Build monodromy and transfer matrices

How does an identity on three small spaces constrain an entire spin chain? One auxiliary spin visits every physical site in a fixed order. The resulting monodromy obeys the same exchange relation as each local factor, and tracing out the auxiliary spin produces a family of operators on the physical chain. You will construct this family and show explicitly why one special spectral value gives translation.

Required background. Use the rational Yang–Baxter identity and tensor-leg notation from Check a rational R-matrix. The XXX model defines the physical spin spaces and periodic bonds.

A separate auxiliary spin

Take a finite homogeneous periodic spin-$1/2$ chain with $N\geq3$ and

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

The auxiliary space $V_a=\mathbb C^2$ is an additional tensor factor used to construct operators; it is not an extra physical site. Define

$$L_{an}(\lambda)=R_{an}(\lambda-i/2) =(\lambda-i/2)I+iP_{an}.$$

Every site uses the same spectral shift; this is the homogeneous choice. The monodromy and transfer matrix are

$$T_a(\lambda)=L_{aN}(\lambda)\cdots L_{a1}(\lambda), \qquad \tau(\lambda)=\operatorname{tr}_a T_a(\lambda).$$

The rightmost factor acts first, so the auxiliary spin visits site $1$, then $2$, and so on. The trace closes that auxiliary path. $T_a$ acts on a space of dimension $2^{N+1}$; $\tau$ acts on the physical space of dimension $2^N$.

These definitions follow Faddeev 1996, § 3, equations (31)–(43), PDF. The convention reference fixes their order and spectral origin.

Entry check and repair

An operator $X$ on $V_a\otimes\mathcal H$ can be written as a $2\times2$ block matrix, whose blocks act on $\mathcal H$. What is $\operatorname{tr}_aX$? Is it a scalar?

Repair. If $X=\sum_{r,s=\uparrow,\downarrow}|r\rangle\langle s|_a\otimes X_{rs}$, then

$$\operatorname{tr}_aX=X_{\uparrow\uparrow}+X_{\downarrow\downarrow}.$$

This is still an operator on the full physical Hilbert space. Taking the trace over the physical spins as well would discard the operator we want to study. For example, $\operatorname{tr}_aI_{a\mathcal H}=2I_{\mathcal H}$, while $\operatorname{tr}_aP_{an}=I_{\mathcal H}$.

From local exchange to monodromy exchange

Introduce a second auxiliary space $V_b$. Substitute the three factors $a,b,n$ into the Yang–Baxter identity to get

$$\begin{aligned} &R_{ab}(\lambda-\mu)L_{an}(\lambda)L_{bn}(\mu)\\ &\hspace{1.5em}=L_{bn}(\mu)L_{an}(\lambda)R_{ab}(\lambda-\mu). \end{aligned}$$

The spectral arguments work because $(\lambda-\mu)+(\mu-i/2)=\lambda-i/2$.

The corresponding chain relation is

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

It is often called the RTT relation, after the order of its factors. Here is the two-site step of its proof. Suppress spectral arguments and write $R=R_{ab}(\lambda-\mu)$:

$$\begin{aligned} R L_{a2}L_{a1}L_{b2}L_{b1} &=R L_{a2}L_{b2}L_{a1}L_{b1}\\ &=L_{b2}L_{a2}R L_{a1}L_{b1}\\ &=L_{b2}L_{a2}L_{b1}L_{a1}R\\ &=L_{b2}L_{b1}L_{a2}L_{a1}R. \end{aligned}$$

The first and last equalities exchange operators acting on disjoint pairs of tensor factors. The middle equalities use the local relation at one site each. Repeating the same step moves $R$ through all $N$ sites; no assumption that neighboring physical spins commute inside a common local factor is required. This is the induction of Faddeev 1996, § 3, equations (44)–(48), PDF.

The next lesson will prove carefully why the auxiliary trace of RTT gives commuting transfer matrices. Generic cyclicity of a partial trace would be an invalid shortcut.

Track three swaps through the trace

At $\lambda_0=i/2$, every local factor is $iP_{an}$. For $N=3$, temporarily omit the overall scalar $i^3$ and track an arbitrary auxiliary state $r$ and physical states $s_1,s_2,s_3$.

Step	Auxiliary factor	Physical factors $1,2,3$
Initial tensor	$r$	$(s_1,s_2,s_3)$
After $P_{a1}$	$s_1$	$(r,s_2,s_3)$
After $P_{a2}$	$s_2$	$(r,s_1,s_3)$
After $P_{a3}$	$s_3$	$(r,s_1,s_2)$

In the partial trace, the outgoing auxiliary value must equal its initial value: $r=s_3$. Thus

$$\tau_3(i/2)|s_1,s_2,s_3\rangle =i^3|s_3,s_1,s_2\rangle.$$

The same bookkeeping on $N$ sites gives

$$\tau(i/2)=i^N U, \qquad U|s_1,\ldots,s_N\rangle=|s_N,s_1,\ldots,s_{N-1}\rangle.$$

Equivalently $U=P_{12}P_{23}\cdots P_{N-1,N}$. It is unitary and invertible, since it permutes an orthonormal basis. This agrees with Faddeev 1996, § 3, equations (49)–(60), PDF.

