View Full Version : Teleportation of photons using entanglement
Sci~Girl
Mar3-05, 04:25 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 howstuffworks.com... article on photon teleportation...\nhttp://travel.howstuffworks.com/teleportation2.htm\n\nThe article only goes into a basic description of how entanglement\nworks, and I wasn\'t able to find a more detailed one anywhere else. I\nwas wondering if anyone knew of any other (accurate) sites that give a\ndetailed description of the method. I\'m writing a research paper on it\nfor an independent study project.\n\nThe article also mentions teleportation of a laser beam, and\ninformation on that would be helpful too.\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 howstuffworks.com... article on photon teleportation...
http://travel.howstuffworks.com/teleportation2.htm
The article only goes into a basic description of how entanglement
works, and I wasn't able to find a more detailed one anywhere else. I
was wondering if anyone knew of any other (accurate) sites that give a
detailed description of the method. I'm writing a research paper on it
for an independent study project.
The article also mentions teleportation of a laser beam, and
information on that would be helpful too.
George Jones
Mar6-05, 01:51 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>Sci~Girl wrote:\n> On howstuffworks.com... article on photon teleportation...\n> http://travel.howstuffworks.com/teleportation2.htm\n>\n> The article only goes into a basic description of how entanglement\n> works, and I wasn\'t able to find a more detailed one anywhere else. I\n> was wondering if anyone knew of any other (accurate) sites that give a\n> detailed description of the method. I\'m writing a research paper on it\n> for an independent study project.\n>\n> The article also mentions teleportation of a laser beam, and\n> information on that would be helpful too.\n\nI\'ll give the theory behind an experiemnt described in the 11 December\n1997 issue of Nature. Maybe what I say won\'t be very helpful. In this\nissue, Nature has a technical but readable article describing the\nexperiment and a non-technical article describing quantum entanglement.\n\nConsider photons that are either horizontally (h) or vertically (v)\nlinearly polarized, so that state space is 2-dimensional, and suppose\nthat an EPR apparatus creates photons A and B which are in the entangled\nstate\n\n|s> = 2^(-1/2)(|Ah>|Bv>-|Av>|Bh>),\n\nand that A is sent to some experimental apparatus, while B\npropagates freely. Suppose further that an arbitary photon, call it P,\nwith arbitrary state\n\n|P> = a|Ph> + b|Pv>, a^2 + b^2 = 1\n\narrives (from anywhere) at the experiment apparatus at the same time in\nand the same place as photon A. The state of the 3-photon system is thus\n\n|u> = |P>|s>\n= (a/2)|Ph>|Ah>|Bv> - (a/2)|Ph>|Av>|Bh> + (b/2)|Pv|Ah>|Bv>\n- (b/2)|Pv>|Av>|Bh>.\n\nThe experimental apparatus then effectively performs a quantum\nmeasurement on the photons P and A in such that they end up in the\nentangled state\n\n|q> = 2^(-1/2)(|Ph>|Av>-|Pv>|Ah>).\n\nBut before the measurement, A and B were in an entangled state |s>, so\nthis can\'t be done without affecting photon B, even though B doesn\'t\ninteract physically with the apparatus.\n\nThe effect of the measurement is represented mathematically by acting on\nthe first two components (P an A) of each term of the initial 3 photon\nstate |u> with the projection operator formed from the final combined\n(and entangled) state |q> of P and A\n\nProj = |q><q|\n\n= |Ph>|Av>(<Ph|<Av| - <Pv|<Ah|) - |Pv>|Ah>(<Ph|<Av| - <Pv|<Ah|),\n\nwhile operating on the third component (the B part) of each term of |u> by\nthe identity operator 1, since no measurement was performed on B.\n\nAfter some algebra, one arrives at\n\n(Proj x 1)|u> = -(a/2)|Ph>|Av>|Bh> + (a/2)|Pv>|Ah>|Bh>\n- (b/2)|Ph>|Av>|Bv> + (b/2)|Pv>|Ah>|Bv>,\n\nwhich factors into\n\n(1/2)(-|Ph>|Av> + |Pv>|Ah>)(a|Bh>+b|Bv>)\n\nShazzam! Photon B, which, after creation, never came anywhere near the\nexperimental apparatus, now has exactly the state that the arbitrary\nphoton P had before the measurement (the interactions of photons P and\nA with the apparatus). In effect the identity of photon P has been\nteleported to photon B! Photon P as single entity has been "destroyed",\nas it is now completely entangled with A.\n\nUnfortunately, the measurement that created state |s> has 3 other\nequally probable outcomes, sot this teleportation only happens 25% of\nthe time.\n\nRegards,\nGeorge\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>Sci~Girl wrote:
> On howstuffworks.com... article on photon teleportation...
> http://travel.howstuffworks.com/teleportation2.htm
>
> The article only goes into a basic description of how entanglement
> works, and I wasn't able to find a more detailed one anywhere else. I
> was wondering if anyone knew of any other (accurate) sites that give a
> detailed description of the method. I'm writing a research paper on it
> for an independent study project.
>
> The article also mentions teleportation of a laser beam, and
> information on that would be helpful too.
I'll give the theory behind an experiemnt described in the 11 December
1997 issue of Nature. Maybe what I say won't be very helpful. In this
issue, Nature has a technical but readable article describing the
experiment and a non-technical article describing quantum entanglement.
Consider photons that are either horizontally (h) or vertically (v)
linearly polarized, so that state space is 2-dimensional, and suppose
that an EPR apparatus creates photons A and B which are in the entangled
state
|s> = 2^(-1/2)(|Ah>|Bv>-|Av>|Bh>),
and that A is sent to some experimental apparatus, while B
propagates freely. Suppose further that an arbitary photon, call it P,
with arbitrary state
|P> =[/itex] a|Ph> + b|Pv>, a^2 + b^2 = 1
arrives (from anywhere) at the experiment apparatus at the same time in
and the same place as photon A. The state of the 3-photon system is thus
|u> = |P>|s>= (a/2)|Ph>|Ah>|Bv> - (a/2)|Ph>|Av>|Bh> + (b/2)|Pv|Ah>|Bv>- (b/2)|Pv>|Av>|Bh>.
The experimental apparatus then effectively performs a quantum
measurement on the photons P and A in such that they end up in the
entangled state
|q> = 2^(-1/2)(|Ph>|Av>-|Pv>|Ah>).
But before the measurement, A and B were in an entangled state |s>, so
this can't be done without affecting photon B, even though B doesn't
interact physically with the apparatus.
The effect of the measurement is represented mathematically by acting on
the first two components (P an A) of each term of the initial 3 photon
state |u> with the projection operator formed from the final combined
(and entangled) state |q> of P and A
Proj [itex]= |q><q|= |Ph>|Av>(<Ph|<Av| - <Pv|<Ah|) - |Pv>|Ah>(<Ph|<Av| - <Pv|<Ah|),
while operating on the third component (the B part) of each term of |u> by
the identity operator 1, since no measurement was performed on B.
After some algebra, one arrives at
(Proj x 1)|u> = -(a/2)|Ph>|Av>|Bh> + (a/2)|Pv>|Ah>|Bh>- (b/2)|Ph>|Av>|Bv> + (b/2)|Pv>|Ah>|Bv>,
which factors into
(1/2)(-|Ph>|Av> + |Pv>|Ah>)(a|Bh>+b|Bv>)
Shazzam! Photon B, which, after creation, never came anywhere near the
experimental apparatus, now has exactly the state that the arbitrary
photon P had before the measurement (the interactions of photons P and
A with the apparatus). In effect the identity of photon P has been
teleported to photon B! Photon P as single entity has been "destroyed",
as it is now completely entangled with A.
Unfortunately, the measurement that created state |s> has 3 other
equally probable outcomes, sot this teleportation only happens 25% of
the time.
Regards,
George
Sci~Girl
Mar7-05, 03:49 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>It helps a little, but I\'m really looking for what "entangled" refers\nto. What does it mean to have photons "entangled?"\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>It helps a little, but I'm really looking for what "entangled" refers
to. What does it mean to have photons "entangled?"
Arnold Neumaier
Mar8-05, 12:48 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>Sci~Girl wrote:\n> It helps a little, but I\'m really looking for what "entangled" refers\n> to. What does it mean to have photons "entangled?"\n>\n\nThat the joint wave function of a a pair of photons is not the\ntensor product of two single-particle wave functions.\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>Sci~Girl wrote:
> It helps a little, but I'm really looking for what "entangled" refers
> to. What does it mean to have photons "entangled?"
>
That the joint wave function of a a pair of photons is not the
tensor product of two single-particle wave functions.
Arnold Neumaier
Igor Khavkine
Mar9-05, 02:01 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 Tue, 08 Mar 2005 06:48:11 +0000, Arnold Neumaier wrote:\n\n> Sci~Girl wrote:\n>> It helps a little, but I\'m really looking for what "entangled" refers\n>> to. What does it mean to have photons "entangled?"\n>>\n>>\n> That the joint wave function of a a pair of photons is not the tensor\n> product of two single-particle wave functions.\n\nThat is correct and very succinct. But, perhaps a hands on demonstration\nwould be helpful.\n\nA state of a particle is described by a wave function. We can represent it\nby a function of one variable, say x. If we have two particles, then their\nstate will be described by wave function of two variables, say x and y,\none for each particle.\n\nWhen two particles are independent, their wave function is the product of\ntwo individual wave functions:\n\npsi_togehter(x,y) = psi_single1(x) psi_single2(y).\n\nHere\'s a simple fact: not every possible function of two variables\npsi_together(x,y) can be written as a product of two individual functions\npsi_single1(x) and psi_single1(y). A simple example is\n\npsi_together(x,y) = x^2 + y^2.\n\nReal wave functions look very different from the above form, but it is\nsufficient to demonstrate the point.\n\nWe call a state of two particles _entangled_ if the corresponding wave\nfunction psi_together(x,y) cannot be written as a product of two single\nparticle wave functions. The discussion above shows that there in fact are\nentangled states.\n\nIf you are at all familiar with linear algebra, you can try an example\nthat is more realistic. A two component (row or column) vector describes\nthe state of a two level quantum system (such as a spin or a photon). If\nyou have two such systems, the joint state will be described by a two by\ntwo matrix (four components). Can you write every possible 2x2 matrix as a\ntensor product of two 2-component tensors (a column vector times a row\nvector)? If not, then you\'ve just proven that a system of two coupled two\nlevel systems exhibits entangled states.\n\nHope this helps.\n\nIgor\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 Tue, 08 Mar 2005 06:48:11 +0000, Arnold Neumaier wrote:
> Sci~Girl wrote:
>> It helps a little, but I'm really looking for what "entangled" refers
>> to. What does it mean to have photons "entangled?"
>>
>>
> That the joint wave function of a a pair of photons is not the tensor
> product of two single-particle wave functions.
That is correct and very succinct. But, perhaps a hands on demonstration
would be helpful.
A state of a particle is described by a wave function. We can represent it
by a function of one variable, say x. If we have two particles, then their
state will be described by wave function of two variables, say x and y,
one for each particle.
When two particles are independent, their wave function is the product of
two individual wave functions:
\psi_togehter(x,y) = \psi_single1(x) \psi_single2(y).
Here's a simple fact: not every possible function of two variables
\psi_together(x,y) can be written as a product of two individual functions
\psi_single1(x) and \psi_single1(y). A simple example is
\psi_together(x,y) = x^2 + y^2.
Real wave functions look very different from the above form, but it is
sufficient to demonstrate the point.
We call a state of two particles _entangled_ if the corresponding wave
function \psi_together(x,y) cannot be written as a product of two single
particle wave functions. The discussion above shows that there in fact are
entangled states.
If you are at all familiar with linear algebra, you can try an example
that is more realistic. A two component (row or column) vector describes
the state of a two level quantum system (such as a spin or a photon). If
you have two such systems, the joint state will be described by a two by
two matrix (four components). Can you write every possible 2x2 matrix as a
tensor product of two 2-component tensors (a column vector times a row
vector)? If not, then you've just proven that a system of two coupled two
level systems exhibits entangled states.
