Operator mixing, integrability, and finite N

Can a small operator-mixing matrix tell us when a spin-chain description is useful—and when we have changed the problem? We will reproduce one spectrum in two ways, lengthen the trace while keeping the approximation fixed, and then test a finite-rank substitution that fails for an exact algebraic reason.

Prepare with the first one-loop spin chain and the reproduction-to-research method. This is a four-dimensional entrance. No three-dimensional localization, curved-space construction, or string-theory calculation is needed. The result below is a known one-loop reproduction and a controlled training extension, not a claim of a new spectrum.

Fix the spectral question

Use four-dimensional N=4 Yang–Mills with gauge group $SU(N)$, gauge rank $N-1$, and the conformal vacuum with zero scalar expectation values. The theory has sixteen real Poincaré supercharges; an N=1 description displays four. The conformal vacuum also has sixteen conformal supercharges. The $SU(2)$ below rotates two scalar flavors; it is not the gauge group.

Keep the canonical complex adjoint fields $Z=(X^1+iX^2)/\sqrt2$ and $W=(X^3+iX^4)/\sqrt2$, with ordinary fundamental trace $\operatorname{Tr}(t^at^b)=\delta^{ab}/2$. First take the planar limit $N\to\infty$ at fixed weak $\lambda=g_{\rm YM}^2N$ and fixed trace length $L$. For operators made from $L-2$ copies of $Z$ and two copies of $W$, the input from the one-loop calculation is

$$\Gamma=a\mathsf H+O(\lambda^2),\qquad a=\frac{\lambda}{8\pi^2},\qquad \mathsf H=\sum_{j=1}^{L}(1-\mathbb P_{j,j+1}), \qquad j+L\equiv j. \tag{D10.1}$$

Here $\mathbb P$ exchanges neighboring flavors. An eigenvalue $E$ of $\mathsf H$ gives the scaling dimension $\Delta=L+aE+O(\lambda^2)$. It is an auxiliary spin-chain energy, not a spacetime particle energy. The closing bond is required by the trace. The scalar polarizations are mutually null under the bilinear $SO(6)$ contraction, so the trace operator in Minahan–Zarembo (2.10)–(2.13) vanishes in this sector, leaving (D10.1).

Our question is deliberately finite: does coordinate scattering give the entire cyclic two-$W$ spectrum at lengths six and seven, including the correct number of states and their norms? Increasing $L$ here changes the operator, not $N$, the coupling, or the theory. Accuracy in diagonalizing $\mathsf H$ does not control the omitted higher loops or nonplanar terms.

Reproduce the short answer before extending it

At length four the two trace classes are $A_4=\operatorname{Tr}(ZZWW)$ and $B_4=\operatorname{Tr}(ZWZW)$. Swapping all mixed bonds gives $\mathsf H A_4=2A_4-2B_4$ and $\mathsf H B_4=-4A_4+4B_4$. The raw-trace matrix, free norm, and eigenoperators are

$$H_4=\begin{pmatrix}2&-2\\-4&4\end{pmatrix},\qquad G_4=c_4\begin{pmatrix}1&0\\0&2\end{pmatrix},\qquad \begin{array}{c|c} \text{operator}&E\\\hline 2A_4+B_4&0\\ A_4-B_4&6 \end{array}. \tag{D10.2}$$

The common coefficient is $c_L=N^L/(8\pi^2)^L$, with the correlator also carrying separation $r^{-2L}$. A word’s diagonal Gram entry counts the cyclic shifts that leave it unchanged. $B_4$ has two such shifts. Thus $H_4G_4=G_4H_4^T$ even though $H_4$ is not symmetric. Our convention is $\mathsf H\boldsymbol{\mathcal O}=H\boldsymbol{\mathcal O}$; the coefficient column of an eigenoperator obeys $H^Tc=Ec$. Confusing these two conventions gives incorrect operator combinations even when the eigenvalues happen to be right.

