Stress Tensor, Primaries, and Conformal Transformations

The free-boson OPEs are local short-distance rules. The stress tensor packages those rules into symmetry. In two-dimensional CFT, $T(z)$ is not merely a conserved current in the usual sense; it is a meromorphic object whose contour integrals generate conformal transformations.

For string theory this is doubly important. First, conformal symmetry is the residual gauge symmetry left after fixing the worldsheet metric to conformal gauge. Second, the vanishing of the stress tensor is the quantum descendant of the classical Virasoro constraints. Understanding $T(z)$ is therefore the gateway from free fields to physical string states and scattering amplitudes.

Stress tensor in complex coordinates

For $D$ free target-space coordinates $X^\mu$, the Euclidean conformal-gauge action is

$$S_E = \frac{1}{2\pi\alpha'} \int d^2z\, \partial X^\mu \bar\partial X_\mu .$$

The basic OPE is

$$\partial X^\mu(z)\partial X^\nu(w) \sim -\frac{\alpha'}{2} \frac{\eta^{\mu\nu}}{(z-w)^2}.$$

The holomorphic and antiholomorphic stress tensors are

$$T(z)\equiv T_{zz} = -\frac{1}{\alpha'} :\partial X^\mu\partial X_\mu:,$$

and

$$\bar T(\bar z)\equiv T_{\bar z\bar z} = -\frac{1}{\alpha'} :\bar\partial X^\mu\bar\partial X_\mu: .$$

Classically, conformal invariance implies tracelessness,

$$T_{z\bar z}=0.$$

Conservation then gives

$$\bar\partial T(z)=0, \qquad \partial \bar T(\bar z)=0,$$

away from operator insertions. Thus $T$ is holomorphic and $\bar T$ is antiholomorphic except at singularities caused by local operators.

Conservation plus tracelessness imply a holomorphic stress tensor $T(z)$ and an antiholomorphic stress tensor $\bar T(\bar z)$. Their singularities near insertions encode conformal transformations.

The holomorphic factorization is what makes two-dimensional CFT so powerful. Symmetry transformations become contour integrals.

Residual conformal transformations

Conformal gauge fixes the metric only up to transformations that preserve angles. In complex coordinates these are

$$z\longrightarrow z'=f(z), \qquad \bar z\longrightarrow \bar z'=\bar f(\bar z).$$

Indeed,

$$dz'\,d\bar z' = f'(z)\bar f'(\bar z)\,dz\,d\bar z,$$

so the metric changes only by a local scale factor, which can be absorbed by a Weyl transformation. This is the residual gauge symmetry of the string worldsheet in conformal gauge.

A conformal map preserves angles locally. In conformal gauge the induced scale factor can be absorbed by a Weyl transformation.

Infinitesimally,

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

The corresponding variation of a local operator is generated by contour integrals of $T$ and $\bar T$.

Primary fields

A local field $\phi(z,\bar z)$ is called a primary field of weights $(h,\bar h)$ if under a finite conformal map it transforms as

$$\phi'(z',\bar z') = \left(\frac{\partial z'}{\partial z}\right)^{-h} \left(\frac{\partial \bar z'}{\partial \bar z}\right)^{-\bar h} \phi(z,\bar z).$$

The numbers

$$h,\qquad \bar h$$

are the holomorphic and antiholomorphic conformal weights. Their sum and difference are

$$\Delta=h+\bar h, \qquad s=h-\bar h,$$

where $\Delta$ is the scaling dimension and $s$ is the two-dimensional spin.

A primary field carries definite conformal weights. A local rescaling by $f'(z)$ multiplies the field by a fixed power determined by $h$ and $\bar h$.

For example:

$$\partial X^\mu(z)$$

has weights $(1,0)$, while

$$\bar\partial X^\mu(\bar z)$$

has weights $(0,1)$. A product such as

$$\partial X^\mu(z)\bar\partial X^\nu(\bar z)$$

has weights $(1,1)$.

A primary field is not just a field that scales nicely under dilatations. It must transform covariantly under every local holomorphic coordinate change.

The stress tensor as a contour generator

Let $\epsilon(z)$ be a holomorphic infinitesimal conformal transformation. The associated charge is represented locally by

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

When this contour surrounds a local operator $\phi(w,\bar w)$, the singular part of $T(z)\phi(w,\bar w)$ determines the transformation of $\phi$:

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

The antiholomorphic part is analogous:

$$\delta_{\bar\epsilon} \phi(w,\bar w) = \frac{1}{2\pi i} \oint_w d\bar z\,\bar\epsilon(\bar z)\bar T(\bar z)\phi(w,\bar w).$$

The contour integral of $T$ around $\phi$ extracts the residues of the singular OPE. Those residues are the infinitesimal conformal transformation of $\phi$.