Hope this helps.
Igor
Sci~Girl
Mar15-05, 12:21 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>\nWe call a state of two particles _entangled_ if the corresponding wave\nfunction psi_together(x,y) cannot be written as a product of two single\n\nparticle wave functions. The discussion above shows that there in fact\nare\nentangled states.\n\n\nIf you are at all familiar with linear algebra, you can try an example\nthat is more realistic. A two component (row or column) vector\ndescribes\nthe state of a two level quantum system (such as a spin or a photon).\nIf\nyou have two such systems, the joint state will be described by a two\nby\ntwo matrix (four components). Can you write every possible 2x2 matrix\nas a\ntensor product of two 2-component tensors (a column vector times a row\nvector)? If not, then you\'ve just proven that a system of two coupled\ntwo\nlevel systems exhibits entangled states.\n\n\nHope this helps.\n\n\nYes, it does, thanx.\nAll I know in math is basic algebra though, I\'m 14 and in 8th grade but\nin 9th grade math. What\'s a vector?\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>We call a state of two particles _entangled_ if the corresponding wave
function \psi_together(x,y) cannot be written as a product of two single
particle wave functions. The discussion above shows that there in fact
are
entangled states.
If you are at all familiar with linear algebra, you can try an example
that is more realistic. A two component (row or column) vector
describes
the state of a two level quantum system (such as a spin or a photon).
If
you have two such systems, the joint state will be described by a two
by
two matrix (four components). Can you write every possible 2x2 matrix
as a
tensor product of two 2-component tensors (a column vector times a row
vector)? If not, then you've just proven that a system of two coupled
two
level systems exhibits entangled states.
Hope this helps.
Yes, it does, thanx.
All I know in math is basic algebra though, I'm 14 and in 8th grade but
in 9th grade math. What's a vector?
Mark Palenik
Mar16-05, 02:07 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>"Sci~Girl" <palmtree117@juno.com> wrote in message\nnews:1110579986.612004.227690@l41g2000cwc .googlegroups.com...\n>\n> Yes, it does, thanx.\n> All I know in math is basic algebra though, I\'m 14 and in 8th grade but\n> in 9th grade math. What\'s a vector?\n>\n\nThe most common types of vectors you\'ll run into are basically arrows.\nThey have direction and length. So, for example, if I were to tell you\nthat something is five miles east, that would be a vector. Similarly, a\nset of x,y, and z coordinates has direction and magnitude, like [0,0,1]\n- left zero, forward zero, and up one. This is the definition of a\nvector that you will learn first in school - that vectors are things\nwith direction and magnitude, and my guess is that most of your teachers\nwon\'t even know that this isn\'t actually the *real* definition of a\nvector.\n\nThere is actually a more fundimental definition of a vector space, and\nin quantum mechanics, the vectors we use actually have neither direction\nnor magnitude (at least, not in the traditional sense - they can have\northogonality, but there\'s no need to confuse you with that). For\nexample, the functions Sin(x), Sin(2x), Sin(3x), etc. can be a vectors.\nYou don\'t really need to worry about how or why, just know that vectors\nare things that we can do algebra with.\n\nThe definitions of entanglement you\'ve been given would probably be\ndifficult for you to understand without an ungodly amount of additional\nexplanation. I\'m not even sure if I would have understood them a\nsemester ago (I haven\'t even done multi-particle QM yet, and just\nlearned the operator/matrix notation about four or five weeks ago - of\ncourse, now the 223 students are learning a simplified version of it\ntoo, which makes I\'d had Thaler when I took that course, but I digress.\n.. .).\n\nAt the most simple level (and someone else feel free to jump in if I\'m\nnot saying this well), you can think of entangled particles as particles\nthat share some net property, but individually don\'t have any definite\nvalue of that property.\n\nFor example, electrons have a property called spin. The spin can be up\nor down. If you have one electron that is spin up and one that\'s spin\ndown, your total spin is zero, because up plus down is zero.\nUnderstand?\n\nNow, say you have two electrons, and you know that the total spin\nbetween the two electrons is zero. But you don\'t know which one is spin\nup and which one is spin down. You can actually set up the system so\nthat neither particle is spin up or spin down until it is measured.\nHowever, as soon as you measure one, the other one will have the\nopposite spin. That\'s an example of entanglement, in my own very\nnon-formal, clumsy words.\n\nI\'d like to thanks people for the linear algebra explanation that was\nposted, because if nothing else, *I* learned something :)\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>"Sci~Girl" <palmtree117@juno.com> wrote in message
news:1110579986.612004.227690@l41g2000cwc.googlegr oups.com...
>
> Yes, it does, thanx.
> All I know in math is basic algebra though, I'm 14 and in 8th grade but
> in 9th grade math. What's a vector?
>
The most common types of vectors you'll run into are basically arrows.
They have direction and length. So, for example, if I were to tell you
that something is five miles east, that would be a vector. Similarly, a
set of x,y, and z coordinates has direction and magnitude, like [0,0,1]
- left zero, forward zero, and up one. This is the definition of a
vector that you will learn first in school - that vectors are things
with direction and magnitude, and my guess is that most of your teachers
won't even know that this isn't actually the *real* definition of a
vector.
There is actually a more fundimental definition of a vector space, and
in quantum mechanics, the vectors we use actually have neither direction
nor magnitude (at least, not in the traditional sense - they can have
orthogonality, but there's no need to confuse you with that). For
example, the functions Sin(x), Sin(2x), Sin(3x), etc. can be a vectors.
You don't really need to worry about how or why, just know that vectors
are things that we can do algebra with.
The definitions of entanglement you've been given would probably be
difficult for you to understand without an ungodly amount of additional
explanation. I'm not even sure if I would have understood them a
semester ago (I haven't even done multi-particle QM yet, and just
learned the operator/matrix notation about four or five weeks ago - of
course, now the 223 students are learning a simplified version of it
too, which makes I'd had Thaler when I took that course, but I digress.
.. .).
At the most simple level (and someone else feel free to jump in if I'm
not saying this well), you can think of entangled particles as particles
that share some net property, but individually don't have any definite
value of that property.
For example, electrons have a property called spin. The spin can be up
or down. If you have one electron that is spin up and one that's spin
down, your total spin is zero, because up plus down is zero.
Understand?
Now, say you have two electrons, and you know that the total spin
between the two electrons is zero. But you don't know which one is spin
up and which one is spin down. You can actually set up the system so
that neither particle is spin up or spin down until it is measured.
However, as soon as you measure one, the other one will have the
opposite spin. That's an example of entanglement, in my own very
non-formal, clumsy words.
I'd like to thanks people for the linear algebra explanation that was
posted, because if nothing else, *I* learned something :)
There is a lot of teleportation researches going on. I am not sure about the entanglement stuff...here is something a bit easier to follow...
http://photonics.anu.edu.au/pubtele/pubtele/slideshow.html
These people teleported a beam of laser or something...vary old news...Now they are trying to teleport a simple atom...
Arnold Neumaier
Mar16-05, 10:18 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>\nSci~Girl wrote:\n\n> All I know in math is basic algebra though, I\'m 14 and in 8th grade but\n> in 9th grade math. What\'s a vector?\n\nA list of numbers written below each other. For example the x,y, and z\ncoordinate of a point in a 3-dimensional coordinate system. Physicists\nwrite the three coordinates as x_1, x_2, x_3 and combine it to a vector\nsimply called x.\n- -\n| x_1 |\nx = | x_2 | (The parentheses look a bit awkward in ascii.)\n| x_3 |\n- -\nThe same for a list of n numbers. This gives a vector x with n\ncoordinates x_1,...x_n, and is thought of as a point in a\nspace with n dimensions.\n\nTwo vectors are added or subtracted or multiplied by a number\njust by adding, subtracting or multiplying their entries.\nThen there is the inner product of two vectors\nx dot y = sum x_i*y_i\nwhich is a number and not a vector.\n\nOnce you master vectors you need to understand matrices.\nThese are rectangular arrays of numbers.\n\n\nIf you want to learn more about quantum physics and really understand\nyou need to learn first how to do calculations with vectors and\nmatrices. This is essential for quantum information theory (and\nentanglement is relevant mainly there). Look in your local library\nfor math books, about \'linear algebra\' or \'analytic geometry\'.\nYou may have to try several before you find one suitable at your level.\n\nMaybe at first it is better to get math schoolbooks from your\nolder peers. Good school books are written in a way that they can be\nused for self study. If you are motivated it can be very exciting!\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>Sci~Girl wrote:
> All I know in math is basic algebra though, I'm 14 and in 8th grade but
> in 9th grade math. What's a vector?
A list of numbers written below each other. For example the x,y, and z
coordinate of a point in a 3-dimensional coordinate system. Physicists
write the three coordinates as x_1, x_2, x_3 and combine it to a vector
simply called x.
- -| x_1 |x = | x_2 | (The parentheses look a bit awkward in ascii.)
| x_3 |- -
The same for a list of n numbers. This gives a vector x with n
coordinates x_1,...x_n, and is thought of as a point in a
space with n dimensions.
Two vectors are added or subtracted or multiplied by a number
just by adding, subtracting or multiplying their entries.
Then there is the inner product of two vectors
x dot y = sum x_i*y_i
which is a number and not a vector.
Once you master vectors you need to understand matrices.
These are rectangular arrays of numbers.
If you want to learn more about quantum physics and really understand
you need to learn first how to do calculations with vectors and
matrices. This is essential for quantum information theory (and
entanglement is relevant mainly there). Look in your local library
for math books, about 'linear algebra' or 'analytic geometry'.
You may have to try several before you find one suitable at your level.
Maybe at first it is better to get math schoolbooks from your
older peers. Good school books are written in a way that they can be
used for self study. If you are motivated it can be very exciting!
Arnold Neumaier
<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>\nMark Palenik <markpalenik@wideopenwest.com> writes\n>For example, electrons have a property called spin. The spin can be up\n>or down. If you have one electron that is spin up and one that\'s spin\n>down, your total spin is zero, because up plus down is zero.\n>Understand?\n>\n>Now, say you have two electrons, and you know that the total spin\n>between the two electrons is zero. But you don\'t know which one is spin\n>up and which one is spin down. You can actually set up the system so\n>that neither particle is spin up or spin down until it is measured.\n\nFor a very elementary idea, suitable to carry someone aged 14 onwards\nwithout putting too much misinformation into the concept, how about the\nfollowing:\n\nTake your two particles, one spin up and one spin down.\n\nIndividually we have three classical combinations:\n\nup up\nup down (and down up)\ndown down\n\nNow quantum mechanics (ie the universe) doesn\'t work quite as simply as\nthis. You can have a pair of particles that is in a very real sense BOTH\nspin up and spin down. This is NOT the same as having a pair with one\nspin up and one spin down.\n\nThis is a fourth valid description.\nIt only exists in a quantum mechanical world (that as yet you know\nnothing about).\n\nIn many ways the latter behaves as a single particle containing both\nspin up and spin down (which isn\'t zero). The devious thing is that if\nyou are very delicate and careful (ie don\'t look at it too closely) you\ncan separate the two particles, each of which is BOTH spin up and spin\ndown. BUT as soon as you actually measure (ie destroy) one, the other is\nguaranteed to be the other way up, even if separated by miles.\n\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\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>Mark Palenik <markpalenik@wideopenwest.com> writes
>For example, electrons have a property called spin. The spin can be up
>or down. If you have one electron that is spin up and one that's spin
>down, your total spin is zero, because up plus down is zero.
>Understand?
>
>Now, say you have two electrons, and you know that the total spin
>between the two electrons is zero. But you don't know which one is spin
>up and which one is spin down. You can actually set up the system so
>that neither particle is spin up or spin down until it is measured.
