alistair
May28-04, 03:06 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>\nIn quantum mechanics a particle can exist in a superposition of two\nstates ( eg spin +1/2 and spin – 1/2) and all observers would agree on\nwhat that state is.\nCan a superposition of states exist in relativity theory? If there\nare two observers moving at different speeds with respect to a mass\neach observer would give a unique value for the mass, so the mass has\nneither one value nor the other.\nIn quantum mechanics a superposition can be destroyed by a\nmeasurement.\nBut if two observers in relativity measure the mass of an object are\nthey creating something analagous to a superposition of states by\ndefining\nthe mass with two unique values simultaneously ( presumably no\nsuperposition would exist until two measurements of the mass have been\nmade)?\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 quantum mechanics a particle can exist in a superposition of two
states ( eg spin +1/2 and spin – 1/2) and all observers would agree on
what that state is.
Can a superposition of states exist in relativity theory? If there
are two observers moving at different speeds with respect to a mass
each observer would give a unique value for the mass, so the mass has
neither one value nor the other.
In quantum mechanics a superposition can be destroyed by a
measurement.
But if two observers in relativity measure the mass of an object are
they creating something analagous to a superposition of states by
defining
the mass with two unique values simultaneously ( presumably no
superposition would exist until two measurements of the mass have been
made)?
Arkadiusz Jadczyk
May29-04, 11:59 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>On 28 May 2004 16:06:28 -0400, alistair@goforit64.fsnet.co.uk (alistair)\nwrote:\n\n>But if two observers in relativity measure the mass of an object are\n>they creating something analagous to a superposition of states by\n>defining\n>the mass with two unique values simultaneously ( presumably no\n>superposition would exist until two measurements of the mass have been\n>made)?\n\nMeasurements, as a rule, would create mixtures rather than\nsuperpositions.\n\nark\n--\n\nArkadiusz Jadczyk\nhttp://www.cassiopaea.org/quantum_future/homepage.htm\n\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 28 May 2004 16:06:28 -0400, alistair@goforit64.fsnet.co.uk (alistair)
wrote:
>But if two observers in relativity measure the mass of an object are
>they creating something analagous to a superposition of states by
>defining
>the mass with two unique values simultaneously ( presumably no
>superposition would exist until two measurements of the mass have been
>made)?
Measurements, as a rule, would create mixtures rather than
superpositions.
ark
--
Arkadiusz Jadczyk
http://www.cassiopaea.org/quantum_future/homepage.htm
--
Greg Egan
May31-04, 06:26 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 <861c1b21.0405281149.2c70f6f3@posting.google.com>, \nalistair@goforit64.fsnet.co.uk (alistair) wrote:\n\n[snip]\n> if two observers in relativity measure the mass of an object are\n> they creating something analagous to a superposition of states by\n> defining the mass with two unique values simultaneously ( presumably no\n> superposition would exist until two measurements of the mass have been\n> made)?\n\nI don\'t believe it\'s useful to see this as analogous to a superposition\nin QM.\n\nIn relativity, when two observers give different numerical results for\nsome physical quantity, it\'s usually because they are measuring one\ncomponent of a 4-dimensional vector, while using different coordinate\nsystems. The mass of an object (in the sense you\'re using the word) is\nthe time component of its energy-momentum 4-vector p, and hence its\nnumerical value will depend on the coordinate system. (However, the\nsquared length of the energy-momentum 4-vector, g(p,p), is -m^2, where m\nis the rest mass, and all observers will agree on that; g here is the\nMinkowskian metric.)\n\nIf you want an analogy for this observer dependance in relativity that\nreally sheds some light on it, the best thing to do is to look at 3-d or\n2-d space rather than spacetime, where the analogous effects become pure\ncommon sense. If Alice and Bob are facing in non-parallel directions,\nbut insist on measuring things in personal coordinate frames aligned to\ntheir bodies (i.e. with left-right and forward-back axes), then they will\ngenerally disagree about the "left-right distance" x between two points,\nand also the "forward-back distance" y between those points. This is\nvery closely analogous to the fact that observers with different\nvelocities disagree on the size of the timelike and spacelike components\nof a 4-vector. However, Alice and Bob will agree on the value of the\noverall distance between any two points, sqrt(x^2+y^2), just as any two\nobservers will agree on the length of a 4-vector.