Supspaces versus Projectors in QM foundations [Archive]

Vonny N

Nov29-04, 02:51 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>In foundational treatments of Quantum Mechanics one normally\nidentifies the elementary propositions/questions/events (whatever we\nchoose to call them) with the projectors (i.e. orthogonal projection\noperators on the underlying Hilbert Space). However Gleason\'s Theorem\nis actually stated and proved in terms of Subspaces rather than\nProjectors. Whilst this identification seems compelling, the\nrelationship between Projectors and Subspaces is not one-to-one. A\nprojector uniquely defines a subspace, however a subspace only defines\na projector up to a phase-factor. Now although the phase-factor\namibiguity is treated very carefully in Wigner\'s Theorem on Symmetry\nRepresentations, I have not seen the issue carefully addressed in\nrelation to Gleason\'s Theorem.\n\nMore succinctly, how can we be sure that subspaces can be replaced by\nprojectors in the foundations of QM when the latter are not\nuniquely-determined by the former?\n\nVonny 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>In foundational treatments of Quantum Mechanics one normally
identifies the elementary propositions/questions/events (whatever we
choose to call them) with the projectors (i.e. orthogonal projection
operators on the underlying Hilbert Space). However Gleason's Theorem
is actually stated and proved in terms of Subspaces rather than
Projectors. Whilst this identification seems compelling, the
relationship between Projectors and Subspaces is not one-to-one. A
projector uniquely defines a subspace, however a subspace only defines
a projector up to a phase-factor. Now although the phase-factor
amibiguity is treated very carefully in Wigner's Theorem on Symmetry
Representations, I have not seen the issue carefully addressed in
relation to Gleason's Theorem.

More succinctly, how can we be sure that subspaces can be replaced by
projectors in the foundations of QM when the latter are not
uniquely-determined by the former?

Vonny N

Pierre Asselin

Nov30-04, 12:50 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>Vonny N <vonnyn@hotmail.com> wrote:\n> [ ... ]\n> A projector uniquely defines a subspace, however a subspace only defines\n> a projector up to a phase-factor.\n\nAre you sure ? I would want my projectors to be idempotent,\nP^2 == P. That relation is not preserved if P is multiplied\nby a nontrivial global phase.\n\n--\npa at panix dot com\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>Vonny N <vonnyn@hotmail.com> wrote:
> [ ... ]
> A projector uniquely defines a subspace, however a subspace only defines
> a projector up to a phase-factor.

Are you sure ? I would want my projectors to be idempotent,
P^2 == P. That relation is not preserved if P is multiplied
by a nontrivial global phase.

--
pa at panix dot com

seratend

Dec1-04, 11:07 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>pa@see.signature.invalid (Pierre Asselin) wrote in message news:<coge54\\$4fl\\$1@reader1.panix.com>...\n> Vonny N <vonnyn@hotmail.com> wrote:\n> > [ ... ]\n> > A projector uniquely defines a subspace, however a subspace only defines\n> > a projector up to a phase-factor.\n>\n> Are you sure ? I would want my projectors to be idempotent,\n> P^2 == P. That relation is not preserved if P is multiplied\n> by a nontrivial global phase.\n\nAlways the same mistakes ;)\n\nA projector is any *hermitian* operator P that satisfies the relation\nP^2=P.\nThe hermitian condition is the one that fixes the phase.\n\nNote that the definition I gave is only a theorem coming from the\ndefinition of the projector (orthogonal subspace) and the properties\nof hilbert spaces. See post arnold neumaier: One of the important\nfeature of projectors is the *closed* subspace property.\n\nSeratend.\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>pa@see.signature.invalid (Pierre Asselin) wrote in message news:<coge54$4fl$1@reader1.panix.com>...
> Vonny N <vonnyn@hotmail.com> wrote:
> > [ ... ]
> > A projector uniquely defines a subspace, however a subspace only defines
> > a projector up to a phase-factor.
>
> Are you sure ? I would want my projectors to be idempotent,
> P^2 == P. That relation is not preserved if P is multiplied
> by a nontrivial global phase.

Always the same mistakes ;)

A projector is any *hermitian* operator P that satisfies the relation
P^2=P.
The hermitian condition is the one that fixes the phase.

Note that the definition I gave is only a theorem coming from the
definition of the projector (orthogonal subspace) and the properties
of hilbert spaces. See post arnold neumaier: One of the important
feature of projectors is the *closed* subspace property.

Seratend.

