A short N=4 spin chain

Can a matrix have the right eigenvalues and still give the wrong operator norms? Yes: cyclic traces with different repetition patterns have different free two-point functions. This laboratory makes that distinction executable. Starting from the cyclic spin-chain lesson, you will construct a short basis, find its scaling operators, and deliberately discard the norm factors to see which checks fail.

Fix the theory and the approximation

We use four-dimensional $\mathcal N=4$ $SU(N)$ Yang–Mills in its conformal vacuum. The parent has sixteen real Lorentzian Poincaré supercharges, with four manifest in an $\mathcal N=1$ description; the gauge rank is $N-1$. No curved background or localization charge is introduced. The two auxiliary spin values label scalar flavor, not spacetime spin or a two-dimensional gauge representation.

Keep $L$ fixed, take $N\to\infty$ with weak $\lambda=g_{\rm YM}^2N$ fixed, and retain planar one-loop mixing. With ordinary $\operatorname{Tr}(t^at^b)=\delta^{ab}/2$, the canonical complex scalars are

$$\begin{aligned} Z&=\frac{X^1+iX^2}{\sqrt2},\qquad W=\frac{X^3+iX^4}{\sqrt2},\\ \Gamma&=\frac{\lambda}{8\pi^2}\mathsf H+O(\lambda^2),\qquad \mathsf H=\sum_{j=1}^{L}(1-P_{j,j+1}),\quad L+1\equiv1. \end{aligned} \tag{LAB08.1}$$

$P$ exchanges neighboring letters. The coefficient is imported from the actual one-loop calculation, not inferred by diagonalizing a toy chain: MZ (2.9)–(2.13) reduces to Minahan (8.1) in this holomorphic sector. The prerequisite explains that reduction.

The program fixes two $W$ fields, with $L-2$ fields $Z$, and accepts $4\leq L\leq7$. Its numbers $E$ give anomalous dimensions $\gamma=\lambda E/(8\pi^2)+O(\lambda^2)$ within the planar approximation. They are dimensionless, not particle energies. Increasing $L$ does not implement finite-$N$ interactions.

The leading double-line contraction agrees with the corresponding $U(N)$ bookkeeping term. The chosen theory remains $SU(N)$; a physical $U(N)$ theory would additionally contain a free central $U(1)$ multiplet.

From a trace to a computational basis

Represent $Z$ by zero and $W$ by one. Generate the $\binom L2$ binary words with exactly two ones. For each word, take the lexicographically smallest of its $L$ rotations as the class representative. This removes cyclic duplicates; it does not identify a word with its reversal.

Let $d_i$ be the number of distinct rotations of representative $i$, and $s_i=L/d_i$ its stabilizer: the number of shifts that return the original word. The raw trace $\mathcal O_i$ has leading free norm $s_i$. More precisely, at separation $r_x>0$,

$$\langle\mathcal O_i(x)\mathcal O_j^\dagger(0)\rangle_0 =\left(\frac{N}{8\pi^2r_x^2}\right)^L G_{ij} +\text{subleading color terms},\qquad G_{ij}=s_i\delta_{ij}. \tag{LAB08.2}$$

The common factor will be removed in the output. The conjugate trace reverses matrix order. Counting its allowed cyclic planar contractions gives $s_i$, as in Minahan (5.8)–(5.9); his canonical matrix fields are $\sqrt2$ times ours. Equivalently, a raw trace maps to $\sqrt{s_i}$ times a normalized orbit state. This explains why counting each orbit once does not make the trace basis orthonormal.

Act on every representative with all $L$ bonds, including the last-to-first bond, and collect the swapped words into cyclic classes. Our matrix convention is

$$\mathsf H\boldsymbol{\mathcal O}=H\boldsymbol{\mathcal O},\qquad \mathcal O_c=c^T\boldsymbol{\mathcal O},\qquad H^Tc=Ec,\qquad HG=GH^T. \tag{LAB08.3}$$

Thus eigenoperator coefficients use the transpose. Their norm is $c^\dagger Gc$; our displayed eigenvectors can be chosen real. The ordinary symmetric matrix is $G^{-1/2}HG^{1/2}$, not $H$ itself.

