QM/EPR/relativity [Archive]

daniel@elit.net

Sep15-05, 12:41 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>Consider an EPR experimental set up, with a singlet state of two spin\n1/2 particles. Measure the spin of particle 1 along the z-axis, and\nlet\'s suppose it is spin up. Proceed as follows:\n\n1) Measure the spin of particle 1 along the z-axis (again). You\'ll get\nspin up for certain.\n2) Then measure the spin of particle 2 along the x-axis. You\'ll get\nspin up or spin down with equal probability.\n\nSet up a new experiment. Do everything as above, except reverse the\norder of 1) and 2). For particle 2, you\'ll get the same probabilities.\nFor particle 1, you\'ll get spin up or spin down with equal probability.\n\nWe can change the order of things by a Lorentz boost. For example, we\ncould set it up so that observer A is in the rest frame of particle 1\nand observer B is in the rest frame of particle 2, and BOTH A and B\nthink THEY are the ones making the first measurement. Leading to a\ncontradiction. How is it all resolved?\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>Consider an EPR experimental set up, with a singlet state of two spin
1/2 particles. Measure the spin of particle 1 along the z-axis, and
let's suppose it is spin up. Proceed as follows:

1) Measure the spin of particle 1 along the z-axis (again). You'll get
spin up for certain.
2) Then measure the spin of particle 2 along the x-axis. You'll get
spin up or spin down with equal probability.

Set up a new experiment. Do everything as above, except reverse the
order of 1) and 2). For particle 2, you'll get the same probabilities.
For particle 1, you'll get spin up or spin down with equal probability.

We can change the order of things by a Lorentz boost. For example, we
could set it up so that observer A is in the rest frame of particle 1
and observer B is in the rest frame of particle 2, and BOTH A and B
think THEY are the ones making the first measurement. Leading to a
contradiction. How is it all resolved?

Ilja Schmelzer

Sep15-05, 09:03 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<daniel@elit.net> schrieb\n> Consider an EPR experimental set up, with a singlet state of two spin\n> 1/2 particles. Measure the spin of particle 1 along the z-axis, and\n> let\'s suppose it is spin up. Proceed as follows:\n>\n> 1) Measure the spin of particle 1 along the z-axis (again). You\'ll get\n> spin up for certain.\n> 2) Then measure the spin of particle 2 along the x-axis. You\'ll get\n> spin up or spin down with equal probability.\n>\n> Set up a new experiment. Do everything as above, except reverse the\n> order of 1) and 2). For particle 2, you\'ll get the same probabilities.\n> For particle 1, you\'ll get spin up or spin down with equal probability.\n\nNo. You\'ll get spin up for certain. The EPR state between the\ntwo particles is already destroyed by the first measurement of particle 1.\n\nIlja\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><daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

Andreas Most

Sep17-05, 11:15 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>daniel@elit.net wrote:\n> Consider an EPR experimental set up, with a singlet state of two spin\n> 1/2 particles. Measure the spin of particle 1 along the z-axis, and\n> let\'s suppose it is spin up. Proceed as follows:\n>\n> 1) Measure the spin of particle 1 along the z-axis (again). You\'ll get\n> spin up for certain.\n> 2) Then measure the spin of particle 2 along the x-axis. You\'ll get\n> spin up or spin down with equal probability.\n>\n> Set up a new experiment. Do everything as above, except reverse the\n> order of 1) and 2). For particle 2, you\'ll get the same probabilities.\n> For particle 1, you\'ll get spin up or spin down with equal probability.\n\nNo, you\'ll get spin up for particle 1 along the z-axis regardless of when\nor how you perform the measurement on particle 2.\nQM does not state that one measurement changes the measurement on the\nother particle. Instead, QM tells you that the measurements on the two\nparticles must be consistent.\n\nIf you think of the QM wave function as the set of information you\npossess about a system, it is clear that with the knowledge of\nthe state of one particle in EPR you can deduce the state of the\nsecond particle. But there is no exchange of "information" between the\ntwo particles after you performed a measurement on one particle.\n\n>\n> We can change the order of things by a Lorentz boost. For example, we\n> could set it up so that observer A is in the rest frame of particle 1\n> and observer B is in the rest frame of particle 2, and BOTH A and B\n> think THEY are the ones making the first measurement. Leading to a\n> contradiction. How is it all resolved?\n>\n\nThere is no contradiction. Einstein said that already when the EPR paper\nwas published.\n\nRegards, Andreas.\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>daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

Ralph Hartley

Sep17-05, 01:09 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>Is there a FAQ entry for this? We see it all the time.\n\ndaniel@elit.net wrote:\n> Consider an EPR experimental set up, with a singlet state of two spin\n> 1/2 particles. Measure the spin of particle 1 along the z-axis, and\n> let\'s suppose it is spin up. Proceed as follows:\n>\n> 1) Measure the spin of particle 1 along the z-axis (again). You\'ll get\n> spin up for certain.\n> 2) Then measure the spin of particle 2 along the x-axis. You\'ll get\n> spin up or spin down with equal probability.\n>\n> Set up a new experiment. Do everything as above, except reverse the\n> order of 1) and 2). For particle 2, you\'ll get the same probabilities.\n> For particle 1, you\'ll get spin up or spin down with equal probability.\n\nNo, you get spin up with probability 1.\n\nLets do the calculation. Initial state (ignoring factors of sqrt(2)):\n\n|Up>|Down> - |Down>|Up>\n\nafter the first measurement:\n\n|Up>|Down>\n\nWe want to measure the second particle in a different direction, so we\nwill use\n\n|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))\n|Down> = |Left> - |Right> (They can take care of themselves)\nand\n|Left> = |Up> + |Down>\n|Right> = |Up> - |Down>\n\nto re-write the state as:\n\n|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>\n\nAfter the second measurement it is either |Up>|Left> or |Up>|Right>. In\nboth cases measuring the first particle again gives up.\n\nLets try again without the first measurement:\n\n|Up>|Down> - |Down>|Up>\n= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)\n= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>\n= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>\n= |Right>|Left> - |Left>|Right>\n\nMeasuring the second particle makes the state either\n|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the\nz axis) *does* give 50-50. Of course, if we had made that measurement\nfirst it still would be 50-50. So it was the extra measurement at the\nstart that made the difference.\n\nEntanglement is a one time thing. Once you do a measurement on either\nparticle, you don\'t have an entangled pair any more, you have a pair of\nparticles, both in known states.\n\nYou might think that the loss of entanglement itself could be used to\nsend a signal. It can\'t, because entanglement cannot be detected\nlocally. To tell if the particles are still entangled you have to\ncompare measurements. You can\'t do that instantaneously unless you can\n*already* send an instantaneous signal, by some other means.\n\nRalph Hartley\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>Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of \sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of \sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley