Check a rational R-matrix

Why does the XXX chain possess a whole family of commuting operators? The starting point is a local identity on just three two-dimensional spaces. You will prove that identity for the rational R-matrix, identify the spectral differences that make it work, and test its exceptional parameters. This opens the algebraic sequence leading to commuting charges, the periodic-chain Hamiltonian and regular Bethe states; it needs tensor products, not prior representation theory or a Bethe completeness theorem.

Required background. Act with matrices on tensor products and compose operators from right to left; the linear algebra bridge develops these operations. Construct a small spin-chain Hamiltonian applies them to physical spin operators and the spin permutation.

Helpful background. The XXX course gives the coordinate-wave viewpoint. The R-matrix convention reference distinguishes the related operator normalizations.

Swaps on three tensor factors

Let $V=\mathbb C^2$ with basis $|\uparrow\rangle,|\downarrow\rangle$. All tensor products here are ordinary, with no signs introduced by exchanging factors. On $V\otimes V$, the permutation $P$ is defined by

$$P(v\otimes w)=w\otimes v, \qquad P^2=I.$$

On $V_1\otimes V_2\otimes V_3$, the subscript tells us which two factors are exchanged. Thus

$$P_{13}(v_1\otimes v_2\otimes v_3) =v_3\otimes v_2\otimes v_1.$$

The labels name tensor factors, not matrix entries. Put $A=P_{12}$, $B=P_{13}$, $C=P_{23}$. Acting on an arbitrary elementary tensor proves

$$AB=BC=CA,\qquad BA=CB=AC,\qquad ABC=CBA=B.$$

For example, $AB$ first exchanges factors $1,3$, then $1,2$, sending the triple $(v_1,v_2,v_3)$ to $(v_2,v_3,v_1)$. The products $BC$ and $CA$ do the same. Equality on elementary tensors implies equality on their linear span, hence on the full tensor product. These are the permutation relations used by Faddeev 1996, § 3, equations (33)–(39), PDF.

Entry check and repair

Do $P_{12}$ and $P_{23}$ commute? Apply both orders to $|\downarrow\uparrow\uparrow\rangle$.

Repair. $P_{12}P_{23}$ gives $|\uparrow\downarrow\uparrow\rangle$, whereas $P_{23}P_{12}$ gives $|\uparrow\uparrow\downarrow\rangle$. They share a tensor factor and do not commute. Operators on disjoint pairs, such as $P_{12}$ and $P_{34}$ on four factors, do commute. A drawing of crossing lines does not replace this ordering check.

The rational Yang–Baxter identity

For a dimensionless complex spectral parameter $u$, define

$$R(u)=uI+iP.$$

In the ordered basis $\uparrow\uparrow,\uparrow\downarrow,\downarrow\uparrow,\downarrow\downarrow$,

$$R(u)= \begin{pmatrix} u+i&0&0&0\\ 0&u&i&0\\ 0&i&u&0\\ 0&0&0&u+i \end{pmatrix}.$$

The relation to verify is

$$\begin{aligned} &R_{12}(a)R_{13}(a+b)R_{23}(b)\\ &\hspace{1.5em}=R_{23}(b)R_{13}(a+b)R_{12}(a). \end{aligned}$$

Here $R_{12}$, for example, acts as the identity on factor $3$. The parameters can also be written $a=\lambda-\mu$, $b=\mu-\nu$, and $a+b=\lambda-\nu$. The middle argument is constrained by these differences.

To see exactly where the constraint matters, keep the middle argument independent and call it $z$. Expand

$$\begin{aligned} \mathcal D(a,z,b)={}&(aI+iA)(zI+iB)(bI+iC)\\ &-(bI+iC)(zI+iB)(aI+iA). \end{aligned}$$

The terms with zero or one permutation cancel immediately. The three-permutation terms cancel because $ABC=CBA$. The two-permutation terms are

$$-bAB-zAC-aBC+aCB+zCA+bBA.$$

Use $AB=BC=CA$ and $BA=CB=AC$ to obtain

$$\mathcal D(a,z,b)=(z-a-b)(AB-BA).$$

Setting $z=a+b$ proves the Yang–Baxter identity for every complex $a,b$, including values at which an individual R-matrix is singular. No inverse was used. The Library proof connects this local identity to the chain-wide exchange relation.

