Need the vacuum belong to Hilbert space? [Archive]

Very cryptic

Aug6-04, 03:05 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>A thought just occured to me. Since the Hamiltonion is an unbounded\noperator and we know in general, the eigenstates of an unbounded\noperator (like the Dirac delta function, for example) belong to a\nrigged Hilbert space in general and the vacuum is defined to be the\neigenstate of the lowest eigenvalue of the Hamiltonian, could it be\npossible the vacuum does NOT belong to the Hilbert space? And if it\ndoesn\'t, could it be possible the n-point correlation functions is NOT\ndefined in general?\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>A thought just occured to me. Since the Hamiltonion is an unbounded
operator and we know in general, the eigenstates of an unbounded
operator (like the Dirac \delta function, for example) belong to a
rigged Hilbert space in general and the vacuum is defined to be the
eigenstate of the lowest eigenvalue of the Hamiltonian, could it be
possible the vacuum does NOT belong to the Hilbert space? And if it
doesn't, could it be possible the n-point correlation functions is NOT
defined in general?

mandro

Aug7-04, 05:05 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>This all depends on which Hamiltonian you\'re\ntalking about. Some Hamiltonians have genuine\neigenstates, others do not. For your case though,\njust the "free" Hamiltonian H=p^2 in QM for a\nparticle living on the real line suffices to prove\nyour point. I.e., none of the "formal" eigenstates\nof H, are actual eigenvectors in the Hilbert space.\nFor other cases, like the Harmonic oscillator, the\ngroundstate is an actual vector in Hilbert space,\nyet H is unbounded above. So this settles the first\nhalf of your question:\n\n> is it possible the vacuum does NOT belong to the Hilbert space?\n\nif you were talking ordinary\nQuantum Mechanics. However, your question:\n\n> And if it doesn\'t, could it be possible\n> the n-point correlation functions is NOT\n> defined in general?\n\nsounds like you\'re asking something about QFT,\nand I don\'t know much about this.\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>This all depends on which Hamiltonian you're
talking about. Some Hamiltonians have genuine
eigenstates, others do not. For your case though,
just the "free" Hamiltonian H=p^2 in QM for a
particle living on the real line suffices to prove
your point. I.e., none of the "formal" eigenstates
of H, are actual eigenvectors in the Hilbert space.
For other cases, like the Harmonic oscillator, the
groundstate is an actual vector in Hilbert space,
yet H is unbounded above. So this settles the first
half of your question:

> is it possible the vacuum does NOT belong to the Hilbert space?

if you were talking ordinary
Quantum Mechanics. However, your question:

> And if it doesn't, could it be possible
> the n-point correlation functions is NOT
> defined in general?

sounds like you're asking something about QFT,
and I don't know much about this.

Arkadiusz Jadczyk

Aug12-04, 08:29 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\nOn Fri, 6 Aug 2004 08:05:03 +0000 (UTC), very_cryptic@hotmail.com (Very\ncryptic) wrote:\n\n>A thought just occured to me. Since the Hamiltonion is an unbounded\n>operator and we know in general, the eigenstates of an unbounded\n>operator (like the Dirac delta function, for example) belong to a\n>rigged Hilbert space in general and the vacuum is defined to be the\n>eigenstate of the lowest eigenvalue of the Hamiltonian, could it be\n>possible the vacuum does NOT belong to the Hilbert space? And if it\n>doesn\'t, could it be possible the n-point correlation functions is NOT\n>defined in general?\n\nIt all depends on what you understand by "the vacuum". Sometimes vacuum\nis understood as the state with zero particle number. In this case, even\nif particle number is an unbounded operator, it usually has a purely\ndiscrete spectrum, and there is no reason not to include its zero\neigenspace.\n\nOn the other hand, if you are looking for "possible difficulties", not\nonly Hamiltonian may have no discrete lowest energy eigenstate, but even\na worse situation can happen: there may be time evolution, but no\nHamiltonian! This may happen within somewhat more general (but useful)\nalgebraic formalism, when the very notion of "the Hilbert space" should\nbe approached with care!\n\nark\n--\n\nArkadiusz Jadczyk\nhttp://quantumfuture.net/quantum_future/jadpub.htm\n--\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Fri, 6 Aug 2004 08:05:03 +0000 (UTC), very_cryptic@hotmail.com (Very
cryptic) wrote:

>A thought just occured to me. Since the Hamiltonion is an unbounded
>operator and we know in general, the eigenstates of an unbounded
>operator (like the Dirac \delta function, for example) belong to a
>rigged Hilbert space in general and the vacuum is defined to be the
>eigenstate of the lowest eigenvalue of the Hamiltonian, could it be
>possible the vacuum does NOT belong to the Hilbert space? And if it
>doesn't, could it be possible the n-point correlation functions is NOT
>defined in general?

