan\'t see how this comes about. But it\'s what gives us the niceties of\nthinking about this stuff in terms of wavefunctions on the classical\nconfiguration space.\n\nIf somebody else want\'s to illuminate this, please feel free!!\n\n> >At the end of the day the mathematical structure is what is important\n> >if you find a different way to look at it with two spaces and a single\n> >particle it\'s just as well but might not be particularly usefull.\n>\n> Conceptually it makes a big difference. One can imagine a particle\n> communicating (or appearing to) \'internally\' at ftl, using a variety of\n> plausible arguments. I am uncomfortable about two separate particles\n> doing so.\n>\n\nWell a particle is a point really classically. Elementary particles\nare not supposed to have any internal structure at all.\nAnd if you are feeliung uncomfortable about it you\'re on the right\nway, remember:\n\n"Anyone who is not shocked by quantum theory has not understood it."\n-Niels Bohr\n;)\n\n---\nfrank\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>>In other words (using Gallileo group): This wave function we are
> >talking about here lives in the tensor product of the Hilbert space of
> >functions on 3 Dimensional Euclidean geometry.
> >H(x)H
> >If it were seperable it would live on
> >H(+)H
>
> Hang on though. Haven't you just implied its 'not separable' since it
> doesn't live on H(+)H? That is, its neither 'single' nor 'double'?
>
Yes it's double. Two particles live on the tensor product space, and
transform under the (reducible) tensor product of the irreducible
representation of the relevant group.
Tensor products is how you group degrees of freedom together in QM.
> Or, put another way the pair is neither (using the normal definitions) a
> single nor a double particle? So, one can say its not a double particle,
> even if its not a single particle? Does this make me half right?
>
As mentioned before you are somewhat on the right track since what is
important here is that you need to consider them as one wavefunction,
however this one wave function does properly represent two particles
since it transforms under R(x)R.
I can kind of see what you mean but it's not how tensor products work.
First of all let's keep things simple, take the tensor product of two
three dimensional Hilbert spaces.
The basis of this 9 dimensional tensor space is:
(x1,x2)(x1,y2)(x1,z2)(y1,x2)(y1,y2)(y1,z2)(z1,x2)(z1,y2)(z1,z2)
The basis of the H(+)H space is
(x1)(x2)(y1)(y2)(z1)(z2)
Now keep in mind that this is not your physical 3D space but a 3
dimensional Hilbert space (we can't use 2D since 2+2=2*2, think a Spin
1 particle with the states 1,0,-1).
A General rotation that leaves the z axis invariant will leave only
(z1,z2) invariant in the tensor space, but both (z1) and (z2) in the
sum space.
I'm somewhat out of my experience here, it would take me a while to
writedown the actual formulae. And I don't quite see what you were
doing there except that you seemed to have a slightly skewed
geometrical intuition of the tensor product.
The problem is that we are really talking about infinite dimensional
Hilbert spaces of functions on low/three dimensional vector spaces.
Somehow it is the case that the tensor product of two such hilbert
spaces looks like a Hilbert space on a six dimensional vector space. I
can't see how this comes about. But it's what gives us the niceties of
thinking about this stuff in terms of wavefunctions on the classical
configuration space.
If somebody else want's to illuminate this, please feel free!!
> >At the end of the day the mathematical structure is what is important
> >if you find a different way to look at it with two spaces and a single
> >particle it's just as well but might not be particularly usefull.
>
> Conceptually it makes a big difference. One can imagine a particle
> communicating (or appearing to) 'internally' at ftl, using a variety of
> plausible arguments. I am uncomfortable about two separate particles
> doing so.
>
Well a particle is a point really classically. Elementary particles
are not supposed to have any internal structure at all.
And if you are feeliung uncomfortable about it you're on the right
way, remember:
"Anyone who is not shocked by quantum theory has not understood it."