Mathematics	Program name
Minimal rotation of a binary word	representative(word)
Distinct rotations, $d_i$	rotations(word), sizes
Operator-column matrix, $H$	H
Free planar Gram matrix, $G$	gram
Coefficient eigenproblem	H.T.eigenvects()

The central construction is short. Here index maps representatives to their positions:

for i, word in enumerate(basis):
    for j in range(length):
        H[i, i] += 1
        changed = representative(swap(word, j))
        H[i, index[changed]] -= 1

If a swap stays in the same cyclic class, its two terms cancel. The complete downloadable program also checks the basis size without enumerating classes twice: for a rotation by $r$, there are $g=\gcd(L,r)$ cycles of length $L/g$. An invariant two-$W$ word exists only when $L/g$ divides two, in which case there are $\binom{g}{2g/L}$ choices. Averaging this fixed-word count over the $L$ rotations counts the cyclic classes, because every orbit contributes exactly $L$ pairs of a word and a shift fixing it.

Reproduce the length-four result

Download these three files into one directory:

Download the complete lab (ZIP) with the program, pinned dependencies, reference output and run instructions.

• Standalone Python calculation
• Pinned dependencies
• Complete expected default output

The tested environment is Python 3.13.7, SymPy 1.13.3 and mpmath 1.3.0. In a virtual environment, run:

python -m pip install -r requirements.txt
python spin_chain.py
python spin_chain.py --length 6
python spin_chain.py --wrong-gram

The default calculation uses exact integer and algebraic arithmetic for the cyclic matrix and eigenproblem. Decimal arithmetic is used only in the independent evolution check described below. No network access or repository files are needed once the dependencies and three downloads are present.

For $L=4$, the representatives are $A=\operatorname{Tr}(ZZWW)$ and $B=\operatorname{Tr}(ZWZW)$. The output contains

$$(d_A,d_B)=(4,2),\qquad G=\begin{pmatrix}1&0\\0&2\end{pmatrix},\qquad H=\begin{pmatrix}2&-2\\-4&4\end{pmatrix}. \tag{LAB08.4}$$

Only two bonds of $A$ change its class; all four bonds of $B$ do. The characteristic polynomial is $E(E-6)$, and the coefficient eigenvectors are

$$\begin{array}{c|c|c} \text{operator}&E&\text{free norm after removing the common factor}\\\hline 2A+B&0&6\\ A-B&6&3 \end{array} \tag{LAB08.5}$$

Their cross norm is zero. Hence the second operator has $\Delta=4+3\lambda/(4\pi^2)+O(\lambda^2)$ at planar one loop.

Why call the first one protected? Its coefficients count the words in the $\alpha^{L-2}\beta^2$ term of $\operatorname{Tr}(\alpha Z+\beta W)^L$. At length four that term is $4A+2B$. The null complex scalar polarization identifies the symmetric traceless half-BPS family, whose protection uses supersymmetry; see MZ (3.1) and the prerequisite. A zero eigenvalue alone would not prove all-order protection of an arbitrary operator. In contrast, $\operatorname{Tr}[Z,W]^2=-2(A-B)$ is holomorphic but has a nonzero one-loop shift.

Check the answer without diagonalizing again

A second construction starts with every labeled binary word, before taking cyclic classes. At unit rate for each bond, swap its two letters. Equal letters simply leave the word unchanged. This defines an auxiliary random walk with total event rate $L$; it is a mathematical diagnostic, not real or Euclidean time in Yang–Mills theory.

