<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>\nIgor Khavkine wrote:\n\n> I\'ve been bothered by this for a while. If you are given an operator\n> O on the Fock space of a field theory. How does one determine its\n> normal ordered version :O: without knowing explicity its expression\n> in terms of c/a operators.\n\nThis question does not make sense. Unless O is already normal-ordered,\nit is just a meaningless formal expression and not an operator in\nFock space - the normal ordering is there precisely to turn O into\na well-defined operator :O:\n\nAt least that is what happens if the 1-particle Hilbert space underlying\nthe Fock space is infinite-dimensional. On the other hand, for\nfinite-dimensional 1-particle Hilbert spaces, both O and :O: make\nsense. Therefore normal ordering is usually motivated by starting\nwith a finite-dimensional discretization where integrals become\nfinite sums; then one can do all the formal manipulations rigorously.\nUpon passing to the continuum limit, most oexpressions become infinite\nand hence meaningless, but the normally ordered expressions happen\nto have a well-defined meaning. So these are the relevant operators.\n\n\n> If every operator on the Fock space\n> were expressible in terms of c/a operators\n\nThis is indeed the case. You can find a proof in Scharf\'s book on\nQED. Every operator is expressible as a power series with normally\nordered terms.\n\n\n> and there were an algorithm\n> for obtaining this explicit expression given O, then there would be\n> no problem with defining normal ordering by `Operator commutation\'.\n\nOperator commutation for normal ordering is not a well-defined activity\nsince [a(p),a^*(p)] is meaningless. a(p) and a^*(p) are only densely\ndefined oprators, and a(p)a^*(p) which is needed in the commutation\ncannot be given a meaning since the domains of definition do not match\ncorrectly.\n\n\n> Otherwise, it seems that this prescription is only valid for the\n> class of operators which are explicitly expressible in terms of\n> c/a operators.\n\nYes, that\'s how it is used. It is a way to _define_ operators,\nnot one to transform operators into other operators.\n\n\n> Can normal ordering be defined for O given only the operator itself,\n> the Fock space, the inner product, and the ground state?\n\nAs already mentioned, O is not an opertator but a formal string looking\nlike a would-be operator. :O: is obtained by formal manipulations of this\nstring, expanding all terms into monomials and then swapping neighboring\npairs of formal c/a operators (and for fermions changing signs) until\nall creators are to the left of all annihilators. The resulting, uniquely\ndefined string can then be interpreted as an operator.\n\nThe inner product and the ground state is needed only to define the\nFock space.\n\n\n> How about matrix elements between two basis elements of the Fock space?\n> Even if the operator itself is normal ordered, the matrix element\n> can be expressed as a vacuum expecation value with some annihilation\n> operators on the left and some creation operators on the right which\n> results in a non-normal ordered operator. Why is the matrix element\n> then garanteed to be finite?\n\nOnly matrix elements between states with compact support are guaranteed\nto be finite. This is because the a(p), a^*(p) are only\ndistribution-valued operators. Matrix element between momentum basis states\nor position basis states are generally not finite but are distributions\nin the variables labeling the basis states.\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>Igor Khavkine wrote:
> I've been bothered by this for a while. If you are given an operator
> O on the Fock space of a field theory. How does one determine its
> normal ordered version :O: without knowing explicity its expression
> in terms of c/a operators.
This question does not make sense. Unless O is already normal-ordered,
it is just a meaningless formal expression and not an operator in
Fock space - the normal ordering is there precisely to turn O into
a well-defined operator :O:
At least that is what happens if the 1-particle Hilbert space underlying
the Fock space is infinite-dimensional. On the other hand, for
finite-dimensional 1-particle Hilbert spaces, both O and :O: make
sense. Therefore normal ordering is usually motivated by starting
with a finite-dimensional discretization where integrals become
finite sums; then one can do all the formal manipulations rigorously.
Upon passing to the continuum limit, most oexpressions become infinite
and hence meaningless, but the normally ordered expressions happen
to have a well-defined meaning. So these are the relevant operators.
> If every operator on the Fock space
> were expressible in terms of c/a operators
This is indeed the case. You can find a proof in Scharf's book on
QED. Every operator is expressible as a power series with normally
ordered terms.
> and there were an algorithm
> for obtaining this explicit expression given O, then there would be
> no problem with defining normal ordering by `Operator commutation'.