For a very elementary idea, suitable to carry someone aged 14 onwards
without putting too much misinformation into the concept, how about the
following:
Take your two particles, one spin up and one spin down.
Individually we have three classical combinations:
up up
up down (and down up)
down down
Now quantum mechanics (ie the universe) doesn't work quite as simply as
this. You can have a pair of particles that is in a very real sense BOTH
spin up and spin down. This is NOT the same as having a pair with one
spin up and one spin down.
This is a fourth valid description.
It only exists in a quantum mechanical world (that as yet you know
nothing about).
In many ways the latter behaves as a single particle containing both
spin up and spin down (which isn't zero). The devious thing is that if
you are very delicate and careful (ie don't look at it too closely) you
can separate the two particles, each of which is BOTH spin up and spin
down. BUT as soon as you actually measure (ie destroy) one, the other is
guaranteed to be the other way up, even if separated by miles.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
bjflanagan
Mar18-05, 12:29 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>You might have a look at \'Mathematics of Classical & Quantum Physics,\'\nby Byron & Fuller. Meanwhile, let me try to provide a bit of motivation\nby pointing out that, in quantum theory, every physical thing can be\ndescribed, in principle, by a wave function that is (more or less)\nmathematically equivalent to a vector in what is called Hilbert space,\nafter the profoundly influential David Hilbert, who, along with Felix\nKlein, assembled one of the most dazzling array of minds ever known, in\nGottingen in the late 19th and early 20th centuries. (See the masterful\n\'Hilbert\' by Constance Reid.)\n\nThis is how Byron & Fuller present the standard wisdom: "Any physical\nsystem is completely described by a normalized vector (the state vector\nor wave function) in Hilbert space. All possible information about the\nsystem can be derived from this state vector by rules..."\n\nHere are some more references:\n\nhttp://www.geometry.net/scientists_bk/hilbert_david.html\n\nhttp://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Klein.html)\n\n___________________________\n\nHere \'s an excerpt from a fairly recent article by Asher Peres (who is\none of the elder statesmen of physics today), which ties this issue\ninto "EPR" and the famous Einstein-Bohr debates. Einstein said that the\nentire future history of physics would revolve around the resolution of\nthis issue. In this connection it is quite exciting to note that three\nof our most prominent contemporary theorists -- Hartle, Smolin and \'t\nHooft -- have recently revived the "hidden variables" debate.\n\n"That inequality was originally derived for solving a completely\ndifferent problem: is quantum theory compatible with an underlying\npseudo-classical "subquantum" back-ground (deterministic or possibly\nstochastic)? Such a post-quantum theory would pre-sumably involve\nadditional "hidden" variables, and the statistical predictions of\nordinary quantum theory would be reproduced by performing suitable\naverages over these hidden variables. Bell [4] was the first to show\nthat if the constraint of locality is imposed on the hidden variables\n(namely, if the hidden variables of two distant quantum systems are\nthemselves separable into two distinct subsets), then there is an upper\nbound to the correlations of results of measurements that can be\nperformed on the two distant systems."\n\narXiv:quant-ph/9707026 v1\n\n(JS Bell was a theorist at CERN who also believed that there "must be\nhidden variables" (private communication)).\n\n_____________________\n\n\nEPR :\n\nhttp://plato.stanford.edu/entries/qt-measurement/\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>You might have a look at 'Mathematics of Classical & Quantum Physics,'
by Byron & Fuller. Meanwhile, let me try to provide a bit of motivation
by pointing out that, in quantum theory, every physical thing can be
described, in principle, by a wave function that is (more or less)
mathematically equivalent to a vector in what is called Hilbert space,
after the profoundly influential David Hilbert, who, along with Felix
Klein, assembled one of the most dazzling array of minds ever known, in
Gottingen in the late 19th and early 20th centuries. (See the masterful
'Hilbert' by Constance Reid.)
This is how Byron & Fuller present the standard wisdom: "Any physical
system is completely described by a normalized vector (the state vector
or wave function) in Hilbert space. All possible information about the
system can be derived from this state vector by rules..."
Here are some more references:
http://www.geometry.net/scientists_bk/hilbert_david.html
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Klein.html)
__{_________________________}
Here's an excerpt from a fairly recent article by Asher Peres (who is
one of the elder statesmen of physics today), which ties this issue
into "EPR" and the famous Einstein-Bohr debates. Einstein said that the
entire future history of physics would revolve around the resolution of
this issue. In this connection it is quite exciting to note that three
of our most prominent contemporary theorists -- Hartle, Smolin and 't
Hooft -- have recently revived the "hidden variables" debate.
"That inequality was originally derived for solving a completely
different problem: is quantum theory compatible with an underlying
pseudo-classical "subquantum" back-ground (deterministic or possibly
stochastic)? Such a post-quantum theory would pre-sumably involve
additional "hidden" variables, and the statistical predictions of
ordinary quantum theory would be reproduced by performing suitable
averages over these hidden variables. Bell [4] was the first to show
that if the constraint of locality is imposed on the hidden variables
(namely, if the hidden variables of two distant quantum systems are
themselves separable into two distinct subsets), then there is an upper
bound to the correlations of results of measurements that can be
performed on the two distant systems."
arXiv:http://www.arxiv.org/abs/quant-ph/9707026 v1
(JS Bell was a theorist at CERN who also believed that there "must be
hidden variables" (private communication)).
__{___________________}
EPR:
http://plato.stanford.edu/entries/qt-measurement/
Sci~Girl
Mar18-05, 12:39 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>> "At the most simple level (and someone else feel free to jump in if I\'m\n> not saying this well), you can think of entangled particles as particles\n> that share some net property, but individually don\'t have any definite\n> value of that property."\n\n> The definitions of entanglement you\'ve been given would probably be\n> difficult for you to understand without an ungodly amount of additional\n> explanation.\n\nI was able to pull out that there\'s some similarity between the\nparticles, but that was about it. That helps a lot, although of course\nI have questions as always. Unfortunately I don\'t think I\'d be able to\nunderstand the answers.\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>> "At the most simple level (and someone else feel free to jump in if I'm
> not saying this well), you can think of entangled particles as particles
> that share some net property, but individually don't have any definite
> value of that property."
> The definitions of entanglement you've been given would probably be
> difficult for you to understand without an ungodly amount of additional
> explanation.
I was able to pull out that there's some similarity between the
particles, but that was about it. That helps a lot, although of course
I have questions as always. Unfortunately I don't think I'd be able to
understand the answers.
Sci~Girl
Mar18-05, 12:39 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>>In many ways the latter behaves as a single particle containing both\n>spin up and spin down (which isn\'t zero).\n\n>You can have a pair of particles that is in a very real sense BOTH\n>spin up and spin down. This is NOT the same as having a pair with >one\nspin up and one spin down.\n\nThat\'s... hard to comprehend... both up and down but not zero? If it\nwasn\'t zero, either the up or the down would have to be dominant and\nthen it would be called just up or just down... Well, I know that\'s not\nright, but I have no other way to think until I know more.\n\nIn order to truly understand quantum mechanics I\'d need to understand\ncalculus. I HAVE THREE YEARS TO GO BEFORE CALCULUS. It would be okay to\njust borrow a textbook if it was, say, next year\'s geometry I was\ntrying to get to, but it\'s much farther away. I\'d have to read the\ngeometry book, then the algebra 2 book, then the advanced math and\ntrigonometry book, before I even arrived at differential calculus. And\nthen there\'s integral... and I don\'t even know what that means. I\'m\nsure it would be exciting and I\'ve thought about doing it before, but\nit just doesn\'t seem like something within my capacity to do.\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>>In many ways the latter behaves as a single particle containing both
>spin up and spin down (which isn't zero).
>You can have a pair of particles that is in a very real sense BOTH
>spin up and spin down. This is NOT the same as having a pair with >one
spin up and one spin down.
That's... hard to comprehend... both up and down but not zero? If it
wasn't zero, either the up or the down would have to be dominant and
then it would be called just up or just down... Well, I know that's not
right, but I have no other way to think until I know more.
In order to truly understand quantum mechanics I'd need to understand
calculus. I HAVE THREE YEARS TO GO BEFORE CALCULUS. It would be okay to
just borrow a textbook if it was, say, next year's geometry I was
trying to get to, but it's much farther away. I'd have to read the
geometry book, then the algebra 2 book, then the advanced math and
trigonometry book, before I even arrived at differential calculus. And
then there's integral... and I don't even know what that means. I'm
sure it would be exciting and I've thought about doing it before, but
it just doesn't seem like something within my capacity to do.
a student
Mar18-05, 12:41 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>"Mark Palenik" <markpalenik@wideopenwest.com> wrote in message news:<HdudnRYcqL5oXarfRVn-gw@wideopenwest.com>...\n>\n> At the most simple level (and someone else feel free to jump in if I\'m\n> not saying this well), you can think of entangled particles as particles\n> that share some net property, but individually don\'t have any definite\n> value of that property.\n\nI think you have said it very well - \'entanglement\' means that the\nproperties of a pair of entangled particles cannot be reduced to\nproperties of the individual particles.\n\nIn classical mechanics two non-interacting particles, considered\ntogether, are nothing more than a first particle plus a second\nparticle. Each particle has, for example, its own position and\nvelocity, and this is all that is needed to predict any measurement\noutcome. Moreover, if one has only incomplete knowledge of the\npositions and velocities, then one can only make statistical\npredictions (eg, of the average velocity) for the outcome of any\nparticular measurement on the particles, but these predictions are\nstill formulated in terms of the properties of the individual\nparticles. So there is no entanglement.\n\nBut quantum particles CAN have entanglement! In particular, if there\nwas NO entanglement, then all properties of a pair of non-interacting\nparticles would be determined completely by the properties of the\nindividual particles. Hence the statistics of any measurements on the\npair could be formulated in terms of the properties of the individual\nparticles (eg, their positions and velocities). But, as first pointed\nout by John Bell, this means that groups of statistical predictions\nmust satisfy certain mathematical conditions (called Bell\ninequalities). Remarkably, quantum mechanics allows the existence of\npairs of non-interacting particles which do NOT satisfy these\nconditions!! Hence there must be entanglement (if quantum mechanics\nis correct, and it has never been wrong yet).\n\n\nIt is perhaps worth noting further that, in quantum mechanics, one\nonly ever has incomplete knowledge of the properties of even a SINGLE\nparticle - the mathematics simply doesn\'t allow, for example, values\nto exist for both the position and the velocity of a given particle.\nSo, one can only make statistical predictions for the position and\nvelocity (and the statistical errors satisfy the famous Heisenberg\ninequality).\n\nHence, it is natural to ask: why doesn\'t the maths allow complete\nknowledge of a particle\'s properties? Is it\n(i) because while particles actually do have well defined positions\nand velocities, quantum mechanics is not good enough to describe them\n(i.e., the theory is incomplete); or\n(ii) because particles don\'t have definite positions and velocities\n(i.e., our classical notion of particles is incomplete).\n\nThe generally accepted answer is (ii).\n\nSo what rules out choosing option (i) (for most people)?\nEntanglement! There simply seems little point choosing option (i) -\nbecause even if individual particles "really" do have well defined\nvalues of position, velocity, etc, these values cannot explain the\nstatistics of PAIRS of non-interacting particles, as noted above.\n\nIn fact, the only way to \'save\' option (i) is to claim that either\n(a) \'non-interacting particles\' are, in fact, interacting ! (it turns\nout that these interactions must be instantaneous and hence travel\nfaster than the speed of light - there is a theory along these lines\ncalled Bohmian mechanics after David Bohm); or\n(b) there is a conspiracy of nature which denies experimenter\'s the\nchoice of which property to measure, implying that the measurement\nstatistics are not really random in the sense that is required for the\nBell inequalities (I don\'t think anyone takes this seriously); or\n(c) quantum mechanics and its prediction of entanglement is wrong (but\nall experiments to date suggest the opposite!).\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>"Mark Palenik" <markpalenik@wideopenwest.com> wrote in message news:<HdudnRYcqL5oXarfRVn-gw@wideopenwest.com>...
