Stress-Tensor OPE and Conformal Ward Identities

The previous page reached a central idea of two-dimensional field theory: at a critical point, local fields can be organized by how they transform under holomorphic coordinate changes. A primary field $O(z,\bar z)$ with weights $(h,\bar h)$ transforms, under $z\mapsto f(z)$ and $\bar z\mapsto \bar f(\bar z)$, as

$$O(z,\bar z)\mapsto \left({df\over dz}\right)^h \left({d\bar f\over d\bar z}\right)^{\bar h} O(f(z),\bar f(\bar z)).$$

This page explains why the stress tensor is the operator that generates this transformation law. The slogan is simple:

$$\boxed{\text{insert }T(z)\text{ and take residues }\Longleftrightarrow\text{perform an infinitesimal conformal transformation.}}$$

The price of making this slogan precise is that we must treat singular functions as distributions. The stress tensor is holomorphic away from operator insertions, but at insertions it has contact terms. These contact terms are not annoyances; they are the Ward identities. They encode the conformal transformation law of every operator in the theory.

From conservation to holomorphicity

Stress-tensor and contour normalization

We use Euclidean complex coordinates

$$z=x^1+ix^2, \qquad \bar z=x^1-ix^2, \qquad \partial={\partial\over\partial z}, \qquad \bar\partial={\partial\over\partial\bar z}.$$

We write $T(z)$ and $\bar T(\bar z)$ for the chiral stress-tensor fields in the standard CFT normalization. They are convention-dependent multiples of the corresponding tensor components:

$$T(z)\propto T_{zz}(z), \qquad \bar T(\bar z)\propto T_{\bar z\bar z}(\bar z).$$

We choose the standard CFT normalization in which the OPE of $T$ with a primary of holomorphic weight $h$ is

$$T(z)O(w,\bar w) \sim {h\,O(w,\bar w)\over (z-w)^2} +{\partial_w O(w,\bar w)\over z-w}.$$

With this normalization,

$$Q_\epsilon={1\over 2\pi i}\oint dz\,\epsilon(z)T(z)$$

generates

$$\delta_\epsilon O=\epsilon\partial O+h(\partial\epsilon)O$$

through the contour integral around the insertion. Different books sometimes put factors of $2\pi$ in the definition of $T$ or in the area element $d^2z$. The residue formulas below are the convention-locking statements for this page.

In an ordinary relativistic QFT, translation invariance gives a conserved stress tensor,

$$\partial^\mu T_{\mu\nu}=0.$$

At a conformal fixed point, after possible improvement terms have been chosen, the trace vanishes inside correlators up to contact terms and anomalies:

$$T^\mu_{\ \mu}=0.$$

In two Euclidean dimensions, these two equations are much stronger than they first look. In complex coordinates the stress tensor has three components:

$$T(z)=T_{zz}, \qquad \bar T(\bar z)=T_{\bar z\bar z}, \qquad \Theta=T_{z\bar z}.$$

Conservation becomes schematically

$$\bar\partial T+\partial\Theta=0, \qquad \partial\bar T+\bar\partial\Theta=0.$$

In two dimensions, scale invariance implies this conformal statement only under the usual additional hypotheses, such as unitarity, Poincaré invariance, a discrete spectrum of scaling dimensions, and the absence of a nontrivial virial-current obstruction; we assume the conformal fixed-point conditions here. Then $\Theta=0$ away from insertions, and therefore

$$\boxed{ \bar\partial T=0, \qquad \partial\bar T=0 }$$

away from operator insertions. Thus $T$ is holomorphic and $\bar T$ is antiholomorphic locally on the punctured surface.

This is the first miracle of two-dimensional CFT. The stress tensor is not merely conserved; it becomes a meromorphic object. Meromorphic functions are almost determined by their singularities. The singularities of $T(z)$ at operator insertions will therefore encode the conformal transformation properties of those operators.

Local conformal variations and contact terms

Consider an infinitesimal coordinate deformation

$$z\mapsto z+\epsilon(z,\bar z), \qquad \bar z\mapsto \bar z+\bar\epsilon(z,\bar z).$$

For a purely holomorphic conformal transformation, $\bar\partial\epsilon=0$. To derive a Ward identity, however, it is useful to temporarily allow $\epsilon$ to be a smooth function that is not holomorphic everywhere. The variation of the action has the form

$$\delta S \sim \int d^2z\, \left[(\bar\partial\epsilon)T+(\partial\bar\epsilon)\bar T\right],$$

up to the overall normalization convention fixed above. The key point is structural: $T$ couples to $\bar\partial\epsilon$, while $\bar T$ couples to $\partial\bar\epsilon$.

