View Full Version : unitary irreps of Galilei group
Arnold Neumaier
Apr14-04, 03:17 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>Are all unitary irreducible representations of the\nGalilei group classified? Do they arise by a limiting\nconstruction from those of the Poincare group?\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>Are all unitary irreducible representations of the
Galilei group classified? Do they arise by a limiting
construction from those of the Poincare group?
Arnold Neumaier
Martin Ouwehand
Apr14-04, 08:36 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\nDans l\'article <407BFF25.1070900@univie.ac.at>,\nArnold Neumaier <Arnold.Neumaier@univie.ac.at> écrit:\n\n] Are all unitary irreducible representations of the\n] Galilei group classified?\n\nI think this was done by Inönü, Wigner and Bargmann in the fifties.\nA short introduction with references to the original articles is\nin Varadarajan\'s "Geometry of quantum physics".\n\nIt turns out that the purely unitary representations are physically\nunenlightening, you have to consider projective representations (up to\na phase) to find the usual quantum mechanics of the non-relativistic\nfree particle.\n\n] Do they arise by a limiting\n] construction from those of the Poincare group?\n\nI don\'t know if everything is known about this (what do all the representations\nof the Poincaré group become in the limit ? do all the representations\nof the Galilei group arise from such a limit ?), but the study of this\nlimiting process is mentioned by Inönü:\n\nhttp://www.physics.umd.edu/robot/wigner/inonu.pdf\n\n--\n| ~~~~~~~~ Martin Ouwehand ~ Swiss Federal Institute of Technology ~ Lausanne\n__|_____________ Email/PGP: http://slwww.epfl.ch/info/Martin.html _____________\nIn 10-15 years you\'ll be happy with software that\'s 5 years old [L. Torvalds]\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>Dans l'article <407BFF25.1070900@univie.ac.at>,
Arnold Neumaier <Arnold.Neumaier@univie.ac.at> écrit:
] Are all unitary irreducible representations of the
] Galilei group classified?
I think this was done by Inönü, Wigner and Bargmann in the fifties.
A short introduction with references to the original articles is
in Varadarajan's "Geometry of quantum physics".
It turns out that the purely unitary representations are physically
unenlightening, you have to consider projective representations (up to
a phase) to find the usual quantum mechanics of the non-relativistic
free particle.
] Do they arise by a limiting
] construction from those of the Poincare group?
I don't know if everything is known about this (what do all the representations
of the Poincaré group become in the limit ? do all the representations
of the Galilei group arise from such a limit ?), but the study of this
limiting process is mentioned by Inönü:
http://www.physics.umd.edu/robot/wigner/inonu.pdf
--
| ~~~~~~~~ Martin Ouwehand ~ Swiss Federal Institute of Technology ~ Lausanne
__|__{___________} Email/PGP: http://slwww.epfl.ch/info/Martin.html __{___________}
In 10-15 years you'll be happy with software that's 5 years old [L. Torvalds]
Danny Ross Lunsford
Apr15-04, 02:56 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>Arnold Neumaier wrote:\n\n> Are all unitary irreducible representations of the\n> Galilei group classified? Do they arise by a limiting\n> construction from those of the Poincare group?\n\nHere is a paper with some interesting and little known information:\n\nhttp://arxiv.org/abs/hep-th/0007199\n\nThe appearence of SL(2,R) from the blue as it were will surprise many!\nI\'m sure it shows up as some kind of limit of SL(2,C) as c->inf.\n\n(For the person asking about 2x2 real matrices - here is a physical\napplication :)\n\n-drl\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>Arnold Neumaier wrote:
> Are all unitary irreducible representations of the
> Galilei group classified? Do they arise by a limiting
> construction from those of the Poincare group?
Here is a paper with some interesting and little known information:
http://arxiv.org/abs/hep-th/0007199
The appearence of SL(2,R) from the blue as it were will surprise many!
I'm sure it shows up as some kind of limit of SL(2,C) as c->inf.
(For the person asking about 2x2 real matrices - here is a physical
application :)
-drl
Arnold Neumaier
Apr15-04, 11:30 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>Martin Ouwehand wrote:\n> Dans l\'article <407BFF25.1070900@univie.ac.at>,\n> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> =E9crit:\n>=20\n> ] Are all unitary irreducible representations of the\n> ] Galilei group classified?\n>=20\n> I think this was done by In=F6n=FC, Wigner and Bargmann in the fifties.=\n\n> A short introduction with references to the original articles is\n> in Varadarajan\'s "Geometry of quantum physics".\n>=20\n\nThanks. Two references are\n\nE.P. Wigner and E. In=F6n=FC,\nProc. Natl. Acad . Sci 39, 510 (1953).\n\nE. In=F6n=FC and E. Wigner,\nNuovo cimento 9, 705 (1952).\n\n\n> It turns out that the purely unitary representations are physically\n> unenlightening, you have to consider projective representations (up to\n> a phase) to find the usual quantum mechanics of the non-relativistic\n> free particle.\n>=20\n> ] Do they arise by a limiting\n> ] construction from those of the Poincare group?\n>=20\n> I don\'t know if everything is known about this (what do all the represe=\nntations\n> of the Poincar=E9 group become in the limit ? do all the representation=\ns\n> of the Galilei group arise from such a limit ?), but the study of this\n> limiting process is mentioned by In=F6n=FC:\n>=20\n> http://www.physics.umd.edu/robot/wigner/inonu.pdf\n\n\nArnold Neumaier\n\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Martin Ouwehand wrote:
> Dans l'article <407BFF25.1070900@univie.ac.at>,
> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> =E9crit:
>=20
> ] Are all unitary irreducible representations of the
> ] Galilei group classified?
>=20
> I think this was done by In=F6n=FC, Wigner and Bargmann in the fifties.=
> A short introduction with references to the original articles is
> in Varadarajan's "Geometry of quantum physics".
>=20
Thanks. Two references are
E.P. Wigner and E. In=F6n=FC,
Proc. Natl. Acad . Sci 39, 510 (1953).
E. In=F6n=FC and E. Wigner,
Nuovo cimento 9, 705 (1952).
> It turns out that the purely unitary representations are physically
> unenlightening, you have to consider projective representations (up to
> a phase) to find the usual quantum mechanics of the non-relativistic
> free particle.
>=20
> ] Do they arise by a limiting
> ] construction from those of the Poincare group?
