Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,873 answers , 20,701 comments
1,470 users with positive rep
502 active unimported users
More ...

Identify the coefficients of Operator Product Expansion (OPE)

+ 4 like - 0 dislike
32 views

Sorry I have a stupid question in Polchinski's string theory book vol 1, p46. For a holomorphic function $T(z)$ with a general operator $\mathcal{A}$, there is a Laurent expansion $$T(z) A(0,0) \sim \sum_{n=0}^{\infty} \frac{1}{z^{n+1}} \mathcal{A}^{(n)}(0,0). \tag{2.4.11}$$ Under transformation $\delta \mathcal{A}=-\epsilon v^a \partial_a \mathcal{A}$, why the OPE is determined as $$T(z) A(0,0) \sim \cdots + \frac{h}{z^2}\mathcal{A}(0,0) + \frac{1}{z} \partial \mathcal{A}(0,0)+\cdots? \tag{2.4.14}$$ How to derive this equation?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
asked Jul 17, 2013 in Theoretical Physics by user26143 (360 points) [ no revision ]
@Qmechanic I really wonder why this is called "homework", to me it looks just like a good technical question ...?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Dilaton
@Dilaton I added the "homework" tag. In the beginning I met difficulty of deriving equation in Polchinski's book. At that time I didn't put "homework" tag. Then a homework tag was added by some forum staff (If I remembered correctly). Later, I keep using homework tag... I don't know whether this is a good technical question, I just feel I am stupid =_=

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
Nah, I personally feel not yet so comfortable about OPE and other things in CFT, so it is very useful for me to look at such questions and the answers you get. Other people seem to be interested in the question too, since it has 4 stars ;-). The homework tag is a bit dangerous, since some homework questions get closed. So seeing it on such technical questions about such advanced topics I like generally worries me a bit ...

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Dilaton
I see. The "homework" tag is removed now, although I don't know other people's opinion...

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
Yeah, I am not quite sure how the homework tag pans out for such high level topics. We will see if Qmechanic or somebody else thinks it should be applied... Up to now everything is ok and the question has attracted no closevotes.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Dilaton

3 Answers

+ 5 like - 0 dislike

One derives it from equation (2.3.11) which is $$ \text{Res}_{z \to z_0} j(z) {\cal A}(z_0,{\bar z}_0) + \overline{\text{Res}}_{{\bar z} \to {\bar z}_0} {\tilde j}({\bar z}) {\cal A}(z_0, {\bar z}_0) = \frac{1}{i \epsilon} \delta {\cal A}(z_0, {\bar z}_0) $$ This is the Ward identity. When $j(z) = i v(z) T(z)$ we find (only focussing on the holomorphic part) $$ \text{Res}_{z \to z_0} i v(z) T(z) {\cal A}(0,0) = \frac{1}{i \epsilon} \delta {\cal A}(z_0, {\bar z}_0) = i v(z) \partial {\cal A}(0,0) $$ Let us now focus on translations first. Here $v(z) = v$(constant). Under translations $\delta {\cal A}(0,0) = - \epsilon v \partial {\cal A}(0,0)$. The equation above then reads $$ i v\partial {\cal A}(0,0) = i v \text{Res}_{z \to 0} \sum\limits_{n=0}^\infty \frac{ {\cal A}^{(n)}(0,0)}{z^{n+1}} = i v {\cal A}^{(0)}(0,0) $$ Thus $$ {\cal A}^{(0)}(0,0) = \partial {\cal A}(0,0) $$ Now, consider the transformation under scaling $v(z) = z$. In this case $$ {\cal A}'(z', {\bar z}) = (1 + \epsilon )^{-h} {\cal A}(z,{\bar z}) \implies \delta {\cal A}(z , {\bar z}) = \epsilon \left[ - h {\cal A}(z,{\bar z})- z \partial{\cal A}(z, {\bar z}) \right] $$ Plugging it into the earlier equation, we find $$ h {\cal A}(0,0) = \text{Res}_{z \to z_0} \sum\limits_{n=0}^\infty \frac{ {\cal A}^{(n)}(0,0)}{z^{n}} = {\cal A}^{(1)}(0,0) $$ Plugging this back into the equation, we find $$ T(z) {\cal A}(0,0) = \sum\limits_{n=0}^\infty \frac{ {\cal A}^{(n)}(0,0)}{z^{n}} = \cdots + \frac{h {\cal A}(0,0)}{z^2} + \frac{\partial {\cal A}(0,0) }{z} + \cdots $$

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Prahar
answered Jul 17, 2013 by prahar21 (535 points) [ no revision ]
Most voted comments show all comments
Sorry, how did you get $z \partial \mathcal{A}$ in $\delta \mathcal{A}=\epsilon [-h \mathcal{A}(z,\bar{z})-z \partial \mathcal{A}(z,\bar{z})] $? is $\delta \mathcal{A}:=A'(z',\bar{z})-A(z,\bar{z})$?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
No. $\delta {\cal A} = {\cal A}'(z,{\bar z}) - {\cal A}(z, {\bar z})$. That's where we get the extra $z \partial {\cal A}$ comes from.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Prahar
By the way, if we take $v(z)=z$, did we lose any generality for identifying the expansion coefficients?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
I don't quite follow your question. What do you mean?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Prahar
I mean, $v(z)$ is in-principle any holomorphic function, if we take $v(z)=\mathrm{constant}$ or $v(z)=z$ in two cases, how to make sure the results (coefficients in OPE ) are independent with particular choice of $v(z)$?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
Most recent comments show all comments
I am sticking to the purely holomorphic case wherein $v(z) = v$ and ${\tilde v}({\bar z}) = 0$.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Prahar
I see, thanks a lot.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
+ 3 like - 0 dislike

Lets work in the Hilbert space framework in which all the objects involved are operators acting on some state space.

A CFT involves following objects (here we consider only holomorphic fields):

A field $T(z)$ called (holomorphic component of) energy momentum tensor. Its mode expansion is written as

$T(z)=\displaystyle\sum _{i}L_nz^{-n-2}$

Where $L_n$ satisfy so called Virasoro algebra.

Besides $T(z)$ there may be other conserved currents too but $T(z)$ is necessarily there in any CFT.

Other basic objects are so called Virasoro primary fields. A field $\phi(z)$ is called Virasoro primary of weight $h$ if under a holomorphic change of coordinates

$\omega =f(z)$

we have

$\phi_{new}(\omega)=(\partial_z f(z))^{-h}\phi(z)$

Infinitesimally this means that if we change our coordinates as $z\rightarrow z+\epsilon(z)$ then

$\phi_{new}(z+\epsilon(z))=(1+\partial_z\epsilon(z))^{-h}\phi(z)$

~ $\phi(z)-h(\partial_z\epsilon(z))\phi(z)$

Or in other words

$\delta\phi(z)=\phi_{new}(z)-\phi(z)=-\epsilon(z)\partial_z\phi(z)-h(\partial_z\epsilon(z))\phi(z)\tag 1$

In terms of operator fields the infinitesimal change in a field $\phi(z)$ (whether primary or not) under infinitesimal change of coordinates $z\rightarrow z+\epsilon(z)$ is given by

$\delta\phi(z)=-\frac{1}{2\pi i}\oint_{C_z} dw\epsilon(w)\mathfrak R(T(w)\phi(z))\tag 2$

Where $\mathfrak R(T(w)\phi(z))$ is radial ordered product or so called operator product; and $C_z$ is a contour about $z$.$^{(*)}$

Now to answer your question all you have to do is to verify that the OPE

$\mathfrak R(T(w)\phi(z))=\displaystyle \frac{h}{(w-z)^2}\phi(z)+\frac{1}{(w-z)}\partial_z\phi(z)\tag 3$

when used in (2) gives (1).

Summary: A primary field by definition satisfies (1). Equation (2) holds for all fields primary or not. So requiring that field $\phi(z)$ is primary its OPE with $T$ should necessarily of the form (3) so that (2) gives (1).

