Fixed points, scaling, and conformal symmetry

Why does a massless correlator fall as a power of distance, and what can interactions change about that power? We will answer first for a free real scalar, including its normalization. Then we will separate three ideas that often appear together: a renormalization-group fixed point, a scaling dimension, and conformal symmetry.

The calculation uses ordinary QFT in flat Euclidean dimension $d=3$ or $4$. No supersymmetry or prior conformal-field-theory knowledge is assumed. Correlators are vacuum correlators on infinite flat space, with no boundary, temperature, or other external length scale. A mass or a boundary would change the scaling question.

Start with the massless scalar

Consider the canonically normalized action

$$S_0=\frac12\int_{\mathbb R^d}d^dx\, \partial_\mu\phi\,\partial_\mu\phi.$$

An integration by parts writes its quadratic operator as $-\partial^2$. The Gaussian two-point function $G(x)=\langle\phi(x)\phi(0)\rangle$ is the inverse of that operator:

$$-\partial^2G(x)=\delta^{(d)}(x).$$

This equation includes a contact source at $x=0$. At separated points it is simply $\partial^2G=0$. Translation and rotation invariance imply $G(x)=G(r)$, where $r=|x|$, so

$$\partial^2G=G''(r)+\frac{d-1}{r}G'(r) =\frac1{r^{d-1}}\frac{d}{dr}\left(r^{d-1}G'(r)\right).$$

Away from the source, $r^{d-1}G'$ is constant. For $d>2$, the decaying solution is $G(r)=A r^{2-d}$. Decay discards an additive constant. It does not fix $A$; the delta function does.

Integrate the Green equation over a ball of radius $\epsilon$ and use its outward normal:

$$1=-\int_{|x|=\epsilon}dS\,\partial_rG =-(2-d)A\,\omega_{d-1}, \qquad \omega_{d-1}=\frac{2\pi^{d/2}}{\Gamma(d/2)}.$$

Here $\omega_{d-1}$ is the area of the unit $(d-1)$-sphere, and $\Gamma$ is the Euler gamma function. The powers of $\epsilon$ cancel. Therefore

$$\boxed{ G(r)=\frac{1}{(d-2)\omega_{d-1}\,r^{d-2}} =\frac{\Gamma(d/2-1)}{4\pi^{d/2}\,r^{d-2}}. }$$

In the two dimensions needed here,

$$d=3:\quad G(r)=\frac1{4\pi r}, \qquad d=4:\quad G(r)=\frac1{4\pi^2r^2}.$$

Both the sign and the coefficient follow from the contact source. A function can solve the separated-point Laplace equation and still have the wrong delta-function normalization.

Dimensional analysis supplies a separate check. With $[\partial_\mu]=1$ and a dimensionless action, the canonical field dimension is $\Delta_0=[\phi]=(d-2)/2$. The correlator has dimension $2\Delta_0=d-2$, exactly the inverse power of length above. Dimensional analysis fixes this power for the free action but cannot determine its numerical coefficient.

What a scaling dimension measures

Let $\mathcal O$ be a local scalar operator that scales with a definite dimension $\Delta$. “Definite” matters: if several operators mix under a change of scale, choose an eigenoperator of that mixing before assigning a single $\Delta$.

In a scale-invariant vacuum, its two-point function obeys

$$G_{\mathcal O}(\lambda r)=\lambda^{-2\Delta}G_{\mathcal O}(r), \qquad \lambda>0.$$

Differentiate with respect to $\lambda$ at $\lambda=1$:

$$\left(r\frac{d}{dr}+2\Delta\right)G_{\mathcal O}(r)=0.$$

The solution is

$$\langle\mathcal O(x)\mathcal O(0)\rangle =\frac{C_{\mathcal O}}{|x|^{2\Delta}}, \qquad x\ne0.$$

For a charged complex operator use $\langle\mathcal O(x)\mathcal O^\dagger(0)\rangle$ instead; a same-charge correlator can vanish by charge conservation. The constant depends on how the operator is normalized. Rescaling $\mathcal O$ changes it without changing $\Delta$.

