measurement in quantum optics [Archive]

Hayden McGuinness

Jul30-04, 10:35 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\n\nHello,\n\nI have four related questions concerning quantum optics and\nmeasurement, they can be answered YES/NO, and any answer would be\ngood, but a little elaboration would be nice, if necessary.\n\n1.If I had a beam where the photons where in a superposition state of\nvertical and horizontal polarization, the state 1/SQ[2](|Vert> +\n|Hori>), and incident that beam on a polarizing beam splitter would\nthat definitely collapse the beam, so that all photons in the beam,\nwhatever direction they happen to go, are no longer in a superposition\nstate?\n\n2. If #1 is no, is there any device that would preserve this\nsuperposition when the before mentioned beam is incident on it yet\ndeterministically differentiates (such as makes them go a different\ndirection) when a beam of known polarization is incident on it?\n\n\n3.If I passed the superpositioned beam through a polarization filter\nwhich instead of being polarized in the horizontal or vertical\ndirection was polarized at 45 degrees between the horizontal and\nvertical would the superposition nature of the beam stay the same\n(maybe just diminish in amplitude?)? If a deterministically horizontal\nor vertical beam was passed through the filter would it be in a\nsuperposition?\n\n\n4.Do you know if it is possible to change the relative weights of the\nbeam in the state 1/SQ[2](|Vert> + |Hori>) without collapsing any of\nthe photons? For example, could I make the beam take the state\n1/2|Vert> + SQ[3]/2|Hori> ?\n\n\nMany thanks,\n\nHayden\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>Hello,

I have four related questions concerning quantum optics and
measurement, they can be answered YES/NO, and any answer would be
good, but a little elaboration would be nice, if necessary.

1.If I had a beam where the photons where in a superposition state of
vertical and horizontal polarization, the state 1/SQ[2](|Vert> +
|Hori>), and incident that beam on a polarizing beam splitter would
that definitely collapse the beam, so that all photons in the beam,
whatever direction they happen to go, are no longer in a superposition
state?

2. If #1 is no, is there any device that would preserve this
superposition when the before mentioned beam is incident on it yet
deterministically differentiates (such as makes them go a different
direction) when a beam of known polarization is incident on it?

3.If I passed the superpositioned beam through a polarization filter
which instead of being polarized in the horizontal or vertical
direction was polarized at 45 degrees between the horizontal and
vertical would the superposition nature of the beam stay the same
(maybe just diminish in amplitude?)? If a deterministically horizontal
or vertical beam was passed through the filter would it be in a
superposition?

4.Do you know if it is possible to change the relative weights of the
beam in the state 1/SQ[2](|Vert> + |Hori>) without collapsing any of
the photons? For example, could I make the beam take the state
1/2|Vert> + SQ[3]/2|Hori> ?

Many thanks,

Hayden

Arnold Neumaier

Jul30-04, 11:45 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\nHayden McGuinness wrote:\n>\n> I have four related questions concerning quantum optics and\n> measurement, they can be answered YES/NO, and any answer would be\n> good, but a little elaboration would be nice, if necessary.\n>\n> 1.If I had a beam where the photons where in a superposition state of\n> vertical and horizontal polarization, the state 1/SQ[2](|Vert> +\n> |Hori>), and incident that beam on a polarizing beam splitter would\n> that definitely collapse the beam, so that all photons in the beam,\n> whatever direction they happen to go, are no longer in a superposition\n> state?\n\nNo. A beam splitter is a unitary filter; there is no collapse.\nA collapse happens only where the photons are destroyed.\n\n\n> 2. If #1 is no, is there any device that would preserve this\n> superposition when the before mentioned beam is incident on it yet\n> deterministically differentiates (such as makes them go a different\n> direction) when a beam of known polarization is incident on it?\n\nIn principle you can transform the beam into a pair of entangled beams,\none in the state |vert> the other in the state |hori>.\nThis is analogous to the Stern-Gerlach experiment that I discussed\nin my theoretical physics FAQ at\nhttp://www.mat.univie.ac.at/~neum/physics-faq.txt\n\n\n> 3.If I passed the superpositioned beam through a polarization filter\n> which instead of being polarized in the horizontal or vertical\n> direction was polarized at 45 degrees between the horizontal and\n> vertical would the superposition nature of the beam stay the same\n> (maybe just diminish in amplitude?)?\n\nYes. You would just get a single beam, since its companion has empty\nintensity. Actually, you get a superposition of a light beam with\na dark beam in the vacuum state (containing no photon); such dark beams\nmay have sometimes unexpected effects (if one is careless).\nSee Ulf Leonhardt\'s nice little booklet \'Measuring the quantum state\nof light\'. In an ideal filter, the amplitude would be the same,\nbut of course real filters are not 100% ideal.\n\n> If a deterministically horizontal\n> or vertical beam was passed through the filter would it be in a\n> superposition?\n\nIt would split into tow entangled beams.\n\n\n> 4.Do you know if it is possible to change the relative weights of the\n> beam in the state 1/SQ[2](|Vert> + |Hori>) without collapsing any of\n> the photons? For example, could I make the beam take the state\n> 1/2|Vert> + SQ[3]/2|Hori> ?\n\nYes, by a rotator. This and lots of other optical filters\nare discussed in the quantum optics bible by Mandel and Wolf.\n\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>Hayden McGuinness wrote:
>
> I have four related questions concerning quantum optics and
> measurement, they can be answered YES/NO, and any answer would be
> good, but a little elaboration would be nice, if necessary.
>
> 1.If I had a beam where the photons where in a superposition state of
> vertical and horizontal polarization, the state 1/SQ[2](|Vert> +
> |Hori>), and incident that beam on a polarizing beam splitter would
> that definitely collapse the beam, so that all photons in the beam,
> whatever direction they happen to go, are no longer in a superposition
> state?

