Radial Quantization and the State–Operator Correspondence

The Virasoro algebra turns local conformal symmetry into a representation-theory problem. To make that statement precise, we need a Hilbert space. In two-dimensional CFT the natural Hilbert space is obtained by radial quantization: the radius $|z|$ of the complex plane becomes Euclidean time.

The central idea is beautifully simple:

$$\boxed{ \text{local operator inserted at the origin} \quad\longleftrightarrow\quad \text{state on a surrounding circle}. }$$

For string theory, this dictionary is not optional. A spacetime string state is represented by a worldsheet vertex operator, and the physical-state conditions become statements about conformal weights, primarity, and BRST cohomology.

The global conformal subalgebra

The Virasoro algebra is

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

The modes

$$L_{-1},\qquad L_0,\qquad L_1$$

form a closed subalgebra because the central term vanishes for $m=-1,0,1$. Explicitly,

$$[L_1,L_{-1}]=2L_0, \qquad [L_0,L_1]=-L_1, \qquad [L_0,L_{-1}]=L_{-1}.$$

These are the holomorphic generators of the globally well-defined conformal maps of the Riemann sphere,

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

The globally regular conformal transformations on the sphere are generated by $L_{-1}$, $L_0$, and $L_1$, together with their antiholomorphic partners.

The vacuum on the sphere is invariant under these global transformations:

$$L_{-1}|0\rangle=L_0|0\rangle=L_1|0\rangle=0,$$

and similarly for $\bar L_{-1},\bar L_0,\bar L_1$. Higher $L_n$ generate local conformal transformations that are singular somewhere on the Riemann sphere. They are still essential for the local operator algebra, but they are not global automorphisms of the sphere.

The stress tensor and the Schwarzian derivative

A primary field of weight $h$ transforms under $z'=f(z)$ as

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

The stress tensor is almost a primary field of weight $2$, but quantum mechanically it has an anomalous term. Under $z'=f(z)$,

$$T(z) = \left(\frac{df}{dz}\right)^2T'(f(z)) + \frac{c}{12}\{f,z\},$$

where

$$\{f,z\} = \frac{f^{(3)}(z)}{f'(z)} - \frac{3}{2}\left(\frac{f''(z)}{f'(z)}\right)^2$$

is the Schwarzian derivative. Equivalently,

$$T'(z') = \left(\frac{dz}{dz'}\right)^2 \left[T(z)-\frac{c}{12}\{z',z\}\right].$$

The Schwarzian term is another face of the central charge. The same number $c$ appears in the $T(z)T(w)$ OPE, in the Virasoro algebra, and in the Casimir energy on the cylinder.

Cylinder to plane

Let

$$w=\tau+i\sigma, \qquad \sigma\sim \sigma+2\pi,$$

be the Euclidean cylinder coordinate. The map

$$z=e^w$$

sends the cylinder to the punctured complex plane. Constant $\tau$ slices map to circles,

$$|z|=e^\tau.$$

Thus time evolution on the cylinder is radial evolution on the plane.

Under $z=e^w$, increasing cylinder time $\tau$ corresponds to moving outward in the complex plane. This is the geometric basis of radial quantization.

For $z=e^w$,

$$\{z,w\}=-\frac{1}{2}.$$

Therefore the cylinder stress tensor is

$$T_{\rm cyl}(w) = z^2T_{\rm plane}(z)-\frac{c}{24},$$

and similarly

$$\bar T_{\rm cyl}(\bar w) = \bar z^2\bar T_{\rm plane}(\bar z)-\frac{\bar c}{24}.$$

For a parity-invariant theory with $c=\bar c$, the cylinder Hamiltonian and momentum are

$$H_{\rm cyl} = L_0+\bar L_0-\frac{c}{12}, \qquad P_{\rm cyl} = L_0-\bar L_0.$$

The exponential map produces the shift $-c/24$ in one chiral sector. This is the CFT origin of the Casimir energy on the cylinder.

This shift is the same physics that appears as normal-ordering constants in string quantization. For example, $24$ transverse light-cone bosons give a chiral shift $-24/24=-1$.

Radial quantization

In ordinary Euclidean quantum mechanics, a path integral over a long time interval prepares the lowest-energy state compatible with the imposed boundary conditions. In radial quantization, the logarithmic radius

$$\tau=\log|z|$$

is the Euclidean time. A circle $|z|=R$ is a constant-time slice.

If we perform the path integral inside a circle with no insertion at the origin, we prepare the vacuum. If we insert a local operator $\mathcal O(0,0)$ at the origin, we prepare an excited state on the circle:

$$\boxed{ |\mathcal O\rangle = \lim_{z,\bar z\to0}\mathcal O(z,\bar z)|0\rangle. }$$

The state-operator correspondence identifies an operator insertion at the origin with a state on a circle surrounding that insertion.

The inverse statement is also useful: a state on a circle can be represented as a boundary condition for the path integral inside the circle, and in a CFT such boundary data can be shrunk to a local operator at the origin.

