Derive commuting charges

How does a transfer matrix encode the Hamiltonian and further conserved operators? The exchange relation yields a commuting family, while regularity at one spectral point makes its logarithmic derivative a sum of local spin exchanges. You will prove both steps, recover the closing bond of the ring, and reproduce finite-chain checks. Commutativity is the result established here; independence of every charge and completeness of Bethe eigenstates are separate questions.

Required background. Build monodromy and transfer matrices derives RTT and $\tau(i/2)=i^NU$. Check a rational R-matrix identifies the inverse and exceptional values of $R$.

Helpful background. The XXX model fixes the ferromagnetic energy zero. The Library proof collects the general finite-chain argument.

The periodic XXX transfer family

Keep the finite homogeneous spin-$1/2$ chain with $N\geq3$, $J\gt0$, $\hbar=1$, and

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

All matrices are finite-dimensional. $J$ carries the energy scale; the spectral parameter is dimensionless. We will recover

$$H=\frac J2\sum_{n=1}^N(I-P_{n,n+1}),\qquad P_{N,N+1}=P_{N,1}.$$

For $N=2$, a literal periodic sum contains the same unordered pair twice: $(1,2)$ and $(2,1)$. Its Hamiltonian is $J(I-P_{12})$, whereas a single bond is $J(I-P_{12})/2$. Our examples keep $N\geq3$; a two-site comparison must state which convention it uses.

Entry check and repair

If $L_1(\lambda),L_2(\lambda),L_3(\lambda)$ do not commute, how many terms appear in the derivative of $L_3L_2L_1$, and in what order?

Repair. The product rule keeps every factor in place:

$$(L_3L_2L_1)'=L_3'L_2L_1+L_3L_2'L_1+L_3L_2L_1'.$$

It does not give three copies of the same term. Here each $L_n'=I$, but deleting factors at different sites still leaves different ordered products. Their traces will encode different bonds.

Take the auxiliary trace correctly

The established RTT relation is

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

For $d=\lambda-\mu\ne\pm i$, $R(d)$ is invertible. Take the trace over both auxiliary factors:

$$\begin{aligned} \tau(\lambda)\tau(\mu) &=\operatorname{tr}_{ab}(T_aT_b)\\ &=\operatorname{tr}_{ab}(R^{-1}T_bT_aR)\\ &=\operatorname{tr}_{ab}(T_bT_a) =\tau(\mu)\tau(\lambda). \end{aligned}$$

The product order in the first equality is retained: expanding the auxiliary matrix units gives the physical product $\tau(\lambda)\tau(\mu)$, not a freely reordered expression. The third equality is valid because $R$ and $R^{-1}$ act only on the traced auxiliary spaces. If $S$ is such an auxiliary matrix and $X$ has physical operator entries, then

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

as follows by summing its numerical matrix elements and relabeling the two auxiliary indices. This does not license cyclic permutation of arbitrary factors with noncommuting physical entries.

Each entry of $[\tau(\lambda),\tau(\mu)]$ is a polynomial in the two complex parameters. It vanishes when $\lambda-\mu\ne\pm i$, hence identically, including the exceptional differences. We have proved

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

This supplies the trace step behind Faddeev 1996, § 3, equations (44)–(48), PDF. An inverse-based proof at the exceptional points would be invalid, even though the resulting polynomial identity remains true there.

Differentiate at the regular point

Set $\lambda_0=i/2$, $\tau_0=\tau(\lambda_0)$ and $\tau'_0=\tau'(\lambda_0)$. The previous lesson showed

$$\tau_0=i^NU,\qquad U|s_1,\ldots,s_N\rangle=|s_N,s_1,\ldots,s_{N-1}\rangle.$$

Thus $\tau_0$ is invertible. Differentiating the monodromy at this point replaces one factor $iP_{an}$ by the identity:

$$\tau'_0=i^{N-1}\sum_{n=1}^N \operatorname{tr}_a \left(P_{aN}\cdots\widehat P_{an}\cdots P_{a1}\right).$$

The hat means omission. With site indices modulo $N$,

$$\operatorname{tr}_a \left(P_{aN}\cdots\widehat P_{an}\cdots P_{a1}\right) =UP_{n-1,n}.$$

Why this particular bond? Skipping site $n$ leaves its spin unchanged and sends the preceding visited spin directly to site $n+1$. The same effect is obtained by exchanging input spins at $n-1,n$ before the full shift. For $n=1$, the preceding site is $N$, so the missing factor produces the closing bond.

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

$$\tau_0^{-1}\tau'_0 =\frac1i\sum_nP_{n-1,n} =-i\sum_nP_{n,n+1}.$$

Therefore

$$\begin{aligned} H &=\frac{JN}{2}I-\frac{iJ}{2}\tau_0^{-1}\tau'_0\\ &=\frac J2\sum_n(I-P_{n,n+1}). \end{aligned}$$

Since $P_{n,n+1}=2\mathbf S_n\cdot\mathbf S_{n+1}+I/2$, this is precisely $J\sum_n(1/4-\mathbf S_n\cdot\mathbf S_{n+1})$. The all-up energy is zero. Our Hamiltonian equals $-J$ times the source Hamiltonian in Faddeev 1996, § 3, equations (61)–(65), PDF.