No. A beam splitter is a unitary filter; there is no collapse.
A collapse happens only where the photons are destroyed.

> 2. If #1 is no, is there any device that would preserve this
> superposition when the before mentioned beam is incident on it yet
> deterministically differentiates (such as makes them go a different
> direction) when a beam of known polarization is incident on it?

In principle you can transform the beam into a pair of entangled beams,
one in the state |vert> the other in the state |hori>.
This is analogous to the Stern-Gerlach experiment that I discussed
in my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt

> 3.If I passed the superpositioned beam through a polarization filter
> which instead of being polarized in the horizontal or vertical
> direction was polarized at 45 degrees between the horizontal and
> vertical would the superposition nature of the beam stay the same
> (maybe just diminish in amplitude?)?

Yes. You would just get a single beam, since its companion has empty
intensity. Actually, you get a superposition of a light beam with
a dark beam in the vacuum state (containing no photon); such dark beams
may have sometimes unexpected effects (if one is careless).
See Ulf Leonhardt's nice little booklet 'Measuring the quantum state
of light'. In an ideal filter, the amplitude would be the same,
but of course real filters are not 100% ideal.

> If a deterministically horizontal
> or vertical beam was passed through the filter would it be in a
> superposition?

It would split into tow entangled beams.

> 4.Do you know if it is possible to change the relative weights of the
> beam in the state 1/SQ[2](|Vert> + |Hori>) without collapsing any of
> the photons? For example, could I make the beam take the state
> 1/2|Vert> + SQ[3]/2|Hori> ?

Yes, by a rotator. This and lots of other optical filters
are discussed in the quantum optics bible by Mandel and Wolf.

Arnold Neumaier

Arkadiusz Jadczyk

Jul31-04, 09:15 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>\nOn 30 Jul 2004 12:45:59 -0400, Arnold Neumaier\n<Arnold.Neumaier@univie.ac.at> wrote:\n\n>No. A beam splitter is a unitary filter; there is no collapse.\n>A collapse happens only where the photons are destroyed.\n\nNot necessarily. Photon may be scattered on an electron, for instance.\nChange of electron\'s momentum may be observed right at the very moment\nof the scattering event. Thus collapse of the state of the\nphoton+electron system. Or, if you want to get rid of the electron\naltogether - you have effectively collapse of the photon state without\ndestroying of the photon.\n\nark\n--\n\nArkadiusz Jadczyk\nhttp://quantumfuture.net/quantum_future/homepage.htm\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>On 30 Jul 2004 12:45:59 -0400, Arnold Neumaier
<Arnold.Neumaier@univie.ac.at> wrote:

>No. A beam splitter is a unitary filter; there is no collapse.
>A collapse happens only where the photons are destroyed.

Not necessarily. Photon may be scattered on an electron, for instance.
Change of electron's momentum may be observed right at the very moment
of the scattering event. Thus collapse of the state of the
photon+electron system. Or, if you want to get rid of the electron
altogether - you have effectively collapse of the photon state without
destroying of the photon.

ark
--

Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/homepage.htm

--

Rahul Jain

Aug4-04, 01:25 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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> writes:\n\n> No. A beam splitter is a unitary filter; there is no collapse.\n> A collapse happens only where the photons are destroyed.\n\nIt seems that hot fullerenes collapse their wavefunctions under certain\nconditions having to do with the frequency of photons they emit while\ncooling. There needs to be some other concept of "destruction" in this\ncase, I believe, as the fullerene is most definitely not destroyed\nbefore it hits the detector, yet it does not interfere with itself in\nthe diffraction gratings.\n\n--\nRahul Jain\nrjain@nyct.net\nProfessional Software Developer, Amateur Quantum Mechanicist\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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> writes:

> No. A beam splitter is a unitary filter; there is no collapse.
> A collapse happens only where the photons are destroyed.