For a down spin at site $x$, $U|x\rangle=|x+1\rangle$. Hence the earlier coefficient convention $|k\rangle=N^{-1/2}\sum_xe^{ikx}|x\rangle$ gives $U|k\rangle=e^{-ik}|k\rangle$. The sign of the translation eigenvalue follows from the action just derived.

A complete three-site transfer polynomial

Put $z=\lambda-i/2$ and $S=P_{12}+P_{13}+P_{23}$ on three physical sites. Expanding the three local factors before taking the trace yields

$$\tau_3(\lambda)=(2z^3+3iz^2)I-zS-iU.$$

The terms have a direct interpretation. Zero swaps give $2z^3I$ because the auxiliary identity has trace $2$. One swap gives $iz^2I$ at each of the three sites. Two swaps give the physical exchange of the two visited sites, multiplied by $i^2z=-z$. Three swaps give $i^3U=-iU$.

This polynomial supplies checks at generic complex $\lambda$, not just the regular point. In particular the all-up state has eigenvalue

$$2z^3+3iz^2-3z-i=(z+i)^3+z^3.$$

The equality follows either from this expansion or from the triangular local action on the all-up state.

Exercises

Guided practice: identify the physical shift

For $N=3$, calculate $\tau(i/2)|\uparrow\downarrow\downarrow\rangle$ and $U^3$. For arbitrary $N$, explain why $U^{-1}S_n^zU=S_{n-1}^z$, with site indices taken modulo $N$.

Hint

Use the tuple action of $U$, including the scalar $i^3$ only for $\tau$. The value arriving at site $n$ was previously at site $n-1$.

Solution

The shifted state is $|\downarrow\uparrow\downarrow\rangle$, so $\tau(i/2)$ gives $-i|\downarrow\uparrow\downarrow\rangle$. Three applications of $U$ return each spin to its original site, hence $U^3=I$ on three sites. More generally $S_n^zU$ reads the original spin at $n-1$, so $S_n^zU=US_{n-1}^z$, proving the stated conjugation identity. Replacing $U^{-1}S_n^zU$ by $US_n^zU^{-1}$ reverses the direction.

Independent practice: reconstruct the polynomial

Prove $\operatorname{tr}_a(P_{aj}P_{ai})=P_{ij}$ for distinct sites $i,j$. Use this identity to rederive the two-swap coefficient in $\tau_3(\lambda)$ and check the all-up eigenvalue.

Hint

Starting from auxiliary state $r$, the two swaps leave the auxiliary factor equal to the original state at $j$. The trace equates that state with $r$.

Solution

On the visited factors the sequence is $(r;s_i,s_j)\mapsto(s_i;r,s_j)\mapsto(s_j;r,s_i)$. Tracing sets $r=s_j$, leaving $(s_j,s_i)$ on the physical sites. Thus the trace is $P_{ij}$. The three possible pairs contribute $-z(P_{12}+P_{13}+P_{23})$.

On the all-up state, each permutation and $U$ act as the identity, giving $2z^3+3iz^2-3z-i$. Expanding $(z+i)^3+z^3$ reproduces it. A trace over all four spins instead would give a scalar and could not pass this operator-level test.

Transfer: give the sites different spectral shifts

Replace the local factor by $L_{an}(\lambda)=R_{an}(\lambda-\theta_n-i/2)$ for fixed complex numbers $\theta_n$. Does the same local relation and RTT proof survive? Must there still be a single $\lambda_0$ at which every local factor equals $iP_{an}$?

Hint

Subtract the two spectral arguments at the same site. Then solve $\lambda_0-\theta_n-i/2=0$ simultaneously for all $n$.

Solution

At site $n$, the difference remains $(\lambda-\theta_n-i/2)-(\mu-\theta_n-i/2)=\lambda-\mu$. The same R-matrix therefore intertwines the local factors, and disjoint-factor commutativity still permits the global induction. RTT survives.

A common regular point requires $\lambda_0=\theta_n+i/2$ for every site, so it exists in this form only when all $\theta_n$ agree. Thus the commuting-family argument can remain valid while the simple translation and homogeneous nearest-neighbor Hamiltonian extraction no longer apply. These are separate proof steps.

Extract dynamics from the family

You can now form the monodromy, take the correct trace and determine its shift orientation. Continue to Derive commuting charges to prove transfer commutativity and recover the periodic XXX Hamiltonian, including the closing bond.

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.