Count the directed events from every word in orbit $i$ to orbit $j$. For $i\ne j$, call this count $w_{ij}$. Reversing the swap proves $w_{ij}=w_{ji}$. The matrix with off-diagonal entries $-w_{ij}$ and diagonal entries $\sum_{j\ne i}w_{ij}$ is a symmetric graph Laplacian $B$. Independent counting should give

$$D=\operatorname{diag}(d_i),\qquad B=DH,\qquad f^TBf=\sum_{i<j}w_{ij}(f_i-f_j)^2\geq0. \tag{LAB08.6}$$

Here $f_i$ is any real test value assigned to a class. This confirms positivity and the correct orbit weights with a different data path from the representative action. The stationary probabilities are $d_i/\binom L2$, while the protected operator coefficients are proportional to $d_i$. Neither is generally a uniform list over classes.

There is also an exact spectral check using only a cofactor, the determinant left after deleting one row and the same column from $B$. All such cofactors equal $\tau$ for this connected graph. With $n$ classes, the coefficient of $E$ in $\det(ED-B)$ is $(-1)^{n-1}\tau\sum_i d_i$: choose one diagonal mass and the complementary minor of $-B$. After dividing by $\det D$ and comparing with $\det(EI-H)$, one obtains

$$\prod_{E_a\ne0}E_a=\frac{(\sum_i d_i)\tau}{\prod_i d_i}. \tag{LAB08.7}$$

At length four, $B=\left(\begin{smallmatrix}8&-8\\-8&8\end{smallmatrix}\right)$, so $\tau=8$ and the product is $6$. At length six, the independently counted conductances along the three-class path are $12,12$; thus $\tau=144$ and the product is $20$. The full spectrum must also have the computed trace: these tests can catch a wrong eigenvalue without a second eigensolver.

Finally, sum the actual labeled-word walk over a Poisson-distributed number of swaps. If $P$ is its single-event transition matrix, the evolution is

$$e^{-Lt}\sum_{n=0}^{\infty}\frac{(Lt)^n}{n!}P^n. \tag{LAB08.8}$$

Starting uniformly on one orbit and summing final probabilities by orbit should reproduce the corresponding row of $e^{-tH}$. The code compares these constructions at $t=0.4$. If terms through $n=M$ are retained and $M+2>Lt$, the omitted probability is bounded by

$$\frac{e^{-Lt}(Lt)^{M+1}}{(M+1)!} \frac1{1-Lt/(M+2)}. \tag{LAB08.9}$$

Every entry’s truncation error is at most this tail. At the default 50-digit precision, the length-four comparison differs by about $1.34\times10^{-51}$, with a separate Poisson tail below $3.46\times10^{-57}$. The acceptance tolerance is $10^{-40}$. The tail inequality does not bound floating-point roundoff; raising --dps repeats that numerical check. None of these computational errors estimates omitted higher-loop or finite-color effects.

Deliberately discard the metric

The --wrong-gram mode replaces $G$ by the identity only in the diagnostic. It does not modify $H$, so its eigenvalues remain $0,6$. Instead it finds

$$H-H^T=\begin{pmatrix}0&2\\-2&0\end{pmatrix}\ne0, \qquad (2,1)\begin{pmatrix}1\\-1\end{pmatrix}=1. \tag{LAB08.10}$$

Hermiticity and distinct-eigenvalue orthogonality fail with that false norm. The output says intentional_error_detected: true; successful exit means the program detected the deliberate error. The ordinary checks still use the correct underlying matrix and Gram data. Use this mode at lengths four or six: at odd lengths in the supplied range all stabilizers are one, so replacing the Gram matrix by the identity does not introduce this particular error, and the program rejects an uninformative negative test.

Change an input and interpret the result

1. Six sites, still two impurities

Run python spin_chain.py --length 6. For the ordered basis $A_6=\operatorname{Tr}(ZZZZWW)$, $B_6=\operatorname{Tr}(ZZZWZW)$ and $C_6=\operatorname{Tr}(ZZWZZW)$, determine the stabilizers, protected combination and both nonzero anomalous dimensions. Check the eigenvalue sum and product independently.

Hint 1

The first two words have six distinct rotations. The last repeats a block of length three.

Hint 2

Use the coefficient transpose, and compare the constant term of the nonzero quadratic factor with (LAB08.7).