It seems that hot fullerenes collapse their wavefunctions under certain
conditions having to do with the frequency of photons they emit while
cooling. There needs to be some other concept of "destruction" in this
case, I believe, as the fullerene is most definitely not destroyed
before it hits the detector, yet it does not interfere with itself in
the diffraction gratings.

--
Rahul Jain
rjain@nyct.net
Professional Software Developer, Amateur Quantum Mechanicist

Rahul Jain

Aug4-04, 01:25 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>Arkadiusz Jadczyk <arkREMOVETHIS@ANDTHIScassiopaea.org> writes:\n\n> On 30 Jul 2004 12:45:59 -0400, Arnold Neumaier\n> <Arnold.Neumaier@univie.ac.at> wrote:\n>\n>>No. A beam splitter is a unitary filter; there is no collapse.\n>>A collapse happens only where the photons are destroyed.\n>\n> Not necessarily. Photon may be scattered on an electron, for instance.\n> Change of electron\'s momentum may be observed right at the very moment\n> of the scattering event. Thus collapse of the state of the\n> photon+electron system.\n\nThe photon and electron are destroyed and a new photon and electron are\ncreated. That which was considered equal no longer is, since we\'re\ndelving into physics where these distinctions may have an effect.\n\n> Or, if you want to get rid of the electron\n> altogether - you have effectively collapse of the photon state without\n> destroying of the photon.\n\nEh? The photon never interacted with anything. Do you have an experiment\nin mind where this caused a photon\'s state to collapse?\n\n--\nRahul Jain\nrjain@nyct.net\nProfessional Software Developer, Amateur Quantum Mechanicist\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>Arkadiusz Jadczyk <arkREMOVETHIS@ANDTHIScassiopaea.org> writes:

> On 30 Jul 2004 12:45:59 -0400, Arnold Neumaier
> <Arnold.Neumaier@univie.ac.at> wrote:
>
>>No. A beam splitter is a unitary filter; there is no collapse.
>>A collapse happens only where the photons are destroyed.
>
> Not necessarily. Photon may be scattered on an electron, for instance.
> Change of electron's momentum may be observed right at the very moment
> of the scattering event. Thus collapse of the state of the
> photon+electron system.

The photon and electron are destroyed and a new photon and electron are
created. That which was considered equal no longer is, since we're
delving into physics where these distinctions may have an effect.

> Or, if you want to get rid of the electron
> altogether - you have effectively collapse of the photon state without
> destroying of the photon.

Eh? The photon never interacted with anything. Do you have an experiment
in mind where this caused a photon's state to collapse?

--
Rahul Jain
rjain@nyct.net
Professional Software Developer, Amateur Quantum Mechanicist

Arkadiusz Jadczyk

Aug7-04, 05:04 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>On Wed, 4 Aug 2004 06:25:24 +0000 (UTC), Rahul Jain <rjain@nyct.net>\nwrote:\n\n>The photon and electron are destroyed and a new photon and electron are\n>created. That which was considered equal no longer is, since we\'re\n>delving into physics where these distinctions may have an effect.\n\nYou see, according to books on quantum mechanics, photons can not be\ndestroyed, as the only thing that happens is a unitary evolution that is\ncontinuous in time. Therefore talking about "destroying" particles you\nare throwing into a trash box all textbooks and 90% of papers that have\nbeen published since 1927.\n\nOf course almost everybody does it, so you are not alone, but once we\nare discussing fundamental issues, it is good to point out that the\nemperor has no cloth. Remember: quantum mechanics, including quantum\noptics, knows no formalism to dynamically describe ANY EVENT.\nIn quantum mechanics NOTHING EVER HAPPENS. In particular, according to\nquantum mechanics you have never written anything.\n\nOnly few people have eyes to see it. John Bell saw it. t\'Hooft sees it\non even days and does not want to see it on odd days, I believe.\nBohmians see it to some extent. GRW see it to some extent. But most\nof the physicists simply do not care.\n\n>> Or, if you want to get rid of the electron\n>> altogether - you have effectively collapse of the photon state without\n>> destroying of the photon.\n>\n>Eh? The photon never interacted with anything. Do you have an experiment\n>in mind where this caused a photon\'s state to collapse?\n\nYes, I have an experiment in mind. It just happened a while ago, when I\nsaw sun shining behind my window.\n\nark\n--\n\nArkadiusz Jadczyk\nhttp://quantumfuture.net/quantum_future/jadpub.htm\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>On Wed, 4 Aug 2004 06:25:24 +0000 (UTC), Rahul Jain <rjain@nyct.net>
wrote:

>The photon and electron are destroyed and a new photon and electron are
>created. That which was considered equal no longer is, since we're
>delving into physics where these distinctions may have an effect.

You see, according to books on quantum mechanics, photons can not be
destroyed, as the only thing that happens is a unitary evolution that is
continuous in time. Therefore talking about "destroying" particles you
are throwing into a trash box all textbooks and 90% of papers that have
been published since 1927.