Pierre Asselin

Dec2-04, 06:16 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>seratend <ser_monmail@yahoo.fr> wrote:\n\n> A projector is any *hermitian* operator P that satisfies the relation\n> P^2=P.\n> The hermitian condition is the one that fixes the phase.\n\nOk, I\'ll bite. Do you have an example of a non-hermitian P\nsuch that P^2= P ? It would have to be non-normal as well,\nand possibly use infinite dimensionality in a tricky way.\nFinding one would be, like, work. Better to ask :-)\n\n--\npa at panix dot com\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>seratend <ser_monmail@yahoo.fr> wrote:

> A projector is any *hermitian* operator P that satisfies the relation
> P^2=P.
> The hermitian condition is the one that fixes the phase.

Ok, I'll bite. Do you have an example of a non-hermitian P
such that P^2= P ? It would have to be non-normal as well,
and possibly use infinite dimensionality in a tricky way.
Finding one would be, like, work. Better to ask :-)

--
pa at panix dot com

Arnold Neumaier

Dec3-04, 04:49 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>Pierre Asselin wrote:\n> seratend <ser_monmail@yahoo.fr> wrote:\n>\n>>A projector is any *hermitian* operator P that satisfies the relation\n>>P^2=P.\n>>The hermitian condition is the one that fixes the phase.\n>\n>\n> Ok, I\'ll bite. Do you have an example of a non-hermitian P\n> such that P^2= P ? It would have to be non-normal as well,\n> and possibly use infinite dimensionality in a tricky way.\n> Finding one would be, like, work. Better to ask :-)\n\nNo, better work. You won\'t learn much without. Try to find a 2x2 example\nby specifying one entry and solving the equation P^2=P for the remaining\nones. To save work, fix another entry by assuming that P is singular.\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>Pierre Asselin wrote:
> seratend <ser_monmail@yahoo.fr> wrote:
>
>>A projector is any *hermitian* operator P that satisfies the relation
>>P^2=P.
>>The hermitian condition is the one that fixes the phase.
>
>
> Ok, I'll bite. Do you have an example of a non-hermitian P
> such that P^2= P ? It would have to be non-normal as well,
> and possibly use infinite dimensionality in a tricky way.
> Finding one would be, like, work. Better to ask :-)

No, better work. You won't learn much without. Try to find a 2x2 example
by specifying one entry and solving the equation P^2=P for the remaining
ones. To save work, fix another entry by assuming that P is singular.

Arnold Neumaier

seratend

Dec3-04, 04:55 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>pa@see.signature.invalid (Pierre Asselin) wrote in message news:<colu19\\$164\\$1@reader1.panix.com>...\n> seratend <ser_monmail@yahoo.fr> wrote:\n>\n> > A projector is any *hermitian* operator P that satisfies the relation\n> > P^2=P.\n> > The hermitian condition is the one that fixes the phase.\n>\n> Ok, I\'ll bite. Do you have an example of a non-hermitian P\n> such that P^2= P ? It would have to be non-normal as well,\n> and possibly use infinite dimensionality in a tricky way.\n> Finding one would be, like, work. Better to ask :-)\n\nI do not understant what you call "non-normal". However, take 2\nvectors |a>,|b> from any hilbert space. Build the operator Q=|a><b|\n(<b| is the unique continuous linear form associated to |b>).\nQ^2=<b|a>.(|a><b|)=<b|a>.Q\n\nNow choose |b>, |a> such that <a|b>=1\nexample:|a>,|c> two orthogonal vectors. |b>=|a>+k|c>\n\nSeratend.\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>pa@see.signature.invalid (Pierre Asselin) wrote in message news:<colu19$164$1@reader1.panix.com>...
> seratend <ser_monmail@yahoo.fr> wrote:
>
> > A projector is any *hermitian* operator P that satisfies the relation
> > P^2=P.
> > The hermitian condition is the one that fixes the phase.
>
> Ok, I'll bite. Do you have an example of a non-hermitian P
> such that P^2= P ? It would have to be non-normal as well,
> and possibly use infinite dimensionality in a tricky way.
> Finding one would be, like, work. Better to ask :-)

I do not understant what you call "non-normal". However, take 2
vectors |a>,|b> from any hilbert space. Build the operator Q=|a><b|(<b| is the unique continuous linear form associated to |b>).Q^2=<b|a>.(|a><b|)=<b|a>.Q

Now choose |b>, |a> such that <a|b>=1
example:|a>,|c> two orthogonal vectors. |b>=|a>+k|c>

Seratend.

