View Full Version : [SOLVED] If [A,B]=0, A=f(C) and B=g(C). An example.
John T Lowry
Jun4-04, 02:50 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>If quantum mechanics (QM) observables A and B are "compatible," ie,\n[A,B]=0, then (the saying goes) there must be some observable C, and\nfunctions f and g, such that A=f(C) and B=g(C). [eg, Fano\'s Mathematical\nMethods of QM, 1971, p 405.]\n\n\n\nBut could this poor beleaguered aviation physicist find even a single\napt example anywhere in his library? Hell no. But here\'s one.\n\n\n\nTake the spin-1/2 system, some fermion. S^2 = (Sx)^2 + (Sy)^2 + (Sz)^2 =\n(3((hbar)^2)/4)*1, where the \'1\' is the identity operator on this\n2-dimensional spin Hilbert space. And Sz = {(1 0), (0, -1)}, where I\'m\nsuggesting 2-by-2 matrices by showing the top row first, then the bottom\nrow. So the question is, what is the operator C, and the functions f and\ng, for which S^2 = f(C) and Sz = g(C)?\n\n\n\nAnswer: C = sqrt(hbar/2)*{(1 0), (0, i)}!\n\n\n\nWhere did I get that? If I\'d been smart, I\'d have remembered that the\nsquare of any Pauli spin matrix is the identity, but I didn\'t. Instead,\nI squared Sz, on spec, and saw I got ((hbar)^2/4)*1 and so assumed\n(hoped!) that by taking a square root of Sz something good would turn\nup. Sz is diagonal, so I did the simplest thing - in homology to\nfunctions of operators expressed via the spectral theorem - and took\nsquare roots of each diagonal entry. Something did turn up.\n\n\n\nThe end result is:\n\n\n\nC is as above, and\n\n\n\nS^2 = f(C) = 3C^4,\n\n\n\nSz = g(C) = C^2.\n\n\n\nAnd so those last two are functions of each other, too:\n\n\n\nSz = (1/sqrt(3))*sqrt(S^2)\n\n\n\nS^2 = 3(Sz)^2.\n\n\n\nFano also shows that any positive definite Hermitean operator has a\nsquare root, and how to get it (with difficulty!). So perhaps the above\nscheme can be pushed through for more complicated cases.\n\n\n\nCould it possibly be that QM\'s "need" to allow square root functions, as\nin the above, is responsible for its expression as a vector space over a\nCOMPLEX field of scalars? I wish I knew.\n\n\n--\nJohn T Lowry, PhD\n\nFlight Physics\n5217 Old Spicewood Springs Rd, #312\nAustin, Texas 78731\n(512) 231-9391\njlowry100@earthlink.net\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>If quantum mechanics (QM) observables A and B are "compatible," ie,
[A,B]=0, then (the saying goes) there must be some observable C, and
functions f and g, such that A=f(C) and B=g(C). [eg, Fano's Mathematical
Methods of QM, 1971, p 405.]
But could this poor beleaguered aviation physicist find even a single
apt example anywhere in his library? Hell no. But here's one.
Take the spin-1/2 system, some fermion. S^2 = (Sx)^2 + (Sy)^2 + (Sz)^2 =(3((\hbar)^2)/4)*1, where the '1' is the identity operator on this
2-dimensional spin Hilbert space. And Sz = {(1 0), (0, -1)}, where I'm
suggesting 2-by-2 matrices by showing the top row first, then the bottom
row. So the question is, what is the operator C, and the functions f and
g, for which S^2 = f(C) and Sz = g(C)?
Answer: C = \sqrt(\hbar/2)*{(1 0), (0, i)}!
Where did I get that? If I'd been smart, I'd have remembered that the
square of any Pauli spin matrix is the identity, but I didn't. Instead,
I squared Sz, on spec, and saw I got ((\hbar)^2/4)*1 and so assumed
(hoped!) that by taking a square root of Sz something good would turn
up. Sz is diagonal, so I did the simplest thing - in homology to
functions of operators expressed via the spectral theorem - and took
square roots of each diagonal entry. Something did turn up.
The end result is:
C is as above, and
S^2 = f(C) = 3C^4,Sz = g(C) = C^2.
And so those last two are functions of each other, too:
Sz = (1/\sqrt(3))*\sqrt(S^2)S^2 = 3(Sz)^2.
