Elementary question: On the vertex [Archive]

mandro

Aug9-04, 11:44 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>Can anyone tell me how the vertex operators\nappear in a scattering process with two incoming\nstrings and one outgoing strinq?\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>Can anyone tell me how the vertex operators
appear in a scattering process with two incoming
strings and one outgoing strinq?

Urs Schreiber

Aug12-04, 12:37 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>"mandro" <ultraman2002@hotmail.com> schrieb im Newsbeitrag\nnews:dec722c5.0408091040.483d48b3-100000@posting.google.com...\n\n> Can anyone tell me how the vertex operators\n> appear in a scattering process with two incoming\n> strings and one outgoing strinq?\n\nRoughly, a vertex operator is a local quantum field on the worldsheet which,\nwhen acting at the point z on the worldsheet vaccum state produces a state\nwhich describes an incoming string in some given state, which in the "far\npast" seems to come from the point z.\n\nIt would help if we knew a little more about what you already understand\nabout the quantization of the string.\n\nDo you know what a vertex operator is? If not, why are you asking about the\nparticular case of a scattering with three external legs? Do you understand\nhow to evaluate a string diagram with 2 external legs?\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>"mandro" <ultraman2002@hotmail.com> schrieb im Newsbeitrag
news:dec722c5.0408091040.483d48b3-100000@posting.google.com...

> Can anyone tell me how the vertex operators
> appear in a scattering process with two incoming
> strings and one outgoing strinq?

Roughly, a vertex operator is a local quantum field on the worldsheet which,
when acting at the point z on the worldsheet vaccum state produces a state
which describes an incoming string in some given state, which in the "far
past" seems to come from the point z.

It would help if we knew a little more about what you already understand
about the quantization of the string.

Do you know what a vertex operator is? If not, why are you asking about the
particular case of a scattering with three external legs? Do you understand
how to evaluate a string diagram with 2 external legs?

mandro

Aug13-04, 09:36 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>> Roughly, a vertex operator is a local quantum field on the worldsheet which,\n> when acting at the point z on the worldsheet vaccum state produces a state\n> which describes an incoming string in some given state, which in the "far\n> past" seems to come from the point z.\n\nI "almost" understand this statement, but I don\'t.\nI mean, in the case of a single string coming in\nand a single string going out, i.e., in the case\nof QFT with the cylinder as space time, one could,\nI guess, interpret your statement above as:\n\nLet Phi(x, t) be a field operator (I guess not\nnecessarily the same as the fields of your theory)\nthen, let |0> be the groundstate. Now, take a state\npsi at time 0, and define something like\nlim_{t -> oo} U_t psi = Phi(x_0, t_0)\n\nBut of course, "something like" doesn\'t cut it.\nI need to have "exactly" what is meant by this.\nIf any of these statements are to be truly meaningful.\n\n> It would help if we knew a little more about what you already understand\n> about the quantization of the string.\n>\n> Do you know what a vertex operator is?\n\nNo I don\'t but my question is designed so that I\ncan find out.\n\nIf not, why are you asking about the\n> particular case of a scattering with three external legs? Do you understand\n> how to evaluate a string diagram with 2 external legs?\n\nThe reason why I asked this, is because I see trouble\narising when one has more than just one incoming and\none outgoing. Firstly in that case were no longer\nlooking at a QFT on the cylinder spacetime since in\nthat case, time somehow "bifurcates".\n\nBut, lets concentrate on the case with one ingoing and\none outgoing state first.\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>> Roughly, a vertex operator is a local quantum field on the worldsheet which,
> when acting at the point z on the worldsheet vaccum state produces a state
> which describes an incoming string in some given state, which in the "far
> past" seems to come from the point z.

I "almost" understand this statement, but I don't.
I mean, in the case of a single string coming in
and a single string going out, i.e., in the case
of QFT with the cylinder as space time, one could,
I guess, interpret your statement above as:

Let \Phi(x, t) be a field operator (I guess not
necessarily the same as the fields of your theory)
then, let |0> be the groundstate. Now, take a state
\psi at time 0, and define something like
lim_{t -> oo} U_t \psi = \Phi(x_0, t_0)

But of course, "something like" doesn't cut it.
I need to have "exactly" what is meant by this.
If any of these statements are to be truly meaningful.

> It would help if we knew a little more about what you already understand
> about the quantization of the string.
>
> Do you know what a vertex operator is?

No I don't but my question is designed so that I
can find out.

If not, why are you asking about the
> particular case of a scattering with three external legs? Do you understand
> how to evaluate a string diagram with 2 external legs?

The reason why I asked this, is because I see trouble
arising when one has more than just one incoming and
one outgoing. Firstly in that case were no longer
looking at a QFT on the cylinder spacetime since in
that case, time somehow "bifurcates".

