Scalar Two- and Three-Point Functions

Correlation functions are where CFT becomes concrete. The previous module developed the language of primaries, stress tensors, sources, and Ward identities. We now begin using that machinery to determine actual correlators.

The first miracle is that conformal symmetry almost completely fixes the simplest scalar correlators. For scalar primary operators in flat space,

$$\langle \mathcal O_1(x_1)\mathcal O_2(x_2)\rangle$$

is fixed up to a normalization matrix, and

$$\langle \mathcal O_1(x_1)\mathcal O_2(x_2)\mathcal O_3(x_3)\rangle$$

is fixed up to one constant, or a finite set of constants if global-symmetry tensor structures are present. These constants are not decorative. After choosing a normalization for two-point functions, the three-point constants are the first genuinely dynamical entries of the CFT data.

For AdS/CFT, this page is one of the first places where the dictionary becomes visible:

$$\boxed{ \begin{aligned} \Delta_i &\longleftrightarrow \text{bulk mass of }\phi_i,\\ \langle \mathcal O_i\mathcal O_j\rangle &\longleftrightarrow \text{bulk quadratic action},\\ C_{ijk} &\longleftrightarrow \text{bulk cubic couplings and Witten diagrams}. \end{aligned} }$$

The important lesson is not merely that symmetry fixes powers of distances. The deeper lesson is this: a CFT is an operator algebra, and two- and three-point functions are the first entries of that algebra.

Setup and notation

We work in flat Euclidean space $\mathbb R^d$ unless otherwise stated. Write

$$x_{ij}^\mu=x_i^\mu-x_j^\mu, \qquad x_{ij}^2=\delta_{\mu\nu}x_{ij}^\mu x_{ij}^\nu, \qquad |x_{ij}|=(x_{ij}^2)^{1/2}.$$

The operators $\mathcal O_i$ are scalar primaries with scaling dimensions $\Delta_i$. Under an infinitesimal conformal transformation generated by a conformal Killing vector $\xi^\mu(x)$,

$$\partial_\mu\xi_\nu+\partial_\nu\xi_\mu = 2\sigma_\xi(x)\delta_{\mu\nu}, \qquad \sigma_\xi(x)=\frac1d\partial_\mu\xi^\mu,$$

a scalar primary transforms as

$$\delta_\xi \mathcal O_i(x) = - \left[ \xi^\mu(x)\partial_\mu+ \Delta_i\sigma_\xi(x) \right] \mathcal O_i(x).$$

Equivalently, under a finite conformal map $x\mapsto x'$ satisfying

$$\frac{\partial x'^\rho}{\partial x^\mu} \frac{\partial x'^\sigma}{\partial x^\nu} \delta_{\rho\sigma} = \Omega(x)^2\delta_{\mu\nu},$$

correlators of scalar primaries transform covariantly as

$$\boxed{ \left\langle \prod_{i=1}^n \mathcal O_i(x_i') \right\rangle = \prod_{i=1}^n \Omega(x_i)^{-\Delta_i} \left\langle \prod_{i=1}^n \mathcal O_i(x_i) \right\rangle . }$$

For a dilatation $x'=\lambda x$, one has $\Omega=\lambda$, so this becomes the familiar scaling law

$$\left\langle \prod_{i=1}^n \mathcal O_i(\lambda x_i) \right\rangle = \lambda^{-\sum_i\Delta_i} \left\langle \prod_{i=1}^n \mathcal O_i(x_i) \right\rangle .$$

All formulas below are statements about separated points. Contact terms supported at $x_i=x_j$ are important in Ward identities and holographic renormalization, but they do not affect the separated-point power laws.

Differential constraints from conformal Ward identities

For a scalar $n$-point function

$$G_n(x_1,\ldots,x_n) = \left\langle \mathcal O_1(x_1)\cdots \mathcal O_n(x_n) \right\rangle,$$

conformal invariance gives differential equations. They are often the fastest way to see why the answer is so rigid.

