Commuting transfer matrices for the XXX chain

Why does a local identity involving three tensor factors produce conserved operators for an entire spin chain? For the finite periodic spin-½ XXX chain, the rational Yang–Baxter equation implies an exchange relation for the monodromy matrix. Its auxiliary trace gives commuting transfer matrices, and a derivative at a special spectral parameter recovers the nearest-neighbor Hamiltonian. This article proves that chain of implications, including the periodic closing bond, and checks it explicitly for three sites. It establishes a commuting operator family; independence of charges and completeness of Bethe states are separate questions.

Required background. Multiply matrices and tensor products, distinguish an operator from its matrix entries, and use the XXX Hamiltonian. Helpful background. The lessons Check a rational R-matrix and Build monodromy and transfer matrices develop the tensor notation. The convention reference collects the normalizations used here.

The rational R-matrix and its tensor factors

Let the physical Hilbert space be

$$\mathcal H=V_1\otimes\cdots\otimes V_N, \qquad V_n=\mathbb C^2, \qquad N\geq3.$$

We use ordinary, ungraded tensor products, periodic sites $N+1\equiv1$, and spin operators $S_n^\alpha=\sigma_n^\alpha/2$. Set $\hbar=1$ and the lattice spacing to one. Spectral parameters are dimensionless complex numbers; $J\gt0$ sets the energy scale. The Hamiltonian is

$$\begin{aligned} H&=J\sum_{n=1}^N \left(\frac14I-\mathbf S_n\cdot\mathbf S_{n+1}\right)\\ &=\frac J2\sum_{n=1}^N(I-P_{n,n+1}), \end{aligned}$$

where $P_{ij}$ swaps factors $i,j$ and acts as the identity elsewhere. The equality follows from

$$P_{ij}=\frac12I+2\mathbf S_i\cdot\mathbf S_j.$$

Thus the fully polarized states have zero energy. For $N=2$, the written periodic sum counts the same unordered bond twice; we examine that distinct convention in the exercises.

Introduce two additional copies $V_a,V_b$ of $\mathbb C^2$, called auxiliary spaces. They are temporary tensor factors, not extra physical sites. On two factors define

$$R_{ij}(u)=uI+iP_{ij}.$$

Products act on states from right to left. In a tensor product containing many factors, a subscript specifies the only two factors on which an $R$ or $P$ acts nontrivially. In particular, $R_{ab}$ has numerical matrix entries on $V_a\otimes V_b$ and acts as the identity on $\mathcal H$. This distinction will justify the trace argument.

The local operator and ordered monodromy are

$$\begin{aligned} L_{an}(\lambda)&=R_{an}(\lambda-i/2),\\ T_a(\lambda)&=L_{aN}(\lambda)\cdots L_{a1}(\lambda),\\ \tau(\lambda)&=\operatorname{tr}_a T_a(\lambda). \end{aligned}$$

Thus $T_a$ acts on $V_a\otimes\mathcal H$, while $\tau$ acts on $\mathcal H$. These conventions agree with Faddeev 1996, v1 PDF, § 3, equations (31)–(37) and (42)–(46). Faddeev’s Hamiltonian has the opposite sign and no factor $J$: $H=-JH_{\mathrm F}$.

Proving the rational Yang–Baxter equation

Work first on three copies of $\mathbb C^2$ and abbreviate

$$p=P_{12},\qquad q=P_{13},\qquad r=P_{23}.$$

Each swap squares to the identity. Applying products to $v_1\otimes v_2\otimes v_3$ gives two cyclic permutations:

$$\begin{aligned} pq=qr=rp &: (v_1,v_2,v_3)\longmapsto(v_2,v_3,v_1),\\ qp=rq=pr &: (v_1,v_2,v_3)\longmapsto(v_3,v_1,v_2). \end{aligned}$$

The ordered triple products also obey $pqr=rqp=q$. These identities hold on arbitrary tensor vectors, so they are operator identities. They are the permutation algebra behind the local exchange relation; compare Faddeev, § 3, equations (38)–(39).