>
> At the most simple level (and someone else feel free to jump in if I'm
> not saying this well), you can think of entangled particles as particles
> that share some net property, but individually don't have any definite
> value of that property.
I think you have said it very well - 'entanglement' means that the
properties of a pair of entangled particles cannot be reduced to
properties of the individual particles.
In classical mechanics two non-interacting particles, considered
together, are nothing more than a first particle plus a second
particle. Each particle has, for example, its own position and
velocity, and this is all that is needed to predict any measurement
outcome. Moreover, if one has only incomplete knowledge of the
positions and velocities, then one can only make statistical
predictions (eg, of the average velocity) for the outcome of any
particular measurement on the particles, but these predictions are
still formulated in terms of the properties of the individual
particles. So there is no entanglement.
But quantum particles CAN have entanglement! In particular, if there
was NO entanglement, then all properties of a pair of non-interacting
particles would be determined completely by the properties of the
individual particles. Hence the statistics of any measurements on the
pair could be formulated in terms of the properties of the individual
particles (eg, their positions and velocities). But, as first pointed
out by John Bell, this means that groups of statistical predictions
must satisfy certain mathematical conditions (called Bell
inequalities). Remarkably, quantum mechanics allows the existence of
pairs of non-interacting particles which do NOT satisfy these
conditions!! Hence there must be entanglement (if quantum mechanics
is correct, and it has never been wrong yet).
It is perhaps worth noting further that, in quantum mechanics, one
only ever has incomplete knowledge of the properties of even a SINGLE
particle - the mathematics simply doesn't allow, for example, values
to exist for both the position and the velocity of a given particle.
So, one can only make statistical predictions for the position and
velocity (and the statistical errors satisfy the famous Heisenberg
inequality).
Hence, it is natural to ask: why doesn't the maths allow complete
knowledge of a particle's properties? Is it
(i) because while particles actually do have well defined positions
and velocities, quantum mechanics is not good enough to describe them
(i.e., the theory is incomplete); or
(ii) because particles don't have definite positions and velocities
(i.e., our classical notion of particles is incomplete).
The generally accepted answer is (ii).
So what rules out choosing option (i) (for most people)?
Entanglement! There simply seems little point choosing option (i) -
because even if individual particles "really" do have well defined
values of position, velocity, etc, these values cannot explain the
statistics of PAIRS of non-interacting particles, as noted above.
In fact, the only way to 'save' option (i) is to claim that either
(a) 'non-interacting particles' are, in fact, interacting ! (it turns
out that these interactions must be instantaneous and hence travel
faster than the speed of light - there is a theory along these lines
called Bohmian mechanics after David Bohm); or
(b) there is a conspiracy of nature which denies experimenter's the
choice of which property to measure, implying that the measurement
statistics are not really random in the sense that is required for the
Bell inequalities (I don't think anyone takes this seriously); or
(c) quantum mechanics and its prediction of entanglement is wrong (but
all experiments to date suggest the opposite!).
Igor Khavkine
Mar18-05, 12:42 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 Tue, 15 Mar 2005 18:21:50 +0000, Sci~Girl wrote:\n\n> All I know in math is basic algebra though, I\'m 14 and in 8th grade but in\n> 9th grade math. What\'s a vector?\n\nIn addition to good explanations by Arnold Neumaier and Mark Palenik,\nhere\'s a slightly more high brow one.\n\nIn modern mathematics vectors are an abstract but simple concept. There\nare many objects that can rightfully be called vectors. The only\nrequirement is that they satisfy certain axioms, that is certain\noperations are possible with them.\n\nHere\'s a boost of confidence for you. You say you know basic algebra. Do\nyou know how to add? Do you know how to multiply? Then you can understand\nwhat vectors are. Suppose U and V are vectors and a and b are numbers.\nThen you can do the following:\n\n1) Add: U + V is also a vector\n2) Multiply: a U (should be read "a times U") is also a vector\n3) Combine the two operations:\na U + b V is also a vector (such an expression is called a linear\ncombination of U and V)\na (U + V) = a U + b V is also a vector\n(a + b) U = a U + b U is also a vector\n(both of these are called distributivity)\n\nThe examples that you\'ve seen so far are:\n\n1) Geometric vectors\nThese are arrows of fixed length and direction. You\'ve probably\nalready learned how to add them, just stick the second arrow\nat the tip of the first one and draw an arrow from the base of the\nfirst arrow to the tip of the second one. Multiplying them by\nnumbers is also easy, just extend or shrink the arrow along its\ndirection.\n\n2) Column vectors\nThese are columns of numbers usually written with either square [ ]\nor round ( ) brackets around them. To add two column vectors with\nthe same number of entries, just add them one by one. To multiply\nby a number, just multiply each entry in the column individually.\n\n3) Functions\nIf I have two functions f(x) and f(y), then their sum is also a\nfunction. If I multiply f(x) by a number b, then\nb f(x) (b times f(x)) is still a function.\n\nIt turns out that if you only care about the abstract properties of\nvectors that I described above, then many different kinds of vectors can\nbe treated as identical. For example, working with column vectors are very\nconvenient, so often people translate their vector problems purely into\nthe language of column vectors. In fact, they are so convenient to work\nwith, that some people forget that there are other kinds of vectors. Which\nis unfortunate, because there are some things that are not particularly\nconvenient to do in the language of column vectors.\n\nIn quantum mechanics, almost every classical state is promoted to a\nvector. If X denotes the state of a particle being at point x and Y\ndenotes the state of a particle being at point y, then in quantum\nmechanics any state of the form a X + b Y is also allowed, even though X\nand Y are exclusive classically. If you\'ve ever heard of Schroedinger\'s\ncat, it is an example of so called superposition of states. If D\nrepresents a state of the cat being dead and A represents the state of the\ncat being alive, then in quantum mechanics a state of the form a D + b A\nis as natural as a vector a N + b E that points somewhere in between\nNorth and East. Don\'t pay attention to all the brouhaha surrounding\nSchroedinger\'s cat, remember that it\'s just an illustration.\n\nLinear algebra can be quite fun to study, especially since it\'s so useful.\nI will second other suggestions in this thread that you go to the library\nand pick up a book on elementary linear algebra. There you will learn\nabout things such as vector spaces, linear equations, functions between\nvector spaces (commonly called matrices), linear forms, bilinear forms,\netc.\n\nFinally, I\'ll expand on my explanation of entanglement. First a note about\ntensor products. It is sometimes convenient (or even necessary) to\nintroduce a product between vectors themselves (note that this\noperation was not defined for abstract vectors I described above).\nHowever, usually if you take the tensor product (or just product) of two\nvectors, you don\'t get the same kind of vector back, but a different kind.\nThe example of multiplication of functions from my previous post is a good\none. A function f(x) is one kind of vector, a function g(y) is another\nkind of vector, while their product h(x,y) = f(x) g(y) is yet a third kind\nof vector. The first two were functions of one variable (but different\nvariables), while the third one is a function of both variables at the\nsame time.\n\nBack to quantum mechanics. Consider a light bulb, if classically it only\nhas two states N = on and F = off, then the quantum analog will have\nmany possible states, each of the form a N + b F where a and b are some\nnumbers. If we have two light bulbs let N1 and F1 describe the states of\nthe first one and N2 and F2 describe the sates of the second one. Then to\nconsider possible states of both bulbs together we must have the states\nN1 N2, N1 F2, F1 N2, and F1 F2. You can look at the joint states as\nproducts of individual states. For example, N1 N2 can be read as "N1\ntimes N2" or "N1 tensor N2". Now, the quantum version of these light bulbs\nwill in general be a linear combination of these four sates\n\na N1 N2 + b N1 F2 + c F1 N2 + d F1 F2.\n\nOne way to produce joint quantum states is two take two individual light\nbulb states and multiply them like we did for the classical states (just\nfollowing standard rules of multiplication):\n\n(e N1 + f F1) (g N2 + h F2) = eg N1 N2 + eh N1 F2 + fg F1 N2 + fh F1 F2.\n\nNow, here\'s a puzzle for you. Can you find a joint quantum state described\nby numbers a, b, c, d that *cannot* be written as a product of two\nindividual states, say (e N1 + f F1) and (g N2 + h F2)? If yes, then\nyou\'ve found an entangled state of the two light bulb system.\n\nHope this helps.\n\nIgor\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 Tue, 15 Mar 2005 18:21:50 +0000, Sci~Girl wrote:
> All I know in math is basic algebra though, I'm 14 and in 8th grade but in
> 9th grade math. What's a vector?
In addition to good explanations by Arnold Neumaier and Mark Palenik,
here's a slightly more high brow one.
In modern mathematics vectors are an abstract but simple concept. There
are many objects that can rightfully be called vectors. The only
requirement is that they satisfy certain axioms, that is certain
operations are possible with them.
Here's a boost of confidence for you. You say you know basic algebra. Do
you know how to add? Do you know how to multiply? Then you can understand
what vectors are. Suppose U and V are vectors and a and b are numbers.
Then you can do the following:
1) Add: U + V is also a vector
2) Multiply: a U (should be read "a times U") is also a vector
3) Combine the two operations:
a U + b V is also a vector (such an expression is called a linear
combination of U and V)
a (U + V) = a U + b V is also a vector
(a + b) U = a U + b U is also a vector
(both of these are called distributivity)
The examples that you've seen so far are:
1) Geometric vectors
These are arrows of fixed length and direction. You've probably
already learned how to add them, just stick the second arrow
at the tip of the first one and draw an arrow from the base of the
first arrow to the tip of the second one. Multiplying them by
numbers is also easy, just extend or shrink the arrow along its
direction.
2) Column vectors
These are columns of numbers usually written with either square [ ]
or round ( ) brackets around them. To add two column vectors with
the same number of entries, just add them one by one. To multiply
by a number, just multiply each entry in the column individually.
3) Functions
If I have two functions f(x) and f(y), then their sum is also a
function. If I multiply f(x) by a number b, then
b f(x) (b times f(x)) is still a function.
It turns out that if you only care about the abstract properties of
vectors that I described above, then many different kinds of vectors can
be treated as identical. For example, working with column vectors are very
convenient, so often people translate their vector problems purely into
the language of column vectors. In fact, they are so convenient to work
with, that some people forget that there are other kinds of vectors. Which
is unfortunate, because there are some things that are not particularly
convenient to do in the language of column vectors.
In quantum mechanics, almost every classical state is promoted to a
vector. If X denotes the state of a particle being at point x and Y
denotes the state of a particle being at point y, then in quantum
mechanics any state of the form a X + b Y is also allowed, even though X
and Y are exclusive classically. If you've ever heard of Schroedinger's
cat, it is an example of so called superposition of states. If D
represents a state of the cat being dead and A represents the state of the
cat being alive, then in quantum mechanics a state of the form a D + b A
is as natural as a vector a N + b E that points somewhere in between
North and East. Don't pay attention to all the brouhaha surrounding
Schroedinger's cat, remember that it's just an illustration.
Linear algebra can be quite fun to study, especially since it's so useful.
I will second other suggestions in this thread that you go to the library
and pick up a book on elementary linear algebra. There you will learn
about things such as vector spaces, linear equations, functions between
vector spaces (commonly called matrices), linear forms, bilinear forms,
etc.
Finally, I'll expand on my explanation of entanglement. First a note about
tensor products. It is sometimes convenient (or even necessary) to
introduce a product between vectors themselves (note that this
operation was not defined for abstract vectors I described above).
However, usually if you take the tensor product (or just product) of two
vectors, you don't get the same kind of vector back, but a different kind.
The example of multiplication of functions from my previous post is a good
one. A function f(x) is one kind of vector, a function g(y) is another
kind of vector, while their product h(x,y) = f(x) g(y) is yet a third kind
of vector. The first two were functions of one variable (but different
variables), while the third one is a function of both variables at the
same time.