Now lengthen to six. Classify the two $W$ positions by their shortest cyclic separation $d=1,2,3$:

$$A=\operatorname{Tr}(Z^4W^2),\qquad B=\operatorname{Tr}(Z^3WZW),\qquad C=\operatorname{Tr}(Z^2WZ^2W).$$

For $A$, two mixed bonds lead to $B$. For $B$, two lead to $A$ and two to $C$. For $C$, all four lead to $B$. Therefore

$$H_6=\begin{pmatrix} 2&-2&0\\-2&4&-2\\0&-4&4 \end{pmatrix},\qquad G_6=c_6\operatorname{diag}(1,1,2),\qquad H_6G_6=G_6H_6^T. \tag{D10.3}$$

The repeated block in $C$ accounts for its factor two. On the normalized representatives $(A,B,C/\sqrt2)$ the off-diagonal entries are $-2,-2\sqrt2$, symmetrically placed. Its characteristic polynomial is $E(E^2-10E+20)$. We obtain

$$\begin{aligned} \mathcal P&=2A+2B+C,& E_{\mathcal P}&=0,\\ \mathcal U_E&=2A+(2-E)B+(E-4)C, & E&=5\pm\sqrt5. \end{aligned} \tag{D10.4}$$

Their free norms are $10c_6$ and $10(E-2)c_6$, respectively; all cross-inner-products vanish. Both nonzero norms are positive. The coefficient of $\alpha^4\beta^2$ in $\operatorname{Tr}(\alpha Z+\beta W)^6$ is $3\mathcal P$. This identifies the zero mode as the symmetric component of the half-BPS family. Its exact protection uses supersymmetry, in addition to the one-loop zero found here.

Recover the answer from two moving flavors

A magnon here means one $W$ inserted into the string of $Z$‘s. Let $\psi(x,y)$ be the coefficient of the spin state with its two $W$‘s at sites $x<y$. Away from adjacency, each $W$ can hop in either direction:

$$E\psi(x,y)=4\psi(x,y)-\psi(x-1,y)-\psi(x+1,y) -\psi(x,y-1)-\psi(x,y+1). \tag{D10.5}$$

Adjacent $W$‘s cannot hop onto the same site, and their shared bond does nothing. The equation becomes

$$E\psi(x,x+1)=2\psi(x,x+1)-\psi(x-1,x+1)-\psi(x,x+2). \tag{D10.6}$$

Try two plane waves whose relative coefficient can change when their order is exchanged: $\psi=e^{ip_1x+ip_2y}+S_{12}e^{ip_2x+ip_1y}$. The separated equation fixes $E=4\sin^2(p_1/2)+4\sin^2(p_2/2)$. At $x=0,y=1$, the adjacent equation reads

$$E(e^{ip_2}+S_{12}e^{ip_1}) =2(e^{ip_2}+S_{12}e^{ip_1}) -e^{i(p_2-p_1)}-S_{12}e^{i(p_1-p_2)} -e^{2ip_2}-S_{12}e^{2ip_1}.$$

Solving this linear equation gives the scattering coefficient

$$S_{12}=-\frac{e^{i(p_1+p_2)}-2e^{ip_2}+1} {e^{i(p_1+p_2)}-2e^{ip_1}+1}. \tag{D10.7}$$

This recovers Minahan (8.6)–(8.8) from the local exclusion condition. No spacetime scattering matrix has been introduced.

Translation invariance of a trace requires $p_1+p_2=0$ modulo $2\pi$. Write $(p_1,p_2)=(p,-p)$. For generic $p$, $S_{12}=e^{-ip}$. Taking the first magnon once around the chain requires $e^{ipL}S_{12}=1$. The phase acquired when passing the second magnon changes the denominator from $L$ to $L-1$:

$$p=\frac{2\pi n}{L-1},\qquad f_p(d)=\cos\!\left[p\left(d-\frac12\right)\right],\qquad E_n=8\sin^2\frac{\pi n}{L-1}. \tag{D10.8}$$

Here $f_p$ differs from the two-plane-wave coefficient only by an overall nonzero phase and scale. It satisfies $f_p(L-d)=f_p(d)$, as required for the two choices of separation around the circle. For $p\ne0$, substituting $u=\tfrac12\cot(p/2)$ gives the equivalent two-root equation

$$\left(\frac{u+i/2}{u-i/2}\right)^L =\frac{2u+i}{2u-i},\qquad (u_1,u_2)=(u,-u), \qquad E=\frac{2}{u^2+1/4}. \tag{D10.9}$$

These are the two-impurity relations in Minahan–Zarembo (5.2)–(5.6). The rapidity $u$ is an auxiliary variable parametrizing the phase; it is not a Coulomb-branch coordinate.

At $L=6$, the independent nonconstant waves have $p=2\pi/5,4\pi/5$. Equation (D10.8) gives $5-\sqrt5,5+\sqrt5$, independently of the matrix determinant. To compare the operators as well, convert the wave coefficients to raw traces. A cyclic orbit of size $L/s_d$ contributes $\sqrt{L/s_d}\,f_p(d)$ in the normalized orbit basis, while a raw trace corresponds to $\sqrt{s_d}$ times that basis vector. Thus its raw-trace coefficient is proportional to $f_p(d)/s_d$. Here the vector is $(f_p(1),f_p(2),f_p(3)/2)$, which reproduces $\mathcal U_E$ up to scale.

Two exclusions matter. The uniform protected wave must be added directly: the generic scattering expression has a $0/0$ at $p_1=p_2=0$. Also, a formal momentum is not enough if its wavefunction vanishes. For odd $L$, $p=\pi$ gives $f_\pi(d)=0$ at every integer $d$. After discarding that null wave and identifying the proportional $p$ and $2\pi-p$ waves, the nonconstant choices are $n=1,\ldots,\lfloor(L-2)/2\rfloor$. Together with the uniform state they give exactly $\lfloor L/2\rfloor$ cyclic two-$W$ classes.

Change one input and interpret the check

The length-six matrix, its Gram matrix, and (D10.8) are sufficient data for a reproduction. A useful extension is to generate all two-$W$ words for $4\leq L\leq12$, identify their cyclic orbits, and apply each nearest-neighbor exchange before collecting classes. Compare the complete spectrum and nonzero wavefunctions, not just the lowest eigenvalue. The length-seven calculation is solved below.

For an additional check that does not call an eigensolver, define $K_L(t)=\operatorname{tr}e^{-tH_L}$, $t>0$, on this finite cyclic space. This is an auxiliary spectral diagnostic, not a thermal partition function of Yang–Mills theory. Put $R_L=4I-H_L$. Its entries are nonnegative and its row sums are four, so

$$K_L(t)=e^{-4t}\sum_{j=0}^{\infty} \frac{t^j}{j!}\operatorname{tr}R_L^j, \qquad 0\leq\text{tail after }J \leq\left\lfloor\frac L2\right\rfloor e^{-4t}\sum_{j>J}\frac{(4t)^j}{j!}. \tag{D10.10}$$

Each $\operatorname{tr}R_L^j$ counts weighted closed exchange walks. This positive series can be compared with $1+\sum_n e^{-tE_n}$, using an explicit tail bound. A failed comparison can expose an omitted closing bond, a null Bethe state, or a stabilizer mistake. Agreement verifies this specified one-loop sector. The stronger claim of integrability involves factorized scattering for more magnons and conserved structures; it is not proved by matching three eigenvalues.

A finite-rank substitution that fails

The rank change is a different experiment. At gauge $SU(2)$, both $Z$ and $W$ are traceless two-by-two matrices. The Cayley–Hamilton identity gives $Z^2=\tfrac12\operatorname{Tr}(Z^2)I$ and likewise for $W$. Consequently the length-six traces obey the exact identity $A-C=0$. But the planar map (D10.3) would give

$$\mathsf H_{\rm planar}(A-C)=2A+2B-4C \equiv2(B-A). \tag{D10.11}$$

The right-hand side is generally nonzero. For example, $Z=\operatorname{diag}(1,-1)$ and $W=\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$ give $A=C=2$, $B=-2$, and the proposed image is $-8$. Thus the planar map does not even define an operation on this finite-rank quotient: the zero operator would have a nonzero image. Deleting one basis vector or replacing $\lambda$ by $2g_{\rm YM}^2$ cannot fix it.

A finite-$N$ calculation needs the complete relevant multi-trace space, its finite-color free inner product and null relations, and the dilatation operator including trace splitting and joining. These are new inputs. The full operator must preserve exact trace identities. McLoughlin–Pereira–Spiering, section II supplies a nonplanar operator and explicitly discusses finite-rank basis reduction; we have not substituted its operator for (D10.1) or computed a finite-$N$ anomalous dimension here. In particular, this failure of a planar substitution is not a proof that every finite-rank subsector is nonintegrable.

A bounded literature route

Literature checked 20 October 2026. This is a selected route through established starting points, not an exhaustive frontier survey or a declaration that the next calculation is open.

Read	What it establishes or supplies	What to carry forward
Minahan–Zarembo, v3, sections 2, 4 and 5.1	The planar one-loop scalar Hamiltonian and its Bethe description	Keep the loop order, trace condition and flavor sector fixed
Minahan, v3, section 8	An author-written derivation of two-magnon scattering and the multi-magnon equations	Distinguish periodic spin states from cyclic gauge-invariant operators
Gromov–Kazakov–Leurent–Volin, v2, introduction and Weak Coupling section	A quantum spectral curve formulation of the planar spectral problem, with a weak-coupling check	This advanced continuation uses analytic functions and finite-difference equations; it does not restore finite color rank
McLoughlin–Pereira–Spiering, v1, sections II–III	Nonplanar mixing data and numerical evidence for chaotic spectral statistics in specified perturbative sectors	Resolve symmetry sectors, exact trace relations and approximation order before comparing level statistics
Caputa and collaborators, June 2026 preprint, introduction and section 2.1	Numerical diagnostics of finite-loop planar truncations treated as exact spin Hamiltonians	This changes approximation order rather than color rank; its reported statistics do not establish chaos in the complete planar gauge theory

A realistic first report contains the length-six and length-seven matrices, their cyclic norms, the independent wave check, and the explicit failed finite-rank substitution. The next bounded choice is either more magnons at fixed planarity, or a separately sourced finite-color operator on a complete small multi-trace basis. Those changes test different assumptions. Neither a small matrix nor a failed literature search establishes novelty or a universal claim about quantum chaos.

Practice

1. Match operators, not only eigenvalues

At length six, use (D10.3) to derive the eigenoperators (D10.4), their stated norms, and the orthogonality of all three. Explain why diagonalizing the symmetric matrix on $(A,B,C)$ with all three norms set to one is not an equivalent prescription.

Hint 1

The coefficient vector obeys the transposed eigenproblem. Set its first entry to two for a nonzero eigenvalue.

Hint 2

Use $E^2-10E+20=0$ only after forming the norm $c^TG_6c$. Orthogonality is measured with $G_6$, not with the identity matrix.

Solution

The first row of $H_6^Tc=Ec$, with $c_1=2$, gives $c_2=2-E$. The third gives $-2c_2+4c_3=Ec_3$, hence $c_3=E-4$ after using the quadratic equation. The second row then vanishes by that same equation. At $E=0$, the solution is proportional to $(2,2,1)$.

The zero-mode norm is $c_6(2^2+2^2+2\cdot1^2)=10c_6$. For either nonzero eigenvalue,

$$\|\mathcal U_E\|_0^2/c_6 =4+(2-E)^2+2(E-4)^2 =10(E-2)>0.$$

The overlap with the protected state is $4+2(2-E)+2(E-4)=0$. For $E_++E_-=10$, $E_+E_-=20$, the two unprotected vectors have overlap $4+(2-E_+)(2-E_-)+2(E_+-4)(E_--4)=0$. Their dimensions are $6+\lambda(5\pm\sqrt5)/(8\pi^2)+O(\lambda^2)$.

The raw matrix is not self-adjoint in the identity metric. To use an ordinary symmetric matrix one must actually change basis to $(A,B,C/\sqrt2)$. Assigning unit norms while retaining the raw matrix or symmetrizing it by hand changes the problem. The stabilizer of $C$ is part of the free QFT correlator.

2. Length seven and a null formal solution

At $L=7$, take $A_7=\operatorname{Tr}(Z^5W^2)$, $B_7=\operatorname{Tr}(Z^4WZW)$, $C_7=\operatorname{Tr}(Z^3WZ^2W)$. Compute the cyclic matrix and its three eigenoperators. Compare with (D10.8), including the formal value $n=3$. For the lowest nonzero mode at general $L\geq5$, also bound the error of replacing $E_1$ by $8\pi^2/(L-1)^2$.

Hint 1

All three words have stabilizer one. At the maximal separation, some exchanges give a different representative of the same cyclic class.

Hint 2

Evaluate the wavefunction, not only its formal energy, at $p=\pi$. For the last bound use $x-x^3/6\leq\sin x\leq x$ when $0\leq x\leq1$.

Solution

The two moves from $A_7$ lead to $B_7$. From $B_7$, two lead to $A_7$ and two to $C_7$. From $C_7$, two lead to $B_7$ and two stay in its own cyclic class. The latter cancel between the identity and permutation terms. Thus

$$H_7=\begin{pmatrix}2&-2&0\\-2&4&-2\\0&-2&2\end{pmatrix}, \qquad G_7=c_7I.$$

The eigenoperators are $A_7+B_7+C_7$ at $E=0$, $A_7-C_7$ at $E=2$, and $A_7-2B_7+C_7$ at $E=6$. Their norms are $3c_7,2c_7,6c_7$. The two nonconstant momenta $p=\pi/3,2\pi/3$ reproduce $2,6$. Although $n=3$ formally gives $p=\pi$ and $E=8$, $\cos[\pi(d-1/2)]=0$ for all integer $d$. There is no fourth state. Its coincident rapidities are outside the generic two-root construction.

Finally let $x=\pi/(L-1)$, which lies below one for $L\geq5$. Squaring the positive lower bound on $\sin x$ gives $\sin^2x\geq x^2-x^4/3$. Consequently

$$0\leq\frac{8\pi^2}{(L-1)^2}-E_1 \leq\frac{8\pi^4}{3(L-1)^4}.$$

The relative error measured against the leading approximation is at most $\pi^2/[3(L-1)^2]$. This is a controlled large-length estimate within the planar one-loop model. It says nothing about a simultaneous scaling of $L$ with $N$ or about higher loops.

3. Test whether a map respects an exact relation

For traceless complex two-by-two matrices, define $z_2=\operatorname{Tr}Z^2$, $w_2=\operatorname{Tr}W^2$ and $u=\operatorname{Tr}ZW$. Derive $A,B,C$ in terms of these three invariants. Use them to verify (D10.11). State the minimal consistency test that any proposed finite-$N$ dilatation map must pass before its eigenvalues can be interpreted.

Hint 1

Cayley–Hamilton reduces $Z^4$ to a multiple of the identity. Apply the two-by-two trace identity to $ZW$ as well; this product need not be traceless.

Hint 2

For any two-by-two matrix $M$, $\operatorname{Tr}M^2=(\operatorname{Tr}M)^2-2\det M$. Here $\det Z=-z_2/2$ and $\det W=-w_2/2$.

Solution

Using $Z^2=z_2I/2$ and $W^2=w_2I/2$ gives $A=C=z_2^2w_2/4$. Also $\operatorname{Tr}[(ZW)^2]=u^2-z_2w_2/2$. Therefore

$$B=\frac{z_2u^2}{2}-\frac{z_2^2w_2}{4},\qquad \mathsf H_{\rm planar}(A-C) \equiv z_2u^2-z_2^2w_2.$$

For the explicit matrices in the text, $z_2=w_2=2$ and $u=0$, so this is $-8$. The input is identically zero for every pair of traceless two-by-two matrices, while the proposed image is not.

A linear map on physical operators must preserve the ideal of exact matrix trace relations: equivalent expressions must have equivalent images. Equivalently, it must map free-inner-product null vectors to null vectors on the complete finite-color space. The planar approximation fails this test at the changed rank. The repair requires the omitted finite-color mixing, including the appropriate multi-trace sector, rather than interpreting its spurious eigenvalues as finite-$N$ physics.