Translations give

$$\sum_{i=1}^n \frac{\partial}{\partial x_i^\mu}G_n=0.$$

Rotations give

$$\sum_{i=1}^n \left( x_i^\mu\frac{\partial}{\partial x_i^\nu} - x_i^\nu\frac{\partial}{\partial x_i^\mu} \right)G_n=0.$$

Dilatations give

$$\left[ \sum_{i=1}^n x_i^\mu\frac{\partial}{\partial x_i^\mu} + \sum_{i=1}^n\Delta_i \right]G_n=0.$$

Special conformal transformations give

$$\sum_{i=1}^n \left[ 2x_i^\mu x_i^\nu\frac{\partial}{\partial x_i^\nu} -x_i^2\frac{\partial}{\partial x_{i\mu}} +2\Delta_i x_i^\mu \right]G_n=0.$$

The first two equations say that scalar correlators depend only on relative distances. The last two impose scaling weights and special-conformal covariance.

A useful counting slogan is:

$$\boxed{ \text{there are no conformal cross-ratios for }n\leq 3. }$$

That is why scalar one-, two-, and three-point functions are fixed up to constants. The first truly functional dependence appears at four points.

One-point functions in flat space

Let $\mathcal O$ be a scalar primary of dimension $\Delta$. Translation invariance says

$$\langle \mathcal O(x)\rangle=\text{constant}.$$

Dilatation invariance then requires

$$\langle \mathcal O(\lambda x)\rangle = \lambda^{-\Delta}\langle \mathcal O(x)\rangle .$$

A nonzero constant can obey this only if $\Delta=0$. In a unitary CFT, the only scalar primary of dimension $0$ is usually the identity operator $\mathbb 1$. Thus, for non-identity primaries on flat space,

$$\boxed{ \langle \mathcal O(x)\rangle=0. }$$

This statement is special to the vacuum on flat space. On a sphere, at finite temperature, in the presence of a boundary, or in a state created by another operator, scalar one-point functions can be nonzero.

Two-point functions

Consider two scalar primaries $\mathcal O_1$ and $\mathcal O_2$. Translation and rotation invariance imply

$$\langle \mathcal O_1(x_1)\mathcal O_2(x_2)\rangle =f(|x_{12}|).$$

Dilatation invariance gives

$$f(\lambda r)=\lambda^{-(\Delta_1+\Delta_2)}f(r),$$

so

$$\langle \mathcal O_1(x_1)\mathcal O_2(x_2)\rangle = \frac{C_{12}}{|x_{12}|^{\Delta_1+\Delta_2}}.$$

This is not yet the final answer. Special conformal invariance imposes an extra condition. It is enough to test inversion,

$$x^\mu\mapsto x'^\mu=\frac{x^\mu}{x^2}.$$

Under inversion,

$$|x_{12}'| = \frac{|x_{12}|}{|x_1||x_2|}, \qquad \Omega(x)=\frac1{x^2}.$$

The candidate two-point function transforms as

$$\frac{1}{|x_{12}'|^{\Delta_1+\Delta_2}} = \frac{|x_1|^{\Delta_1+\Delta_2}|x_2|^{\Delta_1+\Delta_2}} {|x_{12}|^{\Delta_1+\Delta_2}}.$$

But conformal covariance demands the factor

$$\Omega(x_1)^{-\Delta_1}\Omega(x_2)^{-\Delta_2} = |x_1|^{2\Delta_1}|x_2|^{2\Delta_2}.$$

These agree for arbitrary $x_1,x_2$ only if

$$\Delta_1=\Delta_2.$$

Therefore

$$\boxed{ \left\langle \mathcal O_1(x_1)\mathcal O_2(x_2) \right\rangle = \begin{cases} \dfrac{C_{12}}{|x_{12}|^{2\Delta_1}}, & \Delta_1=\Delta_2,\\ 0, & \Delta_1\neq \Delta_2. \end{cases} }$$

If there are several scalar primaries with the same dimension and the same global-symmetry quantum numbers, the constants $C_{12}$ form a matrix, often called the two-point metric on the space of operators of that dimension.