But, lets concentrate on the case with one ingoing and
one outgoing state first.

mandro

Aug18-04, 03:32 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>Not that it matters much since it\'s all\nprobably wrong,\n\nbut in my last post I meant\n\nlim_{t -> oo} U_t psi = Phi(x_0, t_0) |0>\n\nbut since the quastion is really: what did\nUrs mean, I\'ll wait for him to respond.\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>Not that it matters much since it's all
probably wrong,

but in my last post I meant

lim_{t ->[/itex] oo} [itex]U_t \psi = \Phi(x_0, t_0) |0>

but since the quastion is really: what did
Urs mean, I'll wait for him to respond.

Urs Schreiber

Aug18-04, 05:23 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>"mandro" <ultraman2002@hotmail.com> schrieb im Newsbeitrag\nnews:dec722c5.0408130600.508716b3-100000@posting.google.com...\n\n> > Do you know what a vertex operator is?\n>\n> No I don\'t but my question is designed so that I\n> can find out.\n\nThe short answer is:\n\nA vertex operator of a state |psi> is a local field Psi such that\n\nPsi(t=-infty,x=0)|vacuum> = |psi>.\n\nPlease have a look at the beginning of section 2.8 in Polchinki\'s book and\nlet us know which specific questions you have.\n\n> The reason why I asked this, is because I see trouble\n> arising when one has more than just one incoming and\n> one outgoing. Firstly in that case were no longer\n> looking at a QFT on the cylinder spacetime since in\n> that case, time somehow "bifurcates".\n\nWhen computing scattering amplitudes on the closed oriented string at tree\nlevel one is using CFT on the sphere. Vertex operators are inserted on the\nsphere to produce correlation functions as usual. Note that in ordinary QFT\non spacetime the correlation function of three fields does not imply\nanything about time "bifurcating".\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>"mandro" <ultraman2002@hotmail.com> schrieb im Newsbeitrag
news:dec722c5.0408130600.508716b3-100000@posting.google.com...

> > Do you know what a vertex operator is?
>
> No I don't but my question is designed so that I
> can find out.

The short answer is:

A vertex operator of a state |\psi> is a local field \Psi such that

\Psi(t=-\infty,x=0)|vacuum> = |\psi>.

Please have a look at the beginning of section 2.8 in Polchinki's book and
let us know which specific questions you have.

> The reason why I asked this, is because I see trouble
> arising when one has more than just one incoming and
> one outgoing. Firstly in that case were no longer
> looking at a QFT on the cylinder spacetime since in
> that case, time somehow "bifurcates".

When computing scattering amplitudes on the closed oriented string at tree
level one is using CFT on the sphere. Vertex operators are inserted on the
sphere to produce correlation functions as usual. Note that in ordinary QFT
on spacetime the correlation function of three fields does not imply
anything about time "bifurcating".

mandro

Aug18-04, 05:29 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>> The short answer is:\n>\n> A vertex operator of a state |psi> is a local field Psi such that\n>\n> Psi(t=-infty,x=0)|vacuum> = |psi>.\n\nWhen you say this, what does it mean\n1) Are you talking in the theory where you are working\non the disk and you just have Psi(0,0)|0> = |psi>.\n\nor 2) Are you working on the cylinder and doing\nlim_{t -->oo} Psi(t, x_0) |0> = |psi>\nand, if you would do this, would it be independent\nof which point x_0 we choose on the circle.\n\nActually one of my problems understanding this stuff\nis that I have no intuition in working with field\ntheories "in space time" i.e., when I think of field\noperators Psi(x, t) I think of having a universal time\nvariable t, and a space manifold M then Psi(x, t) is\njust the time evolution of a field operator Psi(x)\nwhich depends only on points in M. Now when you\'re turning\na cylinder into the disk, I wonder how the time evolution\noperator is expressed on the disk.\n\n> Note that in ordinary QFT\n> on spacetime the correlation function of three fields does not imply\n> anything about time "bifurcating".\n\nI was thinking trying to do a field theory on a\nspacetime that was a pants diagram (looks like Y),\nin that case it would seem to me there\'s a bifurcation\nin time. I.e., the spacetime cannot be written as MxR\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 short answer is:
>
> A vertex operator of a state |\psi> is a local field \Psi such that
>
> \Psi(t=-\infty,x=0)|vacuum> = |\psi>.

When you say this, what does it mean
1) Are you talking in the theory where you are working
on the disk and you just have \Psi(0,0)|0> = |\psi>.

or 2) Are you working on the cylinder and doing
lim_{t -->oo} \Psi(t, x_0) |0> = |\psi>
and, if you would do this, would it be independent
of which point x_0 we choose on the circle.