Let

$$X=O_1(z_1,\bar z_1)\cdots O_n(z_n,\bar z_n).$$

Changing integration variables in the path integral gives

$$0=\delta\langle X\rangle =\langle \delta X\rangle-\langle X\delta S\rangle,$$

where the sign depends on whether one writes the Euclidean weight as $e^{-S}$ or absorbs the sign into $\delta S$. In the residue convention of this page, the local statement is that $\bar\partial\langle T(z)X\rangle$ vanishes away from the insertion points and consists only of contact terms at $z=z_k$.

A local conformal variation can be chosen so that $\bar\partial\epsilon$ is supported only in thin annuli around the operator insertions. The Ward identity is therefore a statement about contact terms of $\bar\partial T$ at the insertions.

For a primary field of holomorphic weight $h_k$, the infinitesimal transformation is

$$\delta_\epsilon O_k(z_k,\bar z_k) = \epsilon(z_k)\partial_{z_k}O_k(z_k,\bar z_k) +h_k(\partial\epsilon)(z_k)O_k(z_k,\bar z_k).$$

Thus the contact-term Ward identity is the distributional statement

$$\bar\partial_z\langle T(z)X\rangle = \pi\sum_{k=1}^n \left[ - h_k\,\partial_z\delta^{(2)}(z-z_k) + \delta^{(2)}(z-z_k)\partial_{z_k} \right] \langle X\rangle,$$

where the sign in the first term follows from

$$\bar\partial_z{1\over (z-z_k)^2} =-\pi\partial_z\delta^{(2)}(z-z_k).$$

This formula looks more complicated than the OPE, but it is the same information. It says that $T$ is holomorphic except at the punctures, and that its singularity at each puncture has just enough strength to translate and rescale the local operator.

Distribution identities behind the OPE

The basic distribution identity in complex analysis is

$$\boxed{ \bar\partial {1\over z} = \pi\delta^{(2)}(z). }$$

One way to check it is to integrate against a smooth test function $f(z,\bar z)$ with compact support. Since $1/z$ is holomorphic away from the origin, the integral localizes on a small circle around the origin:

$$\int d^2z\,f(z,\bar z)\bar\partial {1\over z} = -\int d^2z\,{1\over z}\bar\partial f = \pi f(0),$$

with the usual orientation convention. Differentiating gives

$$\bar\partial {1\over z^{m+1}} = {\pi(-1)^m\over m!}\partial^m\delta^{(2)}(z), \qquad m=0,1,2,\dots.$$

These formulas are the Euclidean two-dimensional version of the familiar Lorentzian rule that boundary values such as $1/(x\pm i0)$ are distributions rather than ordinary functions. In CFT, the singular part of an OPE should always be read in this distributional sense when it is placed inside a Ward identity.

The most important consequence is that the local contact-term equation can be solved by a meromorphic expression. If $X$ is a product of primary fields, then

$$\boxed{ \langle T(z)X\rangle = \sum_{k=1}^n \left[ {h_k\over (z-z_k)^2} +{1\over z-z_k}\partial_{z_k} \right] \langle X\rangle. }$$

This is the holomorphic conformal Ward identity on the plane. The antiholomorphic version is

$$\langle \bar T(\bar z)X\rangle = \sum_{k=1}^n \left[ {\bar h_k\over (\bar z-\bar z_k)^2} +{1\over \bar z-\bar z_k}\partial_{\bar z_k} \right] \langle X\rangle.$$

The Ward identity determines a correlator with one insertion of $T$ in terms of the correlator without $T$. This is why the stress tensor is not an independent mysterious operator in a two-dimensional CFT; it is the local generator of conformal transformations.

The stress-tensor OPE with a primary

The Ward identity for arbitrary correlators is equivalent to the local operator product expansion

$$\boxed{ T(z)O(w,\bar w) \sim {h\,O(w,\bar w)\over (z-w)^2} +{\partial_w O(w,\bar w)\over z-w} +\text{regular}. }$$

The double pole measures the holomorphic weight. The simple pole translates the insertion.