Operator commutation for normal ordering is not a well-defined activity
since [a(p),a^*(p)] is meaningless. a(p) and a^*(p) are only densely
defined oprators, and a(p)a^*(p) which is needed in the commutation
cannot be given a meaning since the domains of definition do not match
correctly.
> Otherwise, it seems that this prescription is only valid for the
> class of operators which are explicitly expressible in terms of
> c/a operators.
Yes, that's how it is used. It is a way to _define_ operators,
not one to transform operators into other operators.
> Can normal ordering be defined for O given only the operator itself,
> the Fock space, the inner product, and the ground state?
As already mentioned, O is not an opertator but a formal string looking
like a would-be operator. :O: is obtained by formal manipulations of this
string, expanding all terms into monomials and then swapping neighboring
pairs of formal c/a operators (and for fermions changing signs) until
all creators are to the left of all annihilators. The resulting, uniquely
defined string can then be interpreted as an operator.
The inner product and the ground state is needed only to define the
Fock space.
> How about matrix elements between two basis elements of the Fock space?
> Even if the operator itself is normal ordered, the matrix element
> can be expressed as a vacuum expecation value with some annihilation
> operators on the left and some creation operators on the right which
> results in a non-normal ordered operator. Why is the matrix element
> then garanteed to be finite?
Only matrix elements between states with compact support are guaranteed
to be finite. This is because the a(p), a^*(p) are only
distribution-valued operators. Matrix element between momentum basis states
or position basis states are generally not finite but are distributions
in the variables labeling the basis states.
Arnold Neumaier
Igor Khavkine
Jul7-04, 09:09 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Tue, 06 Jul 2004 14:47:42 -0400, Arnold Neumaier wrote:\n\n> Igor Khavkine wrote:\n>\n>> I\'ve been bothered by this for a while. If you are given an operator O\n>> on the Fock space of a field theory. How does one determine its normal\n>> ordered version :O: without knowing explicity its expression in terms of\n>> c/a operators.\n>\n> This question does not make sense. Unless O is already normal-ordered, it\n> is just a meaningless formal expression and not an operator in Fock space\n> - the normal ordering is there precisely to turn O into a well-defined\n> operator :O:\n\nGranted.\n\n> At least that is what happens if the 1-particle Hilbert space underlying\n> the Fock space is infinite-dimensional. On the other hand, for\n> finite-dimensional 1-particle Hilbert spaces, both O and :O: make sense.\n> Therefore normal ordering is usually motivated by starting with a\n> finite-dimensional discretization where integrals become finite sums; then\n> one can do all the formal manipulations rigorously. Upon passing to the\n> continuum limit, most oexpressions become infinite and hence meaningless,\n> but the normally ordered expressions happen to have a well-defined\n> meaning. So these are the relevant operators.\n\nI see. I think most of my questions can be reformulated to be more precise\nby using a finite dimensional space and using a limiting procedure as you\nsuggest above.\n\n>> If every operator on the Fock space\n>> were expressible in terms of c/a operators\n>\n> This is indeed the case. You can find a proof in Scharf\'s book on QED.\n> Every operator is expressible as a power series with normally ordered\n> terms.\n\nThanks. I\'ll check this book out.\n\n>> and there were an algorithm\n>> for obtaining this explicit expression given O, then there would be no\n>> problem with defining normal ordering by `Operator commutation\'.\n>\n> Operator commutation for normal ordering is not a well-defined activity\n> since [a(p),a^*(p)] is meaningless. a(p) and a^*(p) are only densely\n> defined operators, and a(p)a^*(p) which is needed in the commutation\n> cannot be given a meaning since the domains of definition do not match\n> correctly.\n\nOk, so let me recap. If the Fock space is built on a finite dimensional\nstate space, then given a well defined operator O (which could be\nspecified in any number of ways) we should be able to construct it\'s,\npossibly distinct, normal ordered version :O:. If the number of dimensions\nof the underlying 1-particle state space is increased to infinity in some\nsort of limiting procedure, then the limit of O may not be a well defined\noperator, while that of :O: should be. Of course, I\'m omitting a bunch of\ndetails that would really make the above precise, but I hope I\'ve got the\nright idea.\n\nActually, I\'ve already seen a version of this procedure in the physics\nliterature. There, any operator that is written down as a formal series in\nc/a operators can be thought of as the same series where the n-th term is\nsuppressed by exp(-an), in the limit a -> 0. In this case, formal\nmanipulations yield finite answers, at least for a > 0.