The OPE as Operator Algebra

The operator product expansion is where CFT stops being only a theory of symmetry constraints and becomes a theory of dynamics.

Conformal symmetry fixes two-point functions and three-point functions up to constants. It strongly constrains four-point functions, but it does not determine them. The OPE is the principle that organizes all higher-point functions in terms of two kinds of data:

$$\boxed{ \text{spectrum of primary operators} \quad + \quad \text{OPE coefficients} }$$

Together with consistency conditions such as crossing symmetry, this is the modern definition of a CFT. A CFT is not primarily a Lagrangian. It is a consistent local operator algebra.

The basic statement is that when two local operators approach each other, their product can be expanded in local operators at one point:

$$\boxed{ \mathcal O_i(x)\mathcal O_j(0) = \sum_k C_{ij}{}^k(x,\partial)\,\mathcal O_k(0). }$$

The coefficient $C_{ij}{}^k(x,\partial)$ is a differential operator. It contains singular powers of $x$ and a series of derivatives acting on $\mathcal O_k$. In a CFT, the sum may be reorganized into conformal families:

$$\mathcal O_i\times \mathcal O_j = \sum_{\mathcal O\;\text{primary}} \lambda_{ij\mathcal O}\,[\mathcal O],$$

where $[\mathcal O]$ denotes the primary $\mathcal O$ together with all of its descendants.

For AdS/CFT, this is the boundary version of the statement that bulk states can be decomposed into particles, multi-particle states, and their interactions. The spectrum of CFT primaries is the spectrum of bulk fields and multi-particle states. The OPE coefficients are the boundary data that, at large $N$, become bulk interaction vertices.

The OPE inside correlation functions

The OPE should first be understood inside correlation functions. Given additional insertions $\mathcal X_a(y_a)$, the statement is

$$\left\langle \mathcal O_i(x)\mathcal O_j(0)\prod_a \mathcal X_a(y_a) \right\rangle = \sum_k C_{ij}{}^k(x,\partial) \left\langle \mathcal O_k(0)\prod_a \mathcal X_a(y_a) \right\rangle.$$

In Euclidean radial quantization, this is not merely an asymptotic expansion. It is a convergent expansion when the two operators being expanded are closer to each other than to every other insertion. If the OPE is centered at the origin, the convergence condition is

$$|x| < \min_a |y_a|.$$

More geometrically, choose a sphere $S_R^{d-1}$ centered at the origin such that

$$|x|<R<|y_a| \qquad \text{for all external insertions }y_a.$$

Then the pair $\mathcal O_i(x)\mathcal O_j(0)$ creates a state on $S_R^{d-1}$. Expanding that state in a complete basis of dilatation eigenstates gives the OPE.

Radial quantization proof idea for the OPE. If $\mathcal O_i(x)$ and $\mathcal O_j(0)$ lie inside a sphere $S_R^{d-1}$ and all other insertions lie outside, the state created on $S_R^{d-1}$ can be expanded in a complete basis of conformal-family states. This gives a convergent expansion of $\mathcal O_i(x)\mathcal O_j(0)$ inside correlators for $|x|<R<|y_a|$.

This convergence is one of the quiet miracles of Euclidean CFT. It is much stronger than the usual short-distance asymptotic expansions in generic QFT. It is the reason the conformal bootstrap can be formulated as an exact consistency problem rather than as a perturbative approximation.

Why radial quantization gives an operator algebra

Recall the state-operator correspondence:

$$\mathcal O(0)|0\rangle \quad \longleftrightarrow \quad |\mathcal O\rangle.$$

Acting with two local operators inside a sphere gives a state:

$$|\Psi_{ij}(x)\rangle = \mathcal O_i(x)\mathcal O_j(0)|0\rangle.$$

Insert a complete set of states in radial quantization:

$$|\Psi_{ij}(x)\rangle = \sum_\alpha |\alpha\rangle\langle \alpha|\mathcal O_i(x)\mathcal O_j(0)|0\rangle.$$

The states $|\alpha\rangle$ decompose into conformal multiplets:

$$|\alpha\rangle \in \bigoplus_{\mathcal O\;\text{primary}} [\mathcal O].$$

Each conformal family is generated by repeatedly acting with $P_\mu$ on a primary:

$$|\mathcal O\rangle, \qquad P_\mu|\mathcal O\rangle, \qquad P_{\mu_1}P_{\mu_2}|\mathcal O\rangle, \qquad \ldots.$$

In position space this becomes a derivative expansion:

$$\mathcal O(0), \qquad \partial_\mu\mathcal O(0), \qquad \partial_{\mu_1}\partial_{\mu_2}\mathcal O(0), \qquad \ldots.$$

Thus the OPE has two nested structures:

$$\boxed{ \text{sum over primaries} \quad + \quad \text{descendant expansion inside each family}. }$$

The first structure is dynamical: which primaries appear, and with which OPE coefficients? The second structure is kinematical: once a primary appears, conformal symmetry fixes the relative contribution of all descendants.

CFT data and normalization conventions

Let $\mathcal O_i$ be primary operators. If the two-point functions are diagonal and normalized as

$$\langle \mathcal O_i(x)\mathcal O_j(0)\rangle = \frac{\delta_{ij}}{|x|^{2\Delta_i}},$$

then the scalar three-point function has the form

$$\langle \mathcal O_i(x_1)\mathcal O_j(x_2)\mathcal O_k(x_3) \rangle = \frac{\lambda_{ijk}} {|x_{12}|^{\Delta_i+\Delta_j-\Delta_k} |x_{23}|^{\Delta_j+\Delta_k-\Delta_i} |x_{13}|^{\Delta_i+\Delta_k-\Delta_j}}.$$

The constants $\lambda_{ijk}$ are the OPE coefficients in an orthonormal basis. In a non-orthonormal basis, the two-point function defines a metric on operator space:

$$\langle \mathcal O_i(x)\mathcal O_j(0)\rangle = \frac{G_{ij}}{|x|^{2\Delta_i}} \qquad \text{when }\Delta_i=\Delta_j,$$

and the OPE coefficient with an upper index is obtained by raising an index:

$$C_{ij}{}^k = \lambda_{ij\ell}G^{\ell k}.$$

This is why one often says that the CFT data are

$$\boxed{ \left\{\Delta_i,\rho_i,G_{ij},\lambda_{ijk}\right\}, }$$

where $\rho_i$ denotes the spin and global-symmetry representation. In a unit-normalized basis, $G_{ij}=\delta_{ij}$, and the data reduce to the spectrum and the three-point coefficients.

The leading scalar OPE

For two scalar primaries $\mathcal O_i$ and $\mathcal O_j$, suppose a scalar primary $\mathcal O$ of dimension $\Delta$ appears in their OPE. The leading terms are

$$\mathcal O_i(x)\mathcal O_j(0) \supset \lambda_{ij\mathcal O} |x|^{\Delta-\Delta_i-\Delta_j} \left[ \mathcal O(0) + \frac{\Delta+\Delta_i-\Delta_j}{2\Delta} \,x^\mu\partial_\mu\mathcal O(0) + \cdots \right].$$

The only dynamical number here is $\lambda_{ij\mathcal O}$. The coefficient of the first descendant is fixed by conformal symmetry. So are all higher descendant coefficients.

For a symmetric traceless spin-$\ell$ primary $\mathcal O_{\mu_1\cdots\mu_\ell}$, the leading tensor structure in the OPE of two scalar primaries is schematically

$$\mathcal O_i(x)\mathcal O_j(0) \supset \lambda_{ij\mathcal O} \frac{x^{\mu_1}\cdots x^{\mu_\ell}} {|x|^{\Delta_i+\Delta_j-\Delta+\ell}} \mathcal O_{\mu_1\cdots\mu_\ell}(0) + \text{descendants}.$$

The numerator supplies the spin. The denominator supplies the correct scaling dimension.

Identity, currents, and the stress tensor

Every unitary CFT has an identity operator $\mathbf 1$ with dimension

$$\Delta_{\mathbf 1}=0.$$

For a unit-normalized scalar primary $\phi$ of dimension $\Delta_\phi$, the identity contribution is

$$\phi(x)\phi(0) \supset \frac{\mathbf 1}{|x|^{2\Delta_\phi}}.$$

This is just the two-point function written as an OPE coefficient.

If the theory has a continuous global symmetry, there is a conserved current $J_\mu^a$ with

$$\Delta_J=d-1.$$

If the theory is local and has a stress tensor, there is a conserved symmetric tensor $T_{\mu\nu}$ with

$$\Delta_T=d.$$

The appearances of $J_\mu^a$ and $T_{\mu\nu}$ in OPEs are not arbitrary. Their coefficients are constrained by Ward identities. For example, the OPE of a current with a charged operator encodes the action of the symmetry generator, while the OPE of the stress tensor with a primary encodes translations, rotations, dilatations, and special conformal transformations.