This derivation requires scaling, translations, and rotations. It has not yet used special conformal transformations. The dilation Ward identity and its solution are derived in Simmons-Duffin, section 4.2, equations (42)–(47).

At an interacting fixed point, a field or composite operator can have

$$\Delta=\Delta_0+\gamma_*.$$

The quantity $\gamma_*$ is its anomalous dimension, the shift from its engineering dimension in the stated normalization and operator identification. It is part of the scaling law, not an error in dimensional analysis of the free action. Composite operators can acquire their own shifts; one cannot generally add constituent anomalous dimensions to obtain the composite’s dimension.

To compare powers without a dimensionful logarithm, normalize at a reference separation $r_0$. Then

$$\frac{G_{\mathcal O}(r)}{G_{\mathcal O}(r_0)} =\left(\frac r{r_0}\right)^{-2\Delta_0} \exp\!\left[-2\gamma_*\log\!\left(\frac r{r_0}\right)\right].$$

For small $\gamma_*$ and $|\gamma_*\log(r/r_0)|\ll1$, this is

$$\left(\frac r{r_0}\right)^{-2\Delta_0} \left[1-2\gamma_*\log\!\left(\frac r{r_0}\right) +O\!\left(\gamma_*^2\log^2(r/r_0)\right)\right].$$

Thus logarithms in a perturbative calculation can be the expansion of a shifted power. Small $\gamma_*$ alone does not justify truncating the logarithmic series over arbitrarily large distance ratios.

As an illustrative scaling law, take $\Delta_0=1/2$ and $\gamma_*=1/10$. The correlator falls as $r^{-6/5}$ rather than $r^{-1}$. Doubling the separation multiplies it by $2^{-6/5}$, not $1/2$. These numbers illustrate the definition; they are not a claimed result for a particular interacting model.

A fixed point has no running dimensionless couplings

A renormalized description uses a reference energy $\mu$. Write its dimensionless couplings as $g^i(\mu)$ and define

$$\beta^i(g)=\mu\frac{d g^i}{d\mu}.$$

A fixed point is a choice $g_*^i$ for which the beta functions vanish, after treating physically redundant descriptions appropriately. Observables at the fixed point can have scaling laws even when the values of $\Delta$ differ from free-field dimensions. Renormalization-group flows organize theories by how they approach or depart from such regimes; the source’s section 1.1 gives the physical motivation.

Near a fixed point, perturb the action by a scalar scaling operator:

$$\delta S=u\int d^dx\,\mathcal O(x),\qquad [u]=d-\Delta\equiv y.$$

Assume this is an independent deformation, rather than a total derivative or a removable field redefinition. Under a dilation to longer distances by a factor $b>1$, the integrated operator scales as $b^{d-\Delta}$. Its perturbation therefore grows as $b^y$ at linear order. Equivalently, the dimensionless coupling $\widehat u=u\mu^{-y}$ has

$$\beta_{\widehat u}=-y\widehat u+O(\widehat u^2)$$

in a basis of scaling deformations. Lower energy means smaller $\mu$, so the minus sign is consistent with growth toward long distances when $y>0$.

Scaling dimension	Name	Linear behavior toward long distances
$\Delta<d$	Relevant	The perturbation grows.
$\Delta>d$	Irrelevant	The perturbation shrinks.
$\Delta=d$	Marginal	Linear scaling alone does not decide.

A marginal interaction may run through nonlinear terms in its beta function. “Classically marginal” is weaker still: it uses engineering dimensions at the Gaussian fixed point. Neither label by itself proves an exact family of fixed points.

For the free scalar, the normal-ordered mass operator $:\!\phi^2\!:$ has dimension $d-2$, so its coupling $m^2$ has $y=2$ in both three and four dimensions. It is relevant. Indeed $\widehat m^2=m^2/\mu^2$ grows as the probing energy decreases. Normal ordering here subtracts self-contractions at the same insertion; it leaves the separated-point free-field scaling dimension of the composite well defined.