Solution

The orbit sizes are $(6,6,3)$ and $G=\operatorname{diag}(1,1,2)$. Directly collecting all six bond swaps gives

$$H_6=\begin{pmatrix}2&-2&0\\-2&4&-2\\0&-4&4\end{pmatrix}, \qquad\det(EI-H_6)=E(E^2-10E+20).$$

Thus $2A_6+2B_6+C_6$ is the protected coefficient combination, with norm $4+4+2=10$. The nonzero values are $E_\pm=5\pm\sqrt5$, so

$$\Delta_\pm=6+\frac{\lambda}{8\pi^2}(5\pm\sqrt5)+O(\lambda^2) \quad\text{at planar order}.$$

For either nonzero $E$, one coefficient vector is $c=(1,(2-E)/2,(E-2)/(E-4))^T$; substitution into $H_6^Tc=Ec$ checks all three rows. Its Gram inner product with $(2,2,1)^T$ vanishes using $E^2-10E+20=0$. The eigenvalue sum is $10=\operatorname{tr}H_6$, and their product is $20$, independently given by $15\times144/(6\times6\times3)$. The irrational answers are exact algebraic numbers, not numerical noise.

2. Why a passed spectrum test is insufficient

Someone reports the correct length-four eigenvalues but assigns unit norm to both $A$ and $B$. Compute the resulting norms and cross norm of the two operators in (LAB08.5). Then repair the basis and write the symmetric mixing matrix. Which conclusion could the original eigenvalue test still support?

Hint 1

Use the same coefficient vectors before and after changing the metric. The eigenproblem itself has not changed.

Hint 2

Divide each trace representative by the square root of its actual norm.

Solution

The false metric gives norms $5$ and $2$, and cross norm $1$. The true values are $6,3,0$. In the orthonormal basis $(A,B/\sqrt2)^T$,

$$H_{\rm orth}=G^{-1/2}HG^{1/2} =\begin{pmatrix}2&-2\sqrt2\\-2\sqrt2&4\end{pmatrix}.$$

It has the same eigenvalues because this is a similarity transformation. The original spectrum check could therefore confirm the two one-loop shifts, while failing to verify normalization, orthogonality or the mapping to physical two-point functions. A finite list of eigenvalues is not a complete operator-mixing calculation.

3. A short chain is not a small gauge group

Now choose actual gauge group $SU(2)$, so $Z,W$ are traceless complex $2\times2$ matrices. Use Cayley–Hamilton to express $\operatorname{Tr}(Z^2W^2)$ as a double trace. Does the existence of the finite two-by-two matrix (LAB08.4) justify treating its eigenvalues as exact finite-$N=2$ anomalous dimensions?

Hint 1

For a traceless $2\times2$ matrix, its characteristic polynomial has no linear term. Relate its determinant to $\operatorname{Tr}(Z^2)$.

Hint 2

The laboratory retained single traces at leading color order. The identity relates that space to products of traces.

Solution

Cayley–Hamilton gives $Z^2=\tfrac12\operatorname{Tr}(Z^2)I_2$, including for complex non-Hermitian $Z$. Consequently

$$\operatorname{Tr}(Z^2W^2) =\frac12\operatorname{Tr}(Z^2)\operatorname{Tr}(W^2).$$

The identity is exact at gauge rank one. It is not imposed by grouping rotations of four abstract letters. At finite color, single and multiple traces can be related and interact through terms absent from this planar Hamiltonian; the finite-color free Gram matrix also differs from (LAB08.2). Therefore the lab has not computed a finite-$N=2$ spectrum. Some protected combinations retain exact dimensions for independent supersymmetric reasons, but a nonzero planar eigenvalue cannot be upgraded by inserting $N=2$ into $\lambda$. The fixed-length limit and the physical gauge group are separate inputs.

For the next controlled step, the spectrum research dossier develops longer-word tests and an explicit finite-rank obstruction. Here the reproducible achievement is narrower: a correctly normalized cyclic basis, a protected family member, nonzero planar one-loop shifts, and checks that can expose a plausible but incomplete calculation.