Actually one of my problems understanding this stuff
is that I have no intuition in working with field
theories "in space time" i.e., when I think of field
operators \Psi(x, t) I think of having a universal time
variable t, and a space manifold M then \Psi(x, t) is
just the time evolution of a field operator \Psi(x)
which depends only on points in M. Now when you're turning
a cylinder into the disk, I wonder how the time evolution
operator is expressed on the disk.

> Note that in ordinary QFT
> on spacetime the correlation function of three fields does not imply
> anything about time "bifurcating".

I was thinking trying to do a field theory on a
spacetime that was a pants diagram (looks like Y),
in that case it would seem to me there's a bifurcation
in time. I.e., the spacetime cannot be written as MxR

Urs Schreiber

Aug19-04, 09:21 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 Wed, 18 Aug 2004, mandro wrote:\n\n> > The short answer is:\n> >\n> > A vertex operator of a state |psi> is a local field Psi such that\n> >\n> > Psi(t=-infty,x=0)|vacuum> = |psi>.\n\n> or 2) Are you working on the cylinder and doing\n> lim_{t -->oo} Psi(t, x_0) |0> = |psi>\n\nWhat I wrote strictly holds only on the cylinder. But if you have any\nstring diagram with an "external leg" which you imagine as a cylinder,\nthis cylinder is topologically equivalent to a disk with the point at\nt=-infty cut out - the "annulus".\nTherefore a string tree diagram,say, with n external\n"cylinder legs" is topologically the same as a sphere with n special\npoints cut out at which you insert the vertex operators in the above\nfashion.\n\nSo that\'s how closed string scattering amplitudes are computed: At m loops\nyou pick the closed genus m surface and insert a vertex operator for each\nof the n "external legs".\n\n> and, if you would do this, would it be independent\n> of which point x_0 we choose on the circle.\n\nThis question is technically a little involved. By using the residual\nreparameterization freedom three vertices can be fixed on the worldsheet,\nthe remaining ones are integrated over all positions. If you are really\ninterested in these questions there is nothing like having a look at\nsection 3 of Polchinki\'s book.\n\n> Actually one of my problems understanding this stuff\n> is that I have no intuition in working with field\n> theories "in space time" i.e., when I think of field\n> operators Psi(x, t) I think of having a universal time\n> variable t, and a space manifold M then Psi(x, t) is\n> just the time evolution of a field operator Psi(x)\n> which depends only on points in M. Now when you\'re turning\n> a cylinder into the disk, I wonder how the time evolution\n> operator is expressed on the disk.\n\nThe easy answer is: It becomes radial evolution (at least if you have the\nsame transformation in mind that I do).\n\nThe longer answer is: String amplitudes are calculated on _Euclidean_\nworld sheets, where the destinction between worldsheet time and worldsheet\nspace sort of disappears. If you want to get intuition for these\namplitudes you should think of correlation functions in statistical\nmechanics. That might help.\n\n> I was thinking trying to do a field theory on a\n> spacetime that was a pants diagram (looks like Y),\n> in that case it would seem to me there\'s a bifurcation\n> in time. I.e., the spacetime cannot be written as MxR\n\nYes, you are quite right. That is the reason why it is hard to compute\nstring scattering amplitudes using a Lorentzian worldsheet theory and why\nit is usual to first Wick-rotate to a Euclidean worldsheet (as described\nin detail on pp. 82-83 of Polchinki\'s book.)\n\nBut I recall having heard Kasper Peeters talk about computations of string\namplitudes using Lorentzian worldsheet theories. It is much harder, but\napparently not completely impossible. But I can\'t tell you much more about\nit. Maybe somebody else can.\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 Wed, 18 Aug 2004, mandro wrote:

> > The short answer is:
> >
> > A vertex operator of a state |\psi> is a local field \Psi such that
> >
> > \Psi(t=-\infty,x=0)|vacuum> = |\psi>.

> or 2) Are you working on the cylinder and doing
> lim_{t -->oo} \Psi(t, x_0) |0> = |\psi>

What I wrote strictly holds only on the cylinder. But if you have any
string diagram with an "external leg" which you imagine as a cylinder,
this cylinder is topologically equivalent to a disk with the point at
t=-\infty cut out - the "annulus".
Therefore a string tree diagram,say, with n external
"cylinder legs" is topologically the same as a sphere with n special
points cut out at which you insert the vertex operators in the above
fashion.

So that's how closed string scattering amplitudes are computed: At m loops
you pick the closed genus m surface and insert a vertex operator for each
of the n "external legs".

> and, if you would do this, would it be independent
> of which point x_0 we choose on the circle.

This question is technically a little involved. By using the residual
reparameterization freedom three vertices can be fixed on the worldsheet,
the remaining ones are integrated over all positions. If you are really
interested in these questions there is nothing like having a look at
section 3 of Polchinki's book.