Back to quantum mechanics. Consider a light bulb, if classically it only
has two states N = on and F = off, then the quantum analog will have
many possible states, each of the form a N + b F where a and b are some
numbers. If we have two light bulbs let N1 and F1 describe the states of
the first one and N2 and F2 describe the sates of the second one. Then to
consider possible states of both bulbs together we must have the states
N1 N2, N1 F2, F1 N2, and F1 F2. You can look at the joint states as
products of individual states. For example, N1 N2 can be read as "N1
times N2" or "N1 tensor N2". Now, the quantum version of these light bulbs
will in general be a linear combination of these four sates
a N1 N2 + b N1 F2 + c F1 N2 + d F1 F2.
One way to produce joint quantum states is two take two individual light
bulb states and multiply them like we did for the classical states (just
following standard rules of multiplication):
(e N1 + f F1) (g N2 + h F2) = eg N1 N2 + eh N1 F2 + fg F1 N2 + fh F1 F2.
Now, here's a puzzle for you. Can you find a joint quantum state described
by numbers a, b, c, d that *cannot* be written as a product of two
individual states, say (e N1 + f F1) and (g N2 + h F2)? If yes, then
you've found an entangled state of the two light bulb system.
Hope this helps.
Igor
<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>Sci~Girl <palmtree117@juno.com> writes\n>>In many ways the latter behaves as a single particle containing both\n>>spin up and spin down (which isn\'t zero).\n>\n>>You can have a pair of particles that is in a very real sense BOTH\n>>spin up and spin down. This is NOT the same as having a pair with >one\n>spin up and one spin down.\n>\n>That\'s... hard to comprehend... both up and down but not zero? If it\n>wasn\'t zero, either the up or the down would have to be dominant and\n>then it would be called just up or just down...\n\nQuite.\nQuantum mechanics is a wonderful thing.\nIts the sort of thing that keeps physics really interesting, even to\npeople who have studied it for a lifetime.\n\nThere is no particular reason why the universe should behave\nintuitively, after all we see only a tiny subset of the universe.\nThis can be roughly described as that of billions of particles going\nvery slowly. One should not be surprised that a few particles going very\nfast behaves rather differently.\n\nLucky you, you will be taught by people who have been taught by a\ngeneration of teachers very familiar with it.\n\n>Well, I know that\'s not\n>right, but I have no other way to think until I know more.\n\nIn many ways its best to accept these (and many other magical) things\njust DO happen that way, and let your head get round it in its own time.\nThe results of a well chosen selection of experiments helps here. Do not\nbe surprised if some seem contradictory, quantum mechanics can handle\nthem. Just be thankful a couple of generations of physicists racked\ntheir brains trying to make sense of it for you.\n\n>In order to truly understand quantum mechanics I\'d need to understand\n>calculus. I HAVE THREE YEARS TO GO BEFORE CALCULUS.\n\nSure. You will find it fun.\n\n>It would be okay to\n>just borrow a textbook if it was, say, next year\'s geometry I was\n>trying to get to, but it\'s much farther away. I\'d have to read the\n>geometry book, then the algebra 2 book, then the advanced math and\n>trigonometry book, before I even arrived at differential calculus. And\n>then there\'s integral... and I don\'t even know what that means.\n\nYup. Loads of really scary stuff designed to put the fear of god into\nthe uninitiated. Not so hard when you get there though. The really good\nthing is that for an ignorant antique like myself it seems to never end.\nI can\'t understand most of what gets posted here, but it sure is\nfascinating. You, at least, have a lifetime to get stuck into these new\nmathematical tools. In ten years (with some work) most of the esoteric\nstuff you read here will seem quite ordinary, accurate and precise.\nI\'m jealous.\n\n>I\'m\n>sure it would be exciting and I\'ve thought about doing it before, but\n>it just doesn\'t seem like something within my capacity to do.\n\nAt 14? Well you could. Unfortunately you could really easily confuse\nyourself and make something easy into something difficult. It would take\nquite a bit of your spare time up. It might be better to spend some of\nthat spare time \'socialising\' or playing football.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\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>Sci~Girl <palmtree117@juno.com> writes
>>In many ways the latter behaves as a single particle containing both
>>spin up and spin down (which isn't zero).
>
>>You can have a pair of particles that is in a very real sense BOTH
>>spin up and spin down. This is NOT the same as having a pair with >one
>spin up and one spin down.
>
>That's... hard to comprehend... both up and down but not zero? If it
>wasn't zero, either the up or the down would have to be dominant and
>then it would be called just up or just down...
Quite.
Quantum mechanics is a wonderful thing.
Its the sort of thing that keeps physics really interesting, even to
people who have studied it for a lifetime.
There is no particular reason why the universe should behave
intuitively, after all we see only a tiny subset of the universe.
This can be roughly described as that of billions of particles going
very slowly. One should not be surprised that a few particles going very
fast behaves rather differently.
Lucky you, you will be taught by people who have been taught by a
generation of teachers very familiar with it.
>Well, I know that's not
>right, but I have no other way to think until I know more.
In many ways its best to accept these (and many other magical) things
just DO happen that way, and let your head get round it in its own time.
The results of a well chosen selection of experiments helps here. Do not
be surprised if some seem contradictory, quantum mechanics can handle
them. Just be thankful a couple of generations of physicists racked
their brains trying to make sense of it for you.
>In order to truly understand quantum mechanics I'd need to understand
>calculus. I HAVE THREE YEARS TO GO BEFORE CALCULUS.
Sure. You will find it fun.
>It would be okay to
>just borrow a textbook if it was, say, next year's geometry I was
>trying to get to, but it's much farther away. I'd have to read the
>geometry book, then the algebra 2 book, then the advanced math and
>trigonometry book, before I even arrived at differential calculus. And
>then there's integral... and I don't even know what that means.
Yup. Loads of really scary stuff designed to put the fear of god into
the uninitiated. Not so hard when you get there though. The really good
thing is that for an ignorant antique like myself it seems to never end.
I can't understand most of what gets posted here, but it sure is
fascinating. You, at least, have a lifetime to get stuck into these new
mathematical tools. In ten years (with some work) most of the esoteric
stuff you read here will seem quite ordinary, accurate and precise.
I'm jealous.
>I'm
>sure it would be exciting and I've thought about doing it before, but
>it just doesn't seem like something within my capacity to do.
At 14? Well you could. Unfortunately you could really easily confuse
yourself and make something easy into something difficult. It would take
quite a bit of your spare time up. It might be better to spend some of
that spare time 'socialising' or playing football.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
Vincent N. Virgilio
Mar22-05, 02:21 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>Sci~Girl wrote:\n> I was able to pull out that there\'s some similarity between the\n> particles, but that was about it. That helps a lot, although of course\n> I have questions as always. Unfortunately I don\'t think I\'d be able to\n> understand the answers.\n>\n\nBeware that I am an amateur in this forum . . .\n\nEntanglement reminds me of correlated random variables. In very plain\nwords: if you learn something about one variable (photon), you learn at\nleast a little about the other. I think this is not so with\nuncorrelated variables (or dis-entangled photons); whatever you know\nabout one of the variables (photons) tells you nothing about the other.\nAt least in this context. (Side note: Dependence and independence is a\nvery important idea in math and science; even in the linear algebra\nothers have referred to in this thread. Example, vectors can be\ndependent or independent with respect to each other.)\n\nIt is as you say, "Similarity between the particles"; and a measureable\none at that.\n\nCorrections are welcome.\n\nVince Virgilio\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>Sci~Girl wrote:
> I was able to pull out that there's some similarity between the
> particles, but that was about it. That helps a lot, although of course
> I have questions as always. Unfortunately I don't think I'd be able to
> understand the answers.
>
Beware that I am an amateur in this forum . . .
Entanglement reminds me of correlated random variables. In very plain
words: if you learn something about one variable (photon), you learn at
least a little about the other. I think this is not so with
uncorrelated variables (or dis-entangled photons); whatever you know
about one of the variables (photons) tells you nothing about the other.
At least in this context. (Side note: Dependence and independence is a
very important idea in math and science; even in the linear algebra
others have referred to in this thread. Example, vectors can be
dependent or independent with respect to each other.)
It is as you say, "Similarity between the particles"; and a measureable
one at that.
Corrections are welcome.
Vince Virgilio
whopkins@csd.uwm.edu
Mar22-05, 06:38 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>>From Sci~Girl <palmtree...@juno.com>\n>All I know in math is basic algebra though, I\'m 14 and in 8th grade\nbut\n>in 9th grade math.\n\nI describe the situation in Bell\'s theorem in clear terms.\nnews://sci.physics\n2005, January 5\nAlain Aspect Non-Locality/Entanglement experiment\nhttp://groups-beta.google.com/groups?q=whopkins+entanglement+DIFFERENT+questions \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>>From Sci~Girl <palmtree...@juno.com>
>All I know in math is basic algebra though, I'm 14 and in 8th grade
but
>in 9th grade math.
I describe the situation in Bell's theorem in clear terms.
news://sci.physics
2005, January 5
Alain Aspect Non-Locality/Entanglement experiment
http://groups-\beta.google.com/groups?q=whopkins+entanglement+DIFFERENT+questions
Ralph Hartley
Mar23-05, 04:33 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>Vincent N. Virgilio wrote:\n> Sci~Girl wrote:\n>>I was able to pull out that there\'s some similarity between the\n>>particles, but that was about it. That helps a lot, although of course\n>>I have questions as always. Unfortunately I don\'t think I\'d be able to\n>>understand the answers.\n\n> Entanglement reminds me of correlated random variables. In very plain\n> words: if you learn something about one variable (photon), you learn at\n> least a little about the other. I think this is not so with\n> uncorrelated variables (or dis-entangled photons); whatever you know\n> about one of the variables (photons) tells you nothing about the other.\n\nThis is exactly correct!\n\nQuantum theory is just the square root of probability theory.\n\nMost people have some idea what a probability is. A quantum "amplitude"\nis very similar, but it can be negative (footnote 1). You square an\namplitude to get a probability, but always do that as the *last* step in\na calculation. All the other rules for calculating probabilities are the\nsame! (footnote 2)\n\nIndependent particles are like independent coin flips. Entangled\nparticles are like coin flips that are not independent.\n\nWe use the word "independent" in both quantum theory and probability,\nbut when it is absent we use two different words "entangled" and\n"correlated". I\'m not sure why (there is *some* logic to it).\n\nJust to make things trickier, the term "linearly independent" uses the\nword "independent" in *another* sense. Sometimes in the same sentence.\n\nEntangled systems can do things that don\'t make sense in terms of\ncorrelated probabilities, because positive and negative amplitudes can\ncancel out when they are added. That can\'t happen with probabilities\nbecause they are all positive.\n\n1) In physics quantum amplitudes are usually complex numbers. That makes\nthe equations simpler, because every number has a square root etc., but\nmakes no difference at this level of detail.\n\n2) The basic rules are:\n\nDivide all the ways things can happen into independent and/or mutually\nexclusive cases (there may be more than one way to do that, but it won\'t\naffect the answer).\n\nFor independent events A and B, to get the probability (amplitude) of A\n*and* B happening, *multiply* the probabilities (amplitudes).\n\nFor mutually exclusive events C and D, to get the probability\n(amplitude) of C *or* D happening, *add* the probabilities (amplitudes).\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>Vincent N. Virgilio wrote:
> Sci~Girl wrote:
>>I was able to pull out that there's some similarity between the
>>particles, but that was about it. That helps a lot, although of course
>>I have questions as always. Unfortunately I don't think I'd be able to
>>understand the answers.
> Entanglement reminds me of correlated random variables. In very plain
> words: if you learn something about one variable (photon), you learn at
> least a little about the other. I think this is not so with
> uncorrelated variables (or dis-entangled photons); whatever you know
> about one of the variables (photons) tells you nothing about the other.