Schematic examples are

$$J_\mu^a(x)\mathcal O_i(0) \sim \frac{x_\mu}{|x|^d}(T^a)_i{}^j\mathcal O_j(0)+\cdots,$$

and

$$T_{\mu\nu}(x)\mathcal O(0) \sim \text{singular terms fixed by }\Delta,\text{ spin, and Ward identities}.$$

The precise numerical coefficients depend on normalization conventions for $J_\mu^a$ and $T_{\mu\nu}$. The invariant statement is that conserved currents and the stress tensor are universal operators whose OPEs implement symmetry.

Selection rules

The OPE is constrained by all symmetries, not only conformal symmetry. If the CFT has a global symmetry group $G$, and operators transform in representations $R_i$, $R_j$, and $R_k$, then $\mathcal O_k$ can appear in $\mathcal O_i\times\mathcal O_j$ only if

$$R_k \subset R_i\otimes R_j.$$

Equivalently, the three-point coefficient must be a $G$-invariant tensor:

$$\lambda_{ijk}\in \operatorname{Hom}_G(R_i\otimes R_j\otimes R_k,\mathbf 1).$$

Spin and parity impose further rules. For example, in a parity-invariant theory, the OPE of two parity-even scalar operators contains only parity-even scalar primaries in its scalar sector. For identical scalar operators, Bose symmetry also constrains the allowed spins. In the OPE

$$\phi\times\phi,$$

only even-spin symmetric traceless primaries can appear when $\phi$ is a real bosonic scalar and no additional antisymmetric internal tensor is present.

These selection rules matter enormously in bootstrap and AdS/CFT applications. They determine which bulk fields are allowed to couple and which exchange diagrams can appear.

The OPE is not an ordinary algebra product

The phrase “operator algebra” is powerful but slightly dangerous. The OPE is not an ordinary finite-dimensional algebra multiplication.

First, the coefficients are functions or distributions of the separation:

$$C_{ij}{}^k=C_{ij}{}^k(x,\partial).$$

They are often singular as $x\to 0$.

Second, the OPE is a statement inside correlation functions, with a domain of convergence in Euclidean signature. The operator product $\mathcal O_i(x)\mathcal O_j(0)$ is not a well-defined local operator at coincident points without renormalization.

Third, contact terms can appear when operators collide under spacetime integrals. These contact terms are invisible in separated-point correlators but important for Ward identities, anomalies, and conformal perturbation theory.

Fourth, in Lorentzian signature the ordering matters. The Euclidean OPE gives radially ordered products. Lorentzian Wightman, time-ordered, retarded, and commutator correlators are obtained by analytic continuation with different $i\epsilon$ prescriptions. The same formal OPE data appear, but their Lorentzian interpretation depends on operator ordering and causal separation.

A safer slogan is therefore:

$$\boxed{ \text{The OPE is a convergent local expansion of products inside Euclidean correlators.} }$$

The operator-algebra interpretation comes from the fact that this local expansion is associative.

Associativity and crossing

The deepest property of the OPE is associativity. Consider four operators. We may first fuse $\mathcal O_1$ with $\mathcal O_2$, or first fuse $\mathcal O_2$ with $\mathcal O_3$, or use another channel. The final correlator must be the same.

Schematically,

$$(\mathcal O_1\mathcal O_2)\mathcal O_3 = \mathcal O_1(\mathcal O_2\mathcal O_3).$$

Inside a four-point function, this becomes crossing symmetry. For identical scalar operators $\phi$, write

$$\langle \phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle = \frac{1}{x_{12}^{2\Delta_\phi}x_{34}^{2\Delta_\phi}}\mathcal G(u,v),$$

where

$$u=\frac{x_{12}^2x_{34}^2}{x_{13}^2x_{24}^2}, \qquad v=\frac{x_{14}^2x_{23}^2}{x_{13}^2x_{24}^2}.$$

The $12\to 34$ OPE channel gives

$$\mathcal G(u,v) = \sum_{\mathcal O} \lambda_{\phi\phi\mathcal O}^2 G_{\Delta,\ell}(u,v),$$

where $G_{\Delta,\ell}(u,v)$ is the conformal block for the family $[\mathcal O]$. Exchanging $x_1\leftrightarrow x_3$ gives the crossing equation

$$\boxed{ \mathcal G(u,v) = \left(\frac{u}{v}\right)^{\Delta_\phi} \mathcal G(v,u). }$$

Equivalently,

$$\boxed{ \sum_{\mathcal O} \lambda_{\phi\phi\mathcal O}^2 G_{\Delta,\ell}(u,v) = \left(\frac{u}{v}\right)^{\Delta_\phi} \sum_{\mathcal O} \lambda_{\phi\phi\mathcal O}^2 G_{\Delta,\ell}(v,u). }$$

This equation is the seed of the conformal bootstrap. The next page explains the conformal blocks $G_{\Delta,\ell}$ that appear in this expansion.