> Actually one of my problems understanding this stuff
> is that I have no intuition in working with field
> theories "in space time" i.e., when I think of field
> operators \Psi(x, t) I think of having a universal time
> variable t, and a space manifold M then \Psi(x, t) is
> just the time evolution of a field operator \Psi(x)
> which depends only on points in M. Now when you're turning
> a cylinder into the disk, I wonder how the time evolution
> operator is expressed on the disk.

The easy answer is: It becomes radial evolution (at least if you have the
same transformation in mind that I do).

The longer answer is: String amplitudes are calculated on _Euclidean_
world sheets, where the destinction between worldsheet time and worldsheet
space sort of disappears. If you want to get intuition for these
amplitudes you should think of correlation functions in statistical
mechanics. That might help.

> I was thinking trying to do a field theory on a
> spacetime that was a pants diagram (looks like Y),
> in that case it would seem to me there's a bifurcation
> in time. I.e., the spacetime cannot be written as MxR

Yes, you are quite right. That is the reason why it is hard to compute
string scattering amplitudes using a Lorentzian worldsheet theory and why
it is usual to first Wick-rotate to a Euclidean worldsheet (as described
in detail on pp. 82-83 of Polchinki's book.)

But I recall having heard Kasper Peeters talk about computations of string
amplitudes using Lorentzian worldsheet theories. It is much harder, but
apparently not completely impossible. But I can't tell you much more about
it. Maybe somebody else can.

mandro

Aug23-04, 01:32 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>Urs Schreiber <Urs.Schreiber@uni-essen.de> wrote\n\n> What I wrote strictly holds only on the cylinder. But if you have any\n> string diagram with an "external leg" which you imagine as a cylinder,\n> this cylinder is topologically equivalent to a disk with the point at\n> t=-infty cut out - the "annulus".\n> Therefore a string tree diagram,say, with n external\n> "cylinder legs" is topologically the same as a sphere with n special\n> points cut out at which you insert the vertex operators in the above\n> fashion.\n\n*** I\'ve known this for years, I.e., upon first reading\nfrom GSW, that\'s the thinking that suggests itself;\nI.e., some kind of a topological deformation from\na n-legged spider to a n-puncture sphere. Then, one\nmentions the words "insert vertex operators at the\nholes " to code the information that will be lost\nif the holes are plugged, and everything\'s dandy.\n\nAgain I\'d known this "layman\'s version for years".\nHowever if I now think of this from the point of\nview of my present understanding of QFT, this does\nnot seem to make sense. (I must admit that part of\nmy thinking was clouded by my seeing the path\nintegrals that appear in these expressions as\nusual Feynman path integrals rather than euclidean\npath integrals, but I don\'t see how using the\nEuclidean path integrals would change my objections)\nFor example, isn\'t the usual expression of path\nintegrals:\n\npsi_t(y) = integral psi_0(x) K_{xy} dx\nwith the integral being over configuration space\nand where K_{xy} denotes integral e^i(S(r)) dr\nwhere this integral is taken over all time\nparametrized paths r that begin at x at time=0\nand end at y at time=t. Now in the case where\nthe configuration space is multistring\nconfiguration space then one of these paths\nr, may be something like a tree diagram\nwhere the initial and final configurations are\nsay x=two strings, y=two strings Now the state\nat time 0 and the state at time t, are represented\nby wavefunctions on the two string sector of\nmultistring configuration space much like the\nwavefunction of two particles in ordinary quantum\nmechanics is a function on space^2 and generally\nthis wavefunction is not just the product of two\nsingle particle wavefunctions. Now it seems to me\nthat but assuming that for each incoming and outgoing\nparticle we can just introduce a vertex operator\nseparately seems to me in contradiction with this\nprinciple, to me it seems that it\'s like they are\nassuming 2 string wavefns are products of 1 string\nwavefunctions.\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>Urs Schreiber <Urs.Schreiber@uni-essen.de> wrote

> What I wrote strictly holds only on the cylinder. But if you have any
> string diagram with an "external leg" which you imagine as a cylinder,
> this cylinder is topologically equivalent to a disk with the point at
> t=-\infty cut out - the "annulus".
> Therefore a string tree diagram,say, with n external
> "cylinder legs" is topologically the same as a sphere with n special
> points cut out at which you insert the vertex operators in the above
> fashion.

*** I've known this for years, I.e., upon first reading
from GSW, that's the thinking that suggests itself;
I.e., some kind of a topological deformation from
a n-legged spider to a n-puncture sphere. Then, one
mentions the words "insert vertex operators at the
holes " to code the information that will be lost
if the holes are plugged, and everything's dandy.