In a unitary theory, for Hermitian scalar primaries, reflection positivity makes this two-point metric positive definite. We can therefore choose an orthonormal basis:

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

This normalization is convenient but not compulsory. The transformation

$$\mathcal O_i\mapsto \lambda_i\mathcal O_i$$

rescales correlators. Consequently, the numerical value of a three-point coefficient is meaningful only after a convention for two-point normalization has been chosen.

Three-point functions

Now consider three scalar primaries. Translation and rotation invariance imply that the answer can depend only on the three pairwise distances:

$$|x_{12}|, \qquad |x_{23}|, \qquad |x_{31}|.$$

Try the most general power-law form compatible with scale invariance:

$$G_3(x_1,x_2,x_3) = \frac{C_{123}} {|x_{12}|^{\alpha_{12}} |x_{23}|^{\alpha_{23}} |x_{31}|^{\alpha_{31}}}.$$

Dilatation invariance gives only one equation,

$$\alpha_{12}+\alpha_{23}+\alpha_{31} = \Delta_1+\Delta_2+\Delta_3.$$

Special conformal invariance supplies the rest. Again use inversion. Since

$$|x_{ij}'|=\frac{|x_{ij}|}{|x_i||x_j|},$$

the power-law ansatz transforms with the factors

$$|x_1|^{\alpha_{12}+\alpha_{31}} |x_2|^{\alpha_{12}+\alpha_{23}} |x_3|^{\alpha_{23}+\alpha_{31}}.$$

Conformal covariance demands instead

$$|x_1|^{2\Delta_1}|x_2|^{2\Delta_2}|x_3|^{2\Delta_3}.$$

Thus

$$\alpha_{12}+\alpha_{31}=2\Delta_1, \qquad \alpha_{12}+\alpha_{23}=2\Delta_2, \qquad \alpha_{23}+\alpha_{31}=2\Delta_3.$$

Solving gives

$$\alpha_{12}=\Delta_1+\Delta_2-\Delta_3,$$ $$\alpha_{23}=\Delta_2+\Delta_3-\Delta_1,$$ $$\alpha_{31}=\Delta_3+\Delta_1-\Delta_2.$$

Therefore the scalar three-point function is

$$\boxed{ \left\langle \mathcal O_1(x_1) \mathcal O_2(x_2) \mathcal O_3(x_3) \right\rangle = \frac{C_{123}} {|x_{12}|^{\Delta_1+\Delta_2-\Delta_3} |x_{23}|^{\Delta_2+\Delta_3-\Delta_1} |x_{31}|^{\Delta_3+\Delta_1-\Delta_2}}. }$$

For three scalar primaries, conformal symmetry fixes the correlator to be a product of powers of the three pairwise distances. The powers are determined by matching the scaling weight $\Delta_i$ at each insertion. The only remaining information is the coefficient $C_{123}$.

A few comments are worth making.

First, the exponents need not all be positive. If, for example, $\Delta_3>\Delta_1+\Delta_2$, then $\Delta_1+\Delta_2-\Delta_3$ is negative, and the corresponding factor appears in the numerator. This is not a problem at separated points.

Second, conformal symmetry fixes the shape but not the coefficient $C_{123}$. Different CFTs can have the same spectrum of dimensions but different three-point coefficients.

Third, for identical scalar primaries $\mathcal O$ of dimension $\Delta$,

$$\left\langle \mathcal O(x_1)\mathcal O(x_2)\mathcal O(x_3) \right\rangle = \frac{C_{\mathcal O\mathcal O\mathcal O}} {|x_{12}|^{\Delta}|x_{23}|^{\Delta}|x_{31}|^{\Delta}}.$$

If the theory has a $\mathbb Z_2$ symmetry under which $\mathcal O\mapsto -\mathcal O$, then $C_{\mathcal O\mathcal O\mathcal O}=0$.

