Vertex Operators, OPEs, and the Virasoro Algebra

The previous page explained how the stress tensor recognizes primary fields and generates conformal transformations. We now apply that machinery to the most important local operators in perturbative string theory: vertex operators. In flat space, a string state with spacetime momentum $k^\mu$ is represented on the worldsheet by a plane wave $:e^{ik\cdot X}:$ multiplied by oscillator factors such as $\partial X^\mu$, $\bar\partial X^\nu$, or their superstring analogues.

This page has two goals. First, we compute the conformal weights of exponential vertex operators. Second, we derive the OPE $T(z)T(w)$ and convert it into the Virasoro algebra. These are the local CFT facts behind the mass-shell conditions and the infinite set of Virasoro constraints.

Plane-wave vertex operators

For the free boson normalized by

$$\langle X^\mu(z,\bar z)X^\nu(w,\bar w)\rangle = -\frac{\alpha'}{2}\eta^{\mu\nu}\ln |z-w|^2,$$

the basic spacetime-momentum operator is

$$V_k(z,\bar z)=:e^{ik\cdot X(z,\bar z)}: .$$

The normal-ordering symbols remove the self-contractions inside a single exponential. Contractions between different operators remain, and they are precisely what determine OPEs.

Differentiating the two-point function gives

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

Therefore

$$\partial X^\mu(z)V_k(w,\bar w) \sim -i\frac{\alpha'}{2}\frac{k^\mu}{z-w}V_k(w,\bar w).$$

The free-boson stress tensor is

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

Applying Wick’s theorem gives

$$T(z)V_k(w,\bar w) \sim \frac{\alpha'k^2}{4}\frac{V_k(w,\bar w)}{(z-w)^2} + \frac{\partial V_k(w,\bar w)}{z-w}.$$

The antiholomorphic OPE is identical with bars. Thus the plane-wave vertex has weights

$$\boxed{ (h,\bar h)=\left(\frac{\alpha'k^2}{4},\frac{\alpha'k^2}{4}\right). }$$

The double pole in $T(z)V_k(w)$ measures the holomorphic weight $h=\alpha'k^2/4$. The simple pole translates the insertion.

This formula is the CFT version of the string mass-shell condition. For a closed-string integrated vertex operator on the sphere, the matter part must have weights $(1,1)$. For the bosonic closed-string tachyon, this gives

$$\frac{\alpha'k^2}{4}=1, \qquad k^2=\frac{4}{\alpha'}.$$

With mostly-plus spacetime metric, $k^2=-m^2$, so

$$m^2=-\frac{4}{\alpha'}.$$

For a massless closed-string vertex,

$$\epsilon_{\mu\nu}\,\partial X^\mu\bar\partial X^\nu e^{ik\cdot X},$$

the derivatives contribute $(1,1)$, so the exponential must contribute $(0,0)$. Hence

$$k^2=0.$$

Normalization checkpoint

With the convention used here, $:e^{ik\cdot X}:$ has weights $h=\bar h=\alpha'k^2/4$. If one temporarily sets $\alpha'=2$, this becomes $h=\bar h=k^2/2$. The notes below keep $\alpha'$ explicit.

Products of exponential operators

The product of two normal-ordered exponentials is not simply the normal-ordered exponential of the sum. One must also include the cross-contraction:

$$:e^{ik\cdot X(z,\bar z)}: :e^{iq\cdot X(w,\bar w)}: \sim |z-w|^{\alpha' k\cdot q} :e^{i(k+q)\cdot X(w,\bar w)}: \left[1+\cdots\right].$$

The omitted terms contain descendants built from derivatives of $X$. Expanding $X(z,\bar z)$ about $(w,\bar w)$ gives

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

Thus the OPE naturally produces a primary operator with momentum $k+q$ together with its derivative descendants. This is the worldsheet origin of the string-theoretic statement that local vertex operators fuse into towers of intermediate string states.

General OPE structure

The operator product expansion is a local completeness relation for operators. In a holomorphic CFT, if $A_i$ and $A_j$ are primary fields of weights $h_i$ and $h_j$, then schematically

$$A_i(z)A_j(w) \sim \sum_k C_{ij}{}^k (z-w)^{h_k-h_i-h_j} \left[A_k(w)+\text{descendants}\right].$$

For a full two-dimensional CFT one also includes the antiholomorphic power,

$$A_i(z,\bar z)A_j(w,\bar w) \sim \sum_k C_{ij}{}^k (z-w)^{h_k-h_i-h_j} (\bar z-\bar w)^{\bar h_k-\bar h_i-\bar h_j} \left[A_k(w,\bar w)+\cdots\right].$$

The OPE is a short-distance expansion. In string amplitudes, the same local expansion becomes factorization onto intermediate string states.