Again I'd known this "layman's version for years".
However if I now think of this from the point of
view of my present understanding of QFT, this does
not seem to make sense. (I must admit that part of
my thinking was clouded by my seeing the path
integrals that appear in these expressions as
usual Feynman path integrals rather than euclidean
path integrals, but I don't see how using the
Euclidean path integrals would change my objections)
For example, isn't the usual expression of path
integrals:

\psi_t(y) = integral \psi_0(x) K_{xy} dx
with the integral being over configuration space
and where K_{xy} denotes integral e^i(S(r)) dr
where this integral is taken over all time
parametrized paths r that begin at x at time=0
and end at y at time=t. Now in the case where
the configuration space is multistring
configuration space then one of these paths
r, may be something like a tree diagram
where the initial and final configurations are
say x=two strings, y=two strings Now the state
at time and the state at time t, are represented
by wavefunctions on the two string sector of
multistring configuration space much like the
wavefunction of two particles in ordinary quantum
mechanics is a function on space^2 and generally
this wavefunction is not just the product of two
single particle wavefunctions. Now it seems to me
that but assuming that for each incoming and outgoing
particle we can just introduce a vertex operator
separately seems to me in contradiction with this
principle, to me it seems that it's like they are
assuming 2 string wavefns are products of 1 string
wavefunctions.

Urs Schreiber

Aug23-04, 01:58 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>On Mon, 23 Aug 2004, mandro wrote:\n\n> For example, isn\'t the usual expression of path\n> integrals:\n>\n> psi_t(y) = integral psi_0(x) K_{xy} dx\n> with the integral being over configuration space\n> and where K_{xy} denotes integral e^i(S(r)) dr\n> where this integral is taken over all time\n> parametrized paths r that begin at x at time=0\n> and end at y at time=t.\n\nThe "path integral" that we are talking about is the integral of the\nexp(iS) term over all _configurations_. What you are thinking about are\nconfigurations of a non-relativistic field theory on the _line_. But what\nyou must be thinking of in order to understand string amplitudes is the\n"path integral" over all field configuration on the worldsheet (the\nsphere with vertex insertions, say). This is still called a path integral,\neven though there are not really any paths involved.\n\n\n> Now in the case where\n> the configuration space is multistring\n> configuration space\n\nNope. While your previous intuition was too simplistic, this one is more\ncomplicated than what you are really looking for. The integral we want to\nperform is that over all field confidurations (of the coordinate fields)\non the worldsheet, i.e. of a "single" string, but with external legs\nattached, being represented by the insertion of vertices in the path\nintegral (i.e. apart from the factor exp(iS) there are the vertices phi in\nthe integrand).\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>On Mon, 23 Aug 2004, mandro wrote:

> For example, isn't the usual expression of path
> integrals:
>
> \psi_t(y) = integral \psi_0(x) K_{xy} dx
> with the integral being over configuration space
> and where K_{xy} denotes integral e^i(S(r)) dr
> where this integral is taken over all time
> parametrized paths r that begin at x at time=0
> and end at y at time=t.

The "path integral" that we are talking about is the integral of the
\exp(iS) term over all _configurations_. What you are thinking about are
configurations of a non-relativistic field theory on the _line_. But what
you must be thinking of in order to understand string amplitudes is the
"path integral" over all field configuration on the worldsheet (the
sphere with vertex insertions, say). This is still called a path integral,
even though there are not really any paths involved.

> Now in the case where
> the configuration space is multistring
> configuration space

Nope. While your previous intuition was too simplistic, this one is more
complicated than what you are really looking for. The integral we want to
perform is that over all field confidurations (of the coordinate fields)
on the worldsheet, i.e. of a "single" string, but with external legs
attached, being represented by the insertion of vertices in the path
integral (i.e. apart from the factor \exp(iS) there are the vertices \phi in
the integrand).