This is exactly correct!
Quantum theory is just the square root of probability theory.
Most people have some idea what a probability is. A quantum "amplitude"
is very similar, but it can be negative (footnote 1). You square an
amplitude to get a probability, but always do that as the *last* step in
a calculation. All the other rules for calculating probabilities are the
same! (footnote 2)
Independent particles are like independent coin flips. Entangled
particles are like coin flips that are not independent.
We use the word "independent" in both quantum theory and probability,
but when it is absent we use two different words "entangled" and
"correlated". I'm not sure why (there is *some* logic to it).
Just to make things trickier, the term "linearly independent" uses the
word "independent" in *another* sense. Sometimes in the same sentence.
Entangled systems can do things that don't make sense in terms of
correlated probabilities, because positive and negative amplitudes can
cancel out when they are added. That can't happen with probabilities
because they are all positive.
1) In physics quantum amplitudes are usually complex numbers. That makes
the equations simpler, because every number has a square root etc., but
makes no difference at this level of detail.
2) The basic rules are:
Divide all the ways things can happen into independent and/or mutually
exclusive cases (there may be more than one way to do that, but it won't
affect the answer).
For independent events A and B, to get the probability (amplitude) of A
*and* B happening, *multiply* the probabilities (amplitudes).
For mutually exclusive events C and D, to get the probability
(amplitude) of C *or* D happening, *add* the probabilities (amplitudes).
Ralph Hartley
Thomas Palm
Mar23-05, 04:34 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>"Sci~Girl" <palmtree117@juno.com> wrote in news:1111019108.240443.118670\n@o13g2000cwo.google groups.com:\n\n> In order to truly understand quantum mechanics I\'d need to understand\n> calculus. I HAVE THREE YEARS TO GO BEFORE CALCULUS. It would be okay to\n> just borrow a textbook if it was, say, next year\'s geometry I was\n> trying to get to, but it\'s much farther away. I\'d have to read the\n> geometry book, then the algebra 2 book, then the advanced math and\n> trigonometry book, before I even arrived at differential calculus. And\n> then there\'s integral... and I don\'t even know what that means. I\'m\n> sure it would be exciting and I\'ve thought about doing it before, but\n> it just doesn\'t seem like something within my capacity to do.\n\nYou could try to read Feynmann\'s "QED The strange theory of light and\nmatter". It was written based on a series of lectures he held trying to\nexplain quantum mechanics to an audience of laymen and contains\nessentially no math and yet outlines quantum mechanics at a rather deep\nlevel. The problem with it is that the level is so deep that it is hard\nto connect it to the elementary stuff you encounter elsewhere. But it\'s\na very short book so give it a try.\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>"Sci~Girl" <palmtree117@juno.com> wrote in news:1111019108.240443.118670
@o13g2000cwo.googlegroups.com:
> In order to truly understand quantum mechanics I'd need to understand
> calculus. I HAVE THREE YEARS TO GO BEFORE CALCULUS. It would be okay to
> just borrow a textbook if it was, say, next year's geometry I was
> trying to get to, but it's much farther away. I'd have to read the
> geometry book, then the algebra 2 book, then the advanced math and
> trigonometry book, before I even arrived at differential calculus. And
> then there's integral... and I don't even know what that means. I'm
> sure it would be exciting and I've thought about doing it before, but
> it just doesn't seem like something within my capacity to do.
You could try to read Feynmann's "QED The strange theory of light and
matter". It was written based on a series of lectures he held trying to
explain quantum mechanics to an audience of laymen and contains
essentially no math and yet outlines quantum mechanics at a rather deep
level. The problem with it is that the level is so deep that it is hard
to connect it to the elementary stuff you encounter elsewhere. But it's
a very short book so give it a try.
Mike Stay
Mar23-05, 04:35 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>"Sci~Girl" <palmtree117@juno.com> wrote in message news:<1111019108.240443.118670@o13g2000cwo.googleg roups.com>...\n> >In many ways the latter behaves as a single particle containing both\n> >spin up and spin down (which isn\'t zero).\n>\n> >You can have a pair of particles that is in a very real sense BOTH\n> >spin up and spin down. This is NOT the same as having a pair with >one\n> spin up and one spin down.\n>\n> That\'s... hard to comprehend... both up and down but not zero? If it\n> wasn\'t zero, either the up or the down would have to be dominant and\n> then it would be called just up or just down... Well, I know that\'s not\n> right, but I have no other way to think until I know more.\n\nThat\'s because in "Hilbert space", "spin up" and "spin down" are at\nright angles to each other! When you measure a quantum system, all\nthe possible answers are at right angles to each other. Here\'s a\nphysical example:\n\nYou can measure the polarization of photons with polarized sunglasses.\nLight that reflects off the surface of a pool ("glare"), for\ninstance, is polarized parallel to the surface. Polarized sunglasses\nare aligned such that only light perpendicular to the surface passes\nthrough (well, assuming you\'re upright when you wear them ;) so you\ncan see down into the pool.\n\nThe two results, parallel and perpendicular, are at right angles.\nThese form what\'s called a "basis". By taking different combinations\nof parallel and perpendicular (the "basis vectors"), you can get\ndiagonally polarized light, and even left- and right-handed\npolarizations. The set of all the different possible polarizations is\ncalled the "Hilbert space".\n\nWhen you have 45-degree polarized light, a photon will be absorbed\nhalf of the time and transmitted half of the time. To figure out what\nthe probability is for other angles, you\'ll need to do some trig. The\nlight itself lays on the hypotenuse of a triangle. The hypotenuse is\none unit long; the other two sides are on the basis vectors. You just\nsquare the length of the sides to figure out what the probability is\nthat it will be transmitted or absorbed.\n\nFor example, if you have light at 36.8 degrees from parallel (a 3-4-5\ntriangle), then the probability that it gets absorbed is (4/5)^2 =\n0.64 and the probability that it gets transmitted is (3/5)^2 = 0.36 .\n\nSpin and polarization are two names for the exact same thing. "Spin\nup" and "spin down" are at right angles, just like parallel and\nperpendicular. To measure the polarization (or "spin") of an\nelectron, you use a magnet instead of a polarized filter. The terms\n"up" and "down" come from the model of an electron as a spinning ball,\nwhich turns out to have some serious problems. You could just as well\ntalk about the electron being polarized parallel or perpendicular to\nthe magnetic field.\n\nTo give an example of entanglement, consider a "polarizing beam\nsplitter" (PBS): this is a mirror that\'s reflective to parallel light,\nbut transparent to perpendicular light.\n\nNow, if we send a diagonally polarized photon into the PBS, then\nthere\'s a chance it will go through the PBS and a chance it\'ll reflect\noff of it. If you detect it behind the PBS, then you also know that\nit\'s perpendicularly polarized without even measuring it. The\nphoton\'s position and polarization are entangled.\n\nNow comes the really wierd part. Consider a pair of photons. There\nare *four* different basis vectors for this pair of photons:\n\npara*para\npara*perp\nperp*para\nperp* perp\n\nYou can combine any of these in the same way that you combined\nparallel and perpendicular to get diagonal. For instance, we can take\n\npara*para + para*perp\n\nin equal amounts, and say that the first photon is polarized parallel,\nwhile the second one is 45-degree diagonal. These two photons are\n"separable", because we can describe them independently, i.e. we could\nwrite them as\n\npara*(para+perp)\n\nTwo 45-degree photons use all of the basis vectors:\n\n(para+perp)*(para+perp) =\npara*para + para*perp + perp*para + perp*perp\n\nBut we can also consider a pair of photons whose polarizations are\nentagled:\n\npara*perp + perp*para\n\nThere\'s no way to write this as (a)*(b), so these are entangled. You\nmight get either polarization when you measure one of them, but\nwhatever you got, you know what the other one will be.\n\nThere\'s a special crystal, called a "down-conversion" crystal that\ntakes an ultraviolet photon and emits two green photons that have\npolarizations related like this.\n--\nMike Stay\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>"Sci~Girl" <palmtree117@juno.com> wrote in message news:<1111019108.240443.118670@o13g2000cwo.googlegroups. com>...
> >In many ways the latter behaves as a single particle containing both
> >spin up and spin down (which isn't zero).
>
> >You can have a pair of particles that is in a very real sense BOTH
> >spin up and spin down. This is NOT the same as having a pair with >one
> spin up and one spin down.
>
> That's... hard to comprehend... both up and down but not zero? If it
> wasn't zero, either the up or the down would have to be dominant and
> then it would be called just up or just down... Well, I know that's not
> right, but I have no other way to think until I know more.
That's because in "Hilbert space", "spin up" and "spin down" are at
right angles to each other! When you measure a quantum system, all
the possible answers are at right angles to each other. Here's a
physical example:
You can measure the polarization of photons with polarized sunglasses.
Light that reflects off the surface of a pool ("glare"), for
instance, is polarized parallel to the surface. Polarized sunglasses
are aligned such that only light perpendicular to the surface passes
through (well, assuming you're upright when you wear them ;) so you
can see down into the pool.
The two results, parallel and perpendicular, are at right angles.
These form what's called a "basis". By taking different combinations
of parallel and perpendicular (the "basis vectors"), you can get
diagonally polarized light, and even left- and right-handed
polarizations. The set of all the different possible polarizations is
called the "Hilbert space".
When you have 45-degree polarized light, a photon will be absorbed
half of the time and transmitted half of the time. To figure out what
the probability is for other angles, you'll need to do some trig. The
light itself lays on the hypotenuse of a triangle. The hypotenuse is
one unit long; the other two sides are on the basis vectors. You just
square the length of the sides to figure out what the probability is
that it will be transmitted or absorbed.
For example, if you have light at 36.8 degrees from parallel (a 3-4-5
triangle), then the probability that it gets absorbed is (4/5)^2 =
.64 and the probability that it gets transmitted is (3/5)^2 = .36 .
Spin and polarization are two names for the exact same thing. "Spin
up" and "spin down" are at right angles, just like parallel and
perpendicular. To measure the polarization (or "spin") of an
electron, you use a magnet instead of a polarized filter. The terms
"up" and "down" come from the model of an electron as a spinning ball,
which turns out to have some serious problems. You could just as well
talk about the electron being polarized parallel or perpendicular to
the magnetic field.
To give an example of entanglement, consider a "polarizing beam
splitter" (PBS): this is a mirror that's reflective to parallel light,
but transparent to perpendicular light.
Now, if we send a diagonally polarized photon into the PBS, then
there's a chance it will go through the PBS and a chance it'll reflect
off of it. If you detect it behind the PBS, then you also know that
it's perpendicularly polarized without even measuring it. The
photon's position and polarization are entangled.
Now comes the really wierd part. Consider a pair of photons. There
are *four* different basis vectors for this pair of photons:
para*parapara*perpperp*paraperp*perp
You can combine any of these in the same way that you combined
parallel and perpendicular to get diagonal. For instance, we can take
para*para + para*perp
in equal amounts, and say that the first photon is polarized parallel,
while the second one is 45-degree diagonal. These two photons are
"separable", because we can describe them independently, i.e. we could
write them as
para*(para+perp)
Two 45-degree photons use all of the basis vectors:
(para+perp)*(para+perp) =para*para + para*perp + perp*para + perp*perp
But we can also consider a pair of photons whose polarizations are
entagled:
para*perp + perp*para
There's no way to write this as (a)*(b), so these are entangled. You
might get either polarization when you measure one of them, but
whatever you got, you know what the other one will be.
There's a special crystal, called a "down-conversion" crystal that
takes an ultraviolet photon and emits two green photons that have
polarizations related like this.
