[SOLVED] CFT Ward identity [Archive]

Joan Estes

Aug7-04, 09:33 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>I have an elementary technical question regarding the familiar\nWard identity in conformal field theory:\n\n\\oint dz v(z) <T(z) phi(z_1) ... phi(z_n)>\n\n= sum_i < phi(z_1) .... (delta_v phi)(z_i) ... phi(z_n)>\n\nwhere phi is a primary operator and (delta_v phi) denotes its\nvariation under the conformal reparametrization generated\nby v(z). This is indeed taken as an axiom in some treatments.\n\nHowever, when I try to verify this for the simple case with\ntwo insertions of a bosonic primary field dX, I get instead:\n\n\\oint dz v(z) <T(z) dX(z_1) dX(z_2)>\n\n= <(delta dX)(z_1) dX(z_2)> + <dX(z_1) (delta dX)(z_2)>\n\n+ 2 \\oint dz v(z) 1 / (z-z_1)^2 (z-z_2)^2\n\nwhere the first two terms are as expected and come from single\ncontractions\n\nT(z) dX(z_i)\n|_______|\n\nbut the last term comes from the double contractions\n\nT(z) dX(z_1) dX(z_2)\n| |_____| |\n|_____________|\n\nsince T is bilinear in X. This contribution is not in the\nWard identity. Am I missing something obvious?\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>I have an elementary technical question regarding the familiar
Ward identity in conformal field theory:

\oint dz v(z) <T(z) \phi(z_1) .[/itex].. \phi(z_n)>= sum_i < \phi(z_1) ..[itex].. (\delta_v \phi)(z_i) ... \phi(z_n)>

where \phi is a primary operator and (\delta_v \phi) denotes its
variation under the conformal reparametrization generated
by v(z). This is indeed taken as an axiom in some treatments.

However, when I try to verify this for the simple case with
two insertions of a bosonic primary field dX, I get instead:

\oint dz v(z) <T(z) dX(z_1) dX(z_2)>= <(\delta dX)(z_1) dX(z_2)> + <dX(z_1) (\delta dX)(z_2)>+ 2 \oint dz v(z) 1 / (z-z_1)^2 (z-z_2)^2

where the first two terms are as expected and come from single
contractions

T(z) dX(z_i)
|__{_____}|

but the last term comes from the double contractions

T(z) dX(z_1) dX(z_2)
| |__{___}| |
|__{___________}|

since T is bilinear in X. This contribution is not in the
Ward identity. Am I missing something obvious?

Rufus Anton

Aug9-04, 12:04 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>Joan Estes <joanestes2000@yahoo.com> wrote in message news:<ab19fe13.0408071657.74782c5e-100000@posting.google.com>...\n\n> I have an elementary technical question regarding the familiar\n> Ward identity in conformal field theory:\n>\n> \\oint dz v(z) <T(z) phi(z_1) ... phi(z_n)>\n> = sum_i < phi(z_1) .... (delta_v phi)(z_i) ... phi(z_n)>\n\nThe contour of this integral encircles all the \\$z_i\\$. You can\ntherefore deform it to a set of contours consisting of little circles\neach encircling exactly one \\$z_i\\$. This gives\n\n\\sum_i < \\phi(z_1) ... ( \\oint dz v(z) T(z) phi(z_i) ) ... \\phi(z_n) >\n\nwhich is equal to the final result you quote because\n\n\\oint dz v(z) T(z) phi(z_i) = (\\delta_v \\phi)(z_i)\n\n> However, when I try to verify this for the simple case with\n> two insertions of a bosonic primary field dX, I get instead:\n>\n> \\oint dz v(z) <T(z) dX(z_1) dX(z_2)>\n>\n> = <(delta dX)(z_1) dX(z_2)> + <dX(z_1) (delta dX)(z_2)>\n>\n> + 2 \\oint dz v(z) 1 / (z-z_1)^2 (z-z_2)^2\n>\n> where the first two terms are as expected and come from single\n> contractions\n>\n> T(z) dX(z_i)\n> |_______|\n>\n> but the last term comes from the double contractions\n>\n> T(z) dX(z_1) dX(z_2)\n> | |_____| |\n> |_____________|\n>\n> since T is bilinear in X. This contribution is not in the\n> Ward identity.\n\nYes it is there. You confuse a derivative with the conformal\nvariation. There is no contradiction if you keep in mind that\n\n\\oint dz v(z) T(z) phi(z_i) = ( v(z_i) \\partial + h \\partial v(z_i) )\n\\phi(z_i)\n\nis the conformal variation. Thus you *need* both terms.\n\nBy the way, T is not bilinear but quadratic in X.\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>Joan Estes <joanestes2000@yahoo.com> wrote in message news:<ab19fe13.0408071657.74782c5e-100000@posting.google.com>...