Momentum states and oscillator states

For one chiral half of a free boson, use

$$\partial X^\mu(z) = -i\sqrt{\frac{\alpha'}{2}} \sum_{n\in\mathbb Z}\alpha_n^\mu z^{-n-1}.$$

The momentum vacuum $|0;k\rangle$ satisfies

$$\alpha_n^\mu|0;k\rangle=0, \qquad n>0,$$

and

$$\alpha_0^\mu|0;k\rangle = \sqrt{\frac{\alpha'}{2}}k^\mu|0;k\rangle.$$

The state-operator map gives

$$:e^{ik\cdot X(0)}:|0\rangle = |0;k\rangle.$$

Since the exponential has holomorphic weight $h=\alpha'k^2/4$, the momentum state has

$$L_0|0;k\rangle = \frac{\alpha'k^2}{4}|0;k\rangle$$

in the matter CFT.

Oscillator excitations come from derivatives of $X$. Expanding near the origin,

$$\partial X^\mu(z)|0;k\rangle = -i\sqrt{\frac{\alpha'}{2}} \left( \frac{\alpha_0^\mu}{z} +\alpha_{-1}^\mu +\alpha_{-2}^\mu z +\alpha_{-3}^\mu z^2 +\cdots \right)|0;k\rangle.$$

Therefore, up to the phase convention in the oscillator expansion,

$$\boxed{ \alpha_{-m}^\mu|0;k\rangle \longleftrightarrow \frac{i}{(m-1)!}\sqrt{\frac{2}{\alpha'}}\, \partial^m X^\mu(0):e^{ik\cdot X(0)}: }$$

for $m\geq1$.

Momentum states are created by exponentials. Oscillator excitations are created by multiplying the exponential by derivatives of $X^\mu$.

For example,

$$\epsilon_\mu\alpha_{-1}^\mu|0;k\rangle \quad\longleftrightarrow\quad \epsilon_\mu\partial X^\mu e^{ik\cdot X}.$$

Its holomorphic weight is

$$h=1+\frac{\alpha'k^2}{4}.$$

For $k^2=0$ this operator has weight one. To be primary, it must also obey a transversality condition. In oscillator language,

$$L_1\epsilon_\mu\alpha_{-1}^\mu|0;k\rangle = \epsilon_\mu\alpha_0^\mu|0;k\rangle = \sqrt{\frac{\alpha'}{2}}(\epsilon\cdot k)|0;k\rangle,$$

so primarity requires

$$\epsilon\cdot k=0.$$

This is the CFT origin of gauge-boson transversality.

Primary operators and highest-weight states

Let $\phi(z,\bar z)$ be a primary field of weights $(h,\bar h)$:

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

and similarly for $\bar T$. Define the state

$$|\phi\rangle=\phi(0,0)|0\rangle.$$

Using

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

we obtain

$$\boxed{ L_0|\phi\rangle=h|\phi\rangle, \qquad L_n|\phi\rangle=0\quad n>0, }$$

and

$$L_{-1}|\phi\rangle=\partial\phi(0)|0\rangle.$$

The antiholomorphic conditions are

$$\bar L_0|\phi\rangle=\bar h|\phi\rangle, \qquad \bar L_n|\phi\rangle=0\quad n>0, \qquad \bar L_{-1}|\phi\rangle=\bar\partial\phi(0)|0\rangle.$$

Under the state-operator map, a primary field becomes a highest-weight state of the left and right Virasoro algebras.

The terminology can be slightly misleading. Modes $L_{-n}$ with $n>0$ raise the $L_0$ eigenvalue and generate descendants. Modes $L_n$ with $n>0$ lower the eigenvalue and annihilate a highest-weight state.

Preview: descendants

The first descendant of $|\phi\rangle$ is

$$L_{-1}|\phi\rangle=\partial\phi(0)|0\rangle.$$

At the next levels one finds states such as

$$L_{-2}|\phi\rangle, \qquad L_{-1}^2|\phi\rangle, \qquad L_{-3}|\phi\rangle, \qquad L_{-2}L_{-1}|\phi\rangle, \qquad L_{-1}^3|\phi\rangle.$$

These states form the beginning of a Virasoro module. The next page develops this representation theory carefully and explains when special combinations of descendants become null states.

What to remember

The essential dictionary is

$$\boxed{ \mathcal O(0,0) \quad\longleftrightarrow\quad |\mathcal O\rangle }$$

and for primary fields,

$$\boxed{ \text{primary operator of weights }(h,\bar h) \quad\longleftrightarrow\quad \text{highest-weight state with }(L_0,\bar L_0)=(h,\bar h). }$$

The cylinder-plane map $z=e^w$ explains why $L_0+\bar L_0-c/12$ is the cylinder Hamiltonian. The oscillator-operator dictionary explains why string excitations can be represented by vertex operators.

