[SOLVED] Re: equivalence between path integral formulation and other QM formulation [Archive]

Arnold Neumaier

Nov14-04, 07:08 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>L.R. wrote:\n>\n> By reading some old posts of this group, I learned that there is a\n> Stone-von Neumann theorem guarantee the equivalence of Heisenberg\n> picture and Schrödinger picture in QM.\n\nThis is not quite correct. The two pictures are equivalent because\nthere is a unitary transformation between them. The Stone-von Neumann\ntheorem guarantees that the canonical commutation relations have a\nunique representation (apart from unitary transformations) and hence\nsuffice to specify the QM of finitely many degrees of freedom, no matter\nwhich picture.\n\nThe Stone-von Neumann theorem fails for systems of infinitely many\ndegrees of freedom, which in a sense \'causes\' the difficulties in\nquantum field theory. Nevertheless, QFT still has a Schroedinger picture\nand a Heisenberg picture, and these are still equivalent.\n\n\n> My question is, if we think\n> that path integral is also some kind of picture, is there a theorem\n> like Stone-von Neumann that say something about the equivalence\n> between PI and other pictures?\n\nThe Feynman-Kac formula makes the equivalence precise, after analytically\ncontinuing the time to purely imaginary. The Osterwalder-Schrader theory\n[see, e.g., math-ph/0001010 or the book by Glimm and Jaffe] shows how to\ngo back in case of relativistic quantum field theory.\n\n\n> By the way, anyone know where can I found the pronunciation of those\n> great figures\' names like von Neumann, Poincare...?\n\nThe simplest is to go to some physicist and ask him or her to pronounce it.\n\n\nArnold Aeumaier\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>L.R. wrote:
>
> By reading some old posts of this group, I learned that there is a
> Stone-von Neumann theorem guarantee the equivalence of Heisenberg
> picture and Schrödinger picture in QM.

This is not quite correct. The two pictures are equivalent because
there is a unitary transformation between them. The Stone-von Neumann
theorem guarantees that the canonical commutation relations have a
unique representation (apart from unitary transformations) and hence
suffice to specify the QM of finitely many degrees of freedom, no matter
which picture.

The Stone-von Neumann theorem fails for systems of infinitely many
degrees of freedom, which in a sense 'causes' the difficulties in
quantum field theory. Nevertheless, QFT still has a Schroedinger picture
and a Heisenberg picture, and these are still equivalent.

> My question is, if we think
> that path integral is also some kind of picture, is there a theorem
> like Stone-von Neumann that say something about the equivalence
> between \PI and other pictures?

The Feynman-Kac formula makes the equivalence precise, after analytically
continuing the time to purely imaginary. The Osterwalder-Schrader theory
[see, e.g., http://www.arxiv.org/abs/math-ph/0001010 or the book by Glimm and Jaffe] shows how to
go back in case of relativistic quantum field theory.

> By the way, anyone know where can I found the pronunciation of those
> great figures' names like von Neumann, Poincare...?

The simplest is to go to some physicist and ask him or her to pronounce it.

Arnold Aeumaier

L.R.