>=20
> I don't know if everything is known about this (what do all the represe=
ntations
> of the Poincar=E9 group become in the limit ? do all the representation=
s
> of the Galilei group arise from such a limit ?), but the study of this
> limiting process is mentioned by In=F6n=FC:
>=20
> http://www.physics.umd.edu/robot/wigner/inonu.pdf
Arnold Neumaier
Alfred Einstead
Apr30-04, 03:02 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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n> Are all unitary irreducible representations of the\n> Galilei group classified? Do they arise by a limiting\n> construction from those of the Poincare group?\n\nActually, what you really want to look at are the representations\ngiven up to complex unit:\npi(A) pi(B) = U(A,B) pi(AB)\n|U(A,B)|^2 = 1.\nI think this is all classified. Look up Prugovecki.\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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
> Are all unitary irreducible representations of the
> Galilei group classified? Do they arise by a limiting
> construction from those of the Poincare group?
Actually, what you really want to look at are the representations
given up to complex unit:
\pi(A) \pi(B) = U(A,B) \pi(AB)|U(A,B)|^2 = 1.
I think this is all classified. Look up Prugovecki.
Alfred Einstead
May6-04, 01:15 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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n> Are all unitary irreducible representations of the\n> Galilei group classified? Do they arise by a limiting\n> construction from those of the Poincare group?\n\nWell, here you go for starters. The Lie algebra for the\nGalilei group can be given as:\n[J_a,J_b] = e^c_{ab} J^c\n[J_a,K_b] = e^c_{ab} K^c\n[J_a,P_b] = e^c_{ab} P^c\n[K_a,E] = P_a\na,b,c range over 1,2,3; summation convention used\ne^1_{23} = 1 = -e^1_{32} with cyclic permutations\ne^a_{bc} = 0 if a=b,b=c or c=a.\n\nThe invariants F(J,K,P,E) are found from:\n[J,F] = -(J x j + K x k + P x p) = 0\n[K,F] = P e - K x j = 0\n[P,F] = -P x j = 0\n[E,F] = -P.k = 0\nusing 3-vector notation, where\ndF = j.dJ + k.dK + p.dP + edE.\nWhen KxP != 0, one gets:\nP x j = 0 --> j = cP\nPe - Kxj = 0 --> Pe = c KxP --> c=e=0 --> j=e=0.\nKxk + Pxp = 0 --> PxK.k = 0, KxP.p = 0\nP.k = 0 --> k = aPx(PxK)\nSo\na Kx(Px(PxK)) + Pxp = 0\n-aK.P PxK + Pxp = 0\np = aK.P K + bP\nand\ndF = a (Px(PxK).dK + K.PK.dP) + bP.dP\n= a/2 d(|KxP|^2) + b/2 d(|P|^2).\nSo\nF = F(|KxP|^2, |P|^2)\nand the two independent invariants are:\n|KxP|^2 and |P|^2.\n\nWhen P != 0, one also finds that\n[E,t] = 1, where t = K.P/|P|^2.\n\nSo, the Heisenberg algebra is included within the overall algebra,\nover the orbits where |P| != 0.\n\nThe other invariants that arise in special cases are:\nwhen |KxP| = 0 --> P.J\nwhen |P| = 0 --> E, |K|^2 and K.J\nwhen |P| = |K| = 0 --> |J|^2.\n\nThe Lie algebra overall is not semisimple and has a maximal solvable\nideal generated by:\n(P,K,E).\nWhen |P| is not 0, one can define W = KxP/|P|^2 and write\nK = PxW + Pt.\nThe algebra generated by (J,P,W) and that generated by (E,t) are\nmutually commuting; the latter being a copy of the Heisenberg\nalgebra.\n\nRepresentation theory gets more complex for algebras that are\nnot semisimple. The 1st and 2nd cohomologies of the Lie algebras\nare non-zero; and so then central extensions are non-trivial.\nThe general representation is then only given up to a U(1) factor,\nas indicated in the previous article.\n\nThis is a substantial point of difference from the Poincare\'\nalgebra, whose corresponding (J,K) elements form the (semisimple)\nalgebra for SO(3,1), the Lorentz group; with (P,E) giving you an\nindirect product ISO(3,1) -- and representation theory for indirect\nproducts is a well-mapped-out field.\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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
> Are all unitary irreducible representations of the
> Galilei group classified? Do they arise by a limiting
> construction from those of the Poincare group?
Well, here you go for starters. The Lie algebra for the
Galilei group can be given as:
[J_a,J_b] = e^{c_}{ab} J^c[J_a,K_b] = e^{c_}{ab} K^c[J_a,P_b] = e^{c_}{ab} P^c[K_a,E] = P_a
a,b,c range over 1,2,3; summation convention used
e^{1_}{23} = 1 = -e^{1_}{32} with cyclic permutations
e^{a_}{bc} = if a=b,b=c or c=a.
The invariants F(J,K,P,E) are found from:
[J,F] = -(J x j + K x k + P x p) =
[K,F] = P e - K x j =
[P,F] = -P x j =
[E,F] = -P.k =
using 3-vector notation, where
dF = j.dJ + k.dK + p.dP + edE.
When KxP != 0, one gets:
P x j = --> j = cPPe - Kxj = --> Pe = c KxP --> c=e=0 --> j=e=0.
Kxk + Pxp = --> PxK.k = 0, KxP.p = P.k = --> k = aPx(PxK)
So
a Kx(Px(PxK)) + Pxp =
-aK.P PxK + Pxp =
p = aK.P K + bP
and
dF = a (Px(PxK).dK + K.PK.dP) + bP.dP= a/2 d(|KxP|^2) + b/2 d(|P|^2).
So
F = F(|KxP|^2, |P|^2)
and the two independent invariants are:
|KxP|^2 and |P|^2.
When P != 0, one also finds that
[E,t] = 1, where t = K.P/|P|^2.
So, the Heisenberg algebra is included within the overall algebra,
over the orbits where |P| != .
The other invariants that arise in special cases are:
when |KxP| = --> P.J
when |P| = --> E, |K|^2 and K.J
when |P| = |K| = --> |J|^2.
The Lie algebra overall is not semisimple and has a maximal solvable
ideal generated by:
(P,K,E).
When |P| is not 0, one can define W = KxP/|P|^2 and write
K = PxW + Pt.
The algebra generated by (J,P,W) and that generated by (E,t) are
mutually commuting; the latter being a copy of the Heisenberg
algebra.