> I have an elementary technical question regarding the familiar
> Ward identity in conformal field theory:
>
> \oint dz v(z) <T(z) \phi(z_1) ... \phi(z_n)>
> = sum_i < \phi(z_1) .... (\delta_v \phi)(z_i) ... \phi(z_n)>

The contour of this integral encircles all the $z_i$. You can
therefore deform it to a set of contours consisting of little circles
each encircling exactly one $z_i$. This gives

\sum_i < \phi(z_1) .[/itex].. ( \oint dz v(z) T(z) \phi(z_i) ) ... \phi(z_n) >

which is equal to the final result you quote because

\oint dz v(z) T(z) \phi(z_i) = (\delta_v \phi)(z_i)

> However, when I try to verify this for the simple case with
> two insertions of a bosonic primary field dX, I get instead:
>
> [itex]\oint dz v(z) <T(z) dX(z_1) dX(z_2)>
>
> = <(\delta dX)(z_1) dX(z_2)> + <dX(z_1) (\delta dX)(z_2)>
>
> + 2 \oint dz v(z) 1 / (z-z_1)^2 (z-z_2)^2
>
> where the first two terms are as expected and come from single
> contractions
>
> T(z) dX(z_i)
> |__{_____}|
>
> but the last term comes from the double contractions
>
> T(z) dX(z_1) dX(z_2)
> | |__{___}| |
> |__{___________}|
>
> since T is bilinear in X. This contribution is not in the
> Ward identity.

Yes it is there. You confuse a derivative with the conformal
variation. There is no contradiction if you keep in mind that

\oint dz v(z) T(z) \phi(z_i) = ( v(z_i) \partial + h \partial v(z_i) )\phi(z_i)

is the conformal variation. Thus you *need* both terms.

By the way, T is not bilinear but quadratic in X.

Joan Estes

Aug9-04, 12:54 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>Rufus Anton <rufusanton@gmx.de> wrote in message news:<a1c70df8.0408081408.62c51b0-100000@posting.google.com>...\n\n> Yes it is there. You confuse a derivative with the conformal\n> variation. There is no contradiction if you keep in mind that\n>\n> \\oint dz v(z) T(z) phi(z_i) = ( v(z_i) \\partial + h \\partial v(z_i) )\n> \\phi(z_i)\n>\n> is the conformal variation. Thus you *need* both terms.\n\nThank you for your response. But I do believe I already get the\nconformal variation from the single contractions. The operator\nproduct gives\n\nT_z \\partial X(z_1) \\partial X(z_2)\n\n\\equiv 1/(z-z_1)^2 \\partial X(z_1) \\partial X(z_2)\n+ 1/(z-z_1) \\partial^2 X(z_1) \\partial X(z_2)\n+ 1/(z-z_2)^2 \\partial X(z_1) \\partial X(z_2)\n+ 1/(z-z_2) \\partial X(z_1) \\partial^2 X(z_2)\n+ 1/(z-z_1)^2 (z-z_2)2\n\nwhere the first 4 terms, coming from single contractions, give\nthe conformal variations under \\oint.\nSo I am still stumped by the presence of the last term coming\nfrom the double contractions.\n\n[Moderator\'s note: the last term arising from the double-contractions\nis proportional to the conformal anomaly (the central charge). The\nmajority of the terms come from the single contractions, and the OPE\ncorresponds to the "naive", classical variations. In other words, these\nterms is what you would get if you replaced the commutators by the\nPoisson brackets multiplied by i\\hbar. However in quantum mechanics, the\ncommutators are not *just* the Poisson brackets. Equivalently, the term\nwith double contractions *is* there. It corresponds to a quantum\ncorrection to the transformation laws of the operator X^2 or whatever you\nhave there under coordinate transformations. LM]\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>Rufus Anton <rufusanton@gmx.de> wrote in message news:<a1c70df8.0408081408.62c51b0-100000@posting.google.com>...

> Yes it is there. You confuse a derivative with the conformal
> variation. There is no contradiction if you keep in mind that
>
> \oint dz v(z) T(z) \phi(z_i) = ( v(z_i) \partial + h \partial v(z_i) )
> \phi(z_i)
>
> is the conformal variation. Thus you *need* both terms.