It all depends on what you understand by "the vacuum". Sometimes vacuum
is understood as the state with zero particle number. In this case, even
if particle number is an unbounded operator, it usually has a purely
discrete spectrum, and there is no reason not to include its zero
eigenspace.

On the other hand, if you are looking for "possible difficulties", not
only Hamiltonian may have no discrete lowest energy eigenstate, but even
a worse situation can happen: there may be time evolution, but no
Hamiltonian! This may happen within somewhat more general (but useful)
algebraic formalism, when the very notion of "the Hilbert space" should
be approached with care!

ark
--

Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
--

Very cryptic

Aug13-04, 05:41 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\nArkadiusz Jadczyk <arkREMOVETHIS@ANDTHIScassiopaea.org> wrote in message news:<tbkfh0t52l86sb3dfto43t5h22junpu6jv@4ax.com>. ..\n\n> It all depends on what you understand by "the vacuum". Sometimes vacuum\n> is understood as the state with zero particle number. In this case, even\n> if particle number is an unbounded operator, it usually has a purely\n> discrete spectrum, and there is no reason not to include its zero\n> eigenspace.\n\nIsn\'t particle number only well-defined in the interaction picture\n(fock space)or the LHZ formalism? In the former case the vacuum is not\nthe state with zero particles not to mention fock space doesn\'t exist\nfor interacting theories and in the latter, particle numbers are\ndefined in terms of the vacuum, not the other way around.\n\n>\n> On the other hand, if you are looking for "possible difficulties", not\n> only Hamiltonian may have no discrete lowest energy eigenstate, but even\n> a worse situation can happen: there may be time evolution, but no\n> Hamiltonian! This may happen within somewhat more general (but useful)\n> algebraic formalism, when the very notion of "the Hilbert space" should\n> be approached with care!\n\nIs there any concrete example of this outside of 0+1 theories?\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>Arkadiusz Jadczyk <arkREMOVETHIS@ANDTHIScassiopaea.org> wrote in message news:<tbkfh0t52l86sb3dfto43t5h22junpu6jv@4ax.com>...

> It all depends on what you understand by "the vacuum". Sometimes vacuum
> is understood as the state with zero particle number. In this case, even
> if particle number is an unbounded operator, it usually has a purely
> discrete spectrum, and there is no reason not to include its zero
> eigenspace.

Isn't particle number only well-defined in the interaction picture
(fock space)or the LHZ formalism? In the former case the vacuum is not
the state with zero particles not to mention fock space doesn't exist
for interacting theories and in the latter, particle numbers are
defined in terms of the vacuum, not the other way around.

>
> On the other hand, if you are looking for "possible difficulties", not
> only Hamiltonian may have no discrete lowest energy eigenstate, but even
> a worse situation can happen: there may be time evolution, but no
> Hamiltonian! This may happen within somewhat more general (but useful)
> algebraic formalism, when the very notion of "the Hilbert space" should
> be approached with care!

Is there any concrete example of this outside of 0+1 theories?