Representation theory gets more complex for algebras that are
not semisimple. The 1st and 2nd cohomologies of the Lie algebras
are non-zero; and so then central extensions are non-trivial.
The general representation is then only given up to a U(1) factor,
as indicated in the previous article.
This is a substantial point of difference from the Poincare'
algebra, whose corresponding (J,K) elements form the (semisimple)
algebra for SO(3,1), the Lorentz group; with (P,E) giving you an
indirect product ISO(3,1) -- and representation theory for indirect
products is a well-mapped-out field.
Arnold Neumaier
May10-04, 06:02 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>\nAlfred Einstead wrote:\n> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n>\n>>Are all unitary irreducible representations of the\n>>Galilei group classified? Do they arise by a limiting\n>>construction from those of the Poincare group?\n>\n>\n> Well, here you go for starters. [...]\n>\n> Representation theory gets more complex for algebras that are\n> not semisimple. The 1st and 2nd cohomologies of the Lie algebras\n> are non-zero; and so then central extensions are non-trivial.\n> The general representation is then only given up to a U(1) factor,\n> as indicated in the previous article.\n>\n> This is a substantial point of difference from the Poincare\'\n> algebra, whose corresponding (J,K) elements form the (semisimple)\n> algebra for SO(3,1), the Lorentz group; with (P,E) giving you an\n> indirect product ISO(3,1) -- and representation theory for indirect\n> products is a well-mapped-out field.\n\nThanks.\n\nIs there a nice way of parameterizing the Poincare group and its physically\nrelevant unitary irreducible representations in such a way that the\ncorresponding representations of the Galilei group result\nas c -> infinity?\n\n\nArnold Neumaier\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Alfred Einstead wrote:
> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
>
>>Are all unitary irreducible representations of the
>>Galilei group classified? Do they arise by a limiting
>>construction from those of the Poincare group?
>
>
> Well, here you go for starters. [...]
>
> Representation theory gets more complex for algebras that are
> not semisimple. The 1st and 2nd cohomologies of the Lie algebras
> are non-zero; and so then central extensions are non-trivial.
> The general representation is then only given up to a U(1) factor,
> as indicated in the previous article.
>
> This is a substantial point of difference from the Poincare'
> algebra, whose corresponding (J,K) elements form the (semisimple)
> algebra for SO(3,1), the Lorentz group; with (P,E) giving you an
> indirect product ISO(3,1) -- and representation theory for indirect
> products is a well-mapped-out field.
Thanks.
Is there a nice way of parameterizing the Poincare group and its physically
relevant unitary irreducible representations in such a way that the
corresponding representations of the Galilei group result
as c -> infinity?
Arnold Neumaier
Alfred Einstead
May12-04, 06:05 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>\nArnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n> Is there a nice way of parameterizing the Poincare group and its physically\n> relevant unitary irreducible representations in such a way that the\n> corresponding representations of the Galilei group result\n> as c -> infinity?\n\nI was already thinking about that. The answer is to just simply\nkeep the c\'s intact in the Poincare\' group generators and commutators\nand work out the representations to see where the c\'s go.\n\nThe J and K commutators will be:\n[J_a, J_b] = e^c_{ab} J_c\n[J_a, K_b] = e^c_{ab} K_c = [K_a, J_b]\nas before, but now\n[K_a, K_b] = -(1/c)^2 e^c_{ab} J_c.\n\nThere\'s also a non-zero commutator between the K\'s and P\'s,\nI believe, and this takes the place of the one posed in the previous\narticle in the context of the projective mass m representation.\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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
> Is there a nice way of parameterizing the Poincare group and its physically
> relevant unitary irreducible representations in such a way that the
> corresponding representations of the Galilei group result
> as c -> infinity?
I was already thinking about that. The answer is to just simply
keep the c's intact in the Poincare' group generators and commutators
and work out the representations to see where the c's go.
The J and K commutators will be:
[J_a, J_b] = e^{c_}{ab} J_c[J_a, K_b] = e^{c_}{ab} K_c = [K_a, J_b]
as before, but now
[K_a, K_b] = -(1/c)^2 e^{c_}{ab} J_c.
There's also a non-zero commutator between the K's and P's,
I believe, and this takes the place of the one posed in the previous
article in the context of the projective mass m representation.
Alfred Einstead
May12-04, 02:49 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>seeURL@end.of.post.ch (Martin Ouwehand) wrote:\n> It turns out that the purely unitary representations are physically\n> unenlightening, you have to consider projective representations (up to\n> a phase) to find the usual quantum mechanics of the non-relativistic\n> free particle.\n\nI\'ll resolve the matter here, following up on both your comments and\nmy other articles in this thread.\n\nA projective representation of a Lie group is equivalent to an\nordinary representation of a central extension of the group. A\ncentral extension is a U(1) bundle over the group, and its Lie\nalgebra is obtained by adding an abelian unit 1 to the algebra.