Nov16-04, 02:17 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>Hi,\n\n> This is not quite correct. The two pictures are equivalent because\n> there is a unitary transformation between them. The Stone-von Neumann\n> theorem guarantees that the canonical commutation relations have a\n> unique representation (apart from unitary transformations) and hence\n> suffice to specify the QM of finitely many degrees of freedom, no matter\n> which picture.\n\nIf every such picture can be represented by some cannonical\ncommutation relations, does that mean all the pictures are\nequivalence? If in some situation, this theorem cannot apply, I think\nwe may construct some non-equivalenct picture that can be identified\nas Schrodinger picture and Heisenberg picture of that system. So\nwithout this theorem we cannot say the equivalence since the unitary\ntransformation may not exist. (actually I think you are talking about\nwe can always use the Hamiltonian to get an exp(-iHt), and I don\'t\nknow how to shoot this down in the case I just mentioned, so, correct\nme if I\'m wrong:))\n\n> The Stone-von Neumann theorem fails for systems of infinitely many\n> degrees of freedom, which in a sense \'causes\' the difficulties in\n> quantum field theory. Nevertheless, QFT still has a Schroedinger picture\n> and a Heisenberg picture, and these are still equivalent.\n\nIs there a proof or we just believe that, Because there is Haag\'s\ntheorem says that the interaction picture does not exist, and I think\ninteraction picture is something in between of the Schroedinger and\nHeisenberg picture. If we cannot get a unitary transformation for\ninteraction picture, How can we be so sure that we can get one between\nSchroedinger and Heisenberg picture?\n\n> The Feynman-Kac formula makes the equivalence precise, after analytically\n> continuing the time to purely imaginary. The Osterwalder-Schrader theory\n> [see, e.g., math-ph/0001010 or the book by Glimm and Jaffe] shows how to\n> go back in case of relativistic quantum field theory.\n\njust read some introduction about Feynman-Kac formula, not dig into\nthe math yet, I think Feynman-Kac formula is mainly about the\nmathematical meaning of path integral. Does it also say something\nabout the equivalence of the physical consequence?\n\nFeynman integral can be handled mathematically only after analytic\ncontinuation, in what kind of situation can we perform this kind of\ncontinuation? I know it will have some problem when we consider QFT in\nthe curved space, does it always be ok when we battle with the\n"ordinary" QM and QFT?\n\n> > By the way, anyone know where can I found the pronunciation of those\n> > great figures\' names like von Neumann, Poincare...?\n>\n> The simplest is to go to some physicist and ask him or her to pronounce it.\n\nthat was the answer I expected ;)\n\n--\nBest Regards,\nLR\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>Hi,

> This is not quite correct. The two pictures are equivalent because
> there is a unitary transformation between them. The Stone-von Neumann
> theorem guarantees that the canonical commutation relations have a
> unique representation (apart from unitary transformations) and hence
> suffice to specify the QM of finitely many degrees of freedom, no matter
> which picture.

If every such picture can be represented by some cannonical
commutation relations, does that mean all the pictures are
equivalence? If in some situation, this theorem cannot apply, I think
we may construct some non-equivalenct picture that can be identified
as Schrodinger picture and Heisenberg picture of that system. So
without this theorem we cannot say the equivalence since the unitary
transformation may not exist. (actually I think you are talking about
we can always use the Hamiltonian to get an \exp(-iHt), and I don't
know how to shoot this down in the case I just mentioned, so, correct
me if I'm wrong:))

> The Stone-von Neumann theorem fails for systems of infinitely many
> degrees of freedom, which in a sense 'causes' the difficulties in
> quantum field theory. Nevertheless, QFT still has a Schroedinger picture
> and a Heisenberg picture, and these are still equivalent.

Is there a proof or we just believe that, Because there is Haag's
theorem says that the interaction picture does not exist, and I think
interaction picture is something in between of the Schroedinger and
Heisenberg picture. If we cannot get a unitary transformation for
interaction picture, How can we be so sure that we can get one between
Schroedinger and Heisenberg picture?

> The Feynman-Kac formula makes the equivalence precise, after analytically
> continuing the time to purely imaginary. The Osterwalder-Schrader theory
> [see, e.g., http://www.arxiv.org/abs/math-ph/0001010 or the book by Glimm and Jaffe] shows how to
> go back in case of relativistic quantum field theory.

just read some introduction about Feynman-Kac formula, not dig into
the math yet, I think Feynman-Kac formula is mainly about the
mathematical meaning of path integral. Does it also say something
about the equivalence of the physical consequence?

Feynman integral can be handled mathematically only after analytic
continuation, in what kind of situation can we perform this kind of
continuation? I know it will have some problem when we consider QFT in
the curved space, does it always be ok when we battle with the
"ordinary" QM and QFT?