The three-point coefficient is OPE data

Once the two-point functions are normalized, the constants $C_{ijk}$ are the same constants that appear in the operator product expansion.

For scalar primaries in an orthonormal basis,

$$\mathcal O_i(x)\mathcal O_j(0) \sim \sum_k \frac{C_{ijk}}{|x|^{\Delta_i+\Delta_j-\Delta_k}} \mathcal O_k(0) +\text{descendants}.$$

The descendant terms are fixed by conformal symmetry once the primary term is known. More precisely, conformal symmetry determines a tower of derivative corrections of the schematic form

$$\mathcal O_i(x)\mathcal O_j(0) \sim \sum_k C_{ijk}|x|^{\Delta_k-\Delta_i-\Delta_j} \left[ \mathcal O_k(0) + \text{fixed powers of }x^\mu\partial_\mu\mathcal O_k(0) +\cdots \right].$$

Taking the expectation value with $\mathcal O_k(x_3)$ and using the two-point function reproduces the three-point function above. This is why the slogan

$$\boxed{ \text{CFT data}= \left\{\Delta_i,\ell_i,\text{global reps},C_{ijk}\right\} }$$

starts with spectra and three-point coefficients.

In a non-orthonormal basis, the OPE coefficient with one index raised is

$$C_{ij}{}^k=C_{ij\ell}G^{\ell k},$$

where

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

inside a subspace of equal dimension and compatible quantum numbers.

Global symmetries and selection rules

If the CFT has a global symmetry group $G$, local operators transform in representations of $G$. Scalar here means scalar under spacetime rotations; it does not mean singlet under internal symmetry.

Suppose $\mathcal O_i^{a_i}$ carries a global-symmetry index $a_i$. The two-point function has the form

$$\left\langle \mathcal O_i^{a}(x)\mathcal O_j^{b}(0) \right\rangle = \frac{G_{ij}\,\mathcal I^{ab}} {|x|^{2\Delta_i}},$$

where $\mathcal I^{ab}$ is an invariant tensor pairing the representation of $\mathcal O_i$ with that of $\mathcal O_j$. If no invariant pairing exists, the two-point function vanishes.

Similarly, the three-point function becomes

$$\left\langle \mathcal O_1^{a_1}(x_1) \mathcal O_2^{a_2}(x_2) \mathcal O_3^{a_3}(x_3) \right\rangle = \sum_A \frac{C_{123}^{(A)}\,\mathcal I_A^{a_1a_2a_3}} {|x_{12}|^{\Delta_1+\Delta_2-\Delta_3} |x_{23}|^{\Delta_2+\Delta_3-\Delta_1} |x_{31}|^{\Delta_3+\Delta_1-\Delta_2}}.$$

Here $\mathcal I_A^{a_1a_2a_3}$ runs over independent invariant tensors in the tensor product of the three representations. The constants $C_{123}^{(A)}$ are independent CFT data.

For example, if a $\mathbb Z_2$ symmetry assigns parities $p_i=\pm1$ to scalar primaries, then

$$C_{123}=0 \qquad \text{unless} \qquad p_1p_2p_3=+1.$$

This is the conformal-field-theory version of an ordinary selection rule.

Momentum-space two-point functions

Position-space formulas are simplest, but AdS/CFT calculations often produce momentum-space correlators after evaluating the on-shell bulk action. The Fourier transform of a scalar two-point function is fixed by scaling.