\n\nIn both these situations, though, pairs of observers who are sufficiently\nthoughtful will find that they don\'t disagree on anything substantial.\nThere are ways of talking about vectors in which not just their length,\nbut the vectors themselves, are seen to be independent of the coordinate\nsystem. For example, you can think of an object\'s 4-velocity in\nrelativity as a differential operator defined along the object\'s world\nline, which takes derivatives of functions on space-time with respect to\ntime shown on a clock that travels with the object (i.e. the object\'s\nproper time, tau). The individual components (t,x,y,z) of the 4-velocity\nvector in a given coordinate system are then the derivatives with respect\nto tau of t,x,y,z as functions on spacetime. The whole 4-vector is\ndefined in an observer-independent way (referring only to the object\nitself), but if you feed the differential operator different coordinate\nfunctions, of course it will give you different results.\n\nIn QM, it\'s true that there are some things you can say about the state\nvector of the system that will depend on the coordinate basis used, and\nothers that are basis-independent. But one of the things that depends on\nyour choice of basis is whether or not a given state is a "superposition"\nat all; for example, a state representing "an electron with a single,\ndefinite value for its z-axis spin of +1/2" will be a superposition of\nspin +1/2 and spin -1/2 states measured along the x- or y-axes. This\njust follows from the fact that the state vector, which lies in a\ntwo-dimensional complex vector space, can be described either with a\nbasis in which it has two non-zero components (the x and y cases), or\nwith a basis where one of the basis vectors is the state vector itself,\nwith respect to which it obviously has only one non-zero component. As\nlong as you have more than 1 dimension, *every* state vector is "a\nsuperposition" relative to some bases, while also being "not a\nsuperposition" relative to other bases.\n\nThe analogous situation in 2-d Euclidean space is when Alice says two\npoints are separated in both her left-right and forward-back directions,\nwhereas Bob says the separation is "purely left-right".\n\nGreg Egan\n\nEmail address (remove name of animal and add standard punctuation):\ngregegan netspace zebra net au\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 <861c1b21.0405281149.2c70f6f3@posting.google.com>,
alistair@goforit64.fsnet.co.uk (alistair) wrote:
[snip]
> if two observers in relativity measure the mass of an object are
> they creating something analagous to a superposition of states by
> defining the mass with two unique values simultaneously ( presumably no
> superposition would exist until two measurements of the mass have been
> made)?
I don't believe it's useful to see this as analogous to a superposition
in QM.
In relativity, when two observers give different numerical results for
some physical quantity, it's usually because they are measuring one
component of a 4-dimensional vector, while using different coordinate
systems. The mass of an object (in the sense you're using the word) is
the time component of its energy-momentum 4-vector p, and hence its
numerical value will depend on the coordinate system. (However, the
squared length of the energy-momentum 4-vector, g(p,p), is -m^2, where m
is the rest mass, and all observers will agree on that; g here is the
Minkowskian metric.)
If you want an analogy for this observer dependance in relativity that
really sheds some light on it, the best thing to do is to look at 3-d or
2-d space rather than spacetime, where the analogous effects become pure
common sense. If Alice and Bob are facing in non-parallel directions,
but insist on measuring things in personal coordinate frames aligned to
their bodies (i.e. with left-right and forward-back axes), then they will
generally disagree about the "left-right distance" x between two points,
and also the "forward-back distance" y between those points. This is
very closely analogous to the fact that observers with different
velocities disagree on the size of the timelike and spacelike components
of a 4-vector. However, Alice and Bob will agree on the value of the
overall distance between any two points, \sqrt(x^2+y^2), just as any two
observers will agree on the length of a 4-vector.
In both these situations, though, pairs of observers who are sufficiently
thoughtful will find that they don't disagree on anything substantial.