When $T(z)$ approaches a primary insertion $O(w,\bar w)$, the singular part has two universal terms. The coefficient of $(z-w)^{-2}$ is the weight $h$, and the coefficient of $(z-w)^{-1}$ is the derivative $\partial_w O$.

To see the transformation law, surround $w$ by a small contour and define

$$Q_\epsilon={1\over 2\pi i}\oint_w dz\,\epsilon(z)T(z).$$

Using the OPE gives

$$\delta_\epsilon O(w,\bar w) = {1\over 2\pi i}\oint_w dz\,\epsilon(z)T(z)O(w,\bar w).$$

Expand $\epsilon$ near $w$:

$$\epsilon(z)=\epsilon(w)+(z-w)\partial\epsilon(w)+\cdots.$$

The residue is

$$\delta_\epsilon O(w,\bar w) = \epsilon(w)\partial_w O(w,\bar w) +h(\partial\epsilon)(w)O(w,\bar w).$$

This is precisely the infinitesimal version of

$$O(w,\bar w) \mapsto \left({df\over dw}\right)^h O(f(w),\bar w)$$

in the holomorphic sector.

The antiholomorphic stress tensor gives the parallel formula

$$\bar T(\bar z)O(w,\bar w) \sim {\bar h\,O(w,\bar w)\over (\bar z-\bar w)^2} +{\partial_{\bar w}O(w,\bar w)\over \bar z-\bar w} +\text{regular}.$$

Together the two OPEs recover the full local conformal transformation law.

Contour deformation and Ward identities

The local OPE can be repackaged into a contour statement. Let $C$ be a contour enclosing all insertions in $X$. If $\epsilon(z)$ is holomorphic in the region swept out by deforming $C$, then Cauchy’s theorem lets us shrink $C$ to a sum of small contours $C_k$ around the insertions:

$${1\over 2\pi i}\oint_C dz\,\epsilon(z)\langle T(z)X\rangle = \sum_{k=1}^n {1\over 2\pi i}\oint_{C_k} dz\,\epsilon(z)\langle T(z)X\rangle.$$

Each small contour evaluates the conformal variation of the operator it surrounds:

$${1\over 2\pi i}\oint_{C_k} dz\,\epsilon(z)T(z)O_k(z_k,\bar z_k) = \left[ \epsilon(z_k)\partial_{z_k} +h_k(\partial\epsilon)(z_k) \right]O_k(z_k,\bar z_k).$$

Therefore

$$\boxed{ {1\over 2\pi i}\oint_C dz\,\epsilon(z)\langle T(z)X\rangle = \sum_{k=1}^n \left[ \epsilon(z_k)\partial_{z_k} +h_k(\partial\epsilon)(z_k) \right] \langle X\rangle. }$$

Because $T(z)$ is holomorphic away from insertions, a contour integral of $\epsilon(z)T(z)$ may be deformed into a sum of small contours around the local fields. Each residue gives the infinitesimal conformal variation of that field.

This is the working form of the conformal Ward identity. It is more flexible than the explicit formula for $\langle T(z)X\rangle$, because the contour version continues to make sense on general Riemann surfaces and in the presence of branch cuts or boundaries, provided the contours and monodromies are treated correctly.

Global conformal transformations on the plane

On the Riemann sphere, the globally well-defined holomorphic vector fields are

$$\epsilon(z)=a+bz+cz^2.$$

They generate translations, dilations/rotations, and special conformal transformations. If the vacuum is invariant under these global transformations, then the contour at infinity contributes nothing. The Ward identity becomes a constraint on the $n$-point function

$$G(z_i,\bar z_i)=\langle O_1(z_1,\bar z_1)\cdots O_n(z_n,\bar z_n)\rangle.$$

For $\epsilon(z)=1$,

$$\boxed{ \sum_{k=1}^n\partial_{z_k}G=0. }$$

For $\epsilon(z)=z$,

$$\boxed{ \sum_{k=1}^n\left(z_k\partial_{z_k}+h_k\right)G=0. }$$

For $\epsilon(z)=z^2$,

$$\boxed{ \sum_{k=1}^n\left(z_k^2\partial_{z_k}+2h_kz_k\right)G=0. }$$

The antiholomorphic sector gives the same three equations with $z_k,h_k$ replaced by $\bar z_k,\bar h_k$.