\n\nIn the second limiting procedure, since operators are only defined using\nformal series of c/a operators, then operator commutation is a valid\nnormal ordering prescription. However, other prescriptions such as\nsubtracting the vacuum expectation value may also be applicable. I\'d like\nto know when these prescriptions are equivalent, especially when O is not\nlinear in the c/a operators.\n\nHowever, these formal manipulations may already be on shaky ground, and I\nthink the first limiting procedure I described can be made more rigorous.\nBut in that case, the operator O may not necessarily be given as an\nexpression involving c/a operators. Then what is the prescription to\ndefine :O:? Does the theorem from Scharf\'s book answer this?\n\n>> How about matrix elements between two basis elements of the Fock space?\n>> Even if the operator itself is normal ordered, the matrix element can\n>> be expressed as a vacuum expectation value with some annihilation\n>> operators on the left and some creation operators on the right which\n>> results in a non-normal ordered operator. Why is the matrix element\n>> then guaranteed to be finite?\n>\n> Only matrix elements between states with compact support are guaranteed\n> to be finite. This is because the a(p), a^*(p) are only\n> distribution-valued operators. Matrix element between momentum basis\n> states or position basis states are generally not finite but are\n> distributions in the variables labeling the basis states.\n\nIf the 1-particle Hilbert space is separable (which I hope is\nequivalent to having a discrete countable basis), which is the case\nI\'m interested in at the moment, then we need not talk about\ndistributions. If, as you suggested, normal ordering is used to *define*\noperators on the Fock space, then it should be granted that matrix\nelements of normal ordered operators are finite. In this case I\'m\ninterested in why/how various normal ordering procedures produce well\ndefined operators. Does Scharf\'s book have details on this topic?\n\nThanks.\n\nIgor\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 Tue, 06 Jul 2004 14:47:42 -0400, Arnold Neumaier wrote:
> Igor Khavkine wrote:
>
>> I've been bothered by this for a while. If you are given an operator O
>> on the Fock space of a field theory. How does one determine its normal
>> ordered version :O: without knowing explicity its expression in terms of
>> c/a operators.
>
> This question does not make sense. Unless O is already normal-ordered, it
> is just a meaningless formal expression and not an operator in Fock space
> - the normal ordering is there precisely to turn O into a well-defined
> operator :O:
Granted.
> At least that is what happens if the 1-particle Hilbert space underlying
> the Fock space is infinite-dimensional. On the other hand, for
> finite-dimensional 1-particle Hilbert spaces, both O and :O: make sense.
> Therefore normal ordering is usually motivated by starting with a
> finite-dimensional discretization where integrals become finite sums; then
> one can do all the formal manipulations rigorously. Upon passing to the
> continuum limit, most oexpressions become infinite and hence meaningless,
> but the normally ordered expressions happen to have a well-defined
> meaning. So these are the relevant operators.
I see. I think most of my questions can be reformulated to be more precise
by using a finite dimensional space and using a limiting procedure as you
suggest above.
>> If every operator on the Fock space
>> were expressible in terms of c/a operators
>
> This is indeed the case. You can find a proof in Scharf's book on QED.
> Every operator is expressible as a power series with normally ordered
> terms.
Thanks. I'll check this book out.
>> and there were an algorithm
>> for obtaining this explicit expression given O, then there would be no
>> problem with defining normal ordering by `Operator commutation'.
>
> Operator commutation for normal ordering is not a well-defined activity
> since [a(p),a^*(p)] is meaningless. a(p) and a^*(p) are only densely
> defined operators, and a(p)a^*(p) which is needed in the commutation
> cannot be given a meaning since the domains of definition do not match
> correctly.
Ok, so let me recap. If the Fock space is built on a finite dimensional
state space, then given a well defined operator O (which could be
specified in any number of ways) we should be able to construct it's,
possibly distinct, normal ordered version :O:. If the number of dimensions
of the underlying 1-particle state space is increased to infinity in some
sort of limiting procedure, then the limit of O may not be a well defined
operator, while that of :O: should be. Of course, I'm omitting a bunch of
details that would really make the above precise, but I hope I've got the
right idea.