-Niels Bohr
;)
---
frank
Tom Trotter
Jul16-04, 09:19 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nOz <oz@farmeroz.port995.com> wrote in message news:<XEaKJ1BSI38AFwoi@farmeroz.port995.com>...\n> Frank Hellmann <Certhas@gmail.com> writes\n\n[... snip]\n\n> >At the end of the day the mathematical structure is what is important\n> >if you find a different way to look at it with two spaces and a single\n> >particle it\'s just as well but might not be particularly usefull.\n>\n> Conceptually it makes a big difference. One can imagine a particle\n> communicating (or appearing to) \'internally\' at ftl, using a variety of\n> plausible arguments. I am uncomfortable about two separate particles\n> doing so.\n\nThe paired photons produced by Aspect via\natomic calcium cascades are moving in opposite\ndirections, have different wavelengths, and\nopposite angular momenta. They\'re two different,\nand separate, \'particles\'.\n\nAnd, it\'s not necessary to be contemplating\nhow they can be \'communicating\' with each\nother. They aren\'t.\n\nThe correlations in the combined context\nare due to the photons of any given\npair being identically polarized via\nthe emission process.\n\n>From Greenstein and Zajonc (The Quantum\nChallange, 1997, p. 133):\n"... the spin angular momentum of a photon\nis related to its polarization state. The\nspin of a photon can only be aligned parallel\nor antiparallel to its direction of motion;\nparallel alignment corresponds to right-handed\ncircular polarization, antiparallel to left-\nhanded circular polarization. The two photons\nneed not be emitted in opposite directions,\nbut if we select those that are, conservation\nof angular momentum now requires that their\nhandedness be the same. Therefore, they must\nhave the same polarization: both right- or\nboth left-circularly polarized."\n\nNow, if you want to make a local hidden\nvariable theory work wrt the combined\ncontext, then, as Bell noted, you\'ll need\nsome sort of mechanism whereby the two\nends of the experimental setup can\ninstantaneously communicate. But, that\nwould be a silly construction, since it\'s\nalready been shown that lambda (the\npolarization of the photons) is irrelevant\nto the determination of coincidental\ndetection.\n\nIn the individual measurement context,\nthe emission-produced *polarization* of\na photon is (along with the orientation\nof the polarizer that it is interacting\nwith) the determining factor.\n\nIn the combined measurement context, the\nemission-produced *relationship* between\npaired photons is (along with the combined\norientations of the polarizers) the\ndetermining factor.\n\nIn the combined context, since the\n*relationship* between paired photons\ndoesn\'t vary from pair to pair (only\nthe polarization does), the only variable\nleft to consider in determining rates\nof coincidental detection is Theta, the\nangular difference in polarizer settings.\n\nDoes any of this make sense, or do you\nthink we should continue to talk about\nftl or instantaneous communication between\nparticles or filters and/or detectors in\nEPRBell experiments?\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz <oz@farmeroz.port995.com> wrote in message news:<XEaKJ1BSI38AFwoi@farmeroz.port995.com>...
> Frank Hellmann <Certhas@gmail.com> writes
[... snip]
> >At the end of the day the mathematical structure is what is important
> >if you find a different way to look at it with two spaces and a single
> >particle it's just as well but might not be particularly usefull.
>
> Conceptually it makes a big difference. One can imagine a particle
> communicating (or appearing to) 'internally' at ftl, using a variety of
> plausible arguments. I am uncomfortable about two separate particles
> doing so.
The paired photons produced by Aspect via
atomic calcium cascades are moving in opposite
directions, have different wavelengths, and
opposite angular momenta. They're two different,
and separate, 'particles'.
And, it's not necessary to be contemplating
how they can be 'communicating' with each
other. They aren't.
The correlations in the combined context
are due to the photons of any given
pair being identically polarized via
the emission process.
>From Greenstein and Zajonc (The Quantum
Challange, 1997, p. 133):
"... the spin angular momentum of a photon
is related to its polarization state. The
spin of a photon can only be aligned parallel
or antiparallel to its direction of motion;
parallel alignment corresponds to right-handed
circular polarization, antiparallel to left-
handed circular polarization. The two photons
need not be emitted in opposite directions,
but if we select those that are, conservation
of angular momentum now requires that their
handedness be the same. Therefore, they must
have the same polarization: both right- or
both left-circularly polarized."
