Why composite operators mix

Color counting found nonzero free overlaps between different trace structures. An overlap is an inner product; it is not yet a change of scaling dimension. To find the latter, turn on an interaction and look for logarithms that remain after ordinary field renormalization. A logarithm can connect two different composite insertions, requiring a matrix of operator counterterms.

We stay in four-dimensional $\mathcal N=4$ $SU(N)$ Yang–Mills in its conformal vacuum, at theta angle zero. The full Lorentzian theory has sixteen real Poincaré supercharges; an $\mathcal N=1$ description makes four manifest. $N$ is the number of fundamental colors and the gauge rank is $N-1$. The dynamical calculation below is planar and one loop, with $\lambda=g_{\rm YM}^2N\ll1$. Correlators are Euclidean and at separated points.

Choose a basis before assigning dimensions

Retain the six canonical real adjoint scalars $X^I=X^{Ia}t^a$ and ordinary trace $\operatorname{Tr}(t^at^b)=\delta^{ab}/2$. Define

$$A=\operatorname{Tr}[(X^1)^2],\qquad B=\sum_{I=2}^6\operatorname{Tr}[(X^I)^2],\qquad \boldsymbol{\mathcal O}=\begin{pmatrix}A\\B\end{pmatrix}. \tag{N09.1}$$

At zero coupling these composites are normal ordered. Vacuum one-point terms are subtracted; the displayed correlators are connected. Both operators have engineering dimension two and are Lorentz and gauge scalars. They are invariant under the $SO(5)$ subgroup rotating $X^2,\ldots,X^6$. Thus this two-dimensional space can close under renormalization. Neither member is separately irreducible under the full $SO(6)_R$ symmetry.

That last qualification matters. Mixing respects the full symmetry: inequivalent irreducible representations do not mix. A basis that has not separated those representations can nevertheless have off-diagonal counterterms. We will find that separation rather than assume it.

Put $r=|x|>0$. Two Wick pairings, the color trace factor and the component propagator $1/(4\pi^2r^2)$ give

$$\langle A(x)A(0)\rangle_0=\frac{c_N}{r^4},\quad \langle B(x)B(0)\rangle_0=\frac{5c_N}{r^4},\quad \langle A(x)B(0)\rangle_0=0,\qquad c_N=\frac{N^2-1}{32\pi^4}.$$

There is no contraction between different scalar flavors. The factor five counts the five independent terms in $B$. For the following planar calculation use $c=N^2/(32\pi^4)$, keeping the leading term of $c_N$. Define the tree-level Gram matrix, the matrix of inner products, by

$$\langle\boldsymbol{\mathcal O}(x) \boldsymbol{\mathcal O}(0)^T\rangle_0=\frac{G_0}{r^4}, \qquad G_0=c\begin{pmatrix}1&0\\0&5\end{pmatrix}. \tag{N09.2}$$

All loop formulas below retain that planar approximation. The exact free coefficient $c_N$ does not supply a finite-$N$ loop calculation.

Identify the logarithm and its flavor matrix

A familiar position-space integral explains the logarithmic dependence:

$$I_\Lambda(r)= \int_{\substack{|z|>\Lambda^{-1}\\|z-x|>\Lambda^{-1}}} \frac{d^4z}{|z|^4|z-x|^4} =\frac{2\pi^2}{r^4}\log(\Lambda^2r^2) +\frac{\text{finite constant}}{r^4}+o(1). \tag{N09.3}$$

Here $\Lambda$ is a large ultraviolet momentum cutoff. Near $z=0$, replace the other denominator by $r^4$: the radial integral is $2\pi^2r^{-4}\int_{\Lambda^{-1}}^{O(r)}d\rho/\rho$. The neighborhood of $z=x$ gives the same term. Their sum gives the displayed coefficient. The distant region is convergent. This is the Euclidean integral in Minahan (5.12)–(5.13); its finite constant depends on the cutoff prescription.

This integral alone does not determine the operator mixing. The complete one-loop calculation also needs gauge exchange, scalar interactions and the residual self-energy contribution after elementary-field renormalization. For one ordered pair, Minahan–Zarembo’s counterterms preceding (2.9) have the common factor $(\lambda/16\pi^2)\log(\Lambda/\mu)$, where $\mu$ is a reference momentum scale. Their respective flavor factors are

$$-\mathbb I,\qquad \mathbb I-2\mathbb P+\mathbb K,\qquad 2\mathbb I.$$

$\mathbb I$ preserves two flavor labels, $\mathbb P$ exchanges them, and $\mathbb K$ replaces a tensor $T_{IJ}$ by $\delta_{IJ}\sum_KT_{KK}$. The sum is $\mathbb K+2\mathbb I-2\mathbb P$. These coefficients are sourced dynamical input; color topology or supersymmetry counting alone would not fix them.