Actually, I've already seen a version of this procedure in the physics
literature. There, any operator that is written down as a formal series in
c/a operators can be thought of as the same series where the n-th term is
suppressed by \exp(-an), in the limit a -> . In this case, formal
manipulations yield finite answers, at least for a > .
In the second limiting procedure, since operators are only defined using
formal series of c/a operators, then operator commutation is a valid
normal ordering prescription. However, other prescriptions such as
subtracting the vacuum expectation value may also be applicable. I'd like
to know when these prescriptions are equivalent, especially when O is not
linear in the c/a operators.
However, these formal manipulations may already be on shaky ground, and I
think the first limiting procedure I described can be made more rigorous.
But in that case, the operator O may not necessarily be given as an
expression involving c/a operators. Then what is the prescription to
define :O:? Does the theorem from Scharf's book answer this?
>> How about matrix elements between two basis elements of the Fock space?
>> Even if the operator itself is normal ordered, the matrix element can
>> be expressed as a vacuum expectation value with some annihilation
>> operators on the left and some creation operators on the right which
>> results in a non-normal ordered operator. Why is the matrix element
>> then guaranteed to be finite?
>
> Only matrix elements between states with compact support are guaranteed
> to be finite. This is because the a(p), a^*(p) are only
> distribution-valued operators. Matrix element between momentum basis
> states or position basis states are generally not finite but are
> distributions in the variables labeling the basis states.
If the 1-particle Hilbert space is separable (which I hope is
equivalent to having a discrete countable basis), which is the case
I'm interested in at the moment, then we need not talk about
distributions. If, as you suggested, normal ordering is used to *define*
operators on the Fock space, then it should be granted that matrix
elements of normal ordered operators are finite. In this case I'm
interested in why/how various normal ordering procedures produce well
defined operators. Does Scharf's book have details on this topic?
Thanks.
Igor
Charles Torre
Jul9-04, 04:49 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> Ok, so let me recap. If the Fock space is built on a finite dimensional\n> state space, then given a well defined operator O (which could be\n> specified in any number of ways) we should be able to construct it\'s,\n> possibly distinct, normal ordered version :O:. If the number of dimensions\n> of the underlying 1-particle state space is increased to infinity in some\n> sort of limiting procedure, then the limit of O may not be a well defined\n> operator, while that of :O: should be. Of course, I\'m omitting a bunch of\n> details that would really make the above precise, but I hope I\'ve got the\n> right idea.\n>\n\nOne pedantic point: Normal ordering does NOT guarantee\nthat you will get a well-defined operator. Local operators are typically\ninfinite sums of C & A operators and these sums need not converge.\nFor example, I seem to recall (but won\'t swear to it off the top of my head)\nthat the (normal-ordered) energy density of a free field is not a well-defined\noperator on the usual free field Fock space in spacetime dimensions 3 and\nabove. (As I recall, it is a well-defined sesquilinear form, though.)\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>Ok, so let me recap. If the Fock space is built on a finite dimensional
> state space, then given a well defined operator O (which could be
> specified in any number of ways) we should be able to construct it's,
> possibly distinct, normal ordered version :O:. If the number of dimensions
> of the underlying 1-particle state space is increased to infinity in some
> sort of limiting procedure, then the limit of O may not be a well defined
> operator, while that of :O: should be. Of course, I'm omitting a bunch of
> details that would really make the above precise, but I hope I've got the
> right idea.
>
One pedantic point: Normal ordering does NOT guarantee
that you will get a well-defined operator. Local operators are typically
infinite sums of C & A operators and these sums need not converge.
For example, I seem to recall (but won't swear to it off the top of my head)
that the (normal-ordered) energy density of a free field is not a well-defined
operator on the usual free field Fock space in spacetime dimensions 3 and
above. (As I recall, it is a well-defined sesquilinear form, though.)