These are exactly the global conformal Ward identities used earlier to fix the forms of two- and three-point functions. Here we see their origin: they are simply the absence of a residue at infinity for $\epsilon(z)T(z)$.

Equivalently, expand the Ward identity for large $z$:

$$\langle T(z)X\rangle = \sum_k \left[ {h_k\over z^2}\left(1+{2z_k\over z}+{3z_k^2\over z^2}+\cdots\right) +{1\over z}\left(1+{z_k\over z}+{z_k^2\over z^2}+\cdots\right)\partial_{z_k} \right]G.$$

The coefficients of $z^{-1}$, $z^{-2}$, and $z^{-3}$ vanish precisely when the three global Ward identities hold. Then

$$\langle T(z)X\rangle=O(z^{-4}) \qquad (z\to\infty).$$

This $z^{-4}$ falloff has a geometric interpretation. The stress tensor is a quadratic differential: $T(z)dz^2$ should be regular at infinity. If $w=1/z$, then $dz=-w^{-2}dw$, so regularity of $T(w)dw^2$ implies

$$T(z)\sim {1\over z^4} \qquad (z\to\infty),$$

for a correlator with no operator inserted at infinity. Quantum mechanically $T$ has a Schwarzian term under a general conformal map, but inversion is Möbius and has zero Schwarzian. Thus the same $z^{-4}$ regularity statement at the point at infinity remains valid after the central charge is included.

On the Riemann sphere, $z=\infty$ is an ordinary point. Since $T(z)dz^2$ is a quadratic differential, regularity at infinity gives $\langle T(z)X\rangle=O(z^{-4})$. The missing $z^{-1}$, $z^{-2}$, and $z^{-3}$ terms are the global conformal Ward identities.

Check: the two-point function

Let $O_1$ and $O_2$ be holomorphic primaries with weights $h_1$ and $h_2$. Translation invariance gives

$$G(z_1,z_2)=F(z_{12}), \qquad z_{12}=z_1-z_2.$$

The dilation Ward identity gives

$$\left(z_1\partial_{z_1}+z_2\partial_{z_2}+h_1+h_2\right)G=0.$$

Since

$$z_1\partial_{z_1}+z_2\partial_{z_2}=z_{12}{d\over dz_{12}},$$

we get

$$\left(z_{12}{d\over dz_{12}}+h_1+h_2\right)F(z_{12})=0,$$

so

$$F(z_{12})={C\over z_{12}^{h_1+h_2}}.$$

Now impose the special conformal Ward identity:

$$\left(z_1^2\partial_{z_1}+z_2^2\partial_{z_2}+2h_1z_1+2h_2z_2\right)G=0.$$

Using $F=Cz_{12}^{-(h_1+h_2)}$, this equation reduces to

$$(h_1-h_2)(z_1-z_2)F=0.$$

Thus $C$ can be nonzero only if

$$h_1=h_2.$$

The holomorphic part of the two-point function is therefore

$$\langle O_1(z_1)O_2(z_2)\rangle ={C_{12}\over z_{12}^{2h}} \qquad \text{if }h_1=h_2=h.$$

Including the antiholomorphic sector gives the familiar result

$$\langle O_1(z_1,\bar z_1)O_2(z_2,\bar z_2)\rangle ={C_{12}\over z_{12}^{2h}\bar z_{12}^{2\bar h}},$$

with nonzero $C_{12}$ only between fields of matching weights, up to possible degeneracies among fields with the same dimensions and quantum numbers. Logarithmic CFTs are an important exception to this diagonalizable-weight argument: Jordan blocks for $L_0$ can produce logarithmic two-point structures.