For separated points,

$$G(x)=\frac{1}{(x^2)^\Delta}.$$

As a distribution, its Fourier transform is

$$\int d^d x\,e^{ip\cdot x}\frac{1}{(x^2)^\Delta} = \pi^{d/2}2^{d-2\Delta} \frac{\Gamma\left(\frac d2-\Delta\right)}{\Gamma(\Delta)} (p^2)^{\Delta-d/2},$$

up to contact terms and analytic continuation in $\Delta$. The scaling is the main point:

$$\widetilde G(p)\propto (p^2)^{\Delta-d/2}.$$

When $\Delta-d/2$ is a nonnegative integer, the transform develops logarithmic behavior after renormalization:

$$\widetilde G(p) \sim (p^2)^{\Delta-d/2}\log p^2 + \text{local terms}.$$

Those local terms are scheme-dependent contact terms. In holography, this is the boundary manifestation of holographic counterterms and, in even dimensions, conformal anomalies.

AdS/CFT interpretation

A scalar primary $\mathcal O_i$ in a $d$-dimensional CFT is dual, in the simplest case, to a scalar bulk field $\phi_i$ in $\mathrm{AdS}_{d+1}$. Near the boundary in Poincare coordinates,

$$ds^2=\frac{L^2}{z^2}\left(dz^2+dx^\mu dx_\mu\right), \qquad z\to0,$$

the scalar behaves schematically as

$$\phi_i(z,x) \sim z^{d-\Delta_i}J_i(x) + z^{\Delta_i}A_i(x).$$

The coefficient $J_i(x)$ is the source for $\mathcal O_i$. The dimension is related to the bulk mass by

$$\boxed{ m_i^2L^2=\Delta_i(\Delta_i-d). }$$

Equivalently,

$$\Delta_\pm =\frac d2\pm \sqrt{\frac{d^2}{4}+m_i^2L^2}.$$

The standard quantization usually takes $\Delta=\Delta_+$. In the Breitenlohner-Freedman window, alternate quantization may take $\Delta=\Delta_-$.

The CFT two-point function comes from the quadratic bulk action. Roughly,

$$S_{\text{bulk}}^{(2)} \sim \frac12\int \phi_i\mathcal K_{ij}\phi_j \quad\Longrightarrow\quad \langle \mathcal O_i\mathcal O_j\rangle.$$

The power $|x|^{-2\Delta_i}$ is fixed by boundary conformal symmetry. The coefficient depends on the normalization of the bulk kinetic term and on the normalization chosen for $\mathcal O_i$.

The CFT three-point function comes from cubic bulk interactions. A term such as

$$S_{\text{bulk}}^{(3)} \supset \int_{\mathrm{AdS}} d^{d+1}X\sqrt g\, g_{ijk}\phi_i\phi_j\phi_k$$

produces a Witten diagram whose boundary dependence is precisely

$$\frac{1} {|x_{12}|^{\Delta_i+\Delta_j-\Delta_k} |x_{23}|^{\Delta_j+\Delta_k-\Delta_i} |x_{31}|^{\Delta_k+\Delta_i-\Delta_j}}.$$

The coefficient is proportional to the bulk cubic coupling $g_{ijk}$, multiplied by known normalization factors from bulk-to-boundary propagators. Thus, for scalar fields,

$$\boxed{ \text{bulk masses determine }\Delta_i, \qquad \text{bulk cubic vertices determine }C_{ijk}. }$$

This is the cleanest first example of how local bulk dynamics is encoded in boundary CFT data.

What conformal symmetry does not fix

It is tempting to think that conformal symmetry solves the whole theory. It does not.

For scalar primaries, conformal symmetry fixes:

$$\langle \mathcal O_i\rangle, \qquad \langle \mathcal O_i\mathcal O_j\rangle, \qquad \langle \mathcal O_i\mathcal O_j\mathcal O_k\rangle$$

up to constants and tensor structures.

But at four points, conformal cross-ratios appear. For four points in $d\geq2$, define

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

A scalar four-point function has the schematic form

$$\langle \mathcal O_1\mathcal O_2\mathcal O_3\mathcal O_4\rangle = \text{fixed powers of }x_{ij} \times \mathcal G(u,v),$$

where the function $\mathcal G(u,v)$ is dynamical. The next pages build toward the OPE and conformal-block decomposition of this function.