Example: free scalar OPE

Let $\phi$ be a canonically normalized free scalar in $d>2$ with

$$\Delta_\phi=\frac{d-2}{2}, \qquad \langle \phi(x)\phi(0)\rangle=\frac{1}{|x|^{d-2}}.$$

Wick’s theorem gives

$$\phi(x)\phi(0) = \frac{\mathbf 1}{|x|^{d-2}} + :\!\phi(x)\phi(0)\!:.$$

Expanding the normal-ordered product near the origin gives

$$\phi(x)\phi(0) = \frac{\mathbf 1}{|x|^{d-2}} + :\!\phi^2\!:(0) + x^\mu :\!\partial_\mu\phi\,\phi\!:(0) +\cdots.$$

This example is simple but instructive. The singular identity term reproduces the two-point function. The regular terms contain composite local operators. In an interacting CFT, the same logic survives, but the composite operators are replaced by scaling operators with definite conformal dimensions.

Holographic interpretation

In AdS/CFT, the OPE is the boundary signature of bulk locality.

A single-trace primary $\mathcal O$ corresponds, at large $N$, to a single-particle bulk field $\Phi_{\mathcal O}$. A double-trace primary schematically of the form

$$[\mathcal O_1\mathcal O_2]_{n,\ell}$$

corresponds to a two-particle bulk state with radial excitation number $n$ and angular momentum $\ell$. At leading order in a large-$N$ CFT, its dimension is approximately

$$\Delta_{n,\ell} = \Delta_1+\Delta_2+2n+\ell +O\left(\frac{1}{N^2}\right).$$

The $O(1/N^2)$ correction is an anomalous dimension, interpreted in the bulk as an interaction energy.

The rough dictionary is

CFT OPE concept	AdS interpretation
primary operator $\mathcal O$	bulk particle species or multi-particle state
conformal family $[\mathcal O]$	descendants generated by boundary translations
OPE coefficient $\lambda_{ijk}$	bulk cubic coupling, after normalization
anomalous dimension	binding energy or interaction effect
crossing symmetry	consistency of bulk scattering, causality, and locality

This dictionary is not exact at finite $N$ or finite gap, but it is the correct intuition for why OPE data are the raw material of holography.

What to remember

The OPE is the central multiplication rule of a CFT:

$$\mathcal O_i\times\mathcal O_j = \sum_{\mathcal O}\lambda_{ij\mathcal O}[\mathcal O].$$

It is local, convergent in Euclidean radial quantization, and controlled by conformal representation theory. The primary spectrum and OPE coefficients determine all correlation functions, provided the OPE is associative. Associativity becomes crossing symmetry, and crossing symmetry becomes the bootstrap.

For AdS/CFT, the same data encode the bulk spectrum and interactions. Learning to read OPE data is learning to read the boundary imprint of quantum gravity in AdS.

Exercises

Exercise 1: Identity contribution

Let $\phi$ be a unit-normalized scalar primary of dimension $\Delta_\phi$:

$$\langle \phi(x)\phi(0)\rangle=\frac{1}{|x|^{2\Delta_\phi}}.$$

Show that the identity contribution to the OPE is

$$\phi(x)\phi(0)\supset \frac{\mathbf 1}{|x|^{2\Delta_\phi}}.$$

Solution

Take the vacuum expectation value of the OPE:

$$\langle \phi(x)\phi(0)\rangle = \sum_k C_{\phi\phi}{}^k(x,\partial)\langle \mathcal O_k(0)\rangle.$$

In the CFT vacuum on flat space, the only operator with nonzero one-point function is the identity:

$$\langle \mathbf 1\rangle=1, \qquad \langle \mathcal O_k\rangle=0 \quad \text{for non-identity primaries.}$$