Now, if you want to make a local hidden
variable theory work wrt the combined
context, then, as Bell noted, you'll need
some sort of mechanism whereby the two
ends of the experimental setup can
instantaneously communicate. But, that
would be a silly construction, since it's
already been shown that \lambda (the
polarization of the photons) is irrelevant
to the determination of coincidental
detection.
In the individual measurement context,
the emission-produced *polarization* of
a photon is (along with the orientation
of the polarizer that it is interacting
with) the determining factor.
In the combined measurement context, the
emission-produced *relationship* between
paired photons is (along with the combined
orientations of the polarizers) the
determining factor.
In the combined context, since the
*relationship* between paired photons
doesn't vary from pair to pair (only
the polarization does), the only variable
left to consider in determining rates
of coincidental detection is \Theta, the
angular difference in polarizer settings.
Does any of this make sense, or do you
think we should continue to talk about
ftl or instantaneous communication between
particles or filters and/or detectors in
EPRBell experiments?
Oz
Jul16-04, 09:19 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nTom Trotter <tom129@juno.com> writes\n>\n>Oz <oz@farmeroz.port995.com> wrote in message news:<8Kpg65ICw77AFwUX@farmeroz.p\n>o\n>rt995.com>...\n>> Tom Trotter <tom129@juno.com> writes\n>>\n>> >It\'s the "sameness" of polarization via emission that\'s\n>> >being measured (filtered through the separated polarizers).\n>> >\n>> >This is why LHV formulations (such as Bell\'s) where Lambda\n>> >is the "angle" of polarization of the photons following\n>> >emission and prior to polarization/detection don\'t work --\n>> >that is, they produce inequalities that will be experimentally\n>> >violated.\n>>\n>> OK. You are getting near to explaining bells inequalities, which nobody\n>> has done here (simply) before.\n>\n>There\'s really nothing to explain wrt Bell inequalities.\n>They\'re just arithmetic relationships wrt quantities\n>of groups of things.\n\n<sigh>\n\nBut which things and what does experiment show under what circumstances?\nPerhaps I should say \'what effect does the experimental results that\ntest bells inequalities imply\'.\n\nSomething to do with \'hidden variables\', but that\'s too broad a brush to\ngain any insight.\n\n>> Are you saying that bell assumed two particles leaving with a set angle\n>> lambda. That is one at lambda+pi/2 and one lambda-pi/2?\n>\n>In terms of light and photons, Bell\'s lambda is the property\n>of the light coming from the emitter, and incident on the\n>polarizers, a (at A) and b (at B), that, if it were known,\n>would allow more accurate predictions of individual results.\n\nRight. So in this example bell assumed that he did (in theory) know\nlambda and found this did NOT agree with experiment?\n\n>So, lambda effectively refers to the *polarization* of the\n>oppositely directed beams of light (in, say, the\n>Aspect experiment) via emission.\n\nSo the assumption (falsely made) was that each particle left with a set\nlambda?\n\n>> From this he derived the appropriate statistics,\n>> which turned out *not* to agree with experiment?\n>\n>Bell\'s theorem is an arithmetic relationship which\n>must be satisfied if the relationship between lambda\n>and a and lambda and b is relevant to the determination\n>of coincidental detection.\n\nIs there such a thing as \'coincidental detection\', given the many frames\nobservers can be in?\n\n>Experiment shows that\n>it isn\'t. (But this can be deduced without\n>referring to experiments.) It\'s the relationship\n>between the emitted photons (that is, it\'s their\n>combined orientation, not their individual orientations)\n>wrt the polarizers that matters in determining\n>coincidental detection.\n\nOK. That\'s how I always read it.\n\n>This *relationship* is\n>a global or nonlocal parameter pertaining to\n>paired photons. It doesn\'t vary. The relationship\n>is that paired photons are polarized identically.\n>\n>In other words, the correlations in the combined\n>context don\'t depend on the same thing that\n>more accurate predictions of results of individual\n>measurements would depend on.\n>\n>The things that are happening to produce individual\n>results are still happening in the combined context.