The OPE is not merely a computational trick. In a two-dimensional CFT, the spectrum of primary fields, the OPE coefficients $C_{ij}{}^k$, and the rules for descendants are essentially the data of the theory.

The $T(z)T(w)$ OPE

The stress tensor itself has a special OPE with itself:

$$\boxed{ T(z)T(w) \sim \frac{c/2}{(z-w)^4} + \frac{2T(w)}{(z-w)^2} + \frac{\partial T(w)}{z-w}. }$$

The double and simple poles say that $T$ transforms like a field of holomorphic weight $2$. The fourth-order pole is new. It is the central charge.

For $D$ free bosons,

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

and Wick contractions give

$$T(z)T(w) \sim \frac{D/2}{(z-w)^4} + \frac{2T(w)}{(z-w)^2} + \frac{\partial T(w)}{z-w}.$$

Therefore

$$\boxed{c=D}$$

for $D$ free scalar fields. The central charge counts degrees of freedom, but it counts them with signs and weights once ghosts or fermions are included. Later the matter contribution $c=D$ will be balanced by the $bc$ ghost contribution $c=-26$, forcing the critical bosonic string to have $D=26$.

The central term in the Virasoro algebra is already visible locally as the fourth-order pole in the $T(z)T(w)$ OPE.

One should not think of the central charge as a small correction. It controls the transformation of $T$ under conformal maps, the Casimir energy on the cylinder, the normal-ordering constants of the string, and the cancellation of the Weyl anomaly.

Virasoro generators

The holomorphic stress tensor is expanded in Laurent modes as

$$T(z) = \sum_{n\in\mathbb Z} L_n z^{-n-2}.$$

Equivalently,

$$\boxed{ L_n = \frac{1}{2\pi i}\oint dz\,z^{n+1}T(z). }$$

The antiholomorphic sector has

$$\bar T(\bar z) = \sum_{n\in\mathbb Z}\bar L_n\bar z^{-n-2}, \qquad \bar L_n = \frac{1}{2\pi i}\oint d\bar z\,\bar z^{n+1}\bar T(\bar z).$$

The mode $L_0$ measures holomorphic scaling dimension on the plane. Acting on a primary state created by a field of weight $h$, it gives eigenvalue $h$. The modes $L_{-n}$ with $n>0$ create descendants, while $L_n$ with $n>0$ annihilate a primary state.

The Virasoro mode $L_n$ is the $z^{n+1}$ moment of $T(z)$. Commutators of modes follow from moving stress-tensor contours through each other.

Deriving the Virasoro algebra from the $TT$ OPE

The Virasoro commutator is computed by a double contour integral:

$$[L_m,L_n] = \frac{1}{(2\pi i)^2} \oint_0 dw\,w^{n+1} \oint_w dz\,z^{m+1}T(z)T(w) - (m\leftrightarrow n).$$

Using the $TT$ OPE, the $z$ contour extracts residues at $z=w$. The useful residue formulas are

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

and

$$\frac{1}{2\pi i}\oint_w dz\, \frac{z^{m+1}}{(z-w)^4} = \frac{(m+1)m(m-1)}{6}w^{m-2}.$$

The result is

$$\boxed{ [L_m,L_n] = (m-n)L_{m+n} + \frac{c}{12}m(m^2-1)\delta_{m+n,0}. }$$

This is the Virasoro algebra. It is the central extension of the algebra of local holomorphic vector fields on the circle.

The Laurent expansion of $T$ turns the local $TT$ OPE into the mode algebra of conformal transformations.

Classically, the central term is absent. Quantum mechanically, normal ordering creates it. In string theory this quantum term is not optional: Lorentz invariance, Weyl invariance, and BRST nilpotency all know about it.

Antiholomorphic copy and closed strings

Closed strings have two commuting copies of the Virasoro algebra:

$$[L_m,L_n] = (m-n)L_{m+n} + \frac{c}{12}m(m^2-1)\delta_{m+n,0},$$ $$[\bar L_m,\bar L_n] = (m-n)\bar L_{m+n} + \frac{\bar c}{12}m(m^2-1)\delta_{m+n,0},$$

and

$$[L_m,\bar L_n]=0.$$

For a parity-invariant flat target-space theory, $c=\bar c=D$ in the matter sector. The condition $L_0=\bar L_0$ becomes level matching. The stronger physical-state constraints will reappear after ghosts and BRST are introduced.

What to remember

The chain of ideas is compact but powerful:

$$\boxed{ \text{free-field contractions} \Rightarrow \text{OPEs} \Rightarrow \text{conformal weights} \Rightarrow \text{Virasoro algebra} }$$

The vertex operator $:e^{ik\cdot X}:$ has weights $h=\bar h=\alpha'k^2/4$. The stress tensor OPE

$$T(z)T(w) \sim \frac{c/2}{(z-w)^4} + \frac{2T(w)}{(z-w)^2} + \frac{\partial T(w)}{z-w}$$

is equivalent to the Virasoro algebra. The central charge $c$ is the number that will later decide whether the string worldsheet quantum theory is consistent.