Of course almost everybody does it, so you are not alone, but once we
are discussing fundamental issues, it is good to point out that the
emperor has no cloth. Remember: quantum mechanics, including quantum
optics, knows no formalism to dynamically describe ANY EVENT.
In quantum mechanics NOTHING EVER HAPPENS. In particular, according to
quantum mechanics you have never written anything.

Only few people have eyes to see it. John Bell saw it. t'Hooft sees it
on even days and does not want to see it on odd days, I believe.
Bohmians see it to some extent. GRW see it to some extent. But most
of the physicists simply do not care.

>> Or, if you want to get rid of the electron
>> altogether - you have effectively collapse of the photon state without
>> destroying of the photon.
>
>Eh? The photon never interacted with anything. Do you have an experiment
>in mind where this caused a photon's state to collapse?

Yes, I have an experiment in mind. It just happened a while ago, when I
saw sun shining behind my window.

ark
--

Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
--

Alex Green

Aug12-04, 08:29 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\n\nArkadiusz Jadczyk <arkREMOVETHIS@ANDTHIScassiopaea.org> wrote in message news:<9lg6h05k81p1mumu1kbvhtlfaag3lr2tcu@4ax.com>. ..\n> On Wed, 4 Aug 2004 06:25:24 +0000 (UTC), Rahul Jain <rjain@nyct.net>\n> wrote:\n>\n[snip]\n> You see, according to books on quantum mechanics, photons can not be\n> destroyed, as the only thing that happens is a unitary evolution that is\n> continuous in time. Therefore talking about "destroying" particles you\n> are throwing into a trash box all textbooks and 90% of papers that have\n> been published since 1927.\n>\n> Of course almost everybody does it, so you are not alone, but once we\n> are discussing fundamental issues, it is good to point out that the\n> emperor has no cloth. Remember: quantum mechanics, including quantum\n> optics, knows no formalism to dynamically describe ANY EVENT.\n> In quantum mechanics NOTHING EVER HAPPENS. In particular, according to\n> quantum mechanics you have never written anything.\n>\n\nPlease could you explain in what sense you mean \'nothing ever happens\'\nin QM? Nothing happens in GR either if we consider time to be part of\na space time continuum (everything has already happened). Nothing\nhappens in Galilean relativity either because only the instant would\nexist and at any instant everything would be frozen still.\n\nPhysicists such as Weyl have been quite explicit, saying that it is\nthe observer that must move in time. This leads to a definition of\ntime as the coordinate that hosts the succession of arrangements of\nthings in a succession of observations. The only thing missing is the\nphysics of the observer.\n\nBest Wishes\n\nAlex Green\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>Arkadiusz Jadczyk <arkREMOVETHIS@ANDTHIScassiopaea.org> wrote in message news:<9lg6h05k81p1mumu1kbvhtlfaag3lr2tcu@4ax.com>...
> On Wed, 4 Aug 2004 06:25:24 +0000 (UTC), Rahul Jain <rjain@nyct.net>
> wrote:
>
[snip]
> You see, according to books on quantum mechanics, photons can not be
> destroyed, as the only thing that happens is a unitary evolution that is
> continuous in time. Therefore talking about "destroying" particles you
> are throwing into a trash box all textbooks and 90% of papers that have
> been published since 1927.
>
> Of course almost everybody does it, so you are not alone, but once we
> are discussing fundamental issues, it is good to point out that the
> emperor has no cloth. Remember: quantum mechanics, including quantum
> optics, knows no formalism to dynamically describe ANY EVENT.
> In quantum mechanics NOTHING EVER HAPPENS. In particular, according to
> quantum mechanics you have never written anything.
>

Please could you explain in what sense you mean 'nothing ever happens'
in QM? Nothing happens in GR either if we consider time to be part of
a space time continuum (everything has already happened). Nothing
happens in Galilean relativity either because only the instant would
exist and at any instant everything would be frozen still.

Physicists such as Weyl have been quite explicit, saying that it is
the observer that must move in time. This leads to a definition of
time as the coordinate that hosts the succession of arrangements of
things in a succession of observations. The only thing missing is the
physics of the observer.

Best Wishes

Alex Green

Arnold Neumaier

Aug12-04, 08:30 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\nRahul Jain wrote:\n> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> writes:\n>\n>\n>>No. A beam splitter is a unitary filter; there is no collapse.\n>>A collapse happens only where the photons are destroyed.\n>\n> It seems that hot fullerenes collapse their wavefunctions under certain\n> conditions having to do with the frequency of photons they emit while\n> cooling.\n\nWhich experiment are you referring to? (arxiv reference?)\n\n\n> There needs to be some other concept of "destruction" in this\n> case, I believe, as the fullerene is most definitely not destroyed\n> before it hits the detector, yet it does not interfere with itself in\n> the diffraction gratings.\n\nThe fact that one can talk about \'hot\' implies temperature and hence a\nthermodynamic model. Whatever has a well-defined temperature is already\na macroscopic dissipative system, and hence has the capacity to cause\ncollapse. Unitarity is lost in any thermodynamic description of a system.\n\n\nArnold Neumaier\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>Rahul Jain wrote:
> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> writes:
>
>
>>No. A beam splitter is a unitary filter; there is no collapse.
>>A collapse happens only where the photons are destroyed.
>
> It seems that hot fullerenes collapse their wavefunctions under certain
> conditions having to do with the frequency of photons they emit while
> cooling.