\n>They just aren\'t relevent when talking about the\n>combined context.\n\nOk. So if we consider the pair as a single particle it must be\ninevitable that if (on \'decay\' - ie detection of one) one is detected,\nthen the other has defined characteristics.\n\nThat\'s it. Nothing else to it.\n\nSo what\'s wrong with the following argument:\n\n1) The particles are one particle until detected.\n2) Because they are separated (to the outside world) only one particle\nwill be detected at any point in global (flat space, right) spacetime.\n3) We cannot force the properties of the detected particle, just measure\nif its up or down.\n4) The waveform of the emitted (double) particle co-evolves. That is it\nconstantly varies its lambda, with \'one half\' being in antiphase with\nthe other. This must be enforced, it seems to me.\n5) We force a detection. We can only detect one particle of the\n\'combined pair\' so the very detection process must break the\nentanglement. Note that the detector interacts with the \'combined pair\'.\n\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com<<\nozacoohdb@despammed.com still functions.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Tom Trotter <tom129@juno.com> writes
>
>Oz <oz@farmeroz.port995.com> wrote in message news:<8Kpg65ICw77AFwUX@farmeroz.p
>o
>rt995.com>...
>> Tom Trotter <tom129@juno.com> writes
>>
>> >It's the "sameness" of polarization via emission that's
>> >being measured (filtered through the separated polarizers).
>> >
>> >This is why LHV formulations (such as Bell's) where \Lambda
>> >is the "angle" of polarization of the photons following
>> >emission and prior to polarization/detection don't work --
>> >that is, they produce inequalities that will be experimentally
>> >violated.
>>
>> OK. You are getting near to explaining bells inequalities, which nobody
>> has done here (simply) before.
>
>There's really nothing to explain wrt Bell inequalities.
>They're just arithmetic relationships wrt quantities
>of groups of things.
<sigh>
But which things and what does experiment show under what circumstances?
Perhaps I should say 'what effect does the experimental results that
test bells inequalities imply'.
Something to do with 'hidden variables', but that's too broad a brush to
gain any insight.
>> Are you saying that bell assumed two particles leaving with a set angle
>> \lambda. That is one at \lambda+\pi/2 and one \lambda-\pi/2?
>
>In terms of light and photons, Bell's \lambda is the property
>of the light coming from the emitter, and incident on the
>polarizers, a (at A) and b (at B), that, if it were known,
>would allow more accurate predictions of individual results.
Right. So in this example bell assumed that he did (in theory) know
\lambda and found this did NOT agree with experiment?
>So, \lambda effectively refers to the *polarization* of the
>oppositely directed beams of light (in, say, the
>Aspect experiment) via emission.
So the assumption (falsely made) was that each particle left with a set
\lambda?
>> From this he derived the appropriate statistics,
>> which turned out *not* to agree with experiment?
>
>Bell's theorem is an arithmetic relationship which
>must be satisfied if the relationship between \lambda
>and a and \lambda and b is relevant to the determination
>of coincidental detection.
Is there such a thing as 'coincidental detection', given the many frames
observers can be in?
>Experiment shows that
>it isn't. (But this can be deduced without
>referring to experiments.) It's the relationship
>between the emitted photons (that is, it's their
>combined orientation, not their individual orientations)
>wrt the polarizers that matters in determining
>coincidental detection.
OK. That's how I always read it.
>This *relationship* is
>a global or nonlocal parameter pertaining to
>paired photons. It doesn't vary. The relationship
>is that paired photons are polarized identically.
>
>In other words, the correlations in the combined
>context don't depend on the same thing that
>more accurate predictions of results of individual
>measurements would depend on.
>
>The things that are happening to produce individual
>results are still happening in the combined context.
>They just aren't relevent when talking about the
>combined context.
Ok. So if we consider the pair as a single particle it must be
inevitable that if (on 'decay' - ie detection of one) one is detected,
then the other has defined characteristics.
That's it. Nothing else to it.