For a quadratic trace, cyclicity makes $T_{IJ}$ symmetric, so $\mathbb P$ acts as the identity. The two positions around the quadratic trace give equal contributions. With

$$a=\frac{\lambda}{8\pi^2},$$

the resulting order-$a$ operator map is therefore

$$\operatorname{Tr}[(X^I)^2]\ \longmapsto\ a\sum_{J=1}^6\operatorname{Tr}[(X^J)^2].$$

The map is the coefficient of the logarithmic counterterm, not a classical transformation of the field. It is the quadratic specialization of MZ (2.9)–(2.13), agreeing with the previously stated quadratic diagnostic. MZ’s action (2.8) uses coupling-absorbed scalars $\Phi=g_{\rm YM}X$; Minahan’s canonical matrix scalar is instead $\sqrt2X$ in our trace normalization. These field conventions give the same $\lambda$ and anomalous dimensions. We only use the two-field restriction here; organizing arbitrary cyclic words is a separate calculation.

Adding the five terms in $B$ gives the component-column matrix

$$J=\begin{pmatrix}1&1\\5&5\end{pmatrix},\qquad A\longmapsto a(A+B),\quad B\longmapsto5a(A+B). \tag{N09.4}$$

This need not be symmetric: the chosen basis has unequal norms. Notice instead that

$$JG_0=G_0J^T =c\begin{pmatrix}1&5\\5&25\end{pmatrix}.$$

Set $L=aJG_0$. A divergent insertion at either endpoint of the two-point function produces the regulated composite correlator

$$C_{\rm B}(r)=\frac1{r^4} \left[G_0-L\log(\Lambda^2r^2)+\text{finite terms}\right] +O(a^2). \tag{N09.5}$$

The two endpoints each contribute $-L\log(\Lambda r)$; the equality above makes their matrices equal. The subscript B denotes composite insertions before their composite counterterms, with the ordinary elementary-field renormalization included consistently. In particular the off-diagonal logarithm is nonzero although the tree-level overlap vanished.

Cancel the divergence with a matrix

A renormalized composite can be a linear combination of regulated ones. Choose the convention

$$\boldsymbol{\mathcal O}_{\rm R} =Z\boldsymbol{\mathcal O}_{\rm B},\qquad Z=1+aJ\log\frac{\Lambda}{\mu}+O(a^2), \tag{N09.6}$$

where $\mu$ is a renormalization momentum scale. Here $Z$ denotes a counterterm matrix, not the complex scalar used in the previous lesson. Both $\Lambda$ and $\mu$ have mass dimension one, and $Z$ is dimensionless. Then $C_{\rm R}=ZC_{\rm B}Z^T$. To first order, the added term is

$$a(JG_0+G_0J^T)\log\frac{\Lambda}{\mu} =2L\log\frac{\Lambda}{\mu}.$$

It cancels the cutoff dependence:

$$C_{\rm R}(r)=\frac1{r^4} \left[G_0-L\log(\mu^2r^2)\right]+O(a^2). \tag{N09.7}$$

Here a finite operator normalization has absorbed the nonlogarithmic order-$a$ constant at $\mu r=1$. Such a finite choice does not change the one-loop scaling eigenvalues. The fixed-order expression is useful when $|a\log(\mu r)|\ll1$ as well as $a\ll1$.

At fixed bare insertions and fixed coupling, differentiate (N09.6). Define

$$\mu\frac{d}{d\mu}\boldsymbol{\mathcal O}_{\rm R} =-\Gamma\boldsymbol{\mathcal O}_{\rm R},\qquad \Gamma=-\left(\mu\frac{dZ}{d\mu}\right)Z^{-1} =aJ+O(a^2). \tag{N09.8}$$

This $\Gamma$ is a composite-operator anomalous-dimension matrix. It is not a beta function or an elementary-field anomalous dimension. The vanishing beta function of this theory does not set $\Gamma$ to zero. MZ (2.4) differentiates $Z$ with respect to $\log\Lambda$ instead; since our counterterm contains $\log(\Lambda/\mu)$, its positive one-loop matrix is the same.

Use the inner product when finding scaling operators

An operator with definite scaling is a combination $\mathcal O_v=v^T\boldsymbol{\mathcal O}_{\rm R}$, with constant real coefficient column $v$. Equation (N09.8) implies $v^T\Gamma=\gamma v^T$. In terms of the symmetric log coefficient $L$ this is

$$\boxed{Lv=\gamma G_0v.} \tag{N09.9}$$

This is a generalized eigenvalue problem: the inner product $G_0$ is part of the equation. Indeed $\Gamma=LG_0^{-1}$, so $G_0^{-1}L=\Gamma^T$ acts on coefficient columns. Stating whether a matrix acts on the operator column or on its coefficient column prevents a transpose error.

For an eigenoperator with unit tree norm, (N09.7) becomes

$$\langle\mathcal O_v(x)\mathcal O_v(0)\rangle =\frac1{r^4}\left[1-\gamma\log(\mu^2r^2)+\cdots\right] =\frac{(\mu r)^{-2\gamma}}{r^4}+\cdots.$$

Thus $\Delta=2+\gamma$; the factor of two comes from the two insertions. The displayed power is understood perturbatively unless higher orders have been supplied.

For a direct matrix check, replace $B$ by $B/\sqrt5$. Its Gram matrix is $c1$, and its anomalous-dimension matrix is

$$a\begin{pmatrix}1&\sqrt5\\\sqrt5&5\end{pmatrix}.$$

It is symmetric, has zero determinant and trace $6a$, and therefore has real eigenvalues $0,6a$. The nonsymmetric matrix in (N09.4) represented the same self-adjoint dilation operation in an unequally normalized basis.

Worked example: separate two scaling behaviors

Solving (N09.9) gives

$$\begin{array}{c|c|c} \text{operator}&\text{tree norm coefficient}&\text{one-loop }\gamma\\ \hline \mathcal P=5A-B&30c&0\\ \mathcal K=A+B&6c&6a \end{array} \qquad \langle\mathcal P(x)\mathcal K(0)\rangle_0=0. \tag{N09.10}$$

$\mathcal P$ has the flavor polarization $\operatorname{diag}(5,-1,-1,-1,-1,-1)$, whose trace vanishes. $\mathcal K$ is the $SO(6)_R$ singlet Konishi scalar. The full symmetry has separated the two sectors, exactly as it should. Normalize them by $\sqrt{30c}$ and $\sqrt{6c}$ respectively. The dimensions are

$$\Delta_{\mathcal P}=2,\qquad \Delta_{\mathcal K} =2+\frac{3\lambda}{4\pi^2}+O(\lambda^2). \tag{N09.11}$$

The exact protection of the first exponent uses the independent shortening argument in protected and unprotected operators; a zero one-loop eigenvalue alone would not prove it. The second agrees with MZ (3.2)–(3.3).

Because $A=(\mathcal P+\mathcal K)/6$, its two-point function combines both exponents. Resumming only the leading logarithms generated by the one-loop matrix gives

$$\langle A(x)A(0)\rangle_{\rm leading\ log} =\frac c{r^4} \left[\frac56+\frac16(\mu r)^{-12a}\right]. \tag{N09.12}$$

Expanding it gives $c[1-a\log(\mu^2r^2)]/r^4$, checking the $AA$ entry of (N09.7). Assigning $A$ the dimension $2+a$ would reproduce that entry at first order, but would incorrectly replace a sum of scaling sectors by a single eigenoperator. The mixed correlator and its scale evolution expose the error.

What a larger mixing problem requires

Choose all independent operators allowed by the relevant quantum numbers and approximation order, impose cyclicity and finite-rank trace relations, then compute their free Gram matrix and logarithmic mixing. If a putative basis has a null vector, first remove the zero operator; $G_0^{-1}$ on that redundant space is not defined.

The general scalar sector closes at one loop in the source calculation, but that statement cannot be extended to arbitrary higher orders: other field contents can share its quantum numbers. At finite $N$, single and multiple traces can also enter a common basis. Conversely, matching some commuting charges is a necessary condition, not a guarantee of nonzero mixing. Full representations and dynamical selection rules still matter. Minahan, section 7.

For a constant nonsingular change $\boldsymbol{\mathcal O}'=R\boldsymbol{\mathcal O}$,

$$G_0'=RG_0R^T,\qquad L'=RLR^T,\qquad \Gamma'=R\Gamma R^{-1}.$$

The first two matrices change by congruence, the third by similarity; generalized eigenvalues are unchanged. This statement assumes a $\mu$-independent change of basis at fixed coupling. Basis-dependent matrix entries are not themselves observable dimensions.