Fano also shows that any positive definite Hermitean operator has a
square root, and how to get it (with difficulty!). So perhaps the above
scheme can be pushed through for more complicated cases.
Could it possibly be that QM's "need" to allow square root functions, as
in the above, is responsible for its expression as a vector space over a
COMPLEX field of scalars? I wish I knew.
--
John T Lowry, PhD
Flight Physics
5217 Old Spicewood Springs Rd, #312
Austin, Texas 78731
(512) 231-9391
jlowry100@earthlink.net
John T Lowry
Jun6-04, 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>Partial Oops! The last part, about Sz and S^2 as functions of each\nother, is not correct because of cavalier root-taking. The fact that\nthere\'s a C such that Sz = f(C) and S^2=f(C), which is the content of\nFano\'s (and many others\') unproved theorem, stands.\n\n"John T Lowry" <jlowry100@earthlink.net> wrote in message\nnews:pLOvc.20734\\$Tn6.14255@newsread1.ne ws.pas.earthlink.net...\n> If quantum mechanics (QM) observables A and B are "compatible," ie,\n> [A,B]=0, then (the saying goes) there must be some observable C, and\n> functions f and g, such that A=f(C) and B=g(C). [eg, Fano\'s\nMathematical\n> Methods of QM, 1971, p 405.]\n>\n>\n>\n> But could this poor beleaguered aviation physicist find even a single\n> apt example anywhere in his library? Hell no. But here\'s one.\n>\n>\n>\n> Take the spin-1/2 system, some fermion. S^2 = (Sx)^2 + (Sy)^2 + (Sz)^2\n=\n> (3((hbar)^2)/4)*1, where the \'1\' is the identity operator on this\n> 2-dimensional spin Hilbert space. And Sz = {(1 0), (0, -1)}, where I\'m\n> suggesting 2-by-2 matrices by showing the top row first, then the\nbottom\n> row. So the question is, what is the operator C, and the functions f\nand\n> g, for which S^2 = f(C) and Sz = g(C)?\n>\n>\n>\n> Answer: C = sqrt(hbar/2)*{(1 0), (0, i)}!\n>\n>\n>\n> Where did I get that? If I\'d been smart, I\'d have remembered that the\n> square of any Pauli spin matrix is the identity, but I didn\'t.\nInstead,\n> I squared Sz, on spec, and saw I got ((hbar)^2/4)*1 and so assumed\n> (hoped!) that by taking a square root of Sz something good would turn\n> up. Sz is diagonal, so I did the simplest thing - in homology to\n> functions of operators expressed via the spectral theorem - and took\n> square roots of each diagonal entry. Something did turn up.\n>\n>\n>\n> The end result is:\n>\n>\n>\n> C is as above, and\n>\n>\n>\n> S^2 = f(C) = 3C^4,\n>\n>\n>\n> Sz = g(C) = C^2.\n>\n>\n>\n> And so those last two are functions of each other, too:\n>\n>\n>\n> Sz = (1/sqrt(3))*sqrt(S^2)\n>\n>\n>\n> S^2 = 3(Sz)^2.\n>\n>\n>\n> Fano also shows that any positive definite Hermitean operator has a\n> square root, and how to get it (with difficulty!). So perhaps the\nabove\n> scheme can be pushed through for more complicated cases.\n>\n>\n>\n> Could it possibly be that QM\'s "need" to allow square root functions,\nas\n> in the above, is responsible for its expression as a vector space over\na\n> COMPLEX field of scalars? I wish I knew.\n>\n>\n> --\n> John T Lowry, PhD\n>\n> Flight Physics\n> 5217 Old Spicewood Springs Rd, #312\n> Austin, Texas 78731\n> (512) 231-9391\n> jlowry100@earthlink.net\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>Partial Oops! The last part, about Sz and S^2 as functions of each
other, is not correct because of cavalier root-taking. The fact that
there's a C such that Sz = f(C) and S^2=f(C), which is the content of
Fano's (and many others') unproved theorem, stands.
"John T Lowry" <jlowry100@earthlink.net> wrote in message
news:pLOvc.20734$Tn6.14255@newsread1.news.pas.eart hlink.net...