To see why the spectral arguments matter, initially take three unrelated complex numbers $\alpha,\zeta,\beta$. Expand

$$\Delta=R_{12}(\alpha)R_{13}(\zeta)R_{23}(\beta) -R_{23}(\beta)R_{13}(\zeta)R_{12}(\alpha).$$

The scalar and one-swap terms cancel directly. The three-swap terms cancel because $pqr=rqp$. The remaining two-swap terms give

$$\begin{aligned} \Delta &=-\beta(pq-qp)-\zeta(pr-rp)-\alpha(qr-rq)\\ &=(\zeta-\alpha-\beta)(pq-qp). \end{aligned}$$

Consequently, setting $\zeta=\alpha+\beta$ proves the Yang–Baxter equation

$$\begin{aligned} &R_{12}(\alpha)R_{13}(\alpha+\beta)R_{23}(\beta)\\ &\qquad=R_{23}(\beta)R_{13}(\alpha+\beta)R_{12}(\alpha). \end{aligned}$$

This is an exact polynomial identity for every complex $\alpha,\beta$, including singular values of individual $R$ matrices. It does not require an inverse. Replacing the middle argument by an unrelated number generally destroys the identity: for example, $(pq-qp)|001\rangle=|010\rangle-|100\rangle\ne0$.

Take the three tensor factors to be $a,b,n$ and choose the spectral arguments

$$\alpha=\lambda-\mu,\qquad \zeta=\lambda-i/2,\qquad \beta=\mu-i/2.$$

Since $\zeta=\alpha+\beta$, the identity becomes

$$\begin{aligned} &R_{ab}(\lambda-\mu)L_{an}(\lambda)L_{bn}(\mu)\\ &\qquad=L_{bn}(\mu)L_{an}(\lambda)R_{ab}(\lambda-\mu). \end{aligned}$$

This is the local exchange relation. The two $L$ factors share physical site $n$; their entries cannot generally be interchanged. The $R$ matrix is exactly what permits the displayed reordering.

From the local relation to RTT

For different physical sites $m\ne n$, the supports of $L_{am}(\lambda)$ and $L_{bn}(\mu)$ are disjoint: they involve $(a,m)$ and $(b,n)$. Therefore these two operators commute. This permits the interleaving

$$T_a(\lambda)T_b(\mu) =\bigl(L_{aN}L_{bN}\bigr)\cdots\bigl(L_{a1}L_{b1}\bigr),$$

where every $a$ factor has parameter $\lambda$ and every $b$ factor has parameter $\mu$. No same-site exchange has been made.

Multiply by $R_{ab}(\lambda-\mu)$ from the left. The local relation moves this $R$ through the block at site $N$ while reversing that block’s two $L$ factors. Repeat at sites $N-1,\ldots,1$. Finally, commute only disjoint pairs to collect all $b$ factors before all $a$ factors. The result is

$$\boxed{ R_{ab}(\lambda-\mu)T_a(\lambda)T_b(\mu) =T_b(\mu)T_a(\lambda)R_{ab}(\lambda-\mu). }$$

This is the RTT relation. The proof explains why a local identity survives a chain of any finite length: different physical sites allow precisely the interchanges needed to concatenate the local relations. It is the argument in Faddeev, § 3, equation (44) and its two-site derivation.

Taking the auxiliary trace correctly

The inverse of $R(u)$ is

$$R(u)^{-1}=\frac{uI-iP}{u^2+1}, \qquad u\ne\pm i,$$

because $P^2=I$. At $u=i$ it vanishes on the antisymmetric subspace; at $u=-i$ it vanishes on the symmetric subspace. We first work at $\lambda-\mu\ne\pm i$, where RTT gives

$$T_b(\mu)T_a(\lambda) =R_{ab}T_a(\lambda)T_b(\mu)R_{ab}^{-1}.$$

For an auxiliary-only numerical matrix $S$ and an arbitrary operator $X$ on $V_a\otimes V_b\otimes\mathcal H$,

$$\operatorname{tr}_{ab}(SX)=\operatorname{tr}_{ab}(XS).$$

Indeed, if $S_{ij}$ are complex numbers and $X_{ij}$ are operators on $\mathcal H$, the two expressions are $\sum_{i,j}S_{ij}X_{ji}$ and $\sum_{i,j}X_{ij}S_{ji}$. Relabel the indices and commute the scalar coefficients. That proof fails if the entries of $S$ are noncommuting physical operators.

Apply this valid cyclicity to $R_{ab}$ and $R_{ab}^{-1}$. Also note, without changing any physical operator order, that

$$\begin{aligned} \operatorname{tr}_{ab}(T_aT_b)&=\tau(\lambda)\tau(\mu),\\ \operatorname{tr}_{ab}(T_bT_a)&=\tau(\mu)\tau(\lambda). \end{aligned}$$

For example, writing $T_a=\sum E_{ij}^{(a)}\otimes T_{ij}(\lambda)$ makes the first equality the sum $\sum_{i,k}T_{ii}(\lambda)T_{kk}(\mu)$. Thus

$$[\tau(\lambda),\tau(\mu)]=0 \qquad\text{when }\lambda-\mu\ne\pm i.$$

Every entry of this commutator is a polynomial in $\lambda,\mu$. Since it vanishes away from the exceptional differences, continuity, or the polynomial identity theorem, extends it to those differences as well:

$$\boxed{[\tau(\lambda),\tau(\mu)]=0 \quad\text{for all }\lambda,\mu\in\mathbb C.}$$

The resulting commuting family is stated in Faddeev, § 3, equations (46)–(48). Here the inverse and trace steps have been made explicit. In particular, neither the singularity of $R(\pm i)$ nor an unjustified cyclic permutation of physical operators leaves a gap in the argument.

