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?

  • $\begingroup$ @Qmechanic I really wonder why this is called "homework", to me it looks just like a good technical question ...? $\endgroup$
    – Dilaton
    Jul 17, 2013 at 22:30
  • $\begingroup$ @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 =_= $\endgroup$
    – user26143
    Jul 17, 2013 at 22:45
  • $\begingroup$ 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 ... $\endgroup$
    – Dilaton
    Jul 17, 2013 at 23:15
  • $\begingroup$ I see. The "homework" tag is removed now, although I don't know other people's opinion... $\endgroup$
    – user26143
    Jul 17, 2013 at 23:20
  • $\begingroup$ 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. $\endgroup$
    – Dilaton
    Jul 17, 2013 at 23:27

3 Answers 3


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 $$ T(z) {\cal A}(0,0) = \sum\limits_{n=0}^\infty \frac{ {\cal A}^{(n)}(0,0)}{z^{n}} = \cdots + \frac{h {\cal A}(0,0)}{z^2} + \frac{\partial {\cal A}(0,0) }{z} + \cdots $$

  • $\begingroup$ 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 $\delta \mathcal{A}:=A'(z',\bar{z})-A(z,\bar{z})$? $\endgroup$
    – user26143
    Jul 17, 2013 at 19:47
  • $\begingroup$ 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. $\endgroup$
    – Prahar
    Jul 17, 2013 at 19:51
  • $\begingroup$ By the way, if we take $v(z)=z$, did we lose any generality for identifying the expansion coefficients? $\endgroup$
    – user26143
    Jul 17, 2013 at 20:01
  • $\begingroup$ I don't quite follow your question. What do you mean? $\endgroup$
    – Prahar
    Jul 17, 2013 at 20:03
  • $\begingroup$ 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 $v(z)$? $\endgroup$
    – user26143
    Jul 17, 2013 at 20:05

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.


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.


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.


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


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

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



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

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

  • $\begingroup$ Thank you very much. I will come back to your solution after I learn the Virasoro algebra... $\endgroup$
    – user26143
    Jul 17, 2013 at 20:42

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