Thank you for your response. But I do believe I already get the
conformal variation from the single contractions. The operator
product gives

T_z \partial X(z_1) \partial X(z_2)\equiv 1/(z-z_1)^2 \partial X(z_1) \partial X(z_2)+ 1/(z-z_1) \partial^2 X(z_1) \partial X(z_2)+ 1/(z-z_2)^2 \partial X(z_1) \partial X(z_2)+ 1/(z-z_2) \partial X(z_1) \partial^2 X(z_2)+ 1/(z-z_1)^2 (z-z_2)2

where the first 4 terms, coming from single contractions, give
the conformal variations under \oint.
So I am still stumped by the presence of the last term coming
from the double contractions.

[Moderator's note: the last term arising from the double-contractions
is proportional to the conformal anomaly (the central charge). The
majority of the terms come from the single contractions, and the OPE
corresponds to the "naive", classical variations. In other words, these
terms is what you would get if you replaced the commutators by the
Poisson brackets multiplied by i\hbar. However in quantum mechanics, the
commutators are not *just* the Poisson brackets. Equivalently, the term
with double contractions *is* there. It corresponds to a quantum
correction to the transformation laws of the operator X^2 or whatever you
have there under coordinate transformations. LM]

Joan Estes

Aug9-04, 11:42 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>> [Moderator\'s note: the last term arising from the double-contractions\n> is proportional to the conformal anomaly (the central charge). The\n> majority of the terms come from the single contractions, and the OPE\n> corresponds to the "naive", classical variations. In other words, these\n> terms is what you would get if you replaced the commutators by the\n> Poisson brackets multiplied by . However in quantum mechanics, the\n> commutators are not *just* the Poisson brackets. Equivalently, the term\n> with double contractions *is* there. It corresponds to a quantum\n> correction to the transformation laws of the operator or whatever you\n> have there under coordinate transformations. LM]\n\nActually, no. There should be no anomalous term in the transformation\nof\n\n<\\partial X(z_1) \\partial X(z_2)>\n\nsince z_1 and z_2 are distinct points and (\\partial X) is a primary\noperator.\n\nI have finally, and trivially, managed to obtain the correct result.\nI was just double counting: in the *expectation value*\n\n\\oint v(z) <T(z) \\partial X(z_1) \\partial X (z_2)>\n\nit is *only* the double contractions that contribute. So one gets\nthe result\n\n\\oint v(z) / (z-z_1)^2 (z-z_2)^2\n\n= \\partial v(z_1)/(z_1-z_2)^2 - 2 v(z_1) / (z_1-z_2)^3\n+ (1 <-> 2)\n\nwhich is then straightforwardly equal to\n\n<(\\delta_v\\partial X)(z_1) \\partial X(z_2)>\n+ <\\partial X(z_1) (\\delta_v\\partial X)(z_2)>\n\nwhich is the anomaly-free transformation of the expectation value\nof the primary operators at different points.\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>> [Moderator's note: the last term arising from the double-contractions
> is proportional to the conformal anomaly (the central charge). The
> majority of the terms come from the single contractions, and the OPE
> corresponds to the "naive", classical variations. In other words, these
> terms is what you would get if you replaced the commutators by the
> Poisson brackets multiplied by . However in quantum mechanics, the
> commutators are not *just* the Poisson brackets. Equivalently, the term
> with double contractions *is* there. It corresponds to a quantum
> correction to the transformation laws of the operator or whatever you
> have there under coordinate transformations. LM]

Actually, no. There should be no anomalous term in the transformation
of

<\partial X(z_1) \partial X(z_2)>

since z_1 and z_2 are distinct points and (\partial X) is a primary
operator.

I have finally, and trivially, managed to obtain the correct result.
I was just double counting: in the *expectation value*

\oint v(z) <T(z) \partial X(z_1) \partial X (z_2)>

it is *only* the double contractions that contribute. So one gets
the result

\oint v(z) / (z-z_1)^2 (z-z_2)^2= \partial v(z_1)/(z_1-z_2)^2 - 2 v(z_1) / (z_1-z_2)^3+ (1 <-> 2)

which is then straightforwardly equal to

<(\delta_v\partial X)(z_1) \partial X(z_2)>+ <\partial X(z_1) (\delta_v\partial X)(z_2)>

which is the anomaly-free transformation of the expectation value
of the primary operators at different points.