mandro

Aug25-04, 11:04 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>Urs Schreiber <Urs.Schreiber@uni-essen.de> wrote\n\n> The "path integral" that we are talking about is the integral of the\n> exp(iS) term over all _configurations_.\n\nI don\'t know what you mean by all "_configurations_ ".\nI *almost* know what I meant by taking the integral\nover all configurations. What I meant was making a\nspace W, of all "fields" on the line then the wave-\nfunctions would be of form psi : W ---> C. Now there\'s\nsome business in ordinary QM that one can take psi\non R^3, or, you can take them on R^4 (i.e., include\ntime) since L^2 (R^3) and L^2(R^4) are equivalent\nboth formalisms would work--I don\'t know if you mean\nthis or not; probably not.\n\n\n> What you are thinking about are\n> configurations of a non-relativistic field theory on the _line_. But what\n> you must be thinking of in order to understand string amplitudes is the\n> "path integral" over all field configuration on the worldsheet (the\n> sphere with vertex insertions, say). This is still called a path integral,\n> even though there are not really any paths involved.\n\nI don\'t know what you mean here (Again, there\'s not\nenough detail) But, hmm, let\'s see, if one does\nintegral e^{iS(r)} dr where the integration\nis over all paths r(t) s.t. r(0)=x in W and\nr(1) = y in W (note here I\'m talking about\nthe configurations of a single string, and by\nconfigurations I mean at one instant in time).\nNow I make a homeomorphism f : S ---> [0,1]xR .\nNote, every path r(t) can be interpreted as a\nfunction F(x, t) = (r(t))(x) s.t.\nF: [0,1]xR ----> T (T is field target space )\nthen F o f : S --> T\nand of course, if S\'(F o f) = S(F), then\nintegral e^{iS\'(r)} dr where now the integral\nis taken over all "G:S--->T" is equal to\nthe integral e^{iS(r)} dr where the integration\nis over all paths r(t) s.t. r(0)=x in W and\nr(1) = y in W.\n\nI guess what I\'m calling G: S ---> T\nis what you call a field configuration on the\nworldsheet (by which I suppose you mean a\n"suitable" map that send S to field target space)\n\n\n> > Now in the case where\n> > the configuration space is multistring\n> > configuration space\n>\n> Nope. While your previous intuition was too simplistic, this one is more\n> complicated than what you are really looking for.\n\nWell\' yes, I understand that the scattering\namplitudes for tree diagrams are just\nintegral e^{iS(r)} F(r) dr where the integral\nis taken over the space Z of all "suitable" maps\nthat send the sphere to \'field target space\' and\nF is a functional on Z which is to be viewed as\nA product of other functionals on Z "apparently"\none for each particle, that are called "vertex\noperators". This however, does not answer my\nquestion fully, since one half of my question\nwas to see if what I think this means is what you\nthink it means, but the second half is to see the\njustification for why they think it should be done\nthis way (and to dispel, for myself, any contra-\ndictions that appear to be there.---which are the result\nnot of real cotradictions, but of inadequate language )\n\n> The integral we want to perform is that over all field\n> confidurations (of the coordinate fields)\n> on the worldsheet, i.e. of a "single" string, but with external legs\n> attached, being represented by the insertion of vertices in the path\n> integral (i.e. apart from the factor exp(iS) there are the vertices phi in\n> the integrand).\n\nYou know if one really thinks logically, this block of\ntext here makes no sense. It seems to be an abbreviation\nstring theorists use for something that does make sense.\nI guess that sensical explanation is what I\'m trying to\nextract.\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>Urs Schreiber <Urs.Schreiber@uni-essen.de> wrote

> The "path integral" that we are talking about is the integral of the
> \exp(iS) term over all _configurations_.

I don't know what you mean by all "_configurations_ ".
I *almost* know what I meant by taking the integral
over all configurations. What I meant was making a
space W, of all "fields" on the line then the wave-
functions would be of form \psi : W ---> C. Now there's
some business in ordinary QM that one can take \psi
on R^3, or, you can take them on R^4 (i.e., include
time) since L^2 (R^3) and L^2(R^4) are equivalent
both formalisms would work--I don't know if you mean
this or not; probably not.

> What you are thinking about are
> configurations of a non-relativistic field theory on the _line_. But what
> you must be thinking of in order to understand string amplitudes is the
> "path integral" over all field configuration on the worldsheet (the
> sphere with vertex insertions, say). This is still called a path integral,
> even though there are not really any paths involved.

I don't know what you mean here (Again, there's not
enough detail) But, hmm, let's see, if one does
integral e^{iS(r)} dr where the integration
is over all paths r(t) s.t. r(0)=x in W and
r(1) = y in W (note here I'm talking about
the configurations of a single string, and by
configurations I mean at one instant in time).
Now I make a homeomorphism f : S ---> [0,1]xR .
Note, every path r(t) can be interpreted as a
function F(x, t) = (r(t))(x) s.t.
F: [0,1]xR ----> T (T is field target space )
then F o f : S --> T
and of course, if S'(F o f) = S(F), then
integral e^{iS'(r)} dr where now the integral
is taken over all "G:S--->T" is equal to
the integral e^{iS(r)} dr where the integration
is over all paths r(t) s.t. r(0)=x in W and
r(1) = y in W.

I guess what I'm calling G: S ---> T
is what you call a field configuration on the
worldsheet (by which I suppose you mean a
"suitable" map that send S to field target space)

> > Now in the case where
> > the configuration space is multistring
> > configuration space
>
> Nope. While your previous intuition was too simplistic, this one is more
> complicated than what you are really looking for.