Arkadiusz Jadczyk

Aug16-04, 12:55 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>\n\n\n\n\nOn 13 Aug 2004 06:41:26 -0400, very_cryptic@hotmail.com (Very cryptic)\nwrote:\n\n>> It all depends on what you understand by "the vacuum". Sometimes vacuum\n>> is understood as the state with zero particle number. In this case, even\n>> if particle number is an unbounded operator, it usually has a purely\n>> discrete spectrum, and there is no reason not to include its zero\n>> eigenspace.\n>\n>Isn\'t particle number only well-defined in the interaction picture\n>(fock space)or the LHZ formalism? In the former case the vacuum is not\n>the state with zero particles not to mention fock space doesn\'t exist\n>for interacting theories and in the latter, particle numbers are\n>defined in terms of the vacuum, not the other way around.\n\nYes, you are right - to some extent. We can always think of an\n"asymptotic Fock space" though, and hope that it has a common vacuum\nwith "the real Hilbert space", when the latter one may or may not even\nexists!\n\n\n>> On the other hand, if you are looking for "possible difficulties", not\n>> only Hamiltonian may have no discrete lowest energy eigenstate, but even\n>> a worse situation can happen: there may be time evolution, but no\n>> Hamiltonian! This may happen within somewhat more general (but useful)\n>> algebraic formalism, when the very notion of "the Hilbert space" should\n>> be approached with care!\n>\n>Is there any concrete example of this outside of 0+1 theories?\n\nThsi can happen in infinite volume limit of quantum lattice systems, or\nwith infinite volume limit of a Bose gas. See, for instance, the last\nchapter of Emch\'s "Algebraic methods".\n\nark\n--\n\nArkadiusz Jadczyk\nhttp://quantumfuture.net/quantum_future/jadpub.htm\n--\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 13 Aug 2004 06:41:26 -0400, very_cryptic@hotmail.com (Very cryptic)
wrote:

>> It all depends on what you understand by "the vacuum". Sometimes vacuum
>> is understood as the state with zero particle number. In this case, even
>> if particle number is an unbounded operator, it usually has a purely
>> discrete spectrum, and there is no reason not to include its zero
>> eigenspace.
>
>Isn't particle number only well-defined in the interaction picture
>(fock space)or the LHZ formalism? In the former case the vacuum is not
>the state with zero particles not to mention fock space doesn't exist
>for interacting theories and in the latter, particle numbers are
>defined in terms of the vacuum, not the other way around.

Yes, you are right - to some extent. We can always think of an
"asymptotic Fock space" though, and hope that it has a common vacuum
with "the real Hilbert space", when the latter one may or may not even
exists!

>> On the other hand, if you are looking for "possible difficulties", not
>> only Hamiltonian may have no discrete lowest energy eigenstate, but even
>> a worse situation can happen: there may be time evolution, but no
>> Hamiltonian! This may happen within somewhat more general (but useful)
>> algebraic formalism, when the very notion of "the Hilbert space" should
>> be approached with care!
>
>Is there any concrete example of this outside of 0+1 theories?

Thsi can happen in infinite volume limit of quantum lattice systems, or
with infinite volume limit of a Bose gas. See, for instance, the last
chapter of Emch's "Algebraic methods".

ark
--

Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
--

Charles Torre

Aug17-04, 01:28 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n> >\n> > On the other hand, if you are looking for "possible difficulties", not\n> > only Hamiltonian may have no discrete lowest energy eigenstate, but even\n> > a worse situation can happen: there may be time evolution, but no\n> > Hamiltonian! This may happen within somewhat more general (but useful)\n> > algebraic formalism, when the very notion of "the Hilbert space" should\n> > be approached with care!\n>\n> Is there any concrete example of this outside of 0+1 theories?\n\nQuantum fields in a non-static spacetime generically have\nthis problem and then some - the dynamical evolution is\ntypically not even implemented by a unitary transformation!\nA similar thing happens with quantum fields in flat\nspacetime when dynamics are defined by transformations\nwhich aren\'t isometries.\n\nIf you want a 1 parameter unitary group for which there is\nno Hamiltonian you need to cook up an example where the\nunitary transformations t -> U(t) do not depend\ncontinuously on t. Otherwise Stone\'s theorem guarantees a\nHamiltonian. I don\'t have a really good example right off\nthe top of my head. Sorry.\n\ncharlie\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On the other hand, if you are looking for "possible difficulties", not
> > only Hamiltonian may have no discrete lowest energy eigenstate, but even
> > a worse situation can happen: there may be time evolution, but no
> > Hamiltonian! This may happen within somewhat more general (but useful)
> > algebraic formalism, when the very notion of "the Hilbert space" should
> > be approached with care!
>
> Is there any concrete example of this outside of 0+1 theories?

Quantum fields in a non-static spacetime generically have
this problem and then some - the dynamical evolution is
typically not even implemented by a unitary transformation!
A similar thing happens with quantum fields in flat
spacetime when dynamics are defined by transformations
which aren't isometries.

If you want a 1 parameter unitary group for which there is
no Hamiltonian you need to cook up an example where the
unitary transformations t -> U(t) do not depend
continuously on t. Otherwise Stone's theorem guarantees a
Hamiltonian. I don't have a really good example right off
the top of my head. Sorry.

charlie