Therefore the coefficient of $\mathbf 1$ must reproduce the two-point function:

$$C_{\phi\phi}{}^{\mathbf 1}(x) = \frac{1}{|x|^{2\Delta_\phi}}.$$

Thus

$$\phi(x)\phi(0) \supset \frac{\mathbf 1}{|x|^{2\Delta_\phi}}.$$

Exercise 2: First descendant coefficient

Suppose two scalar primaries $\mathcal O_i$ and $\mathcal O_j$ have an OPE contribution from a scalar primary $\mathcal O$ of dimension $\Delta$:

$$\mathcal O_i(x)\mathcal O_j(0) \supset \lambda_{ij\mathcal O}|x|^{\Delta-\Delta_i-\Delta_j} \left[\mathcal O(0)+a\,x^\mu\partial_\mu\mathcal O(0)+\cdots\right].$$

Use the scalar three-point function to show that

$$a=\frac{\Delta+\Delta_i-\Delta_j}{2\Delta}.$$

Solution

Take the three-point function with $\mathcal O(y)$ and assume

$$|x|\ll |y|.$$

Using the OPE and the unit-normalized two-point function,

$$\langle \mathcal O(0)\mathcal O(y)\rangle=\frac{1}{|y|^{2\Delta}},$$

we get

$$\langle \mathcal O_i(x)\mathcal O_j(0)\mathcal O(y)\rangle \supset \lambda_{ij\mathcal O}|x|^{\Delta-\Delta_i-\Delta_j} \left[ \frac{1}{|y|^{2\Delta}} +a x^\mu\partial_{0\mu}\frac{1}{|y|^{2\Delta}} +\cdots \right].$$

Since

$$\partial_{0\mu}\frac{1}{|0-y|^{2\Delta}} = \frac{2\Delta y_\mu}{|y|^{2\Delta+2}},$$

this gives

$$\lambda_{ij\mathcal O}|x|^{\Delta-\Delta_i-\Delta_j} \frac{1}{|y|^{2\Delta}} \left[ 1+2\Delta a\frac{x\cdot y}{y^2}+\cdots \right].$$

On the other hand, the exact three-point function is

$$\frac{\lambda_{ij\mathcal O}} {|x|^{\Delta_i+\Delta_j-\Delta} |y|^{\Delta_j+\Delta-\Delta_i} |y-x|^{\Delta_i+\Delta-\Delta_j}}.$$

Expanding

$$|y-x|^{-(\Delta_i+\Delta-\Delta_j)} = |y|^{-(\Delta_i+\Delta-\Delta_j)} \left[ 1+(\Delta_i+\Delta-\Delta_j)\frac{x\cdot y}{y^2}+\cdots \right],$$

we find

$$2\Delta a=\Delta_i+\Delta-\Delta_j.$$

Therefore

$$\boxed{ a=\frac{\Delta+\Delta_i-\Delta_j}{2\Delta}. }$$

Exercise 3: Convergence domain

Consider

$$\left\langle \mathcal O_i(x)\mathcal O_j(0)\prod_a\mathcal X_a(y_a) \right\rangle.$$

Use radial quantization around the origin to explain why the OPE of $\mathcal O_i(x)\mathcal O_j(0)$ converges when

$$|x|<\min_a |y_a|.$$

Solution

Choose a sphere $S_R^{d-1}$ centered at the origin such that

$$|x|<R<|y_a| \qquad \text{for every }a.$$

The two operators $\mathcal O_i(x)$ and $\mathcal O_j(0)$ lie inside the sphere, while all other insertions lie outside. In radial quantization, the inside insertions create a state on the sphere:

$$|\Psi\rangle = \mathcal O_i(x)\mathcal O_j(0)|0\rangle.$$

The Hilbert space on $S_R^{d-1}$ has a complete basis of dilatation eigenstates. Expanding $|\Psi\rangle$ in that basis gives a sum over conformal families. Since the outside insertions only test this state from outside the sphere, the resulting expansion gives the same correlator.

The expansion parameter is controlled by the ratio between the size of the inner configuration and the radius to the nearest outside insertion. Therefore it converges when the inner separation is smaller than the distance to every external insertion:

$$|x|<\min_a |y_a|.$$

Exercise 4: Crossing from OPE associativity

For identical scalar primaries $\phi$ of dimension $\Delta_\phi$, define

$$\langle \phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle = \frac{1}{x_{12}^{2\Delta_\phi}x_{34}^{2\Delta_\phi}}\mathcal G(u,v).$$