--
Mike Stay
<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>whopkins@csd.uwm.edu writes\n>\n>I describe the situation in Bell\'s theorem in clear terms.\n>news://sci.physics\n>2005, January 5\n>Alain Aspect Non-Locality/Entanglement experiment\n>http://groups-beta.google.com/groups?q=whopkins+entanglement+DIFFERENT+questions \n\nGosh, so you do.\n\nI don\'t know how I missed this at the time because its very simple and\nclear.\n\n===========And Hartley said:\nIndependent particles are like independent coin flips. Entangled\nparticles are like coin flips that are not independent.\n\nWe use the word "independent" in both quantum theory and probability,\nbut when it is absent we use two different words "entangled" and\n"correlated". I\'m not sure why (there is *some* logic to it).\n============\n\nSo it would seem to me that my visualisation of an entangled particle as\na *single* particle is not unreasonable. Its just a very extended\nparticle but I have no problem with that since I see extended particles\nall over the place and particularly photons with 2000km wavelengths. An\nentangled pair is thus just a particularly interesting wavefunction with\nTWO localised maxima of increasing separation. And why not?\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\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>whopkins@csd.uwm.edu writes
>
>I describe the situation in Bell's theorem in clear terms.
>news://sci.physics
>2005, January 5
>Alain Aspect Non-Locality/Entanglement experiment
>http://groups-\beta.google.com/groups?q=whopkins+entanglement+DIFFERENT+questions
Gosh, so you do.
I don't know how I missed this at the time because its very simple and
clear.
===========And Hartley said:
Independent particles are like independent coin flips. Entangled
particles are like coin flips that are not independent.
We use the word "independent" in both quantum theory and probability,
but when it is absent we use two different words "entangled" and
"correlated". I'm not sure why (there is *some* logic to it).
============
So it would seem to me that my visualisation of an entangled particle as
a *single* particle is not unreasonable. Its just a very extended
particle but I have no problem with that since I see extended particles
all over the place and particularly photons with 2000km wavelengths. An
entangled pair is thus just a particularly interesting wavefunction with
TWO localised maxima of increasing separation. And why not?
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
Ralph Hartley
Mar24-05, 01:10 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>Oz wrote:\n> So it would seem to me that my visualisation of an entangled particle as\n> a *single* particle is not unreasonable. Its just a very extended\n> particle but I have no problem with that since I see extended particles\n> all over the place and particularly photons with 2000km wavelengths. An\n> entangled pair is thus just a particularly interesting wavefunction with\n> TWO localised maxima of increasing separation. And why not?\n\nNo! Two particles, one distribution (wavefunction).\n\nIf I tossed two coins at the same time, and it was guaranteed that they\neither would land both heads, or both tails (e.g. because I was\ncheating), would you say there was only one coin?\n\nOf course not. You would say that the coin flips were not independent.\n\nTwo photons are two photons, not one.\n\nI can describe the probabilities of a coin flip with a function of one\n(boolean) variable P(X).\n\nI can use a linear table:\n\nTable (1)\nX | P(X)\n-------------\nHeads | 0.5\nTails | 0.5\n\nSuppose I toss a second coin and call the result Y. I can describe that\nthe same way:\n\nTable (2)\nY | P(Y)\n-------------\nHeads | 0.5\nTails | 0.5\n\nBut does that tell you everything you might want to know? No it doesn\'t.\nYou might want to know about combinations of the two flips. For that I\nwould need a function of two variables P(X,Y).\n\nI would need a square table:\n\nTable (3)\nY\nP(X,Y)| Heads | Tails\n----------------------\nX Heads | 0.25 | 0.25\nTails | 0.25 | 0.25\n\nThe table I just gave is a special case where P(X,Y) = P(X)P(Y). When\nthat is true we call X and Y independent. Instead of table (3) I could\nhave given you tables (1) and (2) and told you that X and Y are independent.\n\nBut suppose instead that the probabilities were given by\n\nTable (4)\nY\nP(X,Y)| Heads | Tails\n----------------------\nX Heads | 0.5 | 0.0\nTails | 0.0 | 0.5\n\nIf you look at X and Y separately, you will see that Tables (1) and (2)\nare still correct, but you would be unwise to bet me that the two coins\nwould be different.\n\nTwo photons are just the same. They *always* have *two* positions, spins\netc. Sometimes they may be independent. Then you can get away with\ndescribing them separately. You still can use the full joint\ndistribution, but you don\'t *have* to.\n\nIf they are entangled, You *do* have to use the full joint distribution,\nbecause describing the particles separately doesn\'t tell you everything\nyou want to know.\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>Oz wrote:
> So it would seem to me that my visualisation of an entangled particle as
> a *single* particle is not unreasonable. Its just a very extended
> particle but I have no problem with that since I see extended particles
> all over the place and particularly photons with 2000km wavelengths. An
> entangled pair is thus just a particularly interesting wavefunction with
> TWO localised maxima of increasing separation. And why not?
No! Two particles, one distribution (wavefunction).
If I tossed two coins at the same time, and it was guaranteed that they
either would land both heads, or both tails (e.g. because I was
cheating), would you say there was only one coin?
Of course not. You would say that the coin flips were not independent.
Two photons are two photons, not one.
I can describe the probabilities of a coin flip with a function of one
(boolean) variable P(X).
I can use a linear table:
Table (1)
X | P(X)
-------------
Heads | .5
Tails | .5
Suppose I toss a second coin and call the result Y. I can describe that
the same way:
Table (2)
Y | P(Y)
-------------
Heads | .5
Tails | .5
But does that tell you everything you might want to know? No it doesn't.
You might want to know about combinations of the two flips. For that I
would need a function of two variables P(X,Y).
I would need a square table:
Table (3)
Y
P(X,Y)| Heads | Tails
----------------------
X Heads | .25 | .25
Tails | .25 | .25
The table I just gave is a special case where P(X,Y) = P(X)P(Y). When
that is true we call X and Y independent. Instead of table (3) I could
have given you tables (1) and (2) and told you that X and Y are independent.
But suppose instead that the probabilities were given by
Table (4)
Y
P(X,Y)| Heads | Tails
----------------------
X Heads | .5 | .
Tails | . | .5
If you look at X and Y separately, you will see that Tables (1) and (2)
are still correct, but you would be unwise to bet me that the two coins
would be different.
Two photons are just the same. They *always* have *two* positions, spins
etc. Sometimes they may be independent. Then you can get away with
describing them separately. You still can use the full joint
distribution, but you don't *have* to.
If they are entangled, You *do* have to use the full joint distribution,
because describing the particles separately doesn't tell you everything
you want to know.
Ralph Hartley
<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>Ralph Hartley <hartley@aic.nrl.navy.mil> writes\n>Oz wrote:\n>> So it would seem to me that my visualisation of an entangled particle as\n>> a *single* particle is not unreasonable. Its just a very extended\n>> particle but I have no problem with that since I see extended particles\n>> all over the place and particularly photons with 2000km wavelengths. An\n>> entangled pair is thus just a particularly interesting wavefunction with\n>> TWO localised maxima of increasing separation. And why not?\n>\n>No! Two particles, one distribution (wavefunction).\n>\n>If I tossed two coins at the same time, and it was guaranteed that they\n>either would land both heads, or both tails (e.g. because I was\n>cheating), would you say there was only one coin?\n\nOf course.\n\n>Of course not. You would say that the coin flips were not independent.\n\nNo, I would claim that that was indistinguishable from a body rigidly\nholding the pennies both facing the same way.\n\n>Two photons are two photons, not one.\n\nNot if they are not independent.\n\n>I can describe the probabilities of a coin flip with a function of one\n>(boolean) variable P(X).\n\n>Table (1)\n> X | P(X)\n>-------------\n>Heads | 0.5\n>Tails | 0.5\n\nsnip\n\n>Table (3)\n> Y\n> P(X,Y)| Heads | Tails\n> ----------------------\n>X Heads | 0.25 | 0.25\n> Tails | 0.25 | 0.25\n\nsnip\n\n>But suppose instead that the probabilities were given by\n>\n>Table (4)\n> Y\n> P(X,Y)| Heads | Tails\n> ----------------------\n>X Heads | 0.5 | 0.0\n> Tails | 0.0 | 0.5\n>\n>If you look at X and Y separately, you will see that Tables (1) and (2)\n>are still correct, but you would be unwise to bet me that the two coins\n>would be different.\n\nQuite. Note though the structural similarity to table (1)\nIts IDENTICAL except I redefine\nhead = (head+head) and\ntail = (tail+tail)\n\n>Table (1)\n> X | P(X)\n>-------------\n>Heads | 0.5\n>Tails | 0.5\n\nThese are thus (in probability terms) identical and so the particle\nnumber (ie divisibility) is the same.\n\nWhat happens after is not important. By chopping the pennies in table\none into two identical halves AFTER the throw, I can have two pairs of\nidentical particles and I could in fact do many such choppings up (for a\ncircularly symmetrical penny).\n\n>Two photons are just the same. They *always* have *two* positions, spins\n>etc. Sometimes they may be independent. Then you can get away with\n>describing them separately. You still can use the full joint\n>distribution, but you don\'t *have* to.\n>\n>If they are entangled, You *do* have to use the full joint distribution,\n>because describing the particles separately doesn\'t tell you everything\n>you want to know.\n\nYes, but the result is a single particle (OK its a TAD more) and behaves\nlike one. A \'double photon\'.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\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>Ralph Hartley <hartley@aic.nrl.navy.mil> writes
>Oz wrote:
>> So it would seem to me that my visualisation of an entangled particle as
>> a *single* particle is not unreasonable. Its just a very extended
>> particle but I have no problem with that since I see extended particles
>> all over the place and particularly photons with 2000km wavelengths. An
>> entangled pair is thus just a particularly interesting wavefunction with
>> TWO localised maxima of increasing separation. And why not?
>
>No! Two particles, one distribution (wavefunction).
>
>If I tossed two coins at the same time, and it was guaranteed that they
>either would land both heads, or both tails (e.g. because I was
>cheating), would you say there was only one coin?
Of course.
>Of course not. You would say that the coin flips were not independent.
No, I would claim that that was indistinguishable from a body rigidly
holding the pennies both facing the same way.
>Two photons are two photons, not one.
Not if they are not independent.
>I can describe the probabilities of a coin flip with a function of one
>(boolean) variable P(X).
>Table (1)
> X | P(X)
>-------------
>Heads | .5
>Tails | .5
snip
>Table (3)
> Y
> P(X,Y)| Heads | Tails
> ----------------------
>X Heads | .25 | .25
> Tails | .25 | .25
snip
>But suppose instead that the probabilities were given by
>
>Table (4)
> Y
> P(X,Y)| Heads | Tails
> ----------------------
>X Heads | .5 | .
> Tails | . | .5
>
>If you look at X and Y separately, you will see that Tables (1) and (2)
>are still correct, but you would be unwise to bet me that the two coins
>would be different.
Quite. Note though the structural similarity to table (1)
Its IDENTICAL except I redefine
head = (head+head) and
tail = (tail+tail)
>Table (1)
> X | P(X)
>-------------
>Heads | .5
>Tails | .5
These are thus (in probability terms) identical and so the particle
number (ie divisibility) is the same.
What happens after is not important. By chopping the pennies in table
one into two identical halves AFTER the throw, I can have two pairs of
identical particles and I could in fact do many such choppings up (for a
circularly symmetrical penny).
>Two photons are just the same. They *always* have *two* positions, spins
>etc. Sometimes they may be independent. Then you can get away with
>describing them separately. You still can use the full joint
>distribution, but you don't *have* to.
>
>If they are entangled, You *do* have to use the full joint distribution,
>because describing the particles separately doesn't tell you everything
>you want to know.
Yes, but the result is a single particle (OK its a TAD more) and behaves
like one. A 'double photon'.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
Sci~Girl
Mar25-05, 04:37 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>[ Quotation and references modified by moderator.\nPlease quote and attribute properly in the future. -ik ]\n\nMike Stay wrote:\n> That\'s because in "Hilbert space", "spin up" and "spin down" are at\n> right angles to each other! When you measure a quantum system, all\n> the possible answers are at right angles to each other. Here\'s a\n> physical example:\n\nHoly cow. Ok now it makes a little more sense. To go back to the\nentanglement though - so how did they teleport a whole laser beam?\nEntangling all the photons?\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>[ Quotation and references modified by moderator.