Which experiment are you referring to? (arxiv reference?)

> There needs to be some other concept of "destruction" in this
> case, I believe, as the fullerene is most definitely not destroyed
> before it hits the detector, yet it does not interfere with itself in
> the diffraction gratings.

The fact that one can talk about 'hot' implies temperature and hence a
thermodynamic model. Whatever has a well-defined temperature is already
a macroscopic dissipative system, and hence has the capacity to cause
collapse. Unitarity is lost in any thermodynamic description of a system.

Arnold Neumaier

Arnold Neumaier

Aug12-04, 08:30 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\nArkadiusz Jadczyk wrote:\n> On 30 Jul 2004 12:45:59 -0400, Arnold Neumaier\n> <Arnold.Neumaier@univie.ac.at> wrote:\n>\n>>No. A beam splitter is a unitary filter; there is no collapse.\n>>A collapse happens only where the photons are destroyed.\n>\n> Not necessarily. Photon may be scattered on an electron, for instance.\n> Change of electron\'s momentum may be observed right at the very moment\n> of the scattering event. Thus collapse of the state of the\n> photon+electron system.\n\nIn that case, the collapse of the photon-electron system only happens\nwhen the electron gets bound to a macroscopic object.\n\nMy point was that to get collapse, one needs interaction with a\nmacroscopic body capable of creating dissipation.\n\nIndependently of that, beam splitters _are_ modelled in quantum optics as\nunitary transformations. Once this modelling assumption is made, there\nis no longer room for collapse through splitting.\n\n\nArnold Neumaier\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>Arkadiusz Jadczyk wrote:
> On 30 Jul 2004 12:45:59 -0400, Arnold Neumaier
> <Arnold.Neumaier@univie.ac.at> wrote:
>
>>No. A beam splitter is a unitary filter; there is no collapse.
>>A collapse happens only where the photons are destroyed.
>
> Not necessarily. Photon may be scattered on an electron, for instance.
> Change of electron's momentum may be observed right at the very moment
> of the scattering event. Thus collapse of the state of the
> photon+electron system.

In that case, the collapse of the photon-electron system only happens
when the electron gets bound to a macroscopic object.

My point was that to get collapse, one needs interaction with a
macroscopic body capable of creating dissipation.

Independently of that, beam splitters _are_ modelled in quantum optics as
unitary transformations. Once this modelling assumption is made, there
is no longer room for collapse through splitting.