There are ways of talking about vectors in which not just their length,
but the vectors themselves, are seen to be independent of the coordinate
system. For example, you can think of an object's 4-velocity in
relativity as a differential operator defined along the object's world
line, which takes derivatives of functions on space-time with respect to
time shown on a clock that travels with the object (i.e. the object's
proper time, \tau). The individual components (t,x,y,z) of the 4-velocity
vector in a given coordinate system are then the derivatives with respect
to \tau of t,x,y,z as functions on spacetime. The whole 4-vector is
defined in an observer-independent way (referring only to the object
itself), but if you feed the differential operator different coordinate
functions, of course it will give you different results.
In QM, it's true that there are some things you can say about the state
vector of the system that will depend on the coordinate basis used, and
others that are basis-independent. But one of the things that depends on
your choice of basis is whether or not a given state is a "superposition"
at all; for example, a state representing "an electron with a single,
definite value for its z-axis spin of +1/2" will be a superposition of
spin +1/2 and spin -1/2 states measured along the x- or y-axes. This
just follows from the fact that the state vector, which lies in a
two-dimensional complex vector space, can be described either with a
basis in which it has two non-zero components (the x and y cases), or
with a basis where one of the basis vectors is the state vector itself,
with respect to which it obviously has only one non-zero component. As
long as you have more than 1 dimension, *every* state vector is "a
superposition" relative to some bases, while also being "not a
superposition" relative to other bases.
The analogous situation in 2-d Euclidean space is when Alice says two
points are separated in both her left-right and forward-back directions,
whereas Bob says the separation is "purely left-right".
Greg Egan
Email address (remove name of animal and add standard punctuation):
gregegan netspace zebra net au
alistair
Jun12-04, 07:09 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>When a quantum mechanical measurement is made it is assumed that the\nmeasurement process is a one way process.But do measured masses make\ntheir own measurement of the measuring apparatus - can there be\nfeedback between the measuring apparatus and the measured mass?\nFor example if a photon is in one of two polarization states,\nand its polarization is determined,will this change the state of the\nwavefunction of the measuring apparatus?\nAnd is there some quantity of the system as a whole that remains\nconstant?\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>When a quantum mechanical measurement is made it is assumed that the
measurement process is a one way process.But do measured masses make
their own measurement of the measuring apparatus - can there be
feedback between the measuring apparatus and the measured mass?
For example if a photon is in one of two polarization states,
and its polarization is determined,will this change the state of the
wavefunction of the measuring apparatus?
And is there some quantity of the system as a whole that remains
constant?
Rahul Jain
Jun16-04, 03:43 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>alistair@goforit64.fsnet.co.uk (alistair) writes:\n\n> When a quantum mechanical measurement is made it is assumed that the\n> measurement process is a one way process.But do measured masses make\n> their own measurement of the measuring apparatus - can there be\n> feedback between the measuring apparatus and the measured mass?\n> For example if a photon is in one of two polarization states,\n> and its polarization is determined,will this change the state of the\n> wavefunction of the measuring apparatus?\n\nOf course. If the wavefunction of the measuring apparatus was unchanged,\nthen it didn\'t measure anything. As far as I understand, the\nwavefunction is what describes the total state of the object. Therefore,\na measurement must result in a change in the state, because otherwise\nthere wouldn\'t be any measurement registered...\n\n> And is there some quantity of the system as a whole that remains\n> constant?\n\nSure, those that are covered by the various conservation laws.\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>alistair@goforit64.fsnet.co.uk (alistair) writes:
> When a quantum mechanical measurement is made it is assumed that the
> measurement process is a one way process.But do measured masses make
> their own measurement of the measuring apparatus - can there be
> feedback between the measuring apparatus and the measured mass?
> For example if a photon is in one of two polarization states,
> and its polarization is determined,will this change the state of the
> wavefunction of the measuring apparatus?
Of course. If the wavefunction of the measuring apparatus was unchanged,
then it didn't measure anything. As far as I understand, the
wavefunction is what describes the total state of the object. Therefore,
a measurement must result in a change in the state, because otherwise
there wouldn't be any measurement registered...
> And is there some quantity of the system as a whole that remains
> constant?
Sure, those that are covered by the various conservation laws.
--
Rahul Jain
rjain@nyct.net
Professional Software Developer, Amateur Quantum Mechanicist
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