Common pitfalls

The most common mistake is to forget normalization dependence. If

$$\mathcal O_i\mapsto \lambda_i\mathcal O_i,$$

then

$$C_{ijk}\mapsto \lambda_i\lambda_j\lambda_k C_{ijk}.$$

So the raw number $C_{ijk}$ is meaningful only after two-point functions have been fixed.

A second common mistake is to confuse vanishing by conformal symmetry with vanishing by internal symmetry. Conformal symmetry says two scalar primaries of different dimensions have zero two-point function. Internal symmetry can force additional vanishings, including three-point selection rules.

A third common mistake is to ignore contact terms. At separated points,

$$\langle \mathcal O(x)\mathcal O(0)\rangle \propto |x|^{-2\Delta}.$$

As a distribution, this expression may require regularization at $x=0$, and local contact terms can be added. These contact terms matter in momentum space and in curved-space Ward identities.

Exercises

Exercise 1. Derive the scalar two-point function

Let $\mathcal O_1$ and $\mathcal O_2$ be scalar primaries of dimensions $\Delta_1$ and $\Delta_2$. Use translations, rotations, dilatations, and inversion to derive the separated-point two-point function.

Solution

Translations and rotations imply

$$G_2(x_1,x_2)=f(r), \qquad r=|x_{12}|.$$

Dilatations imply

$$f(\lambda r)=\lambda^{-(\Delta_1+\Delta_2)}f(r),$$

hence

$$f(r)=\frac{C_{12}}{r^{\Delta_1+\Delta_2}}.$$

Under inversion,

$$r'=\frac{r}{|x_1||x_2|}, \qquad \Omega(x_i)=\frac1{x_i^2}.$$

The candidate transforms as

$$G_2(x_1',x_2') = \frac{C_{12}|x_1|^{\Delta_1+\Delta_2}|x_2|^{\Delta_1+\Delta_2}} {r^{\Delta_1+\Delta_2}}.$$

Conformal covariance demands

$$G_2(x_1',x_2') = |x_1|^{2\Delta_1}|x_2|^{2\Delta_2}G_2(x_1,x_2).$$

This requires $\Delta_1=\Delta_2$ unless $C_{12}=0$. Therefore

$$G_2(x_1,x_2) = \begin{cases} \dfrac{C_{12}}{|x_{12}|^{2\Delta_1}}, & \Delta_1=\Delta_2,\\ 0, & \Delta_1\neq\Delta_2. \end{cases}$$

Exercise 2. Fix the scalar three-point exponents

Assume

$$G_3= \frac{C_{123}} {|x_{12}|^{\alpha_{12}}|x_{23}|^{\alpha_{23}}|x_{31}|^{\alpha_{31}}}.$$

Use inversion to show that

$$\alpha_{12}=\Delta_1+\Delta_2-\Delta_3, \qquad \alpha_{23}=\Delta_2+\Delta_3-\Delta_1, \qquad \alpha_{31}=\Delta_3+\Delta_1-\Delta_2.$$

Solution

Under inversion,

$$|x_{ij}'|=\frac{|x_{ij}|}{|x_i||x_j|}.$$

Therefore the ansatz transforms with the factor

$$|x_1|^{\alpha_{12}+\alpha_{31}} |x_2|^{\alpha_{12}+\alpha_{23}} |x_3|^{\alpha_{23}+\alpha_{31}}.$$

Conformal covariance requires

$$|x_1|^{2\Delta_1}|x_2|^{2\Delta_2}|x_3|^{2\Delta_3}.$$

Thus

$$\alpha_{12}+\alpha_{31}=2\Delta_1,$$ $$\alpha_{12}+\alpha_{23}=2\Delta_2,$$ $$\alpha_{23}+\alpha_{31}=2\Delta_3.$$

Solving these three linear equations gives

$$\alpha_{12}=\Delta_1+\Delta_2-\Delta_3,$$ $$\alpha_{23}=\Delta_2+\Delta_3-\Delta_1,$$ $$\alpha_{31}=\Delta_3+\Delta_1-\Delta_2.$$

Exercise 3. Recover the leading scalar OPE coefficient

Assume an orthonormal scalar basis and the leading OPE

$$\mathcal O_i(x)\mathcal O_j(0) \sim \sum_k A_{ij}{}^k |x|^{\Delta_k-\Delta_i-\Delta_j}\mathcal O_k(0).$$

Show that $A_{ij}{}^k=C_{ijk}$.