charlie
Igor Khavkine
Jul13-04, 03:51 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>\ntorre@cc.usu.edu (Charles Torre) wrote in message news:<aaec8787.0407071239.38aea43b@posting.google.com>...\n> > Ok, so let me recap. If the Fock space is built on a finite dimensional\n> > state space, then given a well defined operator O (which could be\n> > specified in any number of ways) we should be able to construct it\'s,\n> > possibly distinct, normal ordered version :O:. If the number of dimensions\n> > of the underlying 1-particle state space is increased to infinity in some\n> > sort of limiting procedure, then the limit of O may not be a well defined\n> > operator, while that of :O: should be. Of course, I\'m omitting a bunch of\n> > details that would really make the above precise, but I hope I\'ve got the\n> > right idea.\n> >\n>\n> One pedantic point: Normal ordering does NOT guarantee\n> that you will get a well-defined operator. Local operators are typically\n> infinite sums of C & A operators and these sums need not converge.\n> For example, I seem to recall (but won\'t swear to it off the top of my head)\n> that the (normal-ordered) energy density of a free field is not a well-defined\n> operator on the usual free field Fock space in spacetime dimensions 3 and\n> above. (As I recall, it is a well-defined sesquilinear form, though.)\n\nUnfortunately I don\'t know enough functional analysis to understand the\ndifference. Could you elaborate on that a little bit?\n\nThanks in advance.\n\nIgor\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>torre@cc.usu.edu (Charles Torre) wrote in message news:<aaec8787.0407071239.38aea43b@posting.google.com>...
> > Ok, so let me recap. If the Fock space is built on a finite dimensional
> > state space, then given a well defined operator O (which could be
> > specified in any number of ways) we should be able to construct it's,
> > possibly distinct, normal ordered version :O:. If the number of dimensions
> > of the underlying 1-particle state space is increased to infinity in some
> > sort of limiting procedure, then the limit of O may not be a well defined
> > operator, while that of :O: should be. Of course, I'm omitting a bunch of
> > details that would really make the above precise, but I hope I've got the
> > right idea.
> >
>
> One pedantic point: Normal ordering does NOT guarantee
> that you will get a well-defined operator. Local operators are typically
> infinite sums of C & A operators and these sums need not converge.
> For example, I seem to recall (but won't swear to it off the top of my head)
> that the (normal-ordered) energy density of a free field is not a well-defined
> operator on the usual free field Fock space in spacetime dimensions 3 and
> above. (As I recall, it is a well-defined sesquilinear form, though.)
Unfortunately I don't know enough functional analysis to understand the
difference. Could you elaborate on that a little bit?
Thanks in advance.
Igor
Charles Torre
Jul15-04, 04:57 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>\nIgor Khavkine)> >\n> > One pedantic point: Normal ordering does NOT guarantee\n> > that you will get a well-defined operator. Local operators are typically\n> > infinite sums of C & A operators and these sums need not converge.\n> > For example, I seem to recall (but won\'t swear to it off the top of my head)\n> > that the (normal-ordered) energy density of a free field is not a well-defined\n> > operator on the usual free field Fock space in spacetime dimensions 3 and\n> > above. (As I recall, it is a well-defined sesquilinear form, though.)\n>\n> Unfortunately I don\'t know enough functional analysis to understand the\n> difference. Could you elaborate on that a little bit?\n>\n\nSure. All I mean is the following.\n\nAn operator should take vectors (from some chosen domain)\nand turn them into vectors. Assuming we are using Hilbert\nspace, the vectors must have norms. The point I was trying\nto make is that one can have an "operator", e.g., a formal\ninfinite sum of C & A operators such that, when acting on\nany non-zero vector, yields a vector of "infinite norm".\nSaying that you only have a "sesquilinear form" is just a\nfancy way of saying that you can define matrix elements of\nthe (would-be) "operator", which is a weaker requirement\nthan insisting the "operator" really be an operator in the\nabove sense.\n\nDoes this help?\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>Igor Khavkine)> >
> > One pedantic point: Normal ordering does NOT guarantee
> > that you will get a well-defined operator. Local operators are typically
> > infinite sums of C & A operators and these sums need not converge.
> > For example, I seem to recall (but won't swear to it off the top of my head)
> > that the (normal-ordered) energy density of a free field is not a well-defined
> > operator on the usual free field Fock space in spacetime dimensions 3 and
> > above. (As I recall, it is a well-defined sesquilinear form, though.)
>
> Unfortunately I don't know enough functional analysis to understand the
> difference. Could you elaborate on that a little bit?
>
Sure. All I mean is the following.
An operator should take vectors (from some chosen domain)
and turn them into vectors. Assuming we are using Hilbert
space, the vectors must have norms. The point I was trying
to make is that one can have an "operator", e.g., a formal
infinite sum of C & A operators such that, when acting on
any non-zero vector, yields a vector of "infinite norm".