> If quantum mechanics (QM) observables A and B are "compatible," ie,
> [A,B]=0, then (the saying goes) there must be some observable C, and
> functions f and g, such that A=f(C) and B=g(C). [eg, Fano's
Mathematical
> Methods of QM, 1971, p 405.]
>
>
>
> But could this poor beleaguered aviation physicist find even a single
> apt example anywhere in his library? Hell no. But here's one.
>
>
>
> Take the spin-1/2 system, some fermion. S^2 = (Sx)^2 + (Sy)^2 + (Sz)^2
=
> (3((\hbar)^2)/4)*1, where the '1' is the identity operator on this
> 2-dimensional spin Hilbert space. And Sz = {(1 0), (0, -1)}, where I'm
> suggesting 2-by-2 matrices by showing the top row first, then the
bottom
> row. So the question is, what is the operator C, and the functions f
and
> g, for which S^2 = f(C) and Sz = g(C)?
>
>
>
> Answer: C = \sqrt(\hbar/2)*{(1 0), (0, i)}!
>
>
>
> Where did I get that? If I'd been smart, I'd have remembered that the
> square of any Pauli spin matrix is the identity, but I didn't.
Instead,
> I squared Sz, on spec, and saw I got ((\hbar)^2/4)*1 and so assumed
> (hoped!) that by taking a square root of Sz something good would turn
> up. Sz is diagonal, so I did the simplest thing - in homology to
> functions of operators expressed via the spectral theorem - and took
> square roots of each diagonal entry. Something did turn up.
>
>
>
> The end result is:
>
>
>
> C is as above, and
>
>
>
> S^2 = f(C) = 3C^4,
>
>
>
> Sz = g(C) = C^2.
>
>
>
> And so those last two are functions of each other, too:
>
>
>
> Sz = (1/\sqrt(3))*\sqrt(S^2)
>
>
>
> S^2 = 3(Sz)^2.
>
>
>
> Fano also shows that any positive definite Hermitean operator has a
> square root, and how to get it (with difficulty!). So perhaps the
above
> scheme can be pushed through for more complicated cases.
>
>
>
> Could it possibly be that QM's "need" to allow square root functions,
as
> in the above, is responsible for its expression as a vector space over
a
> COMPLEX field of scalars? I wish I knew.
>
>
> --
> John T Lowry, PhD
>
> Flight Physics
> 5217 Old Spicewood Springs Rd, #312
> Austin, Texas 78731
> (512) 231-9391
> jlowry100@earthlink.net
Aaron Bergman
Jun6-04, 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>In article <pLOvc.20734\\$Tn6.14255@newsread1.news.pas.earthl ink.net>,\n"John T Lowry" <jlowry100@earthlink.net> wrote:\n\n> If quantum mechanics (QM) observables A and B are "compatible," ie,\n> [A,B]=0, then (the saying goes) there must be some observable C, and\n> functions f and g, such that A=f(C) and B=g(C). [eg, Fano\'s Mathematical\n> Methods of QM, 1971, p 405.]\n\nI\'m pretty sure this is just the statement that A and B are\nsimultaneously diagonalizable.\n\nAaron\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 article <pLOvc.20734$Tn6.14255@newsread1.news.pas.earthlink .net>,
"John T Lowry" <jlowry100@earthlink.net> wrote:
> If quantum mechanics (QM) observables A and B are "compatible," ie,
> [A,B]=0, then (the saying goes) there must be some observable C, and
> functions f and g, such that A=f(C) and B=g(C). [eg, Fano's Mathematical
> Methods of QM, 1971, p 405.]
I'm pretty sure this is just the statement that A and B are
simultaneously diagonalizable.
Aaron
John T Lowry
Jun7-04, 12:36 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"Aaron Bergman" <abergman@physics.utexas.edu> wrote in message\nnews:abergman-5BB496.20374404062004@localhost...\n> In article <pLOvc.20734\\$Tn6.14255@newsread1.news.pas.earthl ink.net>,\n> "John T Lowry" <jlowry100@earthlink.net> wrote:\n>\n> > If quantum mechanics (QM) observables A and B are "compatible," ie,\n> > [A,B]=0, then (the saying goes) there must be some observable C, and\n> > functions f and g, such that A=f(C) and B=g(C). [eg, Fano\'s\nMathematical\n> > Methods of QM, 1971, p 405.]\n>\n> I\'m pretty sure this is just the statement that A and B are\n> simultaneously diagonalizable.\n>\n> Aaron\n>\nTrue, but I was looking for a concrete example of the two commuting\nobservables being both functions of a single third observable.\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>"Aaron Bergman" <abergman@physics.utexas.edu> wrote in message
news:abergman-5BB496.20374404062004@localhost...