> > By the way, anyone know where can I found the pronunciation of those
> > great figures' names like von Neumann, Poincare...?
>
> The simplest is to go to some physicist and ask him or her to pronounce it.

that was the answer I expected ;)

--
Best Regards,
LR

Cl.Massé

Nov16-04, 02:18 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"L.R." <hephooey@hotmail.com> a écrit dans le message de\nnews:231bb620.0411110407.70fbc1f6@posting.goog le.com...\n\n> By reading some old posts of this group, I learned that there is a\n> Stone-von Neumann theorem guarantee the equivalence of Heisenberg\n> picture and Schrödinger picture in QM. My question is, if we think\n> that path integral is also some kind of picture, is there a theorem\n> like Stone-von Neumann that say something about the equivalence\n> between PI and other pictures?\n\nPath integral is no more than a revisiting of the way Schödringer\ndiscovered his equation, in the converse direction.\n\n> By the way, anyone know where can I found the pronunciation of those\n> great figures\' names like von Neumann, Poincare...?\n\nVon Neumann : German [fOn \'nOiman]\nPoincaré : French (with an accent) [pwe~ka\'Re]\nSchrödinger : German [SR2\'diN@R]\nHeisenberg : German [\'haiz@n\'bERg]\nDe Broglie : French [d@ \'bROi]\nBohr : Danish [bO:R]\n\nTranscribed in SAMPA (Speech Assessment Methods Phonetic Alphabet.)\n\n--\n~~~~ clmasse on free dot F-country\nLiberty, Equality, Profitability.\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>"L.R." <hephooey@hotmail.com> a écrit dans le message de
news:231bb620.0411110407.70fbc1f6@posting.google.c om...

> By reading some old posts of this group, I learned that there is a
> Stone-von Neumann theorem guarantee the equivalence of Heisenberg
> picture and Schrödinger picture in QM. My question is, if we think
> that path integral is also some kind of picture, is there a theorem
> like Stone-von Neumann that say something about the equivalence
> between \PI and other pictures?

Path integral is no more than a revisiting of the way Schödringer
discovered his equation, in the converse direction.

> By the way, anyone know where can I found the pronunciation of those
> great figures' names like von Neumann, Poincare...?

Von Neumann : German [fOn 'nOiman]
Poincaré : French (with an accent) [pwe~ka'Re]
Schrödinger : German [SR2'diN@R]
Heisenberg : German ['haiz@n'bERg]
De Broglie : French [d@ 'bROi]
Bohr : Danish [bO:R]

Transcribed in SAMPA (Speech Assessment Methods Phonetic Alphabet.)