Because $\tau_0$, $\tau'_0$ and $\tau(\lambda)$ commute, $[H,\tau(\lambda)]=0$. This establishes conserved operators for the stated Hamiltonian, beyond the earlier construction of particular one- and two-magnon states.

Define higher charges near the regular point

To avoid choosing a global logarithm of $\tau$, normalize the family locally:

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

For sufficiently small complex $s$, $\|B(s)-I\|\lt1$ in an operator norm. The convergent series

$$\log B(s)=\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m}[B(s)-I]^m$$

defines the logarithm near the identity. The commuting transfer family makes these logarithms commute at different small arguments; their Taylor coefficients therefore commute. With $Q_r=\left.d^r\log B(s)/ds^r\right|_{s=0}$, the first two are

$$\begin{aligned} Q_1&=\tau_0^{-1}\tau'_0,\\ Q_2&=\tau_0^{-1}\tau''_0-(\tau_0^{-1}\tau'_0)^2. \end{aligned}$$

The absence of noncommutative ordering corrections here follows from the already proved commutativity of the transfer family and its derivatives. Each $Q_r$ commutes with $H$. The first charge $Q_1=-i\sum P$ is anti-Hermitian; $iQ_1$ is Hermitian. A conserved algebraic operator need not itself be a Hermitian observable without a suitable normalization or combination.

No argument above establishes that every coefficient is independent, that every coefficient has a fixed finite interaction range, or that a Bethe parametrization reaches every eigenstate. For fixed $N$, arbitrarily many Taylor coefficients cannot all be linearly independent in a finite-dimensional operator space.

Reproduce finite-chain checks

The three-site result from the preceding lesson provides an exact target. With $z=\lambda-i/2$, $S=P_{12}+P_{13}+P_{23}$,

$$\tau_3(\lambda)=(2z^3+3iz^2)I-zS-iU, \qquad H/J=\frac{3I-S}{2}.$$

On three spin-$1/2$ sites, $S=\mathbf S_{\mathrm{tot}}^2-3I/4$. The total-spin $3/2$ quartet therefore has $H=0$, and the two total-spin $1/2$ doublets have $H=3J/2$. Each energy has multiplicity four. This gives an analytic target independent of differentiating a transfer matrix numerically.

Download and extract the complete transfer-matrix experiment (ZIP), then open its xxx-algebra folder. Individual files are also available: Python experiment, inputs, saved results, requirements, and instructions. With the dependencies installed, run:

python3 experiment.py --check

Within the website checkout the corresponding command is:

python3 public/computations/xxx-algebra/experiment.py --check

The experiment builds swaps from tensor-basis actions and compares the extracted Hamiltonian against an independently constructed spin Hamiltonian on $N=3,4,5$. It checks local Yang–Baxter and exchange identities, RTT, commuting traces, the shift direction and analytic product differentiation. Generic complex spectral values and the exceptional differences $\lambda-\mu=\pm i$ are separate cases. Negative controls use a wrong middle spectral argument, an omitted closing bond and an invalid partial-trace interchange.

For matrix identities the normalized residual is

$$\rho(A,B)=\frac{\|A-B\|_{\mathrm F}} {\max(1,\|A\|_{\mathrm F},\|B\|_{\mathrm F})},$$

where $\|\cdot\|_{\mathrm F}$ is the Frobenius norm. The Hamiltonian discrepancy is scaled by $\max(J,\|H_{\mathrm{spin}}\|_{\mathrm F})$; translation on a normalized plane wave uses the Euclidean vector norm. The saved report records the inputs, environment and individual checks. These finite calculations test the implementation of the formulas; the general proof comes from the permutation and trace arguments above.

The recorded Python 3.9.6 / NumPy 2.0.2 run uses binary64 arithmetic, $J=1$, and two generic pairs $(\lambda,\mu)=(0.37+0.21i,-0.42+0.13i)$ and $(1.2-0.4i,-0.2+0.6i)$. Exceptional cases use $\mu=0.25+0.125i$ and $\lambda=\mu\pm i$. The largest accepted residual, including plane-wave phases, is below $8.64\times10^{-16}$. Regularity, the logarithmic derivative and the independent Hamiltonian agree exactly in this floating-point run. Every negative control exceeds $0.157$. These zeros describe the recorded computation, not additional mathematical proofs; inspect the saved report for each norm and parameter. The instructions specify Python 3.9–3.12 with the pinned NumPy version, and --check compares fresh calculations with the saved inputs and results without rewriting them.

Exercises

Guided practice: find the forbidden trace step

On one auxiliary spