A first cyclic spin chain

The mixing lesson explained what a logarithmic matrix means. For a longer scalar trace, how can we construct that matrix without listing every composite counterterm separately? At planar one loop, only neighboring fields along the trace interact. We can encode their flavors as a short quantum spin system and calculate its spectrum.

We work in four-dimensional $\mathcal N=4$ $SU(N)$ Yang–Mills in its conformal vacuum, with sixteen real Lorentzian Poincaré supercharges and four manifest in an $\mathcal N=1$ description. The gauge rank is $N-1$. Take $N$ large at fixed weak $\lambda=g_{\rm YM}^2N$, keep the trace length fixed, and retain planar one-loop mixing. The spins below describe scalar flavor, not spacetime angular momentum or the gauge group.

Two complex scalars make a two-state site

Use the same canonical adjoint components and ordinary trace as before:

$$Z=\frac{X^1+iX^2}{\sqrt2},\qquad W=\frac{X^3+iX^4}{\sqrt2},\qquad \operatorname{Tr}(t^at^b)=\frac12\delta^{ab}.$$

We consider words containing only $Z$ and $W$, with neither conjugates nor derivatives. For instance, $\operatorname{Tr}(ZZWW)$ contains four fields and has classical dimension four. The R-symmetry subgroup rotating $(Z,W)$ is called the scalar-sector $SU(2)$. It is present even though the gauge group is $SU(N)$ for general $N$.

Introduce an auxiliary two-dimensional vector space with basis $|Z\rangle=|\uparrow\rangle$, $|W\rangle=|\downarrow\rangle$. A word of length $L$ becomes a tensor-product vector $|s_1\cdots s_L\rangle$. This is a bookkeeping space for local operators. It is not a chain of particles at spatially separated positions in the original QFT.

The permutation $\mathbb P_{\ell,\ell+1}$ exchanges two adjacent letters. On equal letters it acts trivially; on a mixed pair, $(1-\mathbb P)|ZW\rangle=|ZW\rangle-|WZ\rangle$. The complete scalar kernel derived in the prerequisite contains $\mathbb K+2-2\mathbb P$ with coefficient $\lambda/(16\pi^2)$. Here $\mathbb K$ vanishes. Indeed the complex flavor polarizations of $Z,W$ obey $z\cdot z=w\cdot w=z\cdot w=0$, where the dot is the complex bilinear $SO(6)$ contraction, without conjugation.

Consequently the dimensionless mixing operator is

$$\Gamma=a\mathsf H+O(\lambda^2),\qquad a=\frac{\lambda}{8\pi^2},\qquad \boxed{\mathsf H=\sum_{\ell=1}^{L} (1-\mathbb P_{\ell,\ell+1}),\quad L+1\equiv1.} \tag{N10.1}$$

The coefficient is dynamical input from the full one-loop calculation, not something fixed by the spin analogy. MZ (2.9)–(2.13), Minahan (8.1). The two-flavor restriction preserves the required charges; the source explains its closed-sector meaning in section 7. We use only its one-loop Hamiltonian here.

Periodicity and trace cyclicity impose different conditions

The bond from site $L$ to site $1$ makes (N10.1) periodic. But a single trace has a further property:

$$\operatorname{Tr}(s_1s_2\cdots s_L) =\operatorname{Tr}(s_2\cdots s_Ls_1).$$

If $\mathsf T$ shifts every site by one, a physical single-trace state must satisfy $\mathsf T|\psi\rangle=|\psi\rangle$. The projector onto this subspace is

$$\Pi_{\rm cyc}=\frac1L\sum_{j=0}^{L-1}\mathsf T^j. \tag{N10.2}$$

It obeys $\Pi_{\rm cyc}^2=\Pi_{\rm cyc}$ and commutes with $\mathsf H$, because a shift merely relabels the sum of bonds. Periodicity of the Hamiltonian alone would still allow nonzero translation momentum. Such states need not represent single traces.

Cyclicity also affects normalization. Write $v=1/(8\pi^2r^2)$ for the leading matrix propagator at separation $r$. At leading order in $N$, the free correlator of two length-$L$ words is

$$\langle\mathcal O_w(x)\mathcal O_{w'}^\dagger(0)\rangle_0 =(Nv)^L\sum_{j=0}^{L-1} \langle w'|\mathsf T^j|w\rangle +\text{subleading color terms}. \tag{N10.3}$$

The conjugate trace has reversed matrix order, as required by Hermitian conjugation. Each allowed cyclic planar contraction contributes $N^Lv^L$. Different flavor sequences contract only when their cyclic classes coincide. This is the trace version of Minahan (5.8)–(5.9); his canonical matrix fields are $\sqrt2$ times ours, explaining our $v$ rather than $1/(4\pi^2r^2)$.