--
~~~~ clmasse on free dot F-country
Liberty, Equality, Profitability.

Chris Oakley

Nov17-04, 11:14 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\n\n> > The Stone-von Neumann theorem fails for systems of infinitely many\n> > degrees of freedom, which in a sense \'causes\' the difficulties in\n> > quantum field theory. Nevertheless, QFT still has a Schroedinger picture\n> > and a Heisenberg picture, and these are still equivalent.\n>\n> Is there a proof or we just believe that, Because there is Haag\'s\n> theorem says that the interaction picture does not exist, and I think\n> interaction picture is something in between of the Schroedinger and\n> Heisenberg picture. If we cannot get a unitary transformation for\n> interaction picture, How can we be so sure that we can get one between\n> Schroedinger and Heisenberg picture?\n\nThere must always be a time displacement operator U(t) on the vector space\nof quantum mechanics and this can be used to translate states and operators\ndefined for a given time to those defined at a reference time (t = 0\ncommonly). This is the transformation that takes us from the Heisenberg to\nthe Schrödinger picture.\n\nThe "interaction" or "Dirac" picture OTOH requires one to be able to write\nthe time displacement generator (the Hamiltonian) as a sum of a "free" part\nand an "interaction".\n\nIt may well be that there is no "free" Hamiltonian at all: i.e. when\ninteractions exist it may well be that no operator can be constructed that\ncauses time displacement of interacting states & operators that is the same\nas when they were free. In this case there can be no Interaction Picture.\n\nHaag\'s theorem (Hall/Wightman: Mat. Fys. Medd. Dansk. Vid. Selskapet 31, p.\n5 (1957)) says, amongst other things, that there is no Interaction Picture\nwhen the theory is relativistic.\n\nIntuitively, one can see that relativity would cause problems because here\ntime loses its special status, being mixed in with space in different\nreference frames.\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>The Stone-von Neumann theorem fails for systems of infinitely many
> > degrees of freedom, which in a sense 'causes' the difficulties in
> > quantum field theory. Nevertheless, QFT still has a Schroedinger picture
> > and a Heisenberg picture, and these are still equivalent.
>
> Is there a proof or we just believe that, Because there is Haag's
> theorem says that the interaction picture does not exist, and I think
> interaction picture is something in between of the Schroedinger and
> Heisenberg picture. If we cannot get a unitary transformation for
> interaction picture, How can we be so sure that we can get one between
> Schroedinger and Heisenberg picture?

There must always be a time displacement operator U(t) on the vector space
of quantum mechanics and this can be used to translate states and operators
defined for a given time to those defined at a reference time (t =
commonly). This is the transformation that takes us from the Heisenberg to
the Schrödinger picture.

The "interaction" or "Dirac" picture OTOH requires one to be able to write
the time displacement generator (the Hamiltonian) as a sum of a "free" part
and an "interaction".

It may well be that there is no "free" Hamiltonian at all: i.e. when
interactions exist it may well be that no operator can be constructed that
causes time displacement of interacting states & operators that is the same
as when they were free. In this case there can be no Interaction Picture.

Haag's theorem (Hall/Wightman: Mat. Fys. Medd. Dansk. Vid. Selskapet 31, p.
5 (1957)) says, amongst other things, that there is no Interaction Picture
when the theory is relativistic.

Intuitively, one can see that relativity would cause problems because here
time loses its special status, being mixed in with space in different
reference frames.

Arnold Neumaier

Nov17-04, 11:14 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\nL.R. wrote:\n> Hi,\n>\n>\n>>This is not quite correct. The two pictures are equivalent because\n>>there is a unitary transformation between them. The Stone-von Neumann\n>>theorem guarantees that the canonical commutation relations have a\n>>unique representation (apart from unitary transformations) and hence\n>>suffice to specify the QM of finitely many degrees of freedom, no matter\n>>which picture.\n>\n>\n> If every such picture can be represented by some canonical\n> commutation relations, does that mean all the pictures are\n> equivalent?\n\nFor a finite number of degrees of freedom, yes.\n\n\n>>The Stone-von Neumann theorem fails for systems of infinitely many\n>>degrees of freedom, which in a sense \'causes\' the difficulties in\n>>quantum field theory. Nevertheless, QFT still has a Schroedinger picture\n>>and a Heisenberg picture, and these are still equivalent.\n>\n> Is there a proof or we just believe that\n\nCounterexamples to the Stone-von Neumann theorem fails for systems of\ninfinitely many degrees of freedom are known.\n\nSince there is no known mathematically consistent definition of an\ninteracting QFT in 4D, the statement about QFT is of course not proved.