Please quote and attribute properly in the future. -ik ]
Mike Stay wrote:
> That's because in "Hilbert space", "spin up" and "spin down" are at
> right angles to each other! When you measure a quantum system, all
> the possible answers are at right angles to each other. Here's a
> physical example:
Holy cow. Ok now it makes a little more sense. To go back to the
entanglement though - so how did they teleport a whole laser beam?
Entangling all the photons?
Sci~Girl
Mar26-05, 02:42 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>At 14? Well you could. Unfortunately you could really easily confuse\nyourself and make something easy into something difficult. It would\ntake\nquite a bit of your spare time up. It might be better to spend some of\nthat spare time \'socialising\' or playing football.\n\n\nMake something easy into something difficult? Ugh, I\'ve done that so\nmany times it drives me crazy. I already have independent study\nprojects to do in a couple classes so I don\'t know if I will have the\ntime.\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>At 14? Well you could. Unfortunately you could really easily confuse
yourself and make something easy into something difficult. It would
take
quite a bit of your spare time up. It might be better to spend some of
that spare time 'socialising' or playing football.
Make something easy into something difficult? Ugh, I've done that so
many times it drives me crazy. I already have independent study
projects to do in a couple classes so I don't know if I will have the
time.
whopkins@csd.uwm.edu
Mar28-05, 06:49 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>Sci~Girl wrote:\n> All I know in math is basic algebra though, I\'m 14 and in 8th grade\nbut\n> in 9th grade math. What\'s a vector?\n\n"Mark\'s Elements"\nMe\nnews://sci.math.research\nI. June 22, 1996\nII. June 24, 1996\nIII. June 25, 1996\nhttp://groups-beta.google.com/groups?q=%22Mark%27s+Elements%22\n\n(Part III is an axiomatization of Minkowski space, the earlier parts\ndeal with the formal definition of a vector space)\n\nAn updated (and corrected and formatted, the definition of Hilbert\nspace needs to be redone) copy of this will appear in the archive I\nmentioned previously ... along with more accessible writeups which show\nhow vector algebra and vector calculus works on the context of\nformulating and solving geometric or trigonometric problems.\n\n(For those already familiar with Calculus: the development of Complete\nMetric spaces in Mark\'s Elements uses NO EPSILON-DELTAS. Instead, an\nalternative concept -- the "Zeno sequence" is employed, which is more\ntransparent and even constructive.)\n\nFor those interested in gauge theory, there will also be material which\nprovides a substantial reworking of the theory if principal fibre\nbundles, using a new concept (the quotient operation) to significantly\nsimplify the subject and render it transparent.\n\nAlso of interest:\n"The Absolute: Intro to Relativity"\nMe\nJune 13 - August 24, 1995\nnews://sci.physics\nhttp://groups-beta.google.com/groups?q=%22The+Absolute%22+%22Intro+To+Relativity %22\n\n"TIME IS SPACE -- Part 2/5: The Hudgin Axioms"\nMe\nJune 13, 1995\nnews://sci.physics\nhttp://groups-beta.google.com/groups?q=%22The+Hudgin+Axioms%22\n\nEarly precursor to the material in Part III of Mark\'s Elements.\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>Sci~Girl wrote:
> All I know in math is basic algebra though, I'm 14 and in 8th grade
but
> in 9th grade math. What's a vector?
"Mark's Elements"
Me
news://sci.math.research
I. June 22, 1996
II. June 24, 1996
III. June 25, 1996
http://groups-\beta.google.com/groups?q=%22Mark%27s+Elements%22
(Part III is an axiomatization of Minkowski space, the earlier parts
deal with the formal definition of a vector space)
An updated (and corrected and formatted, the definition of Hilbert
space needs to be redone) copy of this will appear in the archive I
mentioned previously ... along with more accessible writeups which show
how vector algebra and vector calculus works on the context of
formulating and solving geometric or trigonometric problems.
(For those already familiar with Calculus: the development of Complete
Metric spaces in Mark's Elements uses NO \EPSILON-DELTAS. Instead, an
alternative concept -- the "Zeno sequence" is employed, which is more
transparent and even constructive.)
For those interested in gauge theory, there will also be material which
provides a substantial reworking of the theory if principal fibre
bundles, using a new concept (the quotient operation) to significantly
simplify the subject and render it transparent.
Also of interest:
"The Absolute: Intro to Relativity"
Me
June 13 - August 24, 1995
news://sci.physics
http://groups-\beta.google.com/groups?q=%22The+Absolute%22+%22Intro+To+Relativity %22
"TIME IS SPACE -- Part 2/5: The Hudgin Axioms"
Me
June 13, 1995
news://sci.physics
http://groups-\beta.google.com/groups?q=%22The+Hudgin+Axioms%22
Early precursor to the material in Part III of Mark's Elements.
Mike Stay
Mar29-05, 02:21 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>Sci~Girl <palmtree117@juno.com> wrote in message news:<1111776497.654784.22440@g14g2000cwa.googlegr oups.com>...\n> Mike Stay wrote:\n> > That\'s because in "Hilbert space", "spin up" and "spin down" are at\n> > right angles to each other! When you measure a quantum system, all\n> > the possible answers are at right angles to each other. Here\'s a\n> > physical example:\n>\n> Holy cow. Ok now it makes a little more sense. To go back to the\n> entanglement though - so how did they teleport a whole laser beam?\n> Entangling all the photons?\n\nYep. They shone one UV laser through the down-conversion crystal I\nmentioned, and got two entangled lasers coming out.\n\n--\nMike\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>Sci~Girl <palmtree117@juno.com> wrote in message news:<1111776497.654784.22440@g14g2000cwa.googlegroups.c om>...
> Mike Stay wrote:
> > That's because in "Hilbert space", "spin up" and "spin down" are at
> > right angles to each other! When you measure a quantum system, all
> > the possible answers are at right angles to each other. Here's a
> > physical example:
>
> Holy cow. Ok now it makes a little more sense. To go back to the
> entanglement though - so how did they teleport a whole laser beam?
> Entangling all the photons?
Yep. They shone one UV laser through the down-conversion crystal I
mentioned, and got two entangled lasers coming out.
--
Mike
whopkins@csd.uwm.edu
Mar29-05, 02:22 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>Sci~Girl wrote:\n> In order to truly understand quantum mechanics I\'d need to understand\n> calculus. I HAVE THREE YEARS TO GO BEFORE CALCULUS.\n\n"The Untold Story of Calculus"\nMe\nJune 17, 2002\nnews://sci.math\nhttp://groups-beta.google.com/groups?q=%22The+Untold+Story+Of+Calculus%22\n\n"A Trig Primer"\nMe\nJanuary 14, 2002\nnews://sci.math\nhttp://groups-beta.google.com/groups?q=%22A+Trig+Primer%22+whopkins\n\n"A Complete Introduction to Calculus"\nMe\nOctober 27, 1995\nnews://sci.math\nhttp://groups-beta.google.com/groups?q=%22Complete+Introduction+To+Calculus%22\n \n... or, if you\'re willing to wait, I\'m in the process of establishing\nan extensive archive which will have better typeset versions of all\nthese, as well as what will probably end up being upwards of 1000\npapers, monographs and expositories from me, in mathematics and physics\nand other areas. Stay tuned for future developments...\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>Sci~Girl wrote:
> In order to truly understand quantum mechanics I'd need to understand
> calculus. I HAVE THREE YEARS TO GO BEFORE CALCULUS.
"The Untold Story of Calculus"
Me
June 17, 2002
news://sci.math
http://groups-\beta.google.com/groups?q=%22The+Untold+Story+Of+Calculus%22
"A Trig Primer"
Me
January 14, 2002
news://sci.math
http://groups-\beta.google.com/groups?q=%22A+Trig+Primer%22+whopkins
"A Complete Introduction to Calculus"
Me
October 27, 1995
news://sci.math
http://groups-\beta.google.com/groups?q=%22Complete+Introduction+To+Calculus%22
... or, if you're willing to wait, I'm in the process of establishing
an extensive archive which will have better typeset versions of all
these, as well as what will probably end up being upwards of 1000
papers, monographs and expositories from me, in mathematics and physics
and other areas. Stay tuned for future developments...
whopkins@csd.uwm.edu
Mar29-05, 02:25 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>Sci~Girl wrote:\n> All I know in math is basic algebra though, I\'m 14 and in 8th grade but\n> in 9th grade math. What\'s a vector?\n\n"Mark\'s Elements"\nMe\nnews://sci.math.research\nI. June 22, 1996\nII. June 24, 1996\nIII. June 25, 1996\nhttp://groups-beta.google.com/groups?q=%22Mark%27s+Elements%22\n\n(Part III is an axiomatization of Minkowski space, the earlier parts\ndeal with the formal definition of a vector space)\n\nAn updated (and corrected and formatted, the definition of Hilbert\nspace needs to be redone) copy of this will appear in the archive I\nmentioned previously ... along with more accessible writeups which show\nhow vector algebra and vector calculus works on the context of\nformulating and solving geometric or trigonometric problems.\n\n(For those already familiar with Calculus: the development of Complete\nMetric spaces in Mark\'s Elements uses NO EPSILON-DELTAS. Instead, an\nalternative concept -- the "Zeno sequence" is employed, which is more\ntransparent and even constructive.)\n\nFor those interested in gauge theory, there will also be material which\nprovides a substantial reworking of the theory if principal fibre\nbundles, using a new concept (the quotient operation) to significantly\nsimplify the subject and render it transparent.\n\nAlso of interest:\n"The Absolute: Intro to Relativity"\nMe\nJune 13 - August 24, 1995\nnews://sci.physics\nhttp://groups-beta.google.com/groups?q=%22The+Absolute%22+%22Intro+To+Relativity %22\n\n"TIME IS SPACE -- Part 2/5: The Hudgin Axioms"\nMe\nJune 13, 1995\nnews://sci.physics\nhttp://groups-beta.google.com/groups?q=%22The+Hudgin+Axioms%22\n\nEarly precursor to the material in Part III of Mark\'s Elements.\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>Sci~Girl wrote:
> All I know in math is basic algebra though, I'm 14 and in 8th grade but
> in 9th grade math. What's a vector?
"Mark's Elements"
Me
news://sci.math.research
I. June 22, 1996
II. June 24, 1996
III. June 25, 1996
http://groups-\beta.google.com/groups?q=%22Mark%27s+Elements%22
(Part III is an axiomatization of Minkowski space, the earlier parts
deal with the formal definition of a vector space)
An updated (and corrected and formatted, the definition of Hilbert
space needs to be redone) copy of this will appear in the archive I
mentioned previously ... along with more accessible writeups which show
how vector algebra and vector calculus works on the context of
formulating and solving geometric or trigonometric problems.
(For those already familiar with Calculus: the development of Complete
Metric spaces in Mark's Elements uses NO \EPSILON-DELTAS. Instead, an
alternative concept -- the "Zeno sequence" is employed, which is more
transparent and even constructive.)
For those interested in gauge theory, there will also be material which
provides a substantial reworking of the theory if principal fibre
bundles, using a new concept (the quotient operation) to significantly
simplify the subject and render it transparent.
Also of interest:
"The Absolute: Intro to Relativity"
Me
June 13 - August 24, 1995
news://sci.physics
http://groups-\beta.google.com/groups?q=%22The+Absolute%22+%22Intro+To+Relativity %22
"TIME IS SPACE -- Part 2/5: The Hudgin Axioms"
Me
June 13, 1995
news://sci.physics
http://groups-\beta.google.com/groups?q=%22The+Hudgin+Axioms%22
Early precursor to the material in Part III of Mark's Elements.
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