Descendants and why derivatives are not usually primary

If $O$ is a primary of weight $h$, then its derivative

$$V=\partial O$$

has scaling weight $h+1$. But $V$ is not usually primary. The easiest way to see this is to vary it:

$$\delta_\epsilon O=\epsilon\partial O+h(\partial\epsilon)O.$$

Taking $\partial$ gives

$$\begin{aligned} \delta_\epsilon(\partial O) &=\partial\left[\epsilon\partial O+h(\partial\epsilon)O\right]\\ &=\epsilon\partial^2O+(h+1)(\partial\epsilon)\partial O+h(\partial^2\epsilon)O. \end{aligned}$$

A primary of weight $h+1$ would transform as

$$\delta_\epsilon V=\epsilon\partial V+(h+1)(\partial\epsilon)V.$$

The extra term

$$h(\partial^2\epsilon)O$$

is the obstruction. Therefore $\partial O$ is primary only in special cases, for example algebraically when $h=0$, or when the descendant is modified by adding other fields so that the inhomogeneous terms cancel. In an ordinary unitary CFT, a weight-zero primary is often the identity, whose derivative is null.

The same fact appears in the OPE. Differentiate the primary OPE with respect to $w$:

$$T(z)\partial O(w) \sim {2h\,O(w)\over(z-w)^3} +{(h+1)\partial O(w)\over(z-w)^2} +{\partial^2O(w)\over z-w}.$$

The third-order pole is the OPE version of the $h\partial^2\epsilon$ term.

The derivative of a primary has the expected shifted weight $h+1$, but it also acquires an inhomogeneous term proportional to $h\partial^2\epsilon$. In OPE language this is the third-order pole in $T(z)\partial O(w)$.

This is the beginning of the descendant structure of two-dimensional CFT. The identity

$$\partial O=L_{-1}O$$

will later become part of the Virasoro representation theory. The stress-tensor OPE is the bridge between local conformal transformations and that operator algebra.

Primary, quasi-primary, and the stress tensor itself

It is useful to separate three related notions.

A primary field transforms covariantly under all local holomorphic maps:

$$O(z)\mapsto (f'(z))^hO(f(z)).$$

Equivalently, its OPE with $T$ has only the universal double and simple poles.

A quasi-primary field transforms covariantly under the global Möbius group $PSL(2,\mathbb C)$ on the sphere. It may fail to transform as a primary under a general local holomorphic map.

A descendant is obtained by acting with derivatives or, more generally, with negative Virasoro modes. Descendants are essential: they are the local Taylor data of a conformal family. But a generic descendant is not primary.

The stress tensor itself deserves special caution. Classically one might expect $T$ to transform as a field of weight $2$. Quantum mechanically, once the central charge is included, its transformation law acquires an anomalous Schwarzian derivative term:

$$T(z)\mapsto (f'(z))^2T(f(z))+{c\over12}\{f,z\}.$$

That central term is not the topic of this page. It will appear when we study the $T(z)T(w)$ OPE and the Virasoro algebra. For now, the important point is that $T$ is the generator, not just another primary field.

Ward identities as local operator equations

The OPE with $T$ lets us write conformal transformations as operator equations. For any holomorphic test function $\epsilon$,

$$Q_\epsilon={1\over2\pi i}\oint dz\,\epsilon(z)T(z)$$

acts on a primary as

$$[Q_\epsilon,O(w)] = \epsilon(w)\partial O(w)+h(\partial\epsilon)(w)O(w),$$

where the bracket should be interpreted radially in Euclidean signature. For the global choices

$$\epsilon(z)=1, \qquad \epsilon(z)=z, \qquad \epsilon(z)=z^2,$$

we get generators usually denoted $L_{-1}$, $L_0$, and $L_1$:

$$L_{-1}=\oint {dz\over2\pi i}T(z), \qquad L_0=\oint {dz\over2\pi i}zT(z), \qquad L_1=\oint {dz\over2\pi i}z^2T(z).$$

Acting on a primary at the origin,

$$[L_{-1},O(0)]=\partial O(0), \qquad [L_0,O(0)]=hO(0), \qquad [L_1,O(0)]=0.$$

The last equation is one definition of being primary at the origin under the global conformal group. The full local version will be

$$[L_n,O(0)]=0 \qquad n>0,$$

once all Virasoro modes are introduced. This is the algebraic form of the same statement encoded in the OPE.