Let $s_w$ count the shifts that leave a word unchanged. Its cyclic orbit has $d_w=L/s_w$ distinct words. The normalized orbit state is $|[w]\rangle=d_w^{-1/2}\sum_{\text{distinct rotations}}|w\rangle$. Equation (N10.3) identifies the trace, after removing the common factor $(Nv)^{L/2}$, with

$$\sqrt L\,\Pi_{\rm cyc}|w\rangle =\sqrt{s_w}\,|[w]\rangle . \tag{N10.4}$$

Thus raw trace representatives are generally not an orthonormal spin basis. A periodically repeating word has a larger free norm. This factor must accompany any matrix diagonalization.

Work a length-four example

With two $Z$ and two $W$ fields, there are only two cyclic classes:

$$A=\operatorname{Tr}(ZZWW),\qquad B=\operatorname{Tr}(ZWZW).$$

The orbit of $A$ has four members; the orbit of $B$ has two. Their stabilizers are $s_A=1,s_B=2$. Put $c_4=N^4/(8\pi^2)^4$, so the free matrix is

$$\langle\boldsymbol{\mathcal O}(x) \boldsymbol{\mathcal O}(0)^\dagger\rangle_0 =\frac{G_0}{r^8},\qquad \boldsymbol{\mathcal O}=\begin{pmatrix}A\\B\end{pmatrix}, \qquad G_0=c_4\begin{pmatrix}1&0\\0&2\end{pmatrix}. \tag{N10.5}$$

Act on each representative before collecting its cyclic class:

Representative	Bonds with unequal letters	Swapped trace at each such bond
$ZZWW$	$(2,3)$ and $(4,1)$	$B,B$
$ZWZW$	All four bonds	$A,A,A,A$

The other bonds give zero. Hence

$$\mathsf H A=2A-2B,\qquad \mathsf H B=-4A+4B,\qquad \Gamma=a\begin{pmatrix}2&-2\\-4&4\end{pmatrix} \quad\text{on the operator column}. \tag{N10.6}$$

As in the prerequisite, the coefficient column of an eigenoperator obeys the transposed eigenproblem. The apparently nonsymmetric matrix is compatible with the inner product:

$$\Gamma G_0=G_0\Gamma^T =ac_4\begin{pmatrix}2&-4\\-4&8\end{pmatrix}.$$

Equivalently, use the normalized representatives $A$ and $B/\sqrt2$. Their dimensionless matrix is the real symmetric matrix

$$\mathsf H_{\rm orth} =\begin{pmatrix}2&-2\sqrt2\\-2\sqrt2&4\end{pmatrix}. \tag{N10.7}$$

It has zero determinant and trace six. The eigenoperators and free norms are

$$\begin{array}{c|c|c} \text{operator}&\text{norm coefficient}&\text{one-loop }\gamma\\ \hline \mathcal P=2A+B&6c_4&0\\ \mathcal U=A-B&3c_4&6a \end{array} \qquad \langle\mathcal P\mathcal U^\dagger\rangle_0=0 . \tag{N10.8}$$

Therefore $\Delta_{\mathcal U}=4+3\lambda/(4\pi^2)+O(\lambda^2)$. The nonzero eigenvalue is positive and dimensionless, as it should be. It is not a particle energy in spacetime.

There is an independent way to identify the protected combination. The coefficient of $\alpha^2\beta^2$ in $\operatorname{Tr}(\alpha Z+\beta W)^4$ is $4A+2B=2\mathcal P$. The scalar $\alpha Z+\beta W$ has null complex $SO(6)$ polarization, so this is a component of the symmetric traceless half-BPS family. Its exact protection uses shortening, in addition to the one-loop test.

In contrast, $\operatorname{Tr}[Z,W]^2=2(B-A)=-2\mathcal U$. A holomorphic expression need not be protected just because it contains no conjugate fields. The kernel calculation distinguishes these two combinations. The coincidence of its nonzero one-loop shift with another operator’s shift would not, on its own, prove a supermultiplet identification.

Why the spin Hamiltonian is positive

Let $S_\ell^i=\tau^i/2$ act on the two-state flavor space at site $\ell$. On two spins the symmetric triplet has permutation eigenvalue $+1$ and $S_\ell\cdot S_{\ell+1}=1/4$; the antisymmetric singlet has eigenvalues $-1$ and $-3/4$. These two subspaces exhaust the pair, so

$$\mathbb P_{\ell,\ell+1} =2S_\ell\cdot S_{\ell+1}+\frac12,\qquad \mathsf H=\sum_{\ell=1}^L \left(\frac12-2S_\ell\cdot S_{\ell+1}\right). \tag{N10.9}$$

This is the ferromagnetic Heisenberg Hamiltonian: neighboring aligned flavors cost less auxiliary energy. More directly, $1-\mathbb P$ has eigenvalues zero and two, so every bond is positive. For any state,

$$\langle\psi|\mathsf H|\psi\rangle =\frac12\sum_\ell \|(1-\mathbb P_{\ell,\ell+1})|\psi\rangle\|^2\geq0.$$

A zero mode must be symmetric under every adjacent exchange. These exchanges generate all permutations, so at fixed numbers of $Z,W$ the zero mode is the fully symmetric word sum. Its cyclic orbit weights reproduce $2A+B$, not $A+B$.