Conformal symmetry needs an additional step

A fixed point and a power-law two-point function do not, by themselves, establish every conformal Ward identity. We now state the extra input used in this lesson: a symmetric conserved stress tensor whose flat-space trace can be made zero by an allowed improvement. We use its separated-point identities, with contact terms understood when operator insertions coincide.

An infinitesimal coordinate transformation $x^\mu\mapsto x^\mu+\varepsilon^\mu(x)$ is conformal when it preserves angles. Its local change of metric can be a scale factor:

$$\partial_\mu\varepsilon_\nu+\partial_\nu\varepsilon_\mu =\frac2d(\partial\cdot\varepsilon)\delta_{\mu\nu}.$$

This is the conformal Killing equation. Translations and rotations solve it with $\partial\cdot\varepsilon=0$. Dilations use $\varepsilon^\mu=a x^\mu$ with constant $a$. A new solution is

$$\varepsilon^\mu=2(b\cdot x)x^\mu-b^\mu x^2,$$

where $b^\mu$ is constant and has inverse-length units. Differentiation gives

$$\partial_\mu\varepsilon_\nu+\partial_\nu\varepsilon_\mu =4(b\cdot x)\delta_{\mu\nu}, \qquad \partial\cdot\varepsilon=2d(b\cdot x).$$

These are special conformal transformations. Their position-dependent scale factor is why conformal symmetry contains more than a uniform rescaling.

Now form $j^\mu_\varepsilon=T^{\mu\nu}\varepsilon_\nu$. Conservation and symmetry of the stress tensor give

$$\partial_\mu j^\mu_\varepsilon =\frac12T^{\mu\nu} (\partial_\mu\varepsilon_\nu+\partial_\nu\varepsilon_\mu) =\frac{\partial\cdot\varepsilon}{d}\,T^\mu{}_\mu.$$

A vanishing trace makes every conformal Killing vector supply a conserved current. This is the mechanism behind the extra symmetry, as in equations (19) and (22)–(24) of the lectures.

The free massless scalar supplies a concrete example. In flat Euclidean space,

$$T_{\mu\nu} =\partial_\mu\phi\,\partial_\nu\phi -\frac12\delta_{\mu\nu}(\partial\phi)^2 +\xi_c(\delta_{\mu\nu}\partial^2-\partial_\mu\partial_\nu)\phi^2, \qquad \xi_c=\frac{d-2}{4(d-1)}.$$

The improvement has identically zero divergence. Its trace, using $\partial^2\phi=0$ away from insertions, is

$$T^\mu{}_\mu =\left(1-\frac d2+2\xi_c(d-1)\right)(\partial\phi)^2=0.$$

This establishes the required flat-space current check for the free example. Stress tensors and improvement terms derives the coefficient from the curved action and keeps the boundary assumptions explicit. That fuller derivation is optional preparation here.

For a general theory, improving the trace to zero is a substantive condition. A scale current can have the form $j_D^\mu=x_\nu T^{\mu\nu}-V^\mu$, where $V^\mu$ is a local vector called a virial current. Its conservation only gives $T^\mu{}_\mu=\partial_\mu V^\mu$; removing this divergence by an allowed stress-tensor improvement is an additional step. We are not assuming an unrestricted theorem that every fixed point is conformal. Nor does a flat-space trace identity exclude curvature-dependent quantum anomalies on a curved background.

Exercises

1. A mass changes the long-distance question

In three dimensions take $G_m(r)=e^{-mr}/(4\pi r)$ with $m>0$. Check that it solves the massive Green equation with the same unit contact source. Define $\Delta_{\rm eff}(r)=-\tfrac12\,d\log G_m/d\log r$. Does it measure one fixed scaling dimension?

First hint

At $r>0$, apply $-\partial_r^2-2\partial_r/r+m^2$. At the origin, inspect the leading $1/r$ singularity.