\n\nSo the central extension of the Lie algebra I listed before would\nlook like:\n\n[J_a, J_b] = e^c_{ab} J_c + A_{ab} 1\n[J_a, K_b] = e^c_{ab} K_c + B_{ab} 1\n[J_a, P_b] = e^c_{ab} P_c + C_{ab} 1\n[J_a, E] = D_a 1\n[K_a, K_b] = E_{ab} 1\n[K_a, P_b] = F_{ab} 1\n[K_a, E] = P_a + G_a 1\n[P_a, P_b] = H_{ab} 1\n[P_a, E] = I_a 1,\nand of course:\n[X,1] = 0 for all X.\nwhere\ne^1_{23} = 1 = -e^1_{32}, with cyclic permutations\ne^a_{bc} = 0 if a=b,b=c or c=a.\n\nThe A\'s, E\'s and H\'s are all anti-symmetric. In particular,\nthe A\'s can be written as A_{ab} = e^c_{ab} A_c, and can be\nabsorbed into a redefinition of J, without changing the other\ncommutators.\n\nThe Jacobi identity further limits the range of possibilities.\nThe identities on the J-J-K and J-J-P combinations show that the\nB\'s and C\'s are anti-symmetric. So, these likewise can be\nreabsorbed into redefinitions of K, P (and G). The Jacobi\nidentities on the J-J-E combinations show that the D\'s are 0.\nApplying it to the J-K-K and J-P-P combinations shows that\nthe E\'s and H\'s are all 0, too. Applying it to the J-K-E\ncombinations show that the G\'s are all 0; on the J-P-E\ncombinations, you find the I\'s are all 0; and finally on the\nJ-K-P\'s you find that the F\'s are diagonal: F_{ab} = m delta_{ab}.\n\nSo, the most general non-trivial central extension of the\nGalilei group has the Lie algebra given by:\n\n[J_a, J_b] = e^c_{ab} J_c\n[J_a, K_b] = e^c_{ab} K_c\n[J_a, P_b] = e^c_{ab} P_c\n[K_a, P_b] = m delta_{ab}\n[K_a, E] = P_a\nall other commutators 0.\n\nFor non-zero m, neither P^2 nor |KxP|^2 are invariant any longer.\nIn particular, you find that:\n[K_a, P^2] = 2 m P_a,\nso that instead of P^2, you have E0 = E - P^2/(2m) invariant.\n\nDefining X = K/m, and E0 = E - P^2/(2m), you find that:\n\n[J_a, J_b] = e^c_{ab} J_c\n[J_a, X_b] = e^c_{ab} X_c\n[J_a, P_b] = e^c_{ab} P_c\n[X_a, P_b] = delta_{ab}\n[E0,x] = 0 for all x.\n\nSo the algebra splits into the parts <E0> x <J,X,P>.\n\nOne can define L = X x P, and finds that\n[X_a,L_b] = e^c_{ab} X_c\n[P_a,L_b] = e^c_{ab} P_c\n[L_a,L_b] = e^c_{ab} L_c\n[J_a,L_b] = e^c_{ab} L_c.\nSo, defining S = J - L, one gets:\n[X_a,S_b] = [P_a,S_b] = [J_a,S_b] = 0\n[S_a,S_b] = e^c_{ab} S_c.\n\nTherefore, the algebra splits into <S> x <X,P>. The subalgebra\ncorresponding to <X,P> is just the Heisenberg algebra; that\ncorresponding to <S> is just SU(2).\n\nFinally, in place of |KxP|^2 is the invariant\n|S|^2 = |J-(KxP)/m|^2.\nSo, the eigenstates of an irreducible representation have the\nform\n|f,l,m,e> = f_{e m}(x) |l,n>\nwhere\nE |f,l,n,m,e> = e |f,l,n,m,e>\nP |f,l,n,m,e> = d/dx |f,l,n,m,e>\nK |f,l,n,m,e> = m x |f,l,n,m,e>\nJ |f,l,n,m,e> = (x x d/dx) |f,l,n,m,e> + sigma |f,l,n,m,e>\nwhere\nsigma^2 |f,l,n,m,e> = l(l+1) |f,l,n,m,e>\nsigma_3 |f,l,n,m,e> = n |f,l,n,m,e>\nand n = -l, 1-l, 2-l, ..., l-2, l-1, l.\n\nSo, the states are parametrized by a discrete parameter n and a\ncontinuous parameter x; and the state spaces by l, m and e.\n\nThe ordinary representation, in this light, which was discussed\nin a previous article, is then seen clearly to be just the\nrepresentation for 0 mass particles. The ones above are for\nnon-zero mass particles of mass m.\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>seeURL@end.of.post.ch (Martin Ouwehand) wrote:
> It turns out that the purely unitary representations are physically
> unenlightening, you have to consider projective representations (up to
> a phase) to find the usual quantum mechanics of the non-relativistic
> free particle.
I'll resolve the matter here, following up on both your comments and
my other articles in this thread.
A projective representation of a Lie group is equivalent to an
ordinary representation of a central extension of the group. A
central extension is a U(1) bundle over the group, and its Lie
algebra is obtained by adding an abelian unit 1 to the algebra.
So the central extension of the Lie algebra I listed before would
look like:
[J_a, J_b] = e^{c_}{ab} J_c + A_{ab} 1[J_a, K_b] = e^{c_}{ab} K_c + B_{ab} 1[J_a, P_b] = e^{c_}{ab} P_c + C_{ab} 1[J_a, E] = D_a 1[K_a, K_b] = E_{ab} 1[K_a, P_b] = F_{ab} 1[K_a, E] = P_a + G_a 1[P_a, P_b] = H_{ab} 1[P_a, E] = I_a 1,[/itex]
and of course:
[X,1] = for all X.
where
e^{1_}{23} = 1 = -e^{1_}{32}, with cyclic permutations
e^{a_}{bc} = if a=b,b=c or c=a.
The A's, E's and H's are all anti-symmetric. In particular,
the A's can be written as A_{ab} = e^{c_}{ab} A_c, and can be
absorbed into a redefinition of J, without changing the other
commutators.
The Jacobi identity further limits the range of possibilities.
The identities on the J-J-K and J-J-P combinations show that the
B's and C's are anti-symmetric. So, these likewise can be
reabsorbed into redefinitions of K, P (and G). The Jacobi
identities on the J-J-E combinations show that the D's are .
Applying it to the J-K-K and J-P-P combinations shows that
the E's and H's are all 0, too. Applying it to the J-K-E
combinations show that the G's are all 0; on the J-P-E
combinations, you find the I's are all 0; and finally on the
J-K-P's you find that the F's are diagonal: F_{ab} = m \delta_{ab}.
So, the most general non-trivial central extension of the
Galilei group has the Lie algebra given by:
[J_a, J_b] = e^{c_}{ab} J_c[J_a, K_b] = e^{c_}{ab} K_c[J_a, P_b] = e^{c_}{ab} P_c[K_a, P_b] = m \delta_{ab}[K_a, E] = P_a
all other commutators .
For non-zero m, neither P^2 nor |KxP|^2 are invariant any longer.
In particular, you find that:
[K_a, P^2] = 2 m P_a,
so that instead of P^2, you have E0 = E - P^2/(2m) invariant.
Defining X = K/m, and E0 = E - P^2/(2m), you find that:
[J_a, J_b] = e^{c_}{ab} J_c[J_a, X_b] = e^{c_}{ab} X_c[J_a, P_b] = e^{c_}{ab} P_c[X_a, P_b] = \delta_{ab}[E0,x] = for all x.
So the algebra splits into the parts <E0> x <J,X,P>.
One can define L = X x P, and finds that
[X_a,L_b] = e^{c_}{ab} X_c[P_a,L_b] = e^{c_}{ab} P_c[L_a,L_b] = e^{c_}{ab} L_c[J_a,L_b] = e^{c_}{ab} L_c.