Arnold Neumaier

Hayden McGuinness

Aug12-04, 08:30 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\n\nThanks, both, for the response, I\'m currenlty checking out the\nreferences. I just wanted to make sure I understood the responses\ncorrectly. I\'m not sure if the answers given were assuming that all\nmeasurements/filters *were in regard to the same basis*, that is, I\ndon\'t want to rotate the coordinates in which I am measuring (of\ncourse polarizers have to be rotated in regards to the original basis,\nI\'m talking about the final outcome state). the numbers refer to my\nprevious questions.\n\nSo again, Yes/No (+ elaboration if you are feeling generous) would be\ngood:\n\n1. So if I had a beam with intensity A in the state 1/SQ[2](|Vert>\n+|Hori>), and incident that on a POLARIZING beam splitter (not just a\nregular BS) in one direction I would get a beam with intensity A/2 in\nthe state 1/SQ[2](|Vert> +|Hori>) and in the other direction I would\nget a beam with intensity A/2 in the state 1/SQ[2](|Vert> +|Hori>)\n[the same states!]? Of course if I had a beam in either all |Hori> or\nall |Vert> the beam would go in only one direction, no?\n\n4. If I passed my beam in the state 1/SQ[2](|Vert> +|Hori>) through a\npolarizing rotator meant to rotate polarization 45 degrees, I would\nthen have a beam with photons in either completely the |Vert> state or\ncompletely in the |Hori> state *in the original polarization basis*?\nBut if I passed a beam with know polarization with either |Vert> or\n|Hori> polarization into this same filter I would then have a beam in\nthe state 1/SQ[2](|Vert> +|Hori>) in the original basis ?\n\nAgain, thanks,\nHayden\n\nPS sorry about harping about the "orignal basis" but this has been\nconfusing to me in what I\'ve been reading.\n\nArnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<410A7A8D.1010300@univie.ac.at>...\n> Hayden McGuinness wrote:\n> >\n> > I have four related questions concerning quantum optics and\n> > measurement, they can be answered YES/NO, and any answer would be\n> > good, but a little elaboration would be nice, if necessary.\n> >\n> > 1.If I had a beam where the photons where in a superposition state of\n> > vertical and horizontal polarization, the state 1/SQ[2](|Vert> +\n> > |Hori>), and incident that beam on a polarizing beam splitter would\n> > that definitely collapse the beam, so that all photons in the beam,\n> > whatever direction they happen to go, are no longer in a superposition\n> > state?\n>\n> No. A beam splitter is a unitary filter; there is no collapse.\n> A collapse happens only where the photons are destroyed.\n>\n>\n> > 2. If #1 is no, is there any device that would preserve this\n> > superposition when the before mentioned beam is incident on it yet\n> > deterministically differentiates (such as makes them go a different\n> > direction) when a beam of known polarization is incident on it?\n>\n> In principle you can transform the beam into a pair of entangled beams,\n> one in the state |vert> the other in the state |hori>.\n> This is analogous to the Stern-Gerlach experiment that I discussed\n> in my theoretical physics FAQ at\n> http://www.mat.univie.ac.at/~neum/physics-faq.txt\n>\n>\n> > 3.If I passed the superpositioned beam through a polarization filter\n> > which instead of being polarized in the horizontal or vertical\n> > direction was polarized at 45 degrees between the horizontal and\n> > vertical would the superposition nature of the beam stay the same\n> > (maybe just diminish in amplitude?)?\n>\n> Yes. You would just get a single beam, since its companion has empty\n> intensity. Actually, you get a superposition of a light beam with\n> a dark beam in the vacuum state (containing no photon); such dark beams\n> may have sometimes unexpected effects (if one is careless).\n> See Ulf Leonhardt\'s nice little booklet \'Measuring the quantum state\n> of light\'. In an ideal filter, the amplitude would be the same,\n> but of course real filters are not 100% ideal.\n>\n> > If a deterministically horizontal\n> > or vertical beam was passed through the filter would it be in a\n> > superposition?\n>\n> It would split into tow entangled beams.\n>\n>\n> > 4.Do you know if it is possible to change the relative weights of the\n> > beam in the state 1/SQ[2](|Vert> + |Hori>) without collapsing any of\n> > the photons? For example, could I make the beam take the state\n> > 1/2|Vert> + SQ[3]/2|Hori> ?\n>\n> Yes, by a rotator. This and lots of other optical filters\n> are discussed in the quantum optics bible by Mandel and Wolf.\n>\n>\n> Arnold Neumaier\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>Thanks, both, for the response, I'm currenlty checking out the
references. I just wanted to make sure I understood the responses
correctly. I'm not sure if the answers given were assuming that all
measurements/filters *were in regard to the same basis*, that is, I
don't want to rotate the coordinates in which I am measuring (of
course polarizers have to be rotated in regards to the original basis,
I'm talking about the final outcome state). the numbers refer to my
previous questions.

So again, Yes/No (+ elaboration if you are feeling generous) would be
good:

1. So if I had a beam with intensity A in the state 1/SQ[2](|Vert>+|Hori>), and incident that on a POLARIZING beam splitter (not just a
regular BS) in one direction I would get a beam with intensity A/2 in
the state 1/SQ[2](|Vert> +|Hori>) and in the other direction I would
get a beam with intensity A/2 in the state 1/SQ[2](|Vert> +|Hori>)
[the same states!]? Of course if I had a beam in either all |Hori> or
all |Vert> the beam would go in only one direction, no?

4. If I passed my beam in the state 1/SQ[2](|Vert> +|Hori>) through a
polarizing rotator meant to rotate polarization 45 degrees, I would
then have a beam with photons in either completely the |Vert> state or
completely in the |Hori> state *in the original polarization basis*?
But if I passed a beam with know polarization with either |Vert> or
|Hori> polarization into this same filter I would then have a beam in
the state 1/SQ[2](|Vert> +|Hori>) in the original basis ?

Again, thanks,
Hayden

PS sorry about harping about the "orignal basis" but this has been
confusing to me in what I've been reading.

Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<410A7A8D.1010300@univie.ac.at>...
> Hayden McGuinness wrote:
> >
> > I have four related questions concerning quantum optics and
> > measurement, they can be answered YES/NO, and any answer would be
> > good, but a little elaboration would be nice, if necessary.
> >
> > 1.If I had a beam where the photons where in a superposition state of
> > vertical and horizontal polarization, the state 1/SQ[2](|Vert> +
> > |Hori>), and incident that beam on a polarizing beam splitter would
> > that definitely collapse the beam, so that all photons in the beam,
> > whatever direction they happen to go, are no longer in a superposition
> > state?
>
> No. A beam splitter is a unitary filter; there is no collapse.
> A collapse happens only where the photons are destroyed.
>
>
> > 2. If #1 is no, is there any device that would preserve this
> > superposition when the before mentioned beam is incident on it yet
> > deterministically differentiates (such as makes them go a different
> > direction) when a beam of known polarization is incident on it?
>
> In principle you can transform the beam into a pair of entangled beams,
> one in the state |vert> the other in the state |hori>.
> This is analogous to the Stern-Gerlach experiment that I discussed
> in my theoretical physics FAQ at
> http://www.mat.univie.ac.at/~neum/physics-faq.txt
>
>
> > 3.If I passed the superpositioned beam through a polarization filter
> > which instead of being polarized in the horizontal or vertical
> > direction was polarized at 45 degrees between the horizontal and
> > vertical would the superposition nature of the beam stay the same
> > (maybe just diminish in amplitude?)?
>
> Yes. You would just get a single beam, since its companion has empty
> intensity. Actually, you get a superposition of a light beam with
> a dark beam in the vacuum state (containing no photon); such dark beams
> may have sometimes unexpected effects (if one is careless).
> See Ulf Leonhardt's nice little booklet 'Measuring the quantum state
> of light'. In an ideal filter, the amplitude would be the same,
> but of course real filters are not 100% ideal.
>
> > If a deterministically horizontal
> > or vertical beam was passed through the filter would it be in a
> > superposition?
>
> It would split into tow entangled beams.
>
>
> > 4.Do you know if it is possible to change the relative weights of the
> > beam in the state 1/SQ[2](|Vert> + |Hori>) without collapsing any of
> > the photons? For example, could I make the beam take the state
> > 1/2|Vert> + SQ[3]/2|Hori> ?
>
> Yes, by a rotator. This and lots of other optical filters
> are discussed in the quantum optics bible by Mandel and Wolf.
>
>
> Arnold Neumaier

Franz Heymann

Aug12-04, 08:31 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>\n\n\n"Arkadiusz Jadczyk" <arkREMOVETHIS@ANDTHIScassiopaea.org> wrote in\nmessage news:410ba97d@news.sentex.net...\n>\n> On 30 Jul 2004 12:45:59 -0400, Arnold Neumaier\n> <Arnold.Neumaier@univie.ac.at> wrote:\n>\n> >No. A beam splitter is a unitary filter; there is no collapse.\n> >A collapse happens only where the photons are destroyed.\n>\n> Not necessarily. Photon may be scattered on an electron,\n\nIn the process, it is destroyed and an emergent photon is created.\n\n[snip]\n\nFranz\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>"Arkadiusz Jadczyk" <arkREMOVETHIS@ANDTHIScassiopaea.org> wrote in
message news:410ba97d@news.sentex.net...
>
> On 30 Jul 2004 12:45:59 -0400, Arnold Neumaier
> <Arnold.Neumaier@univie.ac.at> wrote:
>
> >No. A beam splitter is a unitary filter; there is no collapse.
> >A collapse happens only where the photons are destroyed.
>
> Not necessarily. Photon may be scattered on an electron,

In the process, it is destroyed and an emergent photon is created.

[snip]

Franz

Arnold Neumaier

Aug12-04, 12:16 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>\nHayden McGuinness wrote:\n> 1. So if I had a beam with intensity A in the state 1/SQ[2](|Vert>\n> +|Hori>), and incident that on a POLARIZING beam splitter (not just a\n> regular BS) in one direction I would get a beam with intensity A/2 in\n> the state 1/SQ[2](|Vert> +|Hori>) and in the other direction I would\n> get a beam with intensity A/2 in the state 1/SQ[2](|Vert> +|Hori>)\n> [the same states!]?\n\nIn a symmetric situation, you\'d get the same intensity and polarization.\nBut not the same states! The state codes not only for the polarization\nbut also the direction. (The actual state is here a tensor product\nof the polarization state and the state defining\nthe photon\'s momentum. of course, people often ignore the factor for\nthe direction, since it is given by the context and does not enter\nmost calculations. But when working at high precision, one needs to\ntake account of the fact that a beam fans out, and the description gets\nsignificantly more involved...)\n\nBut polarization in a beam splitting material is not necessarily symmetric.\n\nWhat you get depends on the detailed properties of the filter you apply.\nIn principle, one can make arrangements that do any unitary linear\ntransformation on the pair of beams. So you can transform psi\n(2 component vector, single beam) into A_1 psi_1 in one direction and\nA_2 psi_2 in the other, provided the A_i are 2x2 matrices satisfying\nA_1^* A_1 + A_2^* A_2 = \\1, the identity matrix, which is required\nbecause of unitarity. Creating a filter of the right sort given the A_i\nby combining standard components with simple behavior is a useful and not\nentirely trivial exercise in quantum optics, that I recommend you to do.\n\n\n> PS sorry about harping about the "orignal basis" but this has been\n> confusing to me in what I\'ve been reading.\n\nOf course, everything can (and should) be done in any _fixed_ basis.\n\n\nArnold Neumaier\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>Hayden McGuinness wrote:
> 1. So if I had a beam with intensity A in the state 1/SQ[2](|Vert>
> +|Hori>), and incident that on a POLARIZING beam splitter (not just a
> regular BS) in one direction I would get a beam with intensity A/2 in
> the state 1/SQ[2](|Vert> +|Hori>) and in the other direction I would
> get a beam with intensity A/2 in the state 1/SQ[2](|Vert> +|Hori>)
> [the same states!]?