This is the local form of the conformal Ward identity.

The primary-field OPE with $T$

The defining local statement for a primary field is

$$\boxed{ T(z)\phi(w,\bar w) \sim \frac{h\,\phi(w,\bar w)}{(z-w)^2} + \frac{\partial\phi(w,\bar w)}{z-w}. }$$

Similarly,

$$\bar T(\bar z)\phi(w,\bar w) \sim \frac{\bar h\,\phi(w,\bar w)}{(\bar z-\bar w)^2} + \frac{\bar\partial\phi(w,\bar w)}{\bar z-\bar w}.$$

To see why this is the correct transformation law, insert the OPE into the contour integral:

$$\frac{1}{2\pi i} \oint_w dz\, \epsilon(z) \left[ \frac{h\phi(w,\bar w)}{(z-w)^2} + \frac{\partial\phi(w,\bar w)}{z-w} \right].$$

Using

$$\frac{1}{2\pi i}\oint_w dz\,\frac{\epsilon(z)}{z-w} = \epsilon(w),$$

and

$$\frac{1}{2\pi i}\oint_w dz\,\frac{\epsilon(z)}{(z-w)^2} = \partial\epsilon(w),$$

we obtain

$$\delta_\epsilon \phi = \epsilon\,\partial\phi + h(\partial\epsilon)\phi .$$

This is precisely the infinitesimal version of primary-field covariance, up to the usual active/passive sign convention for coordinate transformations.

For a primary field, the double pole measures the conformal weight and the simple pole translates the insertion. More singular poles would signal a non-primary operator.

Free-boson examples

Let us verify these statements with the free-boson stress tensor

$$T(z)=-\frac{1}{\alpha'}:\partial X^\mu\partial X_\mu: .$$

First,

$$T(z)X^\mu(w,\bar w) \sim \frac{\partial X^\mu(w)}{z-w}.$$

There is no double pole, so locally $X^\mu$ behaves as a dimension-zero scalar. The logarithmic two-point function makes $X^\mu$ somewhat special globally, but its derivative is a clean primary field.

For $\partial X^\mu$ we find

$$T(z)\partial X^\mu(w) \sim \frac{\partial X^\mu(w)}{(z-w)^2} + \frac{\partial^2 X^\mu(w)}{z-w}.$$

Thus

$$\partial X^\mu \quad \text{has weights} \quad (h,\bar h)=(1,0).$$

Similarly,

$$\bar\partial X^\mu \quad \text{has weights} \quad (h,\bar h)=(0,1).$$

These are the simplest examples of primary fields in the string worldsheet theory.

Ward identities in correlation functions

The stress tensor OPE with a primary also determines how $T$ behaves inside correlation functions. For primary fields $\phi_i(z_i,\bar z_i)$ of weights $(h_i,\bar h_i)$,

$$\left\langle T(z)\prod_i \phi_i(z_i,\bar z_i) \right\rangle = \sum_i \left[ \frac{h_i}{(z-z_i)^2} + \frac{1}{z-z_i}\frac{\partial}{\partial z_i} \right] \left\langle \prod_i \phi_i(z_i,\bar z_i) \right\rangle .$$

This identity is obtained by applying the $T\phi_i$ OPE near each insertion. It is one of the main computational tools of CFT. The antiholomorphic Ward identity is the same with barred quantities.

A useful way to visualize it is contour deformation. A contour integral of $T$ can be moved through the correlator. When it crosses an insertion, the OPE supplies a residue.

Conformal Ward identities follow by deforming a contour of $T(z)$ through a correlation function and summing residues at operator insertions.

Global conformal transformations

On the Riemann sphere, the globally well-defined holomorphic maps are the Möbius transformations

$$z\longrightarrow \frac{az+b}{cz+d}, \qquad ad-bc\neq 0.$$

They form the group $SL(2,\mathbb C)$ modulo its center. Infinitesimally they are generated by

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

These correspond to translation, dilatation/rotation, and special conformal transformation on the complex plane. Later, this three-parameter holomorphic freedom and its antiholomorphic partner will let us fix three insertion points on the sphere when computing closed-string tree amplitudes.