Regularity turns the transfer matrix into translation

At the regular point $\lambda_0=i/2$,

$$L_{an}(\lambda_0)=iP_{an}, \qquad \tau_0\equiv\tau(\lambda_0) =i^N\operatorname{tr}_a(P_{aN}\cdots P_{a1}).$$

Evaluate the trace in a basis. Start with auxiliary state $r$ and physical states $(s_1,\ldots,s_N)$. The rightmost swap acts first:

Operation	Auxiliary state	Physical states
Before any swap	$r$	$(s_1,s_2,\ldots,s_N)$
After $P_{a1}$	$s_1$	$(r,s_2,\ldots,s_N)$
After $P_{a2}$	$s_2$	$(r,s_1,s_3,\ldots,s_N)$
After $P_{aN}$	$s_N$	$(r,s_1,\ldots,s_{N-1})$

The auxiliary trace imposes $r=s_N$ and sums over $r$. Hence

$$\begin{aligned} \tau_0&=i^N U,\\ U|s_1,\ldots,s_N\rangle&=|s_N,s_1,\ldots,s_{N-1}\rangle. \end{aligned}$$

The transfer matrix at this point is invertible because $U$ is a unitary cyclic shift. Equivalently $U=P_{12}P_{23}\cdots P_{N-1,N}$ and $U^N=I$. This is the regularity construction in Faddeev, § 3, equations (49)–(59).

For a state $|x\rangle$ with one down spin at site $x$, $U|x\rangle=|x+1\rangle$. Therefore our Fourier convention gives

$$\begin{aligned} |k\rangle&=\frac1{\sqrt N}\sum_{x=1}^Ne^{ikx}|x\rangle, \qquad k=\frac{2\pi m}{N},\\ U|k\rangle&=e^{-ik}|k\rangle. \end{aligned}$$

The minus sign in the translation eigenvalue follows by changing the summation variable to $x+1$. Reversing the order of the monodromy reverses this shift, so its orientation must be checked before comparing momentum conventions.

The spectral parameter of a creation block also needs a conversion: for this monodromy order, $B(\lambda_B)|0\rangle$ has adjacent one-magnon coefficients in the ratio $(\lambda_B-i/2)/(\lambda_B+i/2)$, so its coordinate rapidity is $\lambda_{\rm coord}=-\lambda_B$. The convention reference derives the ratio and checks a four-site state with translation phase $i$; an energy check or a symmetric pair alone would miss this reversal.

Recovering the local Hamiltonian

Differentiate the ordered product before evaluating the trace. Since $L'_{an}(\lambda)=I$,

$$\tau'_0=i^{N-1}\sum_{n=1}^N Q_n, \qquad Q_n=\operatorname{tr}_a \bigl(P_{aN}\cdots\widehat{P_{an}}\cdots P_{a1}\bigr),$$

where the hat means that the swap at site $n$ is omitted. This is the product differentiation in Faddeev, § 3, equations (61)–(62).

The omitted swap leaves the state at site $n$ untouched and cyclically shifts the states at the other sites. Compare this action with a full cyclic shift: exchanging the two input states at $n-1,n$ first produces exactly the same output. Thus, with $0\equiv N$,

$$Q_n=U P_{n-1,n}.$$

For the endpoint $n=1$, this says $Q_1=UP_{N,1}$: the unchanged first site is obtained by swapping input sites $N,1$ before translating. The boundary term is therefore part of the trace construction, not an extra interaction inserted afterward.

Using $\tau_0^{-1}=i^{-N}U^{-1}$ gives

$$\boxed{ \tau_0^{-1}\tau'_0 =-i\sum_{n=1}^NP_{n,n+1}. }$$

Consequently,

$$\boxed{ H=\frac{JN}{2}I-\frac{iJ}{2}\tau_0^{-1}\tau'_0. }$$

In particular, $(-i)(-i)=-1$ supplies the negative coefficient of the swap sum. The identity term fixes the polarized-state energy to zero. This is Faddeev, § 3, equations (63)–(65), with $H=-JH_{\mathrm F}$ as specified above.

Differentiate the commuting-transfer identity in one parameter. It follows that $\tau'_0$, $\tau_0$ and $\tau_0^{-1}$ commute with every $\tau(\mu)$. Therefore

$$[H,\tau(\mu)]=0.$$

The transfer family consists of conserved operators for this Hamiltonian. Conservation here means that their Heisenberg evolution $e^{itH}\tau(\mu)e^{-itH}$ is constant. A generic complex $\tau(\mu)$ need not itself be a Hermitian observable.

A local logarithm and higher coefficients

To define a logarithmic expansion without choosing a global branch, put

$$B(s)=\tau_0^{-1}\tau(\lambda_0+s), \qquad B(0)=I.$$

In finite dimension, continuity ensures $\|B(s)-I\|\lt1$ for sufficiently small complex $s$. In that neighborhood the matrix series

$$\begin{aligned} \log B(s) &=\sum_{m=1}^\infty \frac{(-1)^{m+1}}m\bigl(B(s)-I\bigr)^m\\ &=\sum_{r=1}^\infty s^r C_r \end{aligned}$$

converges in an operator norm. All $B(s)$ commute, so the logarithms commute; taking Taylor coefficients proves $[C_r,C_q]=0$. They also commute with $H$. The first two coefficients are

$$\begin{aligned} C_1&=\tau_0^{-1}\tau'_0,\\ C_2&=\frac12\left( \tau_0^{-1}\tau''_0-(\tau_0^{-1}\tau'_0)^2\right). \end{aligned}$$

The word “local” in this logarithm refers to its neighborhood in the spectral parameter. It does not by itself prove locality of every $C_r$ as an operator on the lattice. We proved the nearest-neighbor form for $H$ explicitly. Independence, suitable Hermitian normalizations, and support properties of further charges require further arguments. At fixed finite $N$, an infinite list of series coefficients cannot be an infinite linearly independent set of matrices on the finite-dimensional space $\mathcal H$.

A complete three-site check

Set $N=3$ and write $z=\lambda-i/2$. Expand the three factors before tracing. The needed partial traces are

$$\begin{aligned} \operatorname{tr}_a I&=2I, & \operatorname{tr}_aP_{an}&=I,\\ \operatorname{tr}_a(P_{am}P_{an})&=P_{mn}\quad(m\ne n), & \operatorname{tr}_a(P_{a3}P_{a2}P_{a1})&=U. \end{aligned}$$

The pair identity follows by following the two swaps and closing the auxiliary index, just as in the translation table. Define $\Sigma=P_{12}+P_{13}+P_{23}$. The result is the exact polynomial

$$\boxed{ \tau_3(\lambda)=(2z^3+3iz^2)I-z\Sigma-iU. }$$

In particular, $\tau_3(i/2)=-iU$ and $\tau'_3(i/2)=-\Sigma$. To check their logarithmic derivative directly, observe that the totally antisymmetric subspace of three copies of $\mathbb C^2$ is zero: every three-spin basis state repeats at least one of the two spin values, so antisymmetrizing it cancels its terms in pairs. The antisymmetrizer is proportional to

$$I-\Sigma+U+U^{-1}=0.$$

Since $U^3=I$, this implies $U^{-1}\Sigma=\Sigma$. Therefore $\tau_0^{-1}\tau'_0=-i\Sigma$, exactly as the general proof requires, and

$$\frac HJ=\frac12(3I-\Sigma).$$

We can check all eight energies without a Bethe ansatz. In the one-down-spin basis $|1\rangle,|2\rangle,|3\rangle$,

$$\frac HJ= \begin{pmatrix} 1&-1/2&-1/2\\ -1/2&1&-1/2\\ -1/2&-1/2&1 \end{pmatrix}.$$

The vector $(1,1,1)$ has eigenvalue zero, and its two-dimensional orthogonal complement has eigenvalue $3/2$. Flipping every spin gives the same spectrum in the two-down-spin sector. The fully up and fully down states supply two more zero eigenvalues. Thus $H/J$ has eigenvalues $0$ and $3/2$, each with multiplicity four. The count exhausts $\dim\mathcal H=8$ for this example; it is not a general Bethe-completeness argument.

The charge-extraction lesson includes a downloadable short-chain experiment that tests the tensor construction, analytic product derivative, and this polynomial independently. Its numerical residuals check an implementation; the identities above establish the finite-chain result without a floating-point assumption.

Which changes preserve the argument?

A scalar normalization is consequential for extracting an energy. If

$$\widetilde L_{an}(\lambda)=g(\lambda)L_{an}(\lambda), \qquad g(\lambda_0)\ne0,$$

with $g$ analytic near $\lambda_0$, the scalar factors cancel from the exchange relation and $\widetilde\tau=g^N\tau$ still commutes. Its logarithmic derivative is

$$\widetilde\tau_0^{-1}\widetilde\tau'_0 =\tau_0^{-1}\tau'_0 +N\frac{g'(\lambda_0)}{g(\lambda_0)}I.$$

Thus commutativity survives while the identity term in a Hamiltonian extraction must be adjusted. A zero or pole of $g$ at the regular point invalidates this particular inverse formula.

For site-dependent inhomogeneities $\xi_n$, replacing $L_{an}(\lambda)$ by $R_{an}(\lambda-\xi_n-i/2)$ preserves the local exchange relation: both spectral arguments at site $n$ have the same shift, so their difference remains $\lambda-\mu$. The RTT and trace proofs therefore still apply. If the $\xi_n$ differ, there is generally no single spectral point where every factor becomes $iP_{an}$. The homogeneous translation and nearest-neighbor Hamiltonian derivation cannot then be copied unchanged.

The auxiliary trace also closes the chain periodically. Simply deleting $P_{N,1}$ from the Hamiltonian while keeping this transfer matrix does not establish an open-chain commuting family; open boundaries need an appropriate boundary construction. Finally, neither RTT nor the three-site spectrum constructs all Bethe eigenvectors, proves their completeness for general $N$, or settles a thermodynamic limit.

Exercises

Why unrestricted partial-trace cyclicity fails

Let $E_{ij}=|i\rangle\langle j|$ act on a two-dimensional auxiliary space. On its tensor product with a physical spin take

$$X=E_{01}\otimes\sigma_x, \qquad Y=E_{10}\otimes\sigma_z.$$

Calculate $\operatorname{tr}_a(XY)$ and $\operatorname{tr}_a(YX)$. Explain why this does not contradict the RTT trace proof.

Solution

Since $E_{01}E_{10}=E_{00}$ and $E_{10}E_{01}=E_{11}$,

$$\begin{aligned} \operatorname{tr}_a(XY)&=\sigma_x\sigma_z=-i\sigma_y,\\ \operatorname{tr}_a(YX)&=\sigma_z\sigma_x=i\sigma_y. \end{aligned}$$

These are different. Both $X$ and $Y$ have nontrivial, noncommuting physical entries. The proof used cyclicity only for a matrix acting exclusively on the traced auxiliary factors, whose entries are scalars on the physical space.

A harmless scalar factor can change the energy zero

Multiply every $L$ by $g(\lambda)=e^{ic(\lambda-\lambda_0)}$, with real $c$. Keep the same displayed formula $JN I/2-iJ\widetilde\tau_0^{-1}\widetilde\tau'_0/2$ without correcting its constant. What operator results, and what correction recovers the original $H$?

Solution

Here $g'/g=ic$, so the logarithmic derivative changes by $iNcI$. Substitution gives

$$\widetilde H=H+\frac{JNc}{2}I.$$

Subtract $JNcI/2$ to recover the original polarized-state energy zero. Eigenvectors and energy differences are unchanged. This illustrates why specifying $R$ or $L$ only “up to a scalar” is insufficient when quoting absolute energy formulas.

The two-site periodic convention

Use the same ordered trace with $N=2$. Find $\tau_2(\lambda)$ and its logarithmic derivative at $\lambda_0$. Compare the extracted Hamiltonian with a single bond $J(I-P_{12})/2$.

Solution

Expanding two factors, with $z=\lambda-i/2$, gives

$$\begin{aligned} \tau_2(\lambda)&=(2z^2+2iz)I-P_{12},\\ \tau_0&=-P_{12},\qquad \tau'_0=2iI,\\ \tau_0^{-1}\tau'_0&=-2iP_{12}. \end{aligned}$$

The extraction therefore gives $H=J(I-P_{12})$, twice the single-bond operator. This agrees with the periodic sum, which includes both $(1,2)$ and $(2,1)$. Its triplet energy is zero and its singlet energy is $2J$, whereas the single-bond singlet energy is $J$. Specify the bond convention before using a two-site check to judge a many-site formula.

References

• Faddeev, L. D. How Algebraic Bethe Ansatz works for integrable model. arXiv:hep-th/9605187v1 [hep-th], 1996. Version record; open PDF. Section 3, equations (31)–(39), (42)–(48), (49)–(60), and (61)–(65). The site convention is $H=-JH_{\mathrm F}$; citations use equation numbers in this version.