> In article <pLOvc.20734$Tn6.14255@newsread1.news.pas.earthlink .net>,
> "John T Lowry" <jlowry100@earthlink.net> wrote:
>
> > If quantum mechanics (QM) observables A and B are "compatible," ie,
> > [A,B]=0, then (the saying goes) there must be some observable C, and
> > functions f and g, such that A=f(C) and B=g(C). [eg, Fano's
Mathematical
> > Methods of QM, 1971, p 405.]
>
> I'm pretty sure this is just the statement that A and B are
> simultaneously diagonalizable.
>
> Aaron
>
True, but I was looking for a concrete example of the two commuting
observables being both functions of a single third observable.
Robert C. Helling
Jun12-04, 07:11 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Mon, 7 Jun 2004 17:36:37 +0000 (UTC), John T Lowry <jlowry100@earthlink.net> wrote:\n> "Aaron Bergman" <abergman@physics.utexas.edu> wrote in message\n> news:abergman-5BB496.20374404062004@localhost...\n>> In article <pLOvc.20734\\$Tn6.14255@newsread1.news.pas.earthl ink.net>,\n>> "John T Lowry" <jlowry100@earthlink.net> wrote:\n>>\n>> > If quantum mechanics (QM) observables A and B are "compatible," ie,\n>> > [A,B]=0, then (the saying goes) there must be some observable C, and\n>> > functions f and g, such that A=f(C) and B=g(C). [eg, Fano\'s\n> Mathematical\n>> > Methods of QM, 1971, p 405.]\n>>\n>> I\'m pretty sure this is just the statement that A and B are\n>> simultaneously diagonalizable.\n>>\n>> Aaron\n>>\n> True, but I was looking for a concrete example of the two commuting\n> observables being both functions of a single third observable.\n\nI would say, this statement is trivially true but not of much use: For\nsimplicity take finite dimensional matrices instead of operators. Then\nthere is a basis in which A and B are diagonal with eigenvalues a1,\na2,... and b1, b2,... resp. Then define C to be diagonal with\neigenvalues 1,2,3,... and define your functions f and g for example to\nbe the polynomials that have f(1)=a1, f(2)=a2,... and g(1)=b1,\ng(2)=b2,...\n\n\nRobert\n\n\n--\n..oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oO o.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oO\nRobert C. Helling Department of Applied Mathematics and Theoretical Physics\nUniversity of Cambridge\nprint "Just another Phone: +44/1223/766870\nstupid .sig\\n"; http://www.aei-potsdam.mpg.de/~helling\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 Mon, 7 Jun 2004 17:36:37 +0000 (UTC), John T Lowry <jlowry100@earthlink.net> wrote:
> "Aaron Bergman" <abergman@physics.utexas.edu> wrote in message
> news:abergman-5BB496.20374404062004@localhost...
>> In article <pLOvc.20734$Tn6.14255@newsread1.news.pas.earthlink .net>,
>> "John T Lowry" <jlowry100@earthlink.net> wrote:
>>
>> > If quantum mechanics (QM) observables A and B are "compatible," ie,>> > [A,B]=0, then (the saying goes) there must be some observable C, and
>> > functions f and g, such that A=f(C) and B=g(C). [eg, Fano's
> Mathematical
>> > Methods of QM, 1971, p 405.]
>>
>> I'm pretty sure this is just the statement that A and B are
>> simultaneously diagonalizable.
>>
>> Aaron
>>
> True, but I was looking for a concrete example of the two commuting
> observables being both functions of a single third observable.
I would say, this statement is trivially true but not of much use: For
simplicity take finite dimensional matrices instead of operators. Then
there is a basis in which A and B are diagonal with eigenvalues a1,
a2,... and b1, b2,... resp. Then define C to be diagonal with
eigenvalues 1,2,3,... and define your functions f and g for example to
be the polynomials that have f(1)=a1, f(2)=a2,... and g(1)=b1,g(2)=b2,...