Regularity and exceptional parameters

The symmetric and antisymmetric subspaces of $V\otimes V$ have projectors

$$\Pi_+=\frac{I+P}{2},\qquad \Pi_-=\frac{I-P}{2}.$$

They are respectively the three-dimensional triplet and one-dimensional singlet spaces. Therefore

$$R(u)=(u+i)\Pi_++(u-i)\Pi_-.$$

At $u=0$, $R(0)=iP$ is an invertible permutation times a scalar. This property is called regularity and will turn a transfer matrix into translation. At $u=i$, the singlet eigenvalue vanishes; at $u=-i$, the triplet eigenvalue vanishes. Away from those two points,

$$R(u)^{-1}=\frac{uI-iP}{u^2+1}.$$

For a quick worked check, at $u=2$ the two eigenvalues are $2+i$ and $2-i$. The inverse acts with their reciprocals. The matrix itself is not unitary: $R(2)^\dagger R(2)=5I$. Singularity, regularity and unitarity are distinct statements.

This R-matrix is an auxiliary algebraic object. It is not the two-magnon amplitude ratio $S_{12}$ from the coordinate lesson, and its complex spectral parameter is not automatically a physical lattice momentum.

Exercises

Guided practice: complete a permutation proof

Apply $BC$ and $CA$ to $(v_1,v_2,v_3)$ and compare with the stated action of $AB$. Then use $B^2=I$ or a direct action to establish $ABC=B$.

Hint

Keep the rightmost operation first. For $BC$, exchange $2,3$ before exchanging $1,3$.

Solution

The products act as

$$\begin{aligned} BC:(v_1,v_2,v_3)&\mapsto(v_1,v_3,v_2)\mapsto(v_2,v_3,v_1),\\ CA:(v_1,v_2,v_3)&\mapsto(v_2,v_1,v_3)\mapsto(v_2,v_3,v_1). \end{aligned}$$

Thus both equal $AB$. For $ABC$, first $C$ sends the triple to $(v_1,v_3,v_2)$, then $AB$ sends it to $(v_3,v_2,v_1)$, which is exactly $B$. Reversing every factor order similarly gives $CBA=B$. These identities hold for arbitrary vectors, not only spin-basis examples.

Independent practice: reject a false spectral assignment

Choose $a=1$, $b=2$, but use middle argument $z=0$. Compute $\mathcal D(1,0,2)|\uparrow\downarrow\uparrow\rangle$ without multiplying three $8\times8$ matrices. Does checking the Yang–Baxter equation only on $|\uparrow\uparrow\uparrow\rangle$ detect this error?

Hint

Use the derived factor $(z-a-b)(AB-BA)$. Every permutation acts identically on the all-up vector.

Solution

Here $z-a-b=-3$, while

$$(AB-BA)|\uparrow\downarrow\uparrow\rangle =|\downarrow\uparrow\uparrow\rangle-|\uparrow\uparrow\downarrow\rangle.$$

The residual is therefore $-3$ times this nonzero vector, with norm $3\sqrt2$. On the all-up vector, $AB-BA$ vanishes, so that single test misses the incorrect parameter relation. An identity must hold on the whole space; a highly symmetric vector is a weak negative control.

Transfer: normalize a real-parameter R-matrix

For real $u$, show that $\widehat R(u)=R(u)/\sqrt{u^2+1}$ is unitary. Does multiplying $R(u)$ by a scalar function preserve the Yang–Baxter identity? Distinguish this statement from invertibility at every complex parameter.

Hint

Use $P^\dagger=P$ and $P^2=I$. Count the three scalar factors on each side of the identity.

Solution

For real $u$, $R(u)^\dagger=uI-iP$, so $R(u)^\dagger R(u)=(u^2+1)I$ and $\widehat R^\dagger\widehat R=I$. Replacing $R(u)$ by $f(u)R(u)$ multiplies both sides of Yang–Baxter by the same scalar $f(a)f(a+b)f(b)$, wherever these functions are defined. It preserves the identity without requiring cancellation of that factor.

This does not guarantee an invertible matrix: zeros of $f$ can introduce new singularities, and the unscaled $R$ is already singular at $u=\pm i$. The real positive square root used above also does not specify a globally single-valued complex normalization. When constructing a Hamiltonian, a parameter-dependent normalization changes its logarithmic derivative and must be tracked.

Put the local identity on a chain

You have proved the rational Yang–Baxter identity, tested a nontrivial failure and identified its singular parameters. Continue to Build monodromy and transfer matrices to combine one auxiliary space with every spin in a finite ring.

References

• Faddeev, L. D. How Algebraic Bethe Ansatz works for integrable model. Les Houches lectures, arXiv:hep-th/9605187v1, 1996, 59 pp. Version record. Open PDF. Section and equation numbers identify the cited locations in this version.