Summary

The stress tensor is the local generator of conformal transformations. In two dimensions, conservation plus tracelessness imply

$$\bar\partial T=0, \qquad \partial\bar T=0$$

away from insertions. Insertions create contact terms, and those contact terms are equivalent to the universal OPE

$$T(z)O(w,\bar w) \sim {hO(w,\bar w)\over(z-w)^2} +{\partial_wO(w,\bar w)\over z-w}.$$

Contour integrals of $\epsilon(z)T(z)$ implement infinitesimal conformal transformations:

$$\delta_\epsilon O= \epsilon\partial O+h(\partial\epsilon)O.$$

For correlators on the plane, the large-$z$ falloff $\langle T(z)X\rangle=O(z^{-4})$ is equivalent to the global conformal Ward identities. These identities are the residue version of translation, dilation/rotation, and special conformal invariance. Finally, derivatives of primaries are descendants: they have shifted weights, but they are not usually primary because their transformation law contains inhomogeneous higher-derivative terms.

Common pitfalls

The statement that $T$ is holomorphic is not the statement that $\bar\partial T$ is literally zero inside every correlator. Inside correlators, $\bar\partial T$ has contact terms at operator insertions. Those contact terms are the Ward identity.

The double pole in $T(z)O(w)$ does not mean that $T$ creates a new operator of dimension $h+2$. It means that $O$ has weight $h$ under local rescalings and rotations.

The simple pole coefficient is $\partial O$, not an arbitrary descendant. This is forced by translation invariance: $T$ generates motions of the insertion point.

A derivative of a primary has dimension shifted by one, but it is not usually primary. The third-order pole in $T(z)\partial O(w)$ is the fastest way to diagnose this.

The stress tensor itself is not an ordinary primary once the central charge is present. Its anomalous transformation is the origin of the Schwarzian derivative and the Virasoro central extension.

Exercises

Exercise 1: Extracting the conformal variation

Use the OPE

$$T(z)O(w)\sim {hO(w)\over(z-w)^2}+{\partial O(w)\over z-w}$$

to show that

$${1\over2\pi i}\oint_w dz\,\epsilon(z)T(z)O(w) = \epsilon(w)\partial O(w)+h(\partial\epsilon)(w)O(w).$$

Solution

Expand $\epsilon(z)$ around $w$:

$$\epsilon(z)=\epsilon(w)+(z-w)\partial\epsilon(w)+{1\over2}(z-w)^2\partial^2\epsilon(w)+\cdots.$$

Then

$$\epsilon(z)T(z)O(w) \sim \epsilon(z){hO(w)\over(z-w)^2} + \epsilon(z){\partial O(w)\over z-w}.$$

Only the coefficient of $(z-w)^{-1}$ contributes to the contour integral. From the double-pole term, the residue is

$$h(\partial\epsilon)(w)O(w).$$

From the simple-pole term, the residue is

$$\epsilon(w)\partial O(w).$$

Therefore

$${1\over2\pi i}\oint_w dz\,\epsilon(z)T(z)O(w) = \epsilon(w)\partial O(w)+h(\partial\epsilon)(w)O(w).$$

Exercise 2: Two-point functions from global Ward identities

Let $G(z_1,z_2)=\langle O_1(z_1)O_2(z_2)\rangle$ be a holomorphic two-point function of primary fields with weights $h_1$ and $h_2$. Use the three global Ward identities to show that $G$ can be nonzero only if $h_1=h_2$, and then

$$G(z_1,z_2)={C_{12}\over (z_1-z_2)^{2h_1}}.$$

Solution

The translation Ward identity gives

$$(\partial_{z_1}+\partial_{z_2})G=0,$$

so $G$ depends only on $z_{12}=z_1-z_2$:

$$G(z_1,z_2)=F(z_{12}).$$

The dilation Ward identity gives

$$\left(z_1\partial_{z_1}+z_2\partial_{z_2}+h_1+h_2\right)G=0.$$

Since

$$z_1\partial_{z_1}+z_2\partial_{z_2}=z_{12}{d\over dz_{12}},$$

we get

$$\left(z_{12}{d\over dz_{12}}+h_1+h_2\right)F=0.$$

Thus

$$F(z_{12})={C\over z_{12}^{h_1+h_2}}.$$

Now impose the special conformal Ward identity:

$$\left(z_1^2\partial_{z_1}+z_2^2\partial_{z_2}+2h_1z_1+2h_2z_2\right)G=0.$$

Substituting $G=Cz_{12}^{-(h_1+h_2)}$ yields

$$(h_1-h_2)z_{12}G=0.$$

For separated points and nonzero $G$, this requires $h_1=h_2$. Therefore

$$G(z_1,z_2)={C_{12}\over z_{12}^{2h_1}}.$$

Exercise 3: The first descendant is not generally primary

Assume $O$ is primary of weight $h$. Derive the OPE of $T(z)$ with $\partial O(w)$ and identify the term that prevents $\partial O$ from being primary.