So what's wrong with the following argument:
1) The particles are one particle until detected.
2) Because they are separated (to the outside world) only one particle
will be detected at any point in global (flat space, right) spacetime.
3) We cannot force the properties of the detected particle, just measure
if its up or down.
4) The waveform of the emitted (double) particle co-evolves. That is it
constantly varies its \lambda, with 'one half' being in antiphase with
the other. This must be enforced, it seems to me.
5) We force a detection. We can only detect one particle of the
'combined pair' so the very detection process must break the
entanglement. Note that the detector interacts with the 'combined pair'.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com<<
ozacoohdb@despammed.com still functions.
Joe Rongen
Jul19-04, 04:08 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nBRIGHT AND PURE SOURCE OF HIGH-FIDELITY ENTANGLED\nPHOTONS FOR QUANTUM COMPUTATION AND TELEPORTATION,\nJuly 15 Like virtuosos tuning their violins, researchers at the University\nof Illinois at Urbana-Champaign have tuned their instruments and harmonized\nthe production of entangled photons, pushing rates to more than 1 million\npairs per second. The brighter and purer entangled states could assist\nresearchers in applications involving quantum information processing -\nsuch as quantum computation, teleportation and cryptography and help\nscientists better understand the mysterious transition from quantum\nmechanics to classical physics.\nFull story at http://www.physorg.com/news398.html\n\n== End ======================================\n\nAmazingly enough but now entangled photons are\nset up in a "high fidelity" mode... what will be next ?\n\n\n---\nOutgoing mail is certified Virus Free.\nChecked by AVG anti-virus system (http://www.grisoft.com).\nVersion: 6.0.720 / Virus Database: 476 - Release Date: 7/14/04\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>BRIGHT AND PURE SOURCE OF HIGH-FIDELITY ENTANGLED
PHOTONS FOR QUANTUM COMPUTATION AND TELEPORTATION,
July 15 Like virtuosos tuning their violins, researchers at the University
of Illinois at Urbana-Champaign have tuned their instruments and harmonized
the production of entangled photons, pushing rates to more than 1 million
pairs per second. The brighter and purer entangled states could assist
researchers in applications involving quantum information processing -
such as quantum computation, teleportation and cryptography and help
scientists better understand the mysterious transition from quantum
mechanics to classical physics.
Full story at http://www.physorg.com/news398.html
== End ======================================
Amazingly enough but now entangled photons are
set up in a "high fidelity" mode... what will be next ?
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6..720 / Virus Database: 476 - Release Date: 7/14/04
Tom Trotter
Jul19-04, 04:08 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nOz <oz@farmeroz.port995.com> wrote in message news:<L1Rd1nA5A29AFww3@farmeroz.port995.com>...\n> Tom Trotter <tom129@juno.com> writes\n> >\n> >Oz <oz@farmeroz.port995.com> wrote in message news:<8Kpg65ICw77AFwUX@farmeroz.p\n> >o\n> >rt995.com>...\n> >> Tom Trotter <tom129@juno.com> writes\n> >>\n> >> >It\'s the "sameness" of polarization via emission that\'s\n> >> >being measured (filtered through the separated polarizers).\n> >> >\n> >> >This is why LHV formulations (such as Bell\'s) where Lambda\n> >> >is the "angle" of polarization of the photons following\n> >> >emission and prior to polarization/detection don\'t work --\n> >> >that is, they produce inequalities that will be experimentally\n> >> >violated.\n> >>\n> >> OK. You are getting near to explaining bells inequalities, which nobody\n> >> has done here (simply) before.\n> >\n> >There\'s really nothing to explain wrt Bell inequalities.\n> >They\'re just arithmetic relationships wrt quantities\n> >of groups of things.\n>\n> <sigh>\n>\n> But which things ...\n\nAny things. Let\'s say you have a number objects\nthat, among them, have three different, discernable\ncharacteristics, or properties, or parameters,\nA, B, and C.\n\nBell\'s inequality says that the number of objects\nthat have A but not B plus the number of objects\nthat have B but not C is greater than or equal\nto the number of objects that have A but not C.