So, defining S = J - L, one gets:
[itex][X_a,S_b] = [P_a,S_b] = [J_a,S_b] = [S_a,S_b] = e^{c_}{ab} S_c.
Therefore, the algebra splits into <S> x <X,P>. The subalgebra
corresponding to <X,P> is just the Heisenberg algebra; that
corresponding to <S> is just SU(2).
Finally, in place of |KxP|^2 is the invariant
|S|^2 = |J-(KxP)/m|^2.
So, the eigenstates of an irreducible representation have the
form
|f,l,m,e> = f_{e m}(x) |l,n>
where
E |f,l,n,m,e> = e |f,l,n,m,e>P |f,l,n,m,e> = d/dx |f,l,n,m,e>K |f,l,n,m,e> = m x |f,l,n,m,e>J |f,l,n,m,e> = (x x d/dx) |f,l,n,m,e> + \sigma |f,l,n,m,e>
where
\sigma^2 |f,l,n,m,e> = l(l+1) |f,l,n,m,e>\sigma_3 |f,l,n,m,e> = n |f,l,n,m,e>
and n = -l, 1-l, 2-l, ..., l-2, l-1, l.
So, the states are parametrized by a discrete parameter n and a
continuous parameter x; and the state spaces by l, m and e.
The ordinary representation, in this light, which was discussed
in a previous article, is then seen clearly to be just the
representation for mass particles. The ones above are for
non-zero mass particles of mass m.
Arnold Neumaier
May17-04, 07:23 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>Alfred Einstead wrote:\n> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n>\n>>Is there a nice way of parameterizing the Poincare group and its physically\n>>relevant unitary irreducible representations in such a way that the\n>>corresponding representations of the Galilei group result\n>>as c -> infinity?\n>\n>\n> Here\'s a side-by-side comparison of the two Lie algebras:\n>\n> Galilei (with the central extension by mass m included):\n> [J_a,J_b] = e^c_{ab} J_c\n> [J_a,K_b] = e^c_{ab} K_c\n> [J_a,P_b] = e^c_{ab} P_c\n> [K_a,P_b] = m delta_{ab}\n> [K_a,E] = P_a\n> All other commutators are 0.\n>\n> Poincare\' looks like:\n> [J_a,J_b] = e^c_{ab} J_c\n> [J_a,K_b] = e^c_{ab} K_c\n> [J_a,P_b] = e^c_{ab} P_c\n> [K_a,P_b] = E/c^2 delta_{ab}\n> [K_a,E] = P_a\n> [K_a,K_b] = -e^c_{ab} J_c/c^2\n> All other commutators are 0.\n>\n> The [K,P] commutators are the same except for having E/c^2 in place\n> of m. Thus, E = mc^2.\n\nBut this E diverges for c to infinity, while there is an E in the Galilei\nformulas, too. So something must still be incorrect. What about\nusing H=mc^2+E in place of E in the quantum case?\nSince m is central, the Poincare formulas containing E change to\n\n> [K_a,P_b] = (m+E/c^2) delta_{ab}\n> [K_a,E] = P_a\n\nand the limit is apparent. But now one has a central extension\nof Poincare!?\n\n\nBy the way, you\'d change all super- and subscripts c to d to reduce\nconfusion about the double meaning of c.\n\n\nArnold Neumaier\n\n\n\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Alfred Einstead wrote:
> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
>
>>Is there a nice way of parameterizing the Poincare group and its physically
>>relevant unitary irreducible representations in such a way that the
>>corresponding representations of the Galilei group result
>>as c -> infinity?
>
>
> Here's a side-by-side comparison of the two Lie algebras:
>
> Galilei (with the central extension by mass m included):
> [J_a,J_b] = e^{c_}{ab} J_c
> [J_a,K_b] = e^{c_}{ab} K_c
> [J_a,P_b] = e^{c_}{ab} P_c
> [K_a,P_b] = m \delta_{ab}
> [K_a,E] = P_a
> All other commutators are .
>
> Poincare' looks like:
> [J_a,J_b] = e^{c_}{ab} J_c
> [J_a,K_b] = e^{c_}{ab} K_c
> [J_a,P_b] = e^{c_}{ab} P_c
> [K_a,P_b] = E/c^2 \delta_{ab}
> [K_a,E] = P_a
> [K_a,K_b] = -e^{c_}{ab} J_c/c^2
> All other commutators are .
>
> The [K,P] commutators are the same except for having E/c^2 in place
> of m. Thus, E = mc^2.
But this E diverges for c to infinity, while there is an E in the Galilei
formulas, too. So something must still be incorrect. What about
using H=mc^2+E in place of E in the quantum case?
Since m is central, the Poincare formulas containing E change to
> [K_a,P_b] = (m+E/c^2) \delta_{ab}
> [K_a,E] = P_a
and the limit is apparent. But now one has a central extension
of Poincare!?
By the way, you'd change all super- and subscripts c to d to reduce
confusion about the double meaning of c.