A direct check can build the full sixteen-dimensional tensor-spin matrix for four sites from (N10.9), restrict to two down spins and project with (N10.2). It gives the same two-dimensional matrix (N10.7), without choosing raw trace representatives first. The closed chain and the cyclic projection are both essential.

Practice

1. Delete the closing bond

Suppose someone uses the open-chain sum $\mathsf H_{\rm open}=\sum_{\ell=1}^{3}(1-\mathbb P_{\ell,\ell+1})$ for the length-four trace. Compare its action on the two representatives $ZZWW$ and $ZWWZ$, which describe the same trace. Does it define a consistent map on cyclic classes?

Hint 1

List the unequal adjacent letters without allowing the fourth-to-first bond.

Hint 2

Collect the swapped words into $A$ or $B$ only after acting.

Solution

For $ZZWW$, only bond $(2,3)$ contributes, giving $A-B$. For $ZWWZ$, bonds $(1,2)$ and $(3,4)$ contribute, giving $2A-2B$. The two inputs represent the same trace but have different images. Thus the proposed map is not well defined on cyclic classes.

Restoring $(4,1)$ adds $A-B$ for the first representative and zero for the second. Both then give $2A-2B$. The missing term was not a harmless change in an eigenvalue: it broke the trace identification.

2. A single flipped flavor and the trace condition

On a periodic chain of length $L\geq3$, let $|j\rangle$ have one $W$ at site $j$ and $Z$ elsewhere. Derive the eigenvalue on $|p\rangle=L^{-1/2}\sum_j e^{ipj}|j\rangle$. Which of these states actually survive as single-trace operators?

Hint 1

Only the two bonds touching $W$ contribute. Each moves it one site.

Hint 2

Distinguish $\mathsf T^L=1$ from the stronger condition $\mathsf T|\psi\rangle=|\psi\rangle$.

Solution

The local action is $\mathsf H|j\rangle=2|j\rangle-|j-1\rangle-|j+1\rangle$, with site labels understood modulo $L$. Fourier transformation gives

$$\gamma(p)=a(2-e^{ip}-e^{-ip}) =4a\sin^2\frac p2,\qquad p=\frac{2\pi n}{L}.$$

Periodicity permits these discrete momenta. A one-step shift multiplies the state by $e^{ip}$ or $e^{-ip}$, depending on the shift convention; cyclicity requires this eigenvalue to equal one. Only $p=0$ survives. Indeed all $\operatorname{Tr}(Z^{j-1}WZ^{L-j})$ are the same operator, so a nonzero-momentum weighted sum maps to the zero operator.

The surviving sum is a symmetric component of the half-BPS family and has zero shift. The other spin-chain eigenvalues are legitimate energies of the auxiliary periodic chain, but are not dimensions of single-trace one-$W$ operators. This is the distinction in Minahan (8.3)–(8.5) and the following paragraph.

3. Lengthen the trace

With three $Z$ and two $W$ fields, use $A_5=\operatorname{Tr}(ZZZWW)$ and $B_5=\operatorname{Tr}(ZZWZW)$. Determine their cyclic stabilizers, the planar Gram matrix and both one-loop eigenoperators. Is the nonzero shift the same as at length four?

Hint 1

Both words have five distinct rotations. When a swap of $B_5$ gives another representative of $B_5$, its contribution to $1-\mathbb P$ cancels.

Hint 2

There are four mixed bonds in $B_5$, but only two change its cyclic class.

Solution

Both stabilizers equal one, so $G_0^{(5)}=c_5 I$, where $c_5=N^5/(8\pi^2)^5$ and the correlator includes $r^{-10}$. Acting on all five bonds gives

$$\mathsf H A_5=2A_5-2B_5,\qquad \mathsf H B_5=2B_5-2A_5 .$$

For $B_5$, the swaps at bonds $(2,3)$ and $(5,1)$ remain in its own cyclic class; those at $(3,4)$ and $(4,5)$ produce $A_5$. The combinations $A_5+B_5$ and $A_5-B_5$ have norms $2c_5$ and shifts zero and $4a$, respectively. The protected one is proportional to the two-$W$ coefficient of $\operatorname{Tr}(\alpha Z+\beta W)^5$. The other has $\Delta=5+\lambda/(2\pi^2)+O(\lambda^2)$, different from the length-four answer. Trace length and cyclic adjacency are physical inputs to this planar mixing problem.

The calculation describes a fixed-length planar sector. The leading $SU(N)$ propagator agrees with the double-line $U(N)$ bookkeeping term, but this does not change the gauge theory to $U(N)$. Finite-$N$ trace relations, subleading free overlaps and interacting trace splitting or joining require additional information. Likewise, diagonalizing the one-loop chain does not determine its higher-loop corrections or prove general integrability. What it does provide is a concrete bridge from composite renormalization to a finite, cyclic spectral problem with the correct physical inner product.

For a different kind of observable, begin supersymmetric Wilson lines and circles. Its scalar coupling preserves specified combinations of supercharges, while its value still depends on the coupling.

For a controlled continuation, derive the two-magnon spectrum at longer trace length, then test why the planar map fails to respect an exact finite-rank trace identity.