\n\n> ... and what does experiment show under what circumstances?\n> Perhaps I should say \'what effect does the experimental results that\n> test bells inequalities imply\'.\n\nThe experimental results support the qm formulation,\nand the emission model, which says that paired photons\nare entangled via the emission process\n\nA violation of a Bell inequality tells you that the\ninequality is based on a formulation (lhv) that isn\'t\napplicable to the experimental context.\n\nThe lhv formulation is inapplicable because the\nthing (lambda) that determines the results in individual\nmeasurements isn\'t what determines the results in\ncombined contexts. Lambda refers to the angle of\npolarization of the photons incident on the polarizers\nat A and B. This angle of polarization is irrelevant\nin the combined context. What is relevant is that\npaired photons be polarized identically.\n\n>\n> Something to do with \'hidden variables\', but that\'s\n> too broad a brush to gain any insight.\n>\n\nThe EPRBell tests reveal nothing about local hidden variables\nexcept that formulations including them aren\'t applicable\nto these experimental contexts.\n\nThe EPRBell tests don\'t reveal anything about \'reality\',\nor \'nonlocality\' (in the sense that A and B are communicating\nftl or instantaneously), or determinism vs. indeterminism, or\nwhether lhv theories are, in general, possible.\n\nCertainly, lhv formulations are *inapplicable* to certain\ncontexts.\n\n> >> Are you saying that bell assumed two particles leaving with a set angle\n> >> lambda. That is one at lambda+pi/2 and one lambda-pi/2?\n> >\n> >In terms of light and photons, Bell\'s lambda is the property\n> >of the light coming from the emitter, and incident on the\n> >polarizers, a (at A) and b (at B), that, if it were known,\n> >would allow more accurate predictions of individual results.\n>\n> Right. So in this example bell assumed that he did (in theory) know\n> lambda and found this did NOT agree with experiment?\n>\n\nThe subtle but most relevant assumption associated\nwith the inclusion of the lambda term in the formulation\nand combining it with polarizer orientations at a and b,\nis that knowing the polarization of the photons of\na pair would allow for more accurate predictions of\nrates of coincidental detection, ie., that lambda is\nrelevant in the combined context.\n\nBut lambda has nothing to do with determining rates\nof coincidental detection.\n\nSo, violations of Bell inequalities don\'t tell you\nthat there\'s no lambda, but simply that the specific\npolarization of the photons is not the property\nof the photons that you need to know to accurately\npredict rates of coincidental detection.\n\nThe essential knowledge is that the photons\nof any given pair are polarized identically\nvia emission. Since this *relationship* doesn\'t\nvary from pair to pair, then, effectively,\nyou don\'t need to consider anything about\nthe photons in calculating expectation values\nfor rates of coincidental detection wrt\nvarying mutual polarizer orientations.\n\nYou only need to consider the angular difference\nof mutual polarizer settings.\n\n\n> >So, lambda effectively refers to the *polarization* of the\n> >oppositely directed beams of light (in, say, the\n> >Aspect experiment) via emission.\n>\n> So the assumption (falsely made) was that each particle left with a set\n> lambda?\n\nNo, the problematic assumption is that lambda has\nsomething to do with determining rates of\n*coincidental* detection.\n\n>\n> >> From this he derived the appropriate statistics,\n> >> which turned out *not* to agree with experiment?\n> >\n> >Bell\'s theorem is an arithmetic relationship which\n> >must be satisfied if the relationship between lambda\n> >and a and lambda and b is relevant to the determination\n> >of coincidental detection.\n>\n> Is there such a thing as \'coincidental detection\', given the many frames\n> observers can be in?\n\nYes, as I\'ve mentioned before, experimenters expend\ngreat effort in trying to ensure that they\'re dealing\nwith photons emitted from the same atom in their\ncoincidence counting hardware.\n\n>\n> >Experiment shows that\n> >it isn\'t. (But this can be deduced without\n> >referring to experiments.) It\'s the relationship\n> >between the emitted photons (that is, it\'s their\n> >combine