Arnold Neumaier
Hendrik van Hees
May18-04, 04:44 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>Arnold Neumaier wrote:\n\n> Alfred Einstead wrote:\n>> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n>>\n>>>Is there a nice way of parameterizing the Poincare group and its\n>>>physically relevant unitary irreducible representations in such a way\n>>>that the corresponding representations of the Galilei group result\n>>>as c -> infinity?\n>>\n>>\n>> Here\'s a side-by-side comparison of the two Lie algebras:\n>>\n>> Galilei (with the central extension by mass m included):\n>> [J_a,J_b] =3D e^c_{ab} J_c\n>> [J_a,K_b] =3D e^c_{ab} K_c\n>> [J_a,P_b] =3D e^c_{ab} P_c\n>> [K_a,P_b] =3D m delta_{ab}\n>> [K_a,E] =3D P_a\n>> All other commutators are 0.\n>>\n>> Poincare\' looks like:\n>> [J_a,J_b] =3D e^c_{ab} J_c\n>> [J_a,K_b] =3D e^c_{ab} K_c\n>> [J_a,P_b] =3D e^c_{ab} P_c\n>> [K_a,P_b] =3D E/c^2 delta_{ab}\n>> [K_a,E] =3D P_a\n>> [K_a,K_b] =3D -e^c_{ab} J_c/c^2\n>> All other commutators are 0.\n>>\n>> The [K,P] commutators are the same except for having E/c^2 in place\n>> of m. Thus, E =3D mc^2.\n>\n> But this E diverges for c to infinity, while there is an E in the\n> Galilei formulas, too. So something must still be incorrect. What\n> about using H=3Dmc^2+E in place of E in the quantum case?\n> Since m is central, the Poincare formulas containing E change to\n>\n> > [K_a,P_b] =3D (m+E/c^2) delta_{ab}\n> > [K_a,E] =3D P_a\n>\n> and the limit is apparent. But now one has a central extension\n> of Poincare!?\n\nThere is a nice theorem, stated by Weinberg in volume 1 of his book. I\nhope I remember the bottom line of the argument right.\n\nFirst of all one looks at the ray representation of the Lie algebra g of\nthe group G. With hermitean generators, it\'s commutators read\n\n[t_a,t_b]=3Di {f^c}_{ab} t_c + i C_{ab} id,\n\nwhere the {f^c}_{ab} are the structure constants of the group and C_{ab}\nthe central charges.\n\nEach ray representation of a Lie group G (in usual quantum physicisist\'s\nlanguage each unitary representation up to phases) is induced by a\nunitary representation in Hilbert space if\n\n(1) There exist real numbers \\phi_c such that *all* central charges are\ngiven by\n\nC_{ab}=3D{f^c}_{ab} \\phi_c\n\n(2) The group G is simply connected in the sense that two group elements\nare connected by a continuous path.\n\nBoth demands are fulfilled for the proper orthochronous Lorentz group.\n\nThus, all ray representations are induced by Wigner\'s unitary\nrepresentations. To get the usual irrep. representations as single\nparticle states of a Fock space with a Hamilton, which is bounded from\nbelow, the vacuum must have energy and momentum 0.\n\nThe non-relativistic limit of these representations of the Poincare\ngroup, i.e., the corresponding representations of the Galilei group are\nfound in the usual way by the limit |\\vec{p}|<<m. Then we have (with\nc=3D1):\n\nE=3Dsqrt(m^2+\\vec{p}^2)=3Dm+\\v ec{p}^2/(2m)->m for |\\vec{p}|/m->0\n\nand thus you obtain the qm. reps. of the Galileo Lie algebra by setting\n\nH_{gal}=3Dm+T\n\nand taking T=3D\\vec{p}^2/(2m) as negligible against m.\n\nThere you see that here you have a ray representation, leading to a\nsuperselection rule: In Galileian qm, there must be no superpositions\nof state with different mass.\n\nI think the mathematical term of this procedure is called group\ndeformation, although I\'d rather call it a deformation of a\nrepresentation of one group to the (ray) representation of another.\n\nThe Galilei group admits more than one central extension, and the one\nwhich is that chosen by nature is found in this way from the Poincare\ngroup. Wigner and I=F6n=FC have shown, that there exist other central\nextensions of the Galilei group which gives rather strange quantum\ntheories, where no localizable particles exist, etc. Especially the\nunitary representations themselves give such weird quantum theories.\n\nIn this way we unerstand also, why the representation given by the\ninvariance group of the Schr=F6dinger equation leads to a ray\nrepresentation rather than a unitary one.\n\n>\n> By the way, you\'d change all super- and subscripts c to d to reduce\n> confusion about the double meaning of c.\n\nThat\'s true. I\'d rather prefer to set c=3D1 all the time, because that\nmakes even the physical argument clearer: The non-relativistic limit\nmakes only sense for the representations, not for the groups\nthemselves. While the Poincare groups uniquely determines its quantum\ntheory in the way explained above, we have to make further physical\nassumptions than the notion of irreducible unitary representations of\nthe Galilei group to find the correct non-relativistic quantum theory.\n\nAs far as I know, for this reason, there is no deformation of massless\nrepresentations to physically sensible representations of the Galilei\ngroup.\n\n--\nHendrik van Hees Cyclotron Institute\nPhone: +1 979/845-1411 Texas A&M University\nFax: +1 979/845-1899 Cyclotron Institute, MS-3366\nhttp://theory.gsi.de/~vanhees/ College Station, TX 77843-3366\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>Arnold Neumaier wrote:
> Alfred Einstead wrote:
>> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
>>
>>>Is there a nice way of parameterizing the Poincare group and its
>>>physically relevant unitary irreducible representations in such a way
>>>that the corresponding representations of the Galilei group result
>>>as c -> infinity?
>>
>>
>> Here's a side-by-side comparison of the two Lie algebras:
>>
>> Galilei (with the central extension by mass m included):
>> [J_a,J_b] =3D e^{c_}{ab} J_c>> [J_a,K_b] =3D e^{c_}{ab} K_c>> [J_a,P_b] =3D e^{c_}{ab} P_c>> [K_a,P_b] =3D m \delta_{ab}>> [K_a,E] =3D P_a
>> All other commutators are .
>>
>> Poincare' looks like:
>> [J_a,J_b] =3D e^{c_}{ab} J_c>> [J_a,K_b] =3D e^{c_}{ab} K_c>> [J_a,P_b] =3D e^{c_}{ab} P_c>> [K_a,P_b] =3D E/c^2 \delta_{ab}>> [K_a,E] =3D P_a>> [K_a,K_b] =3D -e^{c_}{ab} J_c/c^2
>> All other commutators are .
>>
>> The [K,P] commutators are the same except for having E/c^2 in place
>> of m. Thus, E =3D mc^2.
>
> But this E diverges for c to infinity, while there is an E in the
> Galilei formulas, too. So something must still be incorrect. What
> about using H=3Dmc^2+E in place of E in the quantum case?
> Since m is central, the Poincare formulas containing E change to
>
> > [K_a,P_b] =3D (m+E/c^2) \delta_{ab}
> > [K_a,E] =3D P_a
>
> and the limit is apparent. But now one has a central extension
> of Poincare!?
There is a nice theorem, stated by Weinberg in volume 1 of his book. I
hope I remember the bottom line of the argument right.
First of all one looks at the ray representation of the Lie algebra g of
the group G. With hermitean generators, it's commutators read
[t_a,t_b]=3Di {f^c}_{ab} t_c + i C_{ab}[/itex] id,
where the {f^c}_{ab} are the structure constants of the group and C_{ab}
the central charges.