Well' yes, I understand that the scattering
amplitudes for tree diagrams are just
integral e^{iS(r)} F(r) dr where the integral
is taken over the space Z of all "suitable" maps
that send the sphere to 'field target space' and
F is a functional on Z which is to be viewed as
A product of other functionals on Z "apparently"
one for each particle, that are called "vertex
operators". This however, does not answer my
question fully, since one half of my question
was to see if what I think this means is what you
think it means, but the second half is to see the
justification for why they think it should be done
this way (and to dispel, for myself, any contra-
dictions that appear to be there.---which are the result
not of real cotradictions, but of inadequate language )

> The integral we want to perform is that over all field
> confidurations (of the coordinate fields)
> on the worldsheet, i.e. of a "single" string, but with external legs
> attached, being represented by the insertion of vertices in the path
> integral (i.e. apart from the factor \exp(iS) there are the vertices \phi in
> the integrand).

You know if one really thinks logically, this block of
text here makes no sense. It seems to be an abbreviation
string theorists use for something that does make sense.
I guess that sensical explanation is what I'm trying to
extract.

Urs Schreiber

Aug25-04, 12:01 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>"mandro" <ultraman2002@hotmail.com> schrieb im Newsbeitrag\nnews:dec722c5.0408250719.3999f59a-100000@posting.google.com...\n> Urs Schreiber <Urs.Schreiber@uni-essen.de> wrote\n\n> > The integral we want to perform is that over all field\n> > configurations (of the coordinate fields)\n> > on the worldsheet, i.e. of a "single" string, but with external legs\n> > attached, being represented by the insertion of vertices in the path\n> > integral (i.e. apart from the factor exp(iS) there are the vertices phi\nin\n> > the integrand).\n>\n> You know if one really thinks logically, this block of\n> text here makes no sense. It seems to be an abbreviation\n> string theorists use for something that does make sense.\n> I guess that sensical explanation is what I\'m trying to\n> extract.\n\nInstead of getting angry with me that I don\'t have the time to walk you\nthrough every step of a first course in quantum field theory, you should use\nthe energy contained in that anger to catch up with some of the background\nthat you need in order to understand the answers to the questions that you\nare apparently interested in.\n\nI\'ll be here to give you hints, but I can only show you the door. You\'re\nthe one that has to walk through it. ;-)\n\nFirst of alll we should make sure you know the basic concepts of field\ntheory, like how the path integral in quantum mechanics that you are\napparently familiar with turns into the integral over all field\nconfigurations on parameter space.\n\nYour first task (that\'s homework, I\'ll check it tomorrow!) is to get hold of\na copy of\n\nM. Peskin & D. Schroeder:\nAn Introduction to Quantum Field Theory\nPerseus (1995)\n\nand not let go of it until you at least understand the basic concepts.\n\nIn particular, it seems that all your confusion here in this thread will be\nremoved once you understand equation (9.18) of that book, which shows how\ncorrelation functions are computed using the integral over all field\nconfigurations with certain insertions.\n\nDepending on your background knowledge you\'ll need to read some of the\nprevious material to get to the point that you know where this equation\ncomes from (especially see section 4.2), but just understanding what that\nformula claims will already help you to get to grips with string theory\namplitudes.\n\nIf you have to learn on your own maybe the best thing is to have a look at\nwhat Hendrik van Hees has recently announced:\n\n\n"Hendrik van Hees" <hees@comp.tamu.edu> schrieb im Newsbeitrag\nnews:2p193lFfnhvgU1@uni-berlin.de...\n>\n> We just started an online qft theory course, reading along Zee\'s\n> textbook. Soon, it will be provided as a an online course at the\n> supersymmetry web page:\n>\n> http://www.superstringtheory.com/\n>\n> Since the webmaster of the page is busy at the moment, it will appear\n> there sometimes in October.\n>\n> Since we didn\'t want to loose to much time, we already started as a\n> Yahoo Group. That\'s not as good as on the superstring theory page, but\n> better than nothing:\n>\n> http://groups.yahoo.com/group/qft/\n>\n> In the file folder of this group you find also my own qft script (also\n> downloadable from my home page) and the first set of exercises (dealing\n> with one-particle non-relativistic qt of the harmonic oscillator.\n>\n> If you are interested, you are heartily invited to join us!\n>\n> --\n> Hendrik van Hees Cyclotron Institute\n> Phone: +1 979/845-1411 Texas A&M University\n> Fax: +1 979/845-1899 Cyclotron Institute, MS-3366\n> http://theory.gsi.de/~vanhees/ College Station, TX 77843-3366\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>"mandro" <ultraman2002@hotmail.com> schrieb im Newsbeitrag
news:dec722c5.0408250719.3999f59a-100000@posting.google.com...
> Urs Schreiber <Urs.Schreiber@uni-essen.de> wrote

> > The integral we want to perform is that over all field
> > configurations (of the coordinate fields)
> > on the worldsheet, i.e. of a "single" string, but with external legs
> > attached, being represented by the insertion of vertices in the path
> > integral (i.e. apart from the factor \exp(iS) there are the vertices \phi
in
> > the integrand).
>
> You know if one really thinks logically, this block of
> text here makes no sense. It seems to be an abbreviation
> string theorists use for something that does make sense.
> I guess that sensical explanation is what I'm trying to
> extract.