grubb@math.niu.edu

Dec3-04, 04:56 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>pa@see.signature.invalid (Pierre Asselin) wrote in message news:<colu19\\$164\\$1@reader1.panix.com>...\n> seratend <ser_monmail@yahoo.fr> wrote:\n>\n> > A projector is any *hermitian* operator P that satisfies the relation\n> > P^2=P.\n> > The hermitian condition is the one that fixes the phase.\n>\n> Ok, I\'ll bite. Do you have an example of a non-hermitian P\n> such that P^2= P ? It would have to be non-normal as well,\n> and possibly use infinite dimensionality in a tricky way.\n> Finding one would be, like, work. Better to ask :-)\n\nLet P=[[1 -1][0 0]], a 2x2 matrix. It is a projection, but\nnot an orthogonal projection.\n\n--Dan Grubb\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>pa@see.signature.invalid (Pierre Asselin) wrote in message news:<colu19$164$1@reader1.panix.com>...
> seratend <ser_monmail@yahoo.fr> wrote:
>
> > A projector is any *hermitian* operator P that satisfies the relation
> > P^2=P.
> > The hermitian condition is the one that fixes the phase.
>
> Ok, I'll bite. Do you have an example of a non-hermitian P
> such that P^2= P ? It would have to be non-normal as well,
> and possibly use infinite dimensionality in a tricky way.
> Finding one would be, like, work. Better to ask :-)

Let P=[[1 -1][0 0]], a 2x2 matrix. It is a projection, but
not an orthogonal projection.

--Dan Grubb

P. Gralewicz

Dec4-04, 03:18 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>grubb@math.niu.edu (grubb@math.niu.edu) wrote in message news:<551fc6d2.0412031018.78dcf254@posting.google. com>...\n> pa@see.signature.invalid (Pierre Asselin) wrote:\n> > seratend <ser_monmail@yahoo.fr> wrote:\n> >\n> > > A projector is any *hermitian* operator P that satisfies the relation\n> > > P^2=P.\n> > > The hermitian condition is the one that fixes the phase.\n> >\n> > Ok, I\'ll bite. Do you have an example of a non-hermitian P\n> > such that P^2= P ? It would have to be non-normal as well,\n> > and possibly use infinite dimensionality in a tricky way.\n> > Finding one would be, like, work. Better to ask :-)\n>\n> Let P=[[1 -1][0 0]], a 2x2 matrix. It is a projection, but\n> not an orthogonal projection.\n>\n> --Dan Grubb\n\nIdempotency P^2 = P, is an independent property of Hermiticity. Given\nan operator P:V->V on a linear space V, makes no sense to ask whether\nit is Hermitean or not without additional structure of a metric,\nbecause Hermiticity is defined for bilinear mappings Q:VxV->K. For a\nparticular P one can find a metric g such that Q=g.P and Q^+ = Q is\nHermitean. But then asking what Q^2 is turns nonsense.\n\nIn the above example P=[[1,-1],[0,0]] any metric of the form\ng=[[a,-a],[-a,b]], a<>b would do. (Additionally requiring g itself to\nbe Hermitean, makes a, b real.)\n\nbest\npg\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>grubb@math.niu.edu (grubb@math.niu.edu) wrote in message news:<551fc6d2.0412031018.78dcf254@posting.google.com>...
> pa@see.signature.invalid (Pierre Asselin) wrote:
> > seratend <ser_monmail@yahoo.fr> wrote:
> >
> > > A projector is any *hermitian* operator P that satisfies the relation
> > > P^2=P.
> > > The hermitian condition is the one that fixes the phase.
> >
> > Ok, I'll bite. Do you have an example of a non-hermitian P
> > such that P^2= P ? It would have to be non-normal as well,
> > and possibly use infinite dimensionality in a tricky way.
> > Finding one would be, like, work. Better to ask :-)
>
> Let P=[[1 -1][0 0]], a 2x2 matrix. It is a projection, but
> not an orthogonal projection.
>
> --Dan Grubb

Idempotency P^2 = P, is an independent property of Hermiticity. Given
an operator P:V->V on a linear space V, makes no sense to ask whether
it is Hermitean or not without additional structure of a metric,
because Hermiticity is defined for bilinear mappings Q:VxV->K. For a
particular P one can find a metric g such that Q=g.P and Q^+ = Q is
Hermitean. But then asking what Q^2 is turns nonsense.

In the above example P=[[1,-1],[0,0]] any metric of the form
g=[[a,-a],[-a,b]], a<>b would do. (Additionally requiring g itself to
be Hermitean, makes a, b real.)

best
pg

Vonny N

Dec7-04, 08:06 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\ngralp@poczta.onet.pl (P. Gralewicz) wrote in message news:<d719cdf7.0412040752.33dd61cd@posting.google. com>...\n> Idempotency P^2 = P, is an independent property of Hermiticity. Given\n> an operator P:V->V on a linear space V, makes no sense to ask whether\n> it is Hermitean or not without additional structure of a metric,\n> because Hermiticity is defined for bilinear mappings Q:VxV->K.\n\nIn Quantum Mechanics we are not dealing with an *arbitrary* linear\nspace V but with a *Hilbert* Space H in which the appropriate bilinear\nform is the canonical one (i.e. that which is induced by the inner\nproduct structure of H); thus there is no question as to the\nmeaningfulness of Hermiticity.\n\nVonny 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>gralp@poczta.onet.pl (P. Gralewicz) wrote in message news:<d719cdf7.0412040752.33dd61cd@posting.google.com>...
> Idempotency P^2 = P, is an independent property of Hermiticity. Given
> an operator P:V->V on a linear space V, makes no sense to ask whether
> it is Hermitean or not without additional structure of a metric,
> because Hermiticity is defined for bilinear mappings Q:VxV->K.