Each ray representation of a Lie group G (in usual quantum physicisist's
language each unitary representation up to phases) is induced by a
unitary representation in Hilbert space if
(1) There exist real numbers \phi_c such that *all* central charges are
given by
[itex]C_{ab}=3D{f^c}_{ab} \phi_c
(2) The group G is simply connected in the sense that two group elements
are connected by a continuous path.
Both demands are fulfilled for the proper orthochronous Lorentz group.
Thus, all ray representations are induced by Wigner's unitary
representations. To get the usual irrep. representations as single
particle states of a Fock space with a Hamilton, which is bounded from
below, the vacuum must have energy and momentum .
The non-relativistic limit of these representations of the Poincare
group, i.e., the corresponding representations of the Galilei group are
found in the usual way by the limit |\vec{p}|<<m. Then we have (with
c=3D1):E=3Dsqrt(m^2+\vec{p}^2)=3Dm+\vec{p}^2/(2m)->m for |\vec{p}|/m->0
and thus you obtain the qm. reps. of the Galileo Lie algebra by setting
H_{gal}=3Dm+T
and taking T=3D\vec{p}^2/(2m) as negligible against m.
There you see that here you have a ray representation, leading to a
superselection rule: In Galileian qm, there must be no superpositions
of state with different mass.
I think the mathematical term of this procedure is called group
deformation, although I'd rather call it a deformation of a
representation of one group to the (ray) representation of another.
The Galilei group admits more than one central extension, and the one
which is that chosen by nature is found in this way from the Poincare
group. Wigner and I=F6n=FC have shown, that there exist other central
extensions of the Galilei group which gives rather strange quantum
theories, where no localizable particles exist, etc. Especially the
unitary representations themselves give such weird quantum theories.
In this way we unerstand also, why the representation given by the
invariance group of the Schr=F6dinger equation leads to a ray
representation rather than a unitary one.
>
> By the way, you'd change all super- and subscripts c to d to reduce
> confusion about the double meaning of c.
That's true. I'd rather prefer to set c=3D1 all the time, because that
makes even the physical argument clearer: The non-relativistic limit
makes only sense for the representations, not for the groups
themselves. While the Poincare groups uniquely determines its quantum
theory in the way explained above, we have to make further physical
assumptions than the notion of irreducible unitary representations of
the Galilei group to find the correct non-relativistic quantum theory.
As far as I know, for this reason, there is no deformation of massless
representations to physically sensible representations of the Galilei
group.
--
Hendrik van Hees Cyclotron Institute
Phone: +1 979/845-1411 Texas A&M University
Fax: +1 979/845-1899 Cyclotron Institute, MS-3366
http://theory.gsi.de/~vanhees/ College Station, TX 77843-3366
whopkins@csd.uwm.edu
Jan30-05, 02:01 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\nArnold Neumaier wrote:\n> Alfred Einstead wrote:\n> > Here\'s a side-by-side comparison of the two Lie algebras:\n> > Galilei (with the central extension by mass m included):\n> > [J_a,J_b] = e^c_{ab} J_c\n> > [J_a,K_b] = e^c_{ab} K_c\n> > [J_a,P_b] = e^c_{ab} P_c\n> > [K_a,P_b] = m delta_{ab}\n> > [K_a,E] = P_a\n> > All other commutators are 0.\n> >\n> > Poincare\' looks like:\n> > [J_a,J_b] = e^c_{ab} J_c\n> > [J_a,K_b] = e^c_{ab} K_c\n> > [J_a,P_b] = e^c_{ab} P_c\n> > [K_a,P_b] = E/c^2 delta_{ab}\n> > [K_a,E] = P_a\n> > [K_a,K_b] = -e^c_{ab} J_c/c^2\n> > All other commutators are 0.\n> But this E diverges for c to infinity, while there is an E\n> in the Galilei formulas, too. So something must still\n> be incorrect. What about using H=mc^2+E in place of E in\n> the quantum case?\n\nActually I just came onto that idea, as well. In general it\'s best to\nconsider the E generator, not as the total energy, but only as the\nKINETIC energy. To make the distinction, call the kinetic energy T and\nthe total energy E. Then one can write down a general Lie algebra that\naccomodates the central extensions of all the above:\n\n[J_a,J_b] = e^c_{ab} J_c\n[J_a,K_b] = e^c_{ab} K_c\n[J_a,P_b] = e^c_{ab} P_c\n[K_a,P_b] = M delta_{ab}\n[K_a,T] = P_a\n[K_a,K_b] = -A e^c_{ab} J_c\n[P_a,P_b] = [P_a,T] = [T,T] = 0\nwhere\nM = m + A T\nis the total mass; and the 3 cases for A are:\nPoincare\': A = (1/c)^2 > 0\nGalilei: A = 0\nEuclidean: A = -(1/r)^2 < 0.\n\n> > [K_a,P_b] = (m+E/c^2) delta_{ab}\n> > [K_a,E] = P_a\n> and the limit is apparent. But now one has a central extension\n> of Poincare!?\n\nFor all the cases, except Galilei, the constant m may be reabsorbed\ninto a redefinition: E = M/A = T + m/A.\n\nFor Poincare\' this, of course, gives you E = M c^2.\n\nThis generalizes further, if you consider the case where the 3 groups\nare actually on a hypersphere or hyperbolic geometry. Then you also\nget non-zero commutators for the P\'s and T\'s:\n\n[P_a,P_b] = LA e^c_{ab} J_c\n[P_a,E] = L K_a.\n\nAs is the case for Galilei and Poincare\', you also get 2 invariants in\nthe general case:\np^2 - 2MT + AT^2 + L(A J^2 - K^2)\n|ML + PxK|^2 - A(J.P)^2 - LA(J.K)^2.\n(Check for correct factors and signs, I\'m doing this extemporaneously\nfrom memory).\n\nOne can proceed to write down the Lie-Poisson bracket (which\ntechnically is defined over the space of C^{infinity} functions over\nT*L*, the cotangent space of the dual L* of the Lie algebra L), which\nin ordinary 3-D vector notation becomes:\n\n{f,g} = J.(fJ x gJ - A fK x gK + LA fP x gP\n+ K.(fJ x gK + fK x gJ + L(fP gT - fT gP)\n+ P.