What to remember

The essential local rule is

$$T(z)\phi(w,\bar w) \sim \frac{h\phi(w,\bar w)}{(z-w)^2} + \frac{\partial\phi(w,\bar w)}{z-w}$$

for a primary field. It implies both the infinitesimal conformal transformation law

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

and the conformal Ward identity in correlation functions.

The next page applies this machinery to plane-wave vertex operators and then studies the OPE $T(z)T(w)$, whose central term produces the Virasoro algebra.

Exercises

Exercise 1. Holomorphicity from conservation and tracelessness

In complex coordinates, explain why conservation of the stress tensor together with $T_{z\bar z}=0$ implies $\bar\partial T_{zz}=0$.

Solution

Stress-tensor conservation can be written schematically as

$$\partial_{\bar z}T_{zz}+\partial_z T_{\bar z z}=0.$$

Since $T_{\bar z z}=T_{z\bar z}$, tracelessness in conformal coordinates gives

$$T_{z\bar z}=0.$$

Therefore the second term vanishes and we get

$$\partial_{\bar z}T_{zz}=0.$$

Thus $T_{zz}=T(z)$ is holomorphic away from insertions.

Exercise 2. Residues and infinitesimal transformations

Assume

$$T(z)\phi(w) \sim \frac{h\phi(w)}{(z-w)^2} + \frac{\partial\phi(w)}{z-w}.$$

Evaluate

$$\frac{1}{2\pi i} \oint_w dz\,\epsilon(z)T(z)\phi(w).$$

Solution

Insert the OPE:

$$\frac{1}{2\pi i} \oint_w dz \left[ \frac{h\epsilon(z)\phi(w)}{(z-w)^2} + \frac{\epsilon(z)\partial\phi(w)}{z-w} \right].$$

The simple-pole residue gives

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

The double-pole residue gives

$$h\,\partial\epsilon(w)\phi(w).$$

Thus

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

Exercise 3. Show that $\partial X^\mu$ has weight one

Using

$$T(z)=-\frac{1}{\alpha'}:\partial X^\rho\partial X_\rho:$$

and

$$\partial X^\rho(z)\partial X^\mu(w) \sim -\frac{\alpha'}{2}\frac{\eta^{\rho\mu}}{(z-w)^2},$$

derive

$$T(z)\partial X^\mu(w) \sim \frac{\partial X^\mu(w)}{(z-w)^2} + \frac{\partial^2 X^\mu(w)}{z-w}.$$

Solution

There are two equivalent contractions, one for each $\partial X$ inside $T$. Contracting one field gives

$$-\frac{1}{\alpha'} \cdot 2 \left( -\frac{\alpha'}{2}\frac{1}{(z-w)^2} \right) \partial X^\mu(z).$$

So

$$T(z)\partial X^\mu(w) \sim \frac{\partial X^\mu(z)}{(z-w)^2}.$$

Now expand the uncontracted field around $w$:

$$\partial X^\mu(z) = \partial X^\mu(w) + (z-w)\partial^2X^\mu(w) +\cdots .$$

Therefore

$$T(z)\partial X^\mu(w) \sim \frac{\partial X^\mu(w)}{(z-w)^2} + \frac{\partial^2X^\mu(w)}{z-w}.$$

This is the primary OPE with $h=1$.

Exercise 4. Ward identity for one insertion

Let $\phi$ be a primary of weight $h$. Use the Ward identity to find

$$\left\langle T(z)\phi(w,\bar w)\right\rangle$$

in terms of $\langle \phi(w,\bar w)\rangle$.

Solution

For a single primary insertion, the Ward identity gives

$$\left\langle T(z)\phi(w,\bar w)\right\rangle = \left[ \frac{h}{(z-w)^2} + \frac{1}{z-w}\frac{\partial}{\partial w} \right] \left\langle \phi(w,\bar w)\right\rangle .$$

On the plane, translational invariance usually forces the one-point function of a non-identity primary to vanish. But the local Ward identity itself is the formula above.

Exercise 5. Weights of a tensor product operator

Assume $\partial X^\mu$ has weights $(1,0)$ and $\bar\partial X^\nu$ has weights $(0,1)$. What are the weights, dimension, and spin of

$$\mathcal O^{\mu\nu}=\partial X^\mu\bar\partial X^\nu?$$

Solution

Weights add under products, so

$$(h,\bar h)=(1,0)+(0,1)=(1,1).$$

The scaling dimension is

$$\Delta=h+\bar h=2,$$

and the two-dimensional spin is

$$s=h-\bar h=0.$$

Thus $\partial X^\mu\bar\partial X^\nu$ is a scalar operator on the worldsheet with dimension two.