Exercises

Exercise 1. The global conformal subalgebra

Using the Virasoro algebra, show that $L_{-1}$, $L_0$, and $L_1$ close without a central term.

Solution

The central term is proportional to

$$m(m^2-1).$$

For $m=-1,0,1$, this factor vanishes. Therefore the central term is absent for commutators among $L_{-1},L_0,L_1$.

The nontrivial commutators are

$$[L_1,L_{-1}]=2L_0,$$ $$[L_0,L_1]=-L_1,$$

and

$$[L_0,L_{-1}]=L_{-1}.$$

Thus the three generators form an $sl(2)$ subalgebra.

Exercise 2. The Schwarzian of the exponential map

For $z=e^w$, compute $\{z,w\}$ and derive

$$T_{\rm cyl}(w)=z^2T_{\rm plane}(z)-\frac{c}{24}.$$

Solution

For $z=e^w$,

$$\frac{dz}{dw}=z, \qquad \frac{d^2z}{dw^2}=z, \qquad \frac{d^3z}{dw^3}=z.$$

Hence

$$\{z,w\} = \frac{z^{(3)}}{z'} - \frac{3}{2}\left(\frac{z''}{z'}\right)^2 = 1-\frac{3}{2} =-\frac{1}{2}.$$

Using

$$T_{\rm cyl}(w)=\left(\frac{dz}{dw}\right)^2T_{\rm plane}(z)+\frac{c}{12}\{z,w\},$$

we get

$$T_{\rm cyl}(w)=z^2T_{\rm plane}(z)-\frac{c}{24}.$$

Exercise 3. Primary state conditions

Let $\phi$ be a primary field of holomorphic weight $h$. Show that $|\phi\rangle=\phi(0)|0\rangle$ satisfies

$$L_0|\phi\rangle=h|\phi\rangle, \qquad L_n|\phi\rangle=0\quad n>0.$$

Solution

Use

$$L_n|\phi\rangle = \frac{1}{2\pi i}\oint_0 dz\,z^{n+1}T(z)\phi(0)|0\rangle.$$

The primary OPE is

$$T(z)\phi(0) \sim \frac{h\phi(0)}{z^2}+\frac{\partial\phi(0)}{z}.$$

Thus

$$z^{n+1}T(z)\phi(0) \sim h z^{n-1}\phi(0)+z^n\partial\phi(0).$$

The contour integral extracts the coefficient of $z^{-1}$. For $n=0$, the first term gives $h\phi(0)$, so $L_0|\phi\rangle=h|\phi\rangle$. For $n>0$, neither term contains $z^{-1}$, so $L_n|\phi\rangle=0$.

Exercise 4. Oscillator-operator dictionary

Use

$$\partial X^\mu(z) = -i\sqrt{\frac{\alpha'}{2}} \sum_{n\in\mathbb Z}\alpha_n^\mu z^{-n-1}$$

to identify the operator corresponding to $\alpha_{-m}^\mu|0;k\rangle$.

Solution

Act on the momentum state:

$$\partial X^\mu(z)|0;k\rangle = -i\sqrt{\frac{\alpha'}{2}} \left( \frac{\alpha_0^\mu}{z} +\alpha_{-1}^\mu +\alpha_{-2}^\mu z +\cdots +\alpha_{-m}^\mu z^{m-1} +\cdots \right)|0;k\rangle.$$

The coefficient of $z^{m-1}$ is

$$-i\sqrt{\frac{\alpha'}{2}}\alpha_{-m}^\mu|0;k\rangle.$$

But

$$\partial^m X^\mu(0) = \partial_z^{m-1}\partial X^\mu(z)\big|_{z=0},$$

so the coefficient is multiplied by $(m-1)!$. Therefore

$$\alpha_{-m}^\mu|0;k\rangle \longleftrightarrow \frac{i}{(m-1)!}\sqrt{\frac{2}{\alpha'}}\, \partial^m X^\mu(0):e^{ik\cdot X(0)}:.$$

Exercise 5. Transversality from primarity

Consider the holomorphic operator

$$\epsilon_\mu\partial X^\mu e^{ik\cdot X}.$$

Using the corresponding state $\epsilon_\mu\alpha_{-1}^\mu|0;k\rangle$, show that primarity imposes $\epsilon\cdot k=0$.

Solution

A primary state must be annihilated by $L_n$ for $n>0$, in particular by $L_1$. Using the oscillator Virasoro generator gives

$$L_1\epsilon_\mu\alpha_{-1}^\mu|0;k\rangle = \epsilon_\mu\alpha_0^\mu|0;k\rangle.$$

Since

$$\alpha_0^\mu|0;k\rangle = \sqrt{\frac{\alpha'}{2}}k^\mu|0;k\rangle,$$

we find

$$L_1\epsilon_\mu\alpha_{-1}^\mu|0;k\rangle = \sqrt{\frac{\alpha'}{2}}(\epsilon\cdot k)|0;k\rangle.$$

Primarity requires this to vanish, so

$$\epsilon\cdot k=0.$$