Solution

Start from

$$T(z)O(w)\sim {hO(w)\over(z-w)^2}+{\partial O(w)\over z-w}.$$

Differentiate with respect to $w$. Since

$$\partial_w{1\over z-w}={1\over(z-w)^2}, \qquad \partial_w{1\over(z-w)^2}={2\over(z-w)^3},$$

we find

$$\begin{aligned} T(z)\partial O(w) &\sim {h\partial O(w)\over(z-w)^2} +{2hO(w)\over(z-w)^3} +{\partial^2O(w)\over z-w} +{\partial O(w)\over(z-w)^2}\\ &= {2hO(w)\over(z-w)^3} +{(h+1)\partial O(w)\over(z-w)^2} +{\partial^2O(w)\over z-w}. \end{aligned}$$

A primary of weight $h+1$ would have only the double pole

$${(h+1)\partial O(w)\over(z-w)^2}$$

and the simple pole

$${\partial^2O(w)\over z-w}.$$

The extra third-order pole

$${2hO(w)\over(z-w)^3}$$

is the obstruction. It is equivalent to the inhomogeneous term $h(\partial^2\epsilon)O$ in the infinitesimal transformation law of $\partial O$.

Exercise 4: Regularity at infinity and the three global constraints

Starting from the Ward identity

$$\langle T(z)X\rangle = \sum_{k=1}^n \left[ {h_k\over(z-z_k)^2}+{1\over z-z_k}\partial_{z_k} \right]G,$$

where $G=\langle X\rangle$, expand at large $z$ and show that regularity at infinity gives the three global Ward identities.

Solution

For large $z$,

$${1\over z-z_k} ={1\over z}\left(1+{z_k\over z}+{z_k^2\over z^2}+\cdots\right),$$

and

$${1\over(z-z_k)^2} ={1\over z^2}\left(1+{2z_k\over z}+{3z_k^2\over z^2}+\cdots\right).$$

Therefore

$$\begin{aligned} \langle T(z)X\rangle =&\ {1\over z}\sum_k\partial_{z_k}G\\ &+{1\over z^2}\sum_k\left(z_k\partial_{z_k}+h_k\right)G\\ &+{1\over z^3}\sum_k\left(z_k^2\partial_{z_k}+2h_kz_k\right)G +O(z^{-4}). \end{aligned}$$

Regularity of $T(z)dz^2$ at $z=\infty$ requires

$$\langle T(z)X\rangle=O(z^{-4}).$$

Thus the coefficients of $z^{-1}$, $z^{-2}$, and $z^{-3}$ must vanish:

$$\sum_k\partial_{z_k}G=0,$$ $$\sum_k\left(z_k\partial_{z_k}+h_k\right)G=0,$$

and

$$\sum_k\left(z_k^2\partial_{z_k}+2h_kz_k\right)G=0.$$

These are the Ward identities for translations, dilations/rotations, and special conformal transformations.

References and further reading

• A. A. Belavin, A. M. Polyakov, and A. B. Zamolodchikov, “Infinite conformal symmetry in two-dimensional quantum field theory,” Nuclear Physics B 241 (1984), 333–380.
• P. Di Francesco, P. Mathieu, and D. Sénéchal, Conformal Field Theory, Chapter 5.
• P. Ginsparg, “Applied Conformal Field Theory,” Les Houches lectures, especially the sections on Ward identities and the stress tensor.
• J. Polchinski, String Theory, Vol. 1, Sections 2.4–2.6, for the stress tensor, conformal transformations, and OPE language.
• A. M. Polyakov, Gauge Fields and Strings, Chapter 9, for the stress tensor and conformal methods in the string/random-surface setting.

Further reading

This lesson follows Polyakov’s contact-term-to-contour lecture sequence. For the maintained reference treatment of the self-OPE, modes, and central charge, see The Virasoro algebra and the stress tensor; a momentum-space complement appears in Momentum-space correlators and conformal Ward identities.