Exercises

Exercise 1. Weight of the plane-wave vertex

Using

$$\partial X^\mu(z)V_k(w,\bar w) \sim -i\frac{\alpha'}{2}\frac{k^\mu}{z-w}V_k(w,\bar w),$$

show that $V_k=:e^{ik\cdot X}:$ has holomorphic weight $h=\alpha'k^2/4$.

Solution

Use

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

The double pole comes from contracting both $\partial X$ factors in $T$ with the exponential:

$$-\frac{1}{\alpha'} \left( -i\frac{\alpha'}{2}\frac{k^\mu}{z-w} \right) \left( -i\frac{\alpha'}{2}\frac{k_\mu}{z-w} \right) V_k.$$

Since $(-i)^2=-1$, this becomes

$$\frac{\alpha'k^2}{4}\frac{V_k}{(z-w)^2}.$$

The single contractions produce the simple pole $\partial V_k/(z-w)$. Therefore

$$T(z)V_k(w,\bar w) \sim \frac{\alpha'k^2}{4}\frac{V_k}{(z-w)^2} + \frac{\partial V_k}{z-w},$$

so $h=\alpha'k^2/4$.

Exercise 2. Closed-string tachyon mass

For the bosonic closed-string tachyon vertex $V_k=:e^{ik\cdot X}:$, require matter weights $(1,1)$. Derive the spacetime mass squared using mostly-plus signature.

Solution

The condition $h=\bar h=1$ gives

$$\frac{\alpha'k^2}{4}=1,$$

hence

$$k^2=\frac{4}{\alpha'}.$$

With mostly-plus signature, $k^2=-m^2$, so

$$m^2=-\frac{4}{\alpha'}.$$

This is the closed bosonic string tachyon mass.

Exercise 3. Exponential OPE

Show that

$$:e^{ik\cdot X(z,\bar z)}::e^{iq\cdot X(w,\bar w)}: \sim |z-w|^{\alpha'k\cdot q}:e^{i(k+q)\cdot X(w,\bar w)}:\cdots .$$

Solution

Only cross-contractions contribute. They exponentiate:

$$\exp\left[-k_\mu q_\nu\langle X^\mu(z,\bar z)X^\nu(w,\bar w)\rangle\right].$$

Using

$$\langle X^\mu(z,\bar z)X^\nu(w,\bar w)\rangle =-\frac{\alpha'}{2}\eta^{\mu\nu}\ln|z-w|^2,$$

we obtain

$$\exp\left[\frac{\alpha'}{2}k\cdot q\ln|z-w|^2\right] =|z-w|^{\alpha'k\cdot q}.$$

The remaining normal-ordered exponential is expanded around $w$, giving the displayed primary and its descendants.

Exercise 4. Central charge of $D$ free bosons

Explain why $D$ independent free scalar fields have central charge $c=D$.

Solution

For one free boson,

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

The fourth-order pole in $T(z)T(w)$ comes from the double contraction. There are two Wick pairings, and each contributes a factor proportional to

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

Including the normalization $1/\alpha'^2$, the total coefficient is

$$\frac{1}{2}\frac{1}{(z-w)^4}.$$

Comparing with

$$T(z)T(w)\sim \frac{c/2}{(z-w)^4}+\cdots,$$

one free boson has $c=1$. For $D$ independent bosons, the contributions add, so $c=D$.

Exercise 5. The central term in the Virasoro algebra

Use the fourth-order pole in the $TT$ OPE to derive the central contribution

$$\frac{c}{12}m(m^2-1)\delta_{m+n,0}$$

in $[L_m,L_n]$.

Solution

The central part of the OPE is

$$T(z)T(w)\supset \frac{c/2}{(z-w)^4}.$$

Inside the double contour expression, the inner contour gives

$$\frac{1}{2\pi i}\oint_w dz\, \frac{z^{m+1}}{(z-w)^4} = \frac{1}{3!}\partial_w^3 w^{m+1} = \frac{(m+1)m(m-1)}{6}w^{m-2}.$$

Multiplying by $c/2$ gives

$$\frac{c}{12}m(m^2-1)w^{m-2}.$$

The outer contour includes $w^{n+1}$, so it extracts

$$\frac{1}{2\pi i}\oint_0 dw\,w^{m+n-1}=\delta_{m+n,0}.$$

Therefore the central term is

$$\frac{c}{12}m(m^2-1)\delta_{m+n,0}.$$