Practice

1. Try to renormalize each original operator separately

Use only diagonal counterterms $Z_{\rm diag}=1+a\,\operatorname{diag}(1,5)\log(\Lambda/\mu)$. Show that these cancel the diagonal logarithmic divergences at first order, but fail to renormalize the complete two-point matrix. Explain why a successful calculation of just $\langle AA\rangle$ would be insufficient.

Hint 1

Multiply $Z_{\rm diag}G_0+G_0Z_{\rm diag}^T$ and retain its order-$a$ part.

Hint 2

The tree matrix is diagonal. Diagonal counterterms cannot add an off-diagonal term at this order.

Solution

The added logarithmic coefficient is $2ac\,\operatorname{diag}(1,25)\log(\Lambda/\mu)$. It cancels the $\Lambda$ dependence of the $AA$ and $BB$ entries in (N09.5). But the $AB$ entry still contains

$$-\frac{5ac}{r^4}\log(\Lambda^2r^2),$$

with no compensating term. The matrix fails to have a finite limit. The off-diagonal counterterms in (N09.6) supply precisely the missing $10ac\log(\Lambda/\mu)/r^4$. Renormalizing an insertion requires its correlation functions with the full operator basis, not just fitting a logarithm in its self-correlator.

2. Change an operator normalization

Replace $B$ by $B'=2B$, retaining $A$. Find $G_0'$, $L'$ and $\Gamma'$. Identify the two eigenoperators and show why ordinary eigenvalues of $L'$ are not the anomalous dimensions.

Hint 1

Use $R=\operatorname{diag}(1,2)$ and the three different transformation rules above.

Hint 2

The old operators $\mathcal P$ and $\mathcal K$ are still present; express them in terms of $B'$.

Solution

Direct multiplication gives

$$G_0'=c\begin{pmatrix}1&0\\0&20\end{pmatrix},\quad L'=ac\begin{pmatrix}1&10\\10&100\end{pmatrix},\quad \Gamma'=a\begin{pmatrix}1&1/2\\10&5\end{pmatrix}.$$

The coefficient columns of $5A-B'/2$ and $A+B'/2$ solve $L'v=\gamma G_0'v$ with $\gamma=0,6a$. Their tree norms remain $30c,6c$. In contrast the nonzero ordinary eigenvalue of $L'$ is $101ac$, whereas that of $L$ was $26ac$. Rescaling an operator cannot change a physical scaling dimension; those numbers describe a logarithmic bilinear form without its required metric.

3. Evolve the operators over a finite scale interval

Keep only the one-loop matrix and put $t=\log(\mu/\mu_0)$. Solve $d\boldsymbol{\mathcal O}/dt=-aJ\boldsymbol{\mathcal O}$ for a finite $t$. What happens to $A$, $\mathcal P$ and $\mathcal K$? Does this solution determine the exact anomalous dimension at finite coupling?

Hint 1

Check $J^2=6J$. Hence $J/6$ and $1-J/6$ are complementary projectors.

Hint 2

Exponentiate separately on the two projected subspaces.

Solution

The solution is

$$\boldsymbol{\mathcal O}(\mu)= \left[1+\frac{e^{-6at}-1}{6}J\right] \boldsymbol{\mathcal O}(\mu_0).$$

In particular

$$A(\mu)=\frac{5+e^{-6at}}6A(\mu_0) +\frac{e^{-6at}-1}6B(\mu_0),\quad \mathcal P(\mu)=\mathcal P(\mu_0),\quad \mathcal K(\mu)=e^{-6at}\mathcal K(\mu_0).$$

At $t=0$ this is the identity; differentiating there gives $dA/dt=-a(A+B)$ with the required sign. The evolution matrices multiply correctly when two scale intervals are joined.

This exponentiation sums the leading logarithms supplied by the known one-loop matrix. It does not calculate the order-$a^2$ correction to $\Gamma$. At weak coupling, intervals with $at$ of order one can be useful provided omitted higher-order contributions such as $a^2t$ remain small. The canonical mass dimension two has not disappeared: this equation tracks renormalization-scale dependence at fixed spacetime point.

Continue with a first cyclic spin chain to construct the logarithmic matrix for longer scalar traces. Their cyclic order supplies additional structure. The calculation above supplies the renormalization meaning, inner product and eigenvalue interpretation that this construction must preserve.