In Quantum Mechanics we are not dealing with an *arbitrary* linear
space V but with a *Hilbert* Space H in which the appropriate bilinear
form is the canonical one (i.e. that which is induced by the inner
product structure of H); thus there is no question as to the
meaningfulness of Hermiticity.

Vonny N

grubb@math.niu.edu

Dec7-04, 08:09 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nP. Gralewicz wrote:\n> grubb@math.niu.edu (grubb@math.niu.edu) wrote in message\nnews:<551fc6d2.0412031018.78dcf254@postin g.google.com>...\n> > pa@see.signature.invalid (Pierre Asselin) wrote:\n> > > seratend <ser_monmail@yahoo.fr> wrote:\n> > >\n> > > > A projector is any *hermitian* operator P that satisfies the\nrelation\n> > > > P^2=P.\n> > > > The hermitian condition is the one that fixes the phase.\n> > >\n> > > Ok, I\'ll bite. Do you have an example of a non-hermitian P\n> > > such that P^2= P ? It would have to be non-normal as well,\n> > > and possibly use infinite dimensionality in a tricky way.\n> > > Finding one would be, like, work. Better to ask :-)\n> >\n> > Let P=[[1 -1][0 0]], a 2x2 matrix. It is a projection, but\n> > not an orthogonal projection.\n> >\n> > --Dan Grubb\n>\n> Idempotency P^2 = P, is an independent property of Hermiticity. Given\n> an operator P:V->V on a linear space V, makes no sense to ask whether\n> it is Hermitean or not without additional structure of a metric,\n> because Hermiticity is defined for bilinear mappings Q:VxV->K. For a\n> particular P one can find a metric g such that Q=g.P and Q^+ = Q is\n> Hermitean. But then asking what Q^2 is turns nonsense.\n>\n> In the above example P=[[1,-1],[0,0]] any metric of the form\n> g=[[a,-a],[-a,b]], a<>b would do. (Additionally requiring g itself to\n> be Hermitean, makes a, b real.)\n>\n> best\n> pg\n\n\nYes, for any idempotent operator P, there is a metric for which P\nis Hermitian. However, for any metric, there is an idempotent that\nis not Hermitian (if the dimension is more than 1). I was using the\ndefault metric, of course.\n\n--Dan Grubb\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>P. Gralewicz wrote:
> grubb@math.niu.edu (grubb@math.niu.edu) wrote in message
news:<551fc6d2.0412031018.78dcf254@posting.google.com>...
> > pa@see.signature.invalid (Pierre Asselin) wrote:
> > > seratend <ser_monmail@yahoo.fr> wrote:
> > >
> > > > A projector is any *hermitian* operator P that satisfies the
relation
> > > > P^2=P.
> > > > The hermitian condition is the one that fixes the phase.
> > >
> > > Ok, I'll bite. Do you have an example of a non-hermitian P
> > > such that P^2= P ? It would have to be non-normal as well,
> > > and possibly use infinite dimensionality in a tricky way.
> > > Finding one would be, like, work. Better to ask :-)
> >
> > Let P=[[1 -1][0 0]], a 2x2 matrix. It is a projection, but
> > not an orthogonal projection.
> >
> > --Dan Grubb
>
> Idempotency P^2 = P, is an independent property of Hermiticity. Given
> an operator P:V->V on a linear space V, makes no sense to ask whether
> it is Hermitean or not without additional structure of a metric,
> because Hermiticity is defined for bilinear mappings Q:VxV->K. For a
> particular P one can find a metric g such that Q=g.P and Q^+ = Q is
> Hermitean. But then asking what Q^2 is turns nonsense.
>
> In the above example P=[[1,-1],[0,0]] any metric of the form
> g=[[a,-a],[-a,b]], a<>b would do. (Additionally requiring g itself to
> be Hermitean, makes a, b real.)
>
> best
> pg

Yes, for any idempotent operator P, there is a metric for which P
is Hermitian. However, for any metric, there is an idempotent that
is not Hermitian (if the dimension is more than 1). I was using the
default metric, of course.

--Dan Grubb

Soldalma

Sep14-05, 04:11 PM

The following is easy to prove: A and B are hermitian then AB is Hermitian if and only if A and B commute.

The following is harder, I could not do it: If A and B are Hermitian projectors, then AB is a projector (ABAB = AB) if and only if A and B commute. (The "if" part is easy) Does anyone know of a proof?

Thanks.