Solution

Insert $\mathcal O_k(y)$ with $|x|\ll |y|$:

$$\left\langle \mathcal O_i(x)\mathcal O_j(0)\mathcal O_k(y) \right\rangle \sim A_{ij}{}^k |x|^{\Delta_k-\Delta_i-\Delta_j} \left\langle \mathcal O_k(0)\mathcal O_k(y) \right\rangle.$$

With orthonormal two-point functions,

$$\left\langle \mathcal O_k(0)\mathcal O_k(y) \right\rangle =\frac1{|y|^{2\Delta_k}}.$$

So the OPE predicts

$$\left\langle \mathcal O_i(x)\mathcal O_j(0)\mathcal O_k(y) \right\rangle \sim A_{ij}{}^k \frac{|x|^{\Delta_k-\Delta_i-\Delta_j}} {|y|^{2\Delta_k}}.$$

The exact three-point function is

$$\frac{C_{ijk}} {|x|^{\Delta_i+\Delta_j-\Delta_k} |y|^{\Delta_j+\Delta_k-\Delta_i} |y-x|^{\Delta_k+\Delta_i-\Delta_j}}.$$

For $|x|\ll |y|$, $|y-x|\sim |y|$, hence

$$G_3\sim C_{ijk} \frac{|x|^{\Delta_k-\Delta_i-\Delta_j}} {|y|^{2\Delta_k}}.$$

Comparing gives

$$A_{ij}{}^k=C_{ijk}.$$

Exercise 4. Determine the momentum-space scaling

Suppose

$$G(x)=\frac{1}{|x|^{2\Delta}}.$$

Use dimensional analysis to determine the scaling of its Fourier transform $\widetilde G(p)$.

Solution

The Fourier transform is

$$\widetilde G(p)=\int d^d x\,e^{ip\cdot x}G(x).$$

Under $p\mapsto \lambda p$, change variables $y=\lambda x$:

$$\widetilde G(\lambda p) = \int d^d x\,e^{i\lambda p\cdot x}\frac1{|x|^{2\Delta}} = \lambda^{2\Delta-d} \int d^d y\,e^{ip\cdot y}\frac1{|y|^{2\Delta}}.$$

Thus

$$\widetilde G(\lambda p)=\lambda^{2\Delta-d}\widetilde G(p),$$

so, up to contact terms and logarithms at special values of $\Delta$,

$$\widetilde G(p)\propto (p^2)^{\Delta-d/2}.$$

Takeaway

For scalar primaries on flat space,

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

in an orthonormal basis, and

$$\boxed{ \left\langle \mathcal O_i(x_1) \mathcal O_j(x_2) \mathcal O_k(x_3) \right\rangle = \frac{C_{ijk}} {|x_{12}|^{\Delta_i+\Delta_j-\Delta_k} |x_{23}|^{\Delta_j+\Delta_k-\Delta_i} |x_{31}|^{\Delta_k+\Delta_i-\Delta_j}}. }$$

The dimensions $\Delta_i$ and the constants $C_{ijk}$ are CFT data. In AdS/CFT, they are the boundary fingerprints of bulk masses and cubic interactions.

Further reading

For global conformal constraints on two- and three-point functions, see Di Francesco, Mathieu, and Sénéchal, Chapter 4.3. For a modern higher-dimensional treatment, see Rychkov’s CFT notes and Simmons-Duffin’s TASI lectures on the conformal bootstrap. For the holographic interpretation, these formulas become the boundary limit of bulk two-point and three-point Witten diagrams.