Robert
--
..oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo. oOo.oOo.oOo.oOo.oOo.oOo.oOo.oO
Robert C. Helling Department of Applied Mathematics and Theoretical Physics
University of Cambridge
print "Just another Phone: +44/1223/766870
stupid .sig\n"; http://www.aei-potsdam.mpg.de/~helling
Here's a simpler example. Let C = |1><1| + 2|2><2| + 3|3><3|, let A = |1><1| + 2(|2><2| +|3><3|) and let B = 2(|1><1| + |3><3|) + |2><2|.
Then any function f, which gives
f(1) = 1, f(2) = 2, f(3) = 2
will give A = f(C) and any function g, which gives
g(1) = 2, g(2) = 1, g(3) = 2
will give B = g(C).
I would say, this statement is trivially true but not of much use: For simplicity take finite dimensional matrices instead of operators. Then
there is a basis in which A and B are diagonal with eigenvalues a1,
a2,... and b1, b2,... resp. Then define C to be diagonal with
eigenvalues 1,2,3,... and define your functions f and g for example to
be the polynomials that have ... and
That's almost correct, but you have to take account of the fact that A and B might be coarse-grainings of C, i.e. more than one eigenvalue of C might correspond to the same eigenvalue of A or B. In fact, that's the only case where the statement is not really trivial.
Robert C. Helling
Jun17-04, 04:19 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nslyboy <slyboy@physicsforums.com> wrote in message news:<slyboy.17uvsh@physicsforums.com>...\n> > I would say, this statement is trivially true but not of much use: For\n> > simplicity take finite dimensional matrices instead of operators.\n> > Then\n> > there is a basis in which A and B are diagonal with eigenvalues a1,\n> > a2,... and b1, b2,... resp. Then define C to be diagonal with\n> > eigenvalues 1,2,3,... and define your functions f and g for example\n> > to\n> > be the polynomials that have ... and\n>\n> That\'s almost correct, but you have to take account of the fact that A\n> and B might be coarse-grainings of C, i.e. more than one eigenvalue of\n> C might correspond to the same eigenvalue of A or B. In fact, that\'s\n> the only case where the statement is not really trivial.\n\nIn the original message it said something along the lines of "there is\nan operator C, such..", thus you can choose it an make sure it does\nnot have degenerate eigenvalues.\n\nRobert\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>slyboy <slyboy@physicsforums.com> wrote in message news:<slyboy.17uvsh@physicsforums.com>...
> > I would say, this statement is trivially true but not of much use: For
> > simplicity take finite dimensional matrices instead of operators.
> > Then
> > there is a basis in which A and B are diagonal with eigenvalues a1,
> > a2,... and b1, b2,... resp. Then define C to be diagonal with
> > eigenvalues 1,2,3,... and define your functions f and g for example
> > to
> > be the polynomials that have ... and
>
> That's almost correct, but you have to take account of the fact that A
> and B might be coarse-grainings of C, i.e. more than one eigenvalue of
> C might correspond to the same eigenvalue of A or B. In fact, that's
> the only case where the statement is not really trivial.
In the original message it said something along the lines of "there is
an operator C, such..", thus you can choose it an make sure it does
not have degenerate eigenvalues.
Robert
Matt Leifer
Jun18-04, 02:05 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n> In the original message it said something along the lines of "there is\n> an operator C, such..", thus you can choose it an make sure it does\n> not have degenerate eigenvalues.\n\n\nOf course that is true, but the point is not that C might have\ndegenerate eigenvalues, but rather that A and B might have. Then,\nthere is no 1-1 function from eigenvalues of C to eigenvalues of A or\nto eigenvalues of B. This is where interesting things can start to\nhappen, such as proofs of the Kochen-Spekker theorem.\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 the original message it said something along the lines of "there is
> an operator C, such..", thus you can choose it an make sure it does
> not have degenerate eigenvalues.
Of course that is true, but the point is not that C might have
degenerate eigenvalues, but rather that A and B might have. Then,
there is no 1-1 function from eigenvalues of C to eigenvalues of A or
to eigenvalues of B. This is where interesting things can start to
happen, such as proofs of the Kochen-Spekker theorem.
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