Saying that you only have a "sesquilinear form" is just a
fancy way of saying that you can define matrix elements of
the (would-be) "operator", which is a weaker requirement
than insisting the "operator" really be an operator in the
above sense.
Does this help?
charlie
Arnold Neumaier
Jul29-04, 08:33 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\nCharles Torre wrote:\n> Igor Khavkine)> >\n>\n>>>One pedantic point: Normal ordering does NOT guarantee\n>>>that you will get a well-defined operator. Local operators are typically\n>>>infinite sums of C & A operators and these sums need not converge.\n>>>For example, I seem to recall (but won\'t swear to it off the top of my head)\n>>>that the (normal-ordered) energy density of a free field is not a well-defined\n>>>operator on the usual free field Fock space in spacetime dimensions 3 and\n>>>above. (As I recall, it is a well-defined sesquilinear form, though.)\n>>\n>>Unfortunately I don\'t know enough functional analysis to understand the\n>>difference. Could you elaborate on that a little bit?\n>\n> Sure. All I mean is the following.\n>\n> An operator should take vectors (from some chosen domain)\n> and turn them into vectors. Assuming we are using Hilbert\n> space, the vectors must have norms. The point I was trying\n> to make is that one can have an "operator", e.g., a formal\n> infinite sum of C & A operators such that, when acting on\n> any non-zero vector, yields a vector of "infinite norm".\n> Saying that you only have a "sesquilinear form" is just a\n> fancy way of saying that you can define matrix elements of\n> the (would-be) "operator", which is a weaker requirement\n> than insisting the "operator" really be an operator in the\n> above sense.\n\nA sesquilinear form is essentially a linear mapping f from a\nspace H (the domain; a dense subspace of the Hilbert space)\nto its dual space H^* (which properly contains H), while\nan operator maps H into H. Thus the latter can be iterated\nwhile the former usually cannot. <phi|f|psi> is always defined\nwhen phi,psi in H, but matrix elements <phi|fg|psi> of products fg\nmake sense only for operators f,g.\n\nBut this has nothing to do with infinite sums, since already\nexpressions like f(x)=a^(x)*a(x) are only sesquilinear forms.\nThat\'s why f(x)f(y) is meaningless while\n:f(x)f(y): = a^*(x)a^*(y)a(y)a(x)\nis well defined (again as sesquilinear form only).\n\n\nArnold Neumaier\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>Charles Torre wrote:
> Igor Khavkine)> >
>
>>>One pedantic point: Normal ordering does NOT guarantee
>>>that you will get a well-defined operator. Local operators are typically
>>>infinite sums of C & A operators and these sums need not converge.
>>>For example, I seem to recall (but won't swear to it off the top of my head)
>>>that the (normal-ordered) energy density of a free field is not a well-defined
>>>operator on the usual free field Fock space in spacetime dimensions 3 and
>>>above. (As I recall, it is a well-defined sesquilinear form, though.)
>>
>>Unfortunately I don't know enough functional analysis to understand the
>>difference. Could you elaborate on that a little bit?
>
> Sure. All I mean is the following.
>
> An operator should take vectors (from some chosen domain)
> and turn them into vectors. Assuming we are using Hilbert
> space, the vectors must have norms. The point I was trying
> to make is that one can have an "operator", e.g., a formal
> infinite sum of C & A operators such that, when acting on
> any non-zero vector, yields a vector of "infinite norm".
> Saying that you only have a "sesquilinear form" is just a
> fancy way of saying that you can define matrix elements of
> the (would-be) "operator", which is a weaker requirement
> than insisting the "operator" really be an operator in the
> above sense.
A sesquilinear form is essentially a linear mapping f from a
space H (the domain; a dense subspace of the Hilbert space)
to its dual space H^* (which properly contains H), while
an operator maps H into H. Thus the latter can be iterated
while the former usually cannot. <\phi|f|\psi> is always defined
when \phi,\psi in H, but matrix elements <\phi|fg|\psi> of products fg
make sense only for operators f,g.
But this has nothing to do with infinite sums, since already
expressions like f(x)=a^(x)*a(x) are only sesquilinear forms.
That's why f(x)f(y) is meaningless while
:f(x)f(y): = a^*(x)a^*(y)a(y)a(x)
is well defined (again as sesquilinear form only).