In a symmetric situation, you'd get the same intensity and polarization.
But not the same states! The state codes not only for the polarization
but also the direction. (The actual state is here a tensor product
of the polarization state and the state defining
the photon's momentum. of course, people often ignore the factor for
the direction, since it is given by the context and does not enter
most calculations. But when working at high precision, one needs to
take account of the fact that a beam fans out, and the description gets
significantly more involved...)

But polarization in a beam splitting material is not necessarily symmetric.

What you get depends on the detailed properties of the filter you apply.
In principle, one can make arrangements that do any unitary linear
transformation on the pair of beams. So you can transform \psi
(2 component vector, single beam) into A_1 \psi_1 in one direction and
A_2 \psi_2 in the other, provided the A_i are 2x2 matrices satisfying
A_1^* A_1 + A_2^* A_2 = \1, the identity matrix, which is required
because of unitarity. Creating a filter of the right sort given the A_i
by combining standard components with simple behavior is a useful and not
entirely trivial exercise in quantum optics, that I recommend you to do.

> PS sorry about harping about the "orignal basis" but this has been
> confusing to me in what I've been reading.

Of course, everything can (and should) be done in any _fixed_ basis.

Arnold Neumaier

Arkadiusz Jadczyk

Aug16-04, 12:55 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\n\nOn 12 Aug 2004 09:29:40 -0400, dralexgreen@yahoo.co.uk (Alex Green)\nwrote:\n\n>Please could you explain in what sense you mean \'nothing ever happens\'\n>in QM? Nothing happens in GR either if we consider time to be part of\n>a space time continuum (everything has already happened).\n\nIn GR when particle trajectory intersects a space like hyper surface -\nwe get an "event". Space like hyper surface can represent, for instance\nactivated photographic plate.\n\nIn GR we do not need to take into account the action of the measurement\napparatus on the measured system. In quantum theory we need to, but the\ndynamics is lacking, as there is no such thing as a "measurement\napparatus" in quantum mechanics. The only thing that is allowed to exist\nin standard quantum mechanics is "the whole system". And nothing EVER\nhappens to the whole system. Standard Quantum mechanics has no\nmechanisms to dynamically describe events. Standard quantum mechanics\nhas no mechanism to answer the most simple experimental question: given\na photographic plate, what is the probability the particle, described by\na given wave function, will hit the plate at time t? It is easy to\nmeasure, it is impossible to calculate from standard quantum mechanics\n(though it is possible to calculate from Bohmian theory or GRW or couple\nof other extensions of qm)\n\nP.S. Of course physicists will use tricks to replace this simple\nquestion by other questions. These tricks sometimes work, but most often\nthey do not work. See Mielnik\'s paper:\n\nMielnik, B.: "The Screen Problem", Found. Phys.\n24 (1994) 1113--1129\n\nark\n--\n\nArkadiusz Jadczyk\nhttp://quantumfuture.net/quantum_future/jadpub.htm\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>On 12 Aug 2004 09:29:40 -0400, dralexgreen@yahoo.co.uk (Alex Green)
wrote:

>Please could you explain in what sense you mean 'nothing ever happens'
>in QM? Nothing happens in GR either if we consider time to be part of
>a space time continuum (everything has already happened).

In GR when particle trajectory intersects a space like hyper surface -
we get an "event". Space like hyper surface can represent, for instance
activated photographic plate.

In GR we do not need to take into account the action of the measurement
apparatus on the measured system. In quantum theory we need to, but the
dynamics is lacking, as there is no such thing as a "measurement
apparatus" in quantum mechanics. The only thing that is allowed to exist
in standard quantum mechanics is "the whole system". And nothing EVER
happens to the whole system. Standard Quantum mechanics has no
mechanisms to dynamically describe events. Standard quantum mechanics
has no mechanism to answer the most simple experimental question: given
a photographic plate, what is the probability the particle, described by
a given wave function, will hit the plate at time t? It is easy to
measure, it is impossible to calculate from standard quantum mechanics
(though it is possible to calculate from Bohmian theory or GRW or couple
of other extensions of qm)

P.S. Of course physicists will use tricks to replace this simple
question by other questions. These tricks sometimes work, but most often
they do not work. See Mielnik's paper:

Mielnik, B.: "The Screen Problem", Found. Phys.
24 (1994) 1113--1129

ark
--

Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
--