Second hint

The volume integral of $m^2G_m$ over a shrinking ball tends to zero. The derivative flux therefore fixes the delta function just as in the massless case.

Solution

Direct differentiation gives $(-\partial^2+m^2)G_m=0$ away from the origin. Since $G_m=1/(4\pi r)-m/(4\pi)+O(r)$ near zero, $-4\pi\epsilon^2G_m'(\epsilon)\to1$. Meanwhile $\int_{|x|<\epsilon}m^2G_m\,d^3x=O(m^2\epsilon^2)$ vanishes. Thus the full equation is $(-\partial^2+m^2)G_m=\delta^{(3)}(x)$.

The logarithmic slope is

$$\Delta_{\rm eff}(r)=\frac12(1+mr).$$

It approaches the free scaling dimension $1/2$ for $mr\ll1$, but grows with distance rather than remaining constant. The mass introduces a correlation length $1/m$, and the exponential decay cannot be described by one fixed anomalous power. A measured local slope is not automatically a fixed-point dimension.

2. When does a logarithmic approximation fail?

Suppose $\Delta=\Delta_0+\gamma_*$ with $|\gamma_*|\ll1$. Divide the normalized correlator by its free power and call the result $R(r)$. Compare its exact and first-order values at $r/r_0=e^{1/|\gamma_*|}$. Explain which small quantity a controlled expansion actually requires.

First hint

Use $R(r)=\exp[-2\gamma_*\log(r/r_0)]$.

Second hint

The exponent is not small at the specified separation, even when $\gamma_*$ is small.

Solution

For $\gamma_*>0$, the exact result is $e^{-2}$, whereas first order gives $1-2=-1$. For $\gamma_*<0$, they are $e^2$ and $1+2=3$. The discrepancy remains of order one as $|\gamma_*|$ tends to zero along this prescribed distance ratio.

The expansion parameter is $\gamma_*\log(r/r_0)$, not $\gamma_*$ alone. In the region where its magnitude is much smaller than one, the omitted relative term starts at $2\gamma_*^2\log^2(r/r_0)$. The negative truncated answer in the first case diagnoses misuse of the approximation; it is not a negative exact two-point normalization.

3. Classify deformations at the Gaussian fixed point

For one free scalar, the normal-ordered operator $:\!\phi^n\!:$ has dimension $n(d-2)/2$. Classify the deformations by $:\!\phi^2\!:$, $:\!\phi^4\!:$, and $:\!\phi^6\!:$ in $d=3$ and $d=4$. Which conclusions survive without knowing nonlinear beta functions?

First hint

Calculate $y=d-\Delta$ separately for each operator and dimension.

Second hint

A zero value of $y$ removes only the linear scaling term. It does not set the entire beta function to zero.

Solution

Dimension	$:\!\phi^2\!:$	$:\!\phi^4\!:$	$:\!\phi^6\!:$
$d=3$	$\Delta=1$, $y=2$: relevant	$\Delta=2$, $y=1$: relevant	$\Delta=3$, $y=0$: marginal
$d=4$	$\Delta=2$, $y=2$: relevant	$\Delta=4$, $y=0$: marginal	$\Delta=6$, $y=-2$: irrelevant

These are linear statements near the Gaussian fixed point. The positive and negative eigenvalues determine initial growth or decay in their respective directions. The marginal cases require a beta-function calculation before deciding whether they grow, shrink, or stay exactly marginal. At an interacting fixed point, the composite dimensions must be determined there; the free formula cannot simply be carried over.

The two-point function now answers two different questions: its normalization tests the specified quadratic action, while its power measures scaling. Conformal symmetry adds further relations through the stress tensor and special conformal transformations. Continue with local operators and their products to organize primaries and descendants and interpret scaling as evolution on a cylinder.

For a different use of the conformal two-point form, the first AdS/CFT dictionary derives a scalar response from a bulk boundary problem. Its additional preparation is source calculus and local contact terms; the example assumes no supersymmetry or theory-specific gauge-rank dictionary.