Added later : For a quasi primary field equation (1) holds only for translation, scale transformation and special conformal transformation of coordinates (which are respectively generated by $L_{-1},L_0,L_1$) and thus only two terms of singular part of its OPE with $T(z)$ can be determined.


*) In usual QFT we have to define quantum operators corresponding to generators of Lorentz group transformations of Minkowski space. There the change in a field under infintesimal change of coordinates is given by commutator of the corresponding operators with the given field. Conformal group is infinite dimensional and so its representation on state space is given by field $T(z)$ rather than by a finite set of operators. The integral in (2) is nothing but the commutator

$Q_{\epsilon}^{+}\phi(z)-\phi(z)Q_{\epsilon}^{-}$

where

$Q_{\epsilon}^{\pm}=-\frac{1}{2\pi i}\oint_{C^{\pm}} dw\epsilon(w)T(w)$

where $C^{+}$ is a circle with center 0 and radius $>|z|$ and $C^{-}$ is a circle with center 0 and radius $<|z|$.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user10001
answered Jul 17, 2013 by user10001 (635 points) [ no revision ]
Thank you very much. I will come back to your solution after I learn the Virasoro algebra...

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
+ 2 like - 0 dislike

You have 2 kinds of transformation to help you to find the OPE $(2.4.14)$, dilatations $(2.4.13)$ and translations.

For each of these transformations, we have to identify infinitesimal transformations quantities $v(z)$ defined by: $z' = z+\epsilon v(z)$ and the infinitesimal modification of the fields $\delta A(z,\bar z)$. The current being given by $j(z) = i v(z)T(z)$ $(2.4.5)$, we are going to use Ward identities $(2.3.11)$ :

$$Res_{z \rightarrow z_0} (j(z)A(z_0, \bar z_0)) + \bar Res_{\bar z \rightarrow \bar z_0}(\bar j(z)A(z_0, \bar z_0)) = \frac{1}{i \epsilon} \delta A(z_0, \bar z_0)$$

We Suppose an OPE of the form :

$T(z) A(0,0) \sim \cdots + \frac{a}{z^2}A(0,0) + \frac{b}{z} \partial A(0,0)+\cdots$, where $a$ and $b$ are to be determined.

Dilatations

The infinitesimal transformation corresponding to $z'= \zeta z$, is, using $\zeta = 1 + \epsilon$, $z' = z +\epsilon z$, so here $v(z) = z$; and $\bar v(z) = \bar z$

The transformation of fields is $A(z',\bar z') = \zeta^{-h}\bar \zeta^{- \tilde h} A(z,\bar z)$, this corresponds to a infinitesimal transformation $\delta A(z, \bar z) = - \epsilon h~ A(z,\bar z) - \bar \epsilon \tilde h~ A(z,\bar z)$

So, appying Ward identity, and only keeping the holomorphic part,we see that :

$$Res_{z \rightarrow 0}(i ~z ~T(z)~A(0,0)) = \frac{1}{i ~\epsilon}(- \epsilon h A(0,0))$$

This means that $T(z)~A(0,0)$ has a component $\frac {h}{z^2}A(0,0)$, in order to have a pole with the correct residue.

Translations

Here $v(z) = v$ = Constant; and $\delta A = - \epsilon (v\partial A + \bar v \bar \partial A)$. So, applying the Ward identity, keeping the holomorphic part, we get :

$$Res_{z \rightarrow 0}(i ~v ~T(z)~A(0,0)) = \frac{1}{i ~\epsilon}(- \epsilon v \partial A(0,0))$$

This means that $T(z)~A(0,0)$ has a component $\frac {1}{z}A(0,0)$, in order to have a pole with the correct residue.

So, finally :

$$T(z) A(0,0) \sim \cdots + \frac{h}{z^2}A(0,0) + \frac{1}{z} \partial A(0,0)+\cdots$$

Of course, an equivalent demonstration is valid for the anti-homorphic part.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Trimok
answered Jul 17, 2013 by Trimok (950 points) [ no revision ]
Thank you very much! Favorite solution ^_^

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysi$\varnothing$sOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...