Show that exchanging $x_1\leftrightarrow x_3$ implies

$$\mathcal G(u,v) = \left(\frac{u}{v}\right)^{\Delta_\phi}\mathcal G(v,u).$$

Solution

The same four-point function can also be written after exchanging $x_1$ and $x_3$:

$$\langle \phi(x_3)\phi(x_2)\phi(x_1)\phi(x_4)\rangle = \frac{1}{x_{32}^{2\Delta_\phi}x_{14}^{2\Delta_\phi}}\mathcal G(v,u),$$

because the exchange $x_1\leftrightarrow x_3$ swaps the cross-ratios:

$$u \leftrightarrow v.$$

The fields are identical bosonic scalars, so the correlator is unchanged by the exchange. Therefore

$$\frac{1}{x_{12}^{2\Delta_\phi}x_{34}^{2\Delta_\phi}}\mathcal G(u,v) = \frac{1}{x_{23}^{2\Delta_\phi}x_{14}^{2\Delta_\phi}}\mathcal G(v,u).$$

Multiplying by $x_{12}^{2\Delta_\phi}x_{34}^{2\Delta_\phi}$ gives

$$\mathcal G(u,v) = \left(\frac{x_{12}^2x_{34}^2}{x_{14}^2x_{23}^2}\right)^{\Delta_\phi} \mathcal G(v,u).$$

Using

$$\frac{u}{v} = \frac{x_{12}^2x_{34}^2}{x_{14}^2x_{23}^2},$$

we obtain

$$\boxed{ \mathcal G(u,v) = \left(\frac{u}{v}\right)^{\Delta_\phi}\mathcal G(v,u). }$$

Exercise 5: Generalized free field preview

A generalized free scalar $\mathcal O$ has a two-point function

$$\langle \mathcal O(x)\mathcal O(0)\rangle=\frac{1}{|x|^{2\Delta}},$$

and its four-point function is defined by Wick-like factorization:

$$\langle \mathcal O_1\mathcal O_2\mathcal O_3\mathcal O_4\rangle = \langle \mathcal O_1\mathcal O_2\rangle\langle \mathcal O_3\mathcal O_4\rangle + \langle \mathcal O_1\mathcal O_3\rangle\langle \mathcal O_2\mathcal O_4\rangle + \langle \mathcal O_1\mathcal O_4\rangle\langle \mathcal O_2\mathcal O_3\rangle.$$

Show that

$$\mathcal G(u,v)=1+u^\Delta+\left(\frac{u}{v}\right)^\Delta.$$

What does this suggest about the OPE spectrum?

Solution

By definition,

$$\langle \mathcal O_1\mathcal O_2\mathcal O_3\mathcal O_4\rangle = \frac{1}{x_{12}^{2\Delta}x_{34}^{2\Delta}}\mathcal G(u,v).$$

The three Wick contractions are

$$\frac{1}{x_{12}^{2\Delta}x_{34}^{2\Delta}}, \qquad \frac{1}{x_{13}^{2\Delta}x_{24}^{2\Delta}}, \qquad \frac{1}{x_{14}^{2\Delta}x_{23}^{2\Delta}}.$$

Factoring out $x_{12}^{-2\Delta}x_{34}^{-2\Delta}$ gives

$$\mathcal G(u,v) = 1+ \left(\frac{x_{12}^2x_{34}^2}{x_{13}^2x_{24}^2}\right)^\Delta + \left(\frac{x_{12}^2x_{34}^2}{x_{14}^2x_{23}^2}\right)^\Delta.$$

Using the cross-ratios,

$$\boxed{ \mathcal G(u,v)=1+u^\Delta+\left(\frac{u}{v}\right)^\Delta. }$$

The identity operator accounts for the $1$ in the $12\to 34$ OPE channel. The remaining terms require an infinite tower of double-trace operators, schematically

$$[\mathcal O\mathcal O]_{n,\ell}, \qquad \Delta_{n,\ell}=2\Delta+2n+\ell,$$

with even spin $\ell$ for identical real scalars. In holography, this is the leading large-$N$ spectrum of two-particle states in AdS.

Further reading

For the classic two-dimensional operator-algebra viewpoint, read Di Francesco, Mathieu, and Sénéchal, especially the parts on radial ordering, OPE, conformal families, conformal blocks, and crossing symmetry. For the higher-dimensional bootstrap viewpoint, read Rychkov’s EPFL Lectures on Conformal Field Theory in $D\ge 3$ and Simmons-Duffin’s TASI Lectures on the Conformal Bootstrap. For the AdS/CFT connection, keep this page in mind before studying Witten diagrams, large-$N$ factorization, and double-trace operators.