\n\n\n> Because there is Haag\'s\n> theorem says that the interaction picture does not exist, and I think\n> interaction picture is something in between of the Schroedinger and\n> Heisenberg picture. If we cannot get a unitary transformation for\n> interaction picture, How can we be so sure that we can get one between\n> Schroedinger and Heisenberg picture?\n\nOn the nonrigorous level commonly used by physicists, all three pictures\nexist perturbatively. On a rigorous level in 2D or 3D, the\nHeisenberg picture can be immediately constructed from the Wightman\nfields. Then the canonical procedure - fixing the Heisenberg operators\nat time t=0 and instead defining dynamical states\npsi(t) := exp(-itH)psi\n- produces the Schroedinger picture from it.\n\n\n>>The Feynman-Kac formula makes the equivalence precise, after analytically\n>>continuing the time to purely imaginary. The Osterwalder-Schrader theory\n>>[see, e.g., math-ph/0001010 or the book by Glimm and Jaffe] shows how to\n>>go back in case of relativistic quantum field theory.\n>\n> just read some introduction about Feynman-Kac formula, not dig into\n> the math yet, I think Feynman-Kac formula is mainly about the\n> mathematical meaning of path integral. Does it also say something\n> about the equivalence of the physical consequence?\n\nIt says that time-ordered expectation values are expressible as\nfunctional integrals, thus providing the \'translation key\'.\n\n\n> Feynman integral can be handled mathematically only after analytic\n> continuation, in what kind of situation can we perform this kind of\n> continuation?\n\nThis is precisely the topic of Osterwalder-Schrader theory.\n\n\n> I know it will have some problem when we consider QFT in\n> the curved space, does it always be ok when we battle with the\n> "ordinary" QM and QFT?\n\nCurved space is not covered by axiomatic field theory since\nPoincare invariance is lost. So mathematically one is on more shaky\nground. Nevertheless, people use the path integral.\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>L.R. wrote:
> Hi,
>
>
>>This is not quite correct. The two pictures are equivalent because
>>there is a unitary transformation between them. The Stone-von Neumann
>>theorem guarantees that the canonical commutation relations have a
>>unique representation (apart from unitary transformations) and hence
>>suffice to specify the QM of finitely many degrees of freedom, no matter
>>which picture.
>
>
> If every such picture can be represented by some canonical
> commutation relations, does that mean all the pictures are
> equivalent?

For a finite number of degrees of freedom, yes.

>>The Stone-von Neumann theorem fails for systems of infinitely many
>>degrees of freedom, which in a sense 'causes' the difficulties in
>>quantum field theory. Nevertheless, QFT still has a Schroedinger picture
>>and a Heisenberg picture, and these are still equivalent.
>
> Is there a proof or we just believe that

Counterexamples to the Stone-von Neumann theorem fails for systems of
infinitely many degrees of freedom are known.

Since there is no known mathematically consistent definition of an
interacting QFT in 4D, the statement about QFT is of course not proved.

> Because there is Haag's
> theorem says that the interaction picture does not exist, and I think
> interaction picture is something in between of the Schroedinger and
> Heisenberg picture. If we cannot get a unitary transformation for
> interaction picture, How can we be so sure that we can get one between
> Schroedinger and Heisenberg picture?

On the nonrigorous level commonly used by physicists, all three pictures
exist perturbatively. On a rigorous level in 2D or 3D, the
Heisenberg picture can be immediately constructed from the Wightman
fields. Then the canonical procedure - fixing the Heisenberg operators
at time t=0 and instead defining dynamical states
\psi(t) := \exp(-itH)\psi
- produces the Schroedinger picture from it.

>>The Feynman-Kac formula makes the equivalence precise, after analytically
>>continuing the time to purely imaginary. The Osterwalder-Schrader theory
>>[see, e.g., http://www.arxiv.org/abs/math-ph/0001010 or the book by Glimm and Jaffe] shows how to
>>go back in case of relativistic quantum field theory.
>
> just read some introduction about Feynman-Kac formula, not dig into
> the math yet, I think Feynman-Kac formula is mainly about the
> mathematical meaning of path integral. Does it also say something
> about the equivalence of the physical consequence?

It says that time-ordered expectation values are expressible as
functional integrals, thus providing the 'translation key'.

> Feynman integral can be handled mathematically only after analytic
> continuation, in what kind of situation can we perform this kind of
> continuation?

This is precisely the topic of Osterwalder-Schrader theory.

> I know it will have some problem when we consider QFT in
> the curved space, does it always be ok when we battle with the
> "ordinary" QM and QFT?

Curved space is not covered by axiomatic field theory since
Poincare invariance is lost. So mathematically one is on more shaky
ground. Nevertheless, people use the path integral.

Arnold Neumaier