Instead of getting angry with me that I don't have the time to walk you
through every step of a first course in quantum field theory, you should use
the energy contained in that anger to catch up with some of the background
that you need in order to understand the answers to the questions that you
are apparently interested in.

I'll be here to give you hints, but I can only show you the door. You're
the one that has to walk through it. ;-)

First of alll we should make sure you know the basic concepts of field
theory, like how the path integral in quantum mechanics that you are
apparently familiar with turns into the integral over all field
configurations on parameter space.

Your first task (that's homework, I'll check it tomorrow!) is to get hold of
a copy of

M. Peskin & D. Schroeder:
An Introduction to Quantum Field Theory
Perseus (1995)

and not let go of it until you at least understand the basic concepts.

In particular, it seems that all your confusion here in this thread will be
removed once you understand equation (9.18) of that book, which shows how
correlation functions are computed using the integral over all field
configurations with certain insertions.

Depending on your background knowledge you'll need to read some of the
previous material to get to the point that you know where this equation
comes from (especially see section 4.2), but just understanding what that
formula claims will already help you to get to grips with string theory
amplitudes.

If you have to learn on your own maybe the best thing is to have a look at
what Hendrik van Hees has recently announced:

"Hendrik van Hees" <hees@comp.tamu.edu> schrieb im Newsbeitrag
news:2p193lFfnhvgU1@uni-berlin.de...
>
> We just started an online qft theory course, reading along Zee's
> textbook. Soon, it will be provided as a an online course at the
> supersymmetry web page:
>
> http://www.superstringtheory.com/
>
> Since the webmaster of the page is busy at the moment, it will appear
> there sometimes in October.
>
> Since we didn't want to loose to much time, we already started as a
> Yahoo Group. That's not as good as on the superstring theory page, but
> better than nothing:
>
> http://groups.yahoo.com/group/qft/
>
> In the file folder of this group you find also my own qft script (also
> downloadable from my home page) and the first set of exercises (dealing
> with one-particle non-relativistic qt of the harmonic oscillator.
>
> If you are interested, you are heartily invited to join us!
>
> --
> 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

mandro

Aug26-04, 02:06 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>> Instead of getting angry with me that I don\'t have the time to walk you\n> through every step of a first course in quantum field theory, you should use\n> the energy contained in that anger to catch up with some of the background\n> that you need in order to understand the answers to the questions that you\n> are apparently interested in.\n\nUrs I\'m not angry with you at all. I was\njust expressing my honest opinion regarding\nthat block of text. However, I have posed to\nyou some questions and arguments that are\nbased on logic, and whose answer to anyone\nthat really understood this stuff would not\ntake that long at all, and you seem to have\ntotally ignored answering them. I posted here\nto get those questions answered. I would like\nto hear your opinions on these things.\n\n> I\'ll be here to give you hints, but\n> I can only show you the door. You\'re\n> the one that has to walk through it. ;-)\n\n> First of all we should make sure you know the basic concepts of field\n> theory, like how the path integral in quantum mechanics that you are\n> apparently familiar with turns into the integral over all field\n> configurations on parameter space.\n\nAll field configurations on parameter space?\nI don\'y understand this sentence too well.\n\nOk I\'ll try to understand equation (9.18) of\nPeskin Schroeder. But If I ask you about it,\nare you going to make me read basic QM instead\nof answering on that? ;-)\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>> Instead of getting angry with me that I don't have the time to walk you
> through every step of a first course in quantum field theory, you should use
> the energy contained in that anger to catch up with some of the background
> that you need in order to understand the answers to the questions that you
> are apparently interested in.

Urs I'm not angry with you at all. I was
just expressing my honest opinion regarding
that block of text. However, I have posed to
you some questions and arguments that are
based on logic, and whose answer to anyone
that really understood this stuff would not
take that long at all, and you seem to have
totally ignored answering them. I posted here
to get those questions answered. I would like
to hear your opinions on these things.

> I'll be here to give you hints, but
> I can only show you the door. You're
> the one that has to walk through it. ;-)

> First of all we should make sure you know the basic concepts of field
> theory, like how the path integral in quantum mechanics that you are
> apparently familiar with turns into the integral over all field
> configurations on parameter space.

All field configurations on parameter space?
I don'y understand this sentence too well.

Ok I'll try to understand equation (9.18) of
Peskin Schroeder. But If I ask you about it,
are you going to make me read basic QM instead
of answering on that? ;-)