(fJ x gP + fP x gJ + fK gT - fT gK)\n+ M (fK.gP - fP.gK)\nwhere\ndf = fJ.dJ + fK.dK + fP.dP + fT dT\ndg = gJ.dJ + gK.dK + gP.dP + gT dT\n(the differentials dJ_a, dK_a, dP_a, dT residing in T*L* = L).\n\nAn invariant f is a function which gives you 0 Poisson bracket with all\ng\'s. This yields a condition for each differential coefficient of g:\ngJ: J x fJ + K x fK + P x fP = 0\ngK: -AJ x fK + K x fJ - P fT - M fP = 0\ngP: LAJ x fP - LK fT + P x fJ + M fK = 0\ngT: LK.fP + P.fK = 0.\nParcelling out the cases (where different combinations of the\ncoefficients give you 0 multipliers) you get the full range of\npossibilities for invariants. The two invariants above are for the\ngeneral case, but reduce to trivial constants in the specialized cases.\n\nIf M is non-zero, then the equation for gT becomes redundant, because\nof the quadratic invariant. If M is 0, then its Poisson brackets are\nall also 0, which gives you restrictions on the possibilities for (L,A)\nand (J,K,P) that in turn yields some of the special cases. For\nnon-zero M, you\'ll find that the quantity (M^2 + L (AJ)^2) becomes of\nspecial significance. If 0, the equations for gK and gP do not yield\nunique solutions for fP and fK; otherwise they do. The first condition\n(gJ) essentially implies that the invariant must be scalar polynomials\n[of a constant degree] whose monomials are formed out of the vectors\n(J,K,P) by dot products, triple product; and multiplied by powers of E\nand M. So, adopting that ansatz, merely solving for the gP and gK\nequations is almost enough to get the most general form of the\ndifferential for f; which in turn gives you the function f.\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>Arnold Neumaier wrote:
> Alfred Einstead wrote:
> > Here's a side-by-side comparison of the two Lie algebras:
> > Galilei (with the central extension by mass m included):
> > [J_a,J_b] = e^{c_}{ab} J_c
> > [J_a,K_b] = e^{c_}{ab} K_c
> > [J_a,P_b] = e^{c_}{ab} P_c
> > [K_a,P_b] = m \delta_{ab}
> > [K_a,E] = P_a
> > All other commutators are .
> >
> > Poincare' looks like:
> > [J_a,J_b] = e^{c_}{ab} J_c
> > [J_a,K_b] = e^{c_}{ab} K_c
> > [J_a,P_b] = e^{c_}{ab} P_c
> > [K_a,P_b] = E/c^2 \delta_{ab}
> > [K_a,E] = P_a
> > [K_a,K_b] = -e^{c_}{ab} J_c/c^2
> > All other commutators are .
> But this E diverges for c to infinity, while there is an E
> in the Galilei formulas, too. So something must still
> be incorrect. What about using H=mc^2+E in place of E in
> the quantum case?
Actually I just came onto that idea, as well. In general it's best to
consider the E generator, not as the total energy, but only as the
KINETIC energy. To make the distinction, call the kinetic energy T and
the total energy E. Then one can write down a general Lie algebra that
accomodates the central extensions of all the above:
[J_a,J_b] = e^{c_}{ab} J_c[J_a,K_b] = e^{c_}{ab} K_c[J_a,P_b] = e^{c_}{ab} P_c[K_a,P_b] = M \delta_{ab}[K_a,T] = P_a[K_a,K_b] = -A e^{c_}{ab} J_c[P_a,P_b] = [P_a,T] =[/itex] [T,T] =
where
M = m + A T
is the total mass; and the 3 cases for A are:
Poincare': A = (1/c)^2 >
Galilei: A =
Euclidean: A = -(1/r)^2 < .
> > [K_a,P_b] = (m+E/c^2) \delta_{ab}
> > [K_a,E] = P_a
> and the limit is apparent. But now one has a central extension
> of Poincare!?
For all the cases, except Galilei, the constant m may be reabsorbed
into a redefinition: E = M/A = T + m/A.
For Poincare' this, of course, gives you E = M c^2.
This generalizes further, if you consider the case where the 3 groups
are actually on a hypersphere or hyperbolic geometry. Then you also
get non-zero commutators for the P's and T's:
[P_a,P_b] = LA e^{c_}{ab} J_c[P_a,E] = L K_a.
As is the case for Galilei and Poincare', you also get 2 invariants in
the general case:
p^2 - 2MT + AT^2 + L(A J^2 - K^2)
|ML + PxK|^2 - A(J.P)^2 - LA(J.K)^2.
(Check for correct factors and signs, I'm doing this extemporaneously
from memory).
One can proceed to write down the Lie-Poisson bracket (which
technically is defined over the space of C^{infinity} functions over
T*L*, the cotangent space of the dual L* of the Lie algebra L), which
in ordinary 3-D vector notation becomes:
{f,g} = J.(fJ x gJ - A fK x gK + LA fP x gP
+ K.(fJ x gK + fK x gJ + L(fP gT - fT gP)
+ P.(fJ x gP + fP x gJ + fK gT - fT gK)+ M (fK.gP - fP.gK)
where
df = fJ.dJ + fK.dK + fP.dP + fT dTdg = gJ.dJ + gK.dK + gP.dP + gT dT
(the differentials dJ_a, dK_a, dP_a, dT residing [itex]in T*L* = L).
An invariant f is a function which gives you Poisson bracket with all
g's. This yields a condition for each differential coefficient of g:
gJ: J x fJ + K x fK + P x fP =
gK: -AJ x fK + K x fJ - P fT - M fP =
gP: LAJ x fP - LK fT + P x fJ + M fK =
gT: LK.fP + P.fK = .
Parcelling out the cases (where different combinations of the
coefficients give you multipliers) you get the full range of
possibilities for invariants. The two invariants above are for the
general case, but reduce to trivial constants in the specialized cases.
If M is non-zero, then the equation for gT becomes redundant, because
of the quadratic invariant. If M is 0, then its Poisson brackets are
all also 0, which gives you restrictions on the possibilities for (L,A)
and (J,K,P) that in turn yields some of the special cases. For
non-zero M, you'll find that the quantity (M^2 + L (AJ)^2) becomes of
special significance. If 0, the equations for gK and gP do not yield
unique solutions for fP and fK; otherwise they do. The first condition
(gJ) essentially implies that the invariant must be scalar polynomials
[of a constant degree] whose monomials are formed out of the vectors
(J,K,P) by dot products, triple product; and multiplied by powers of E
and M. So, adopting that ansatz, merely solving for the gP and gK
equations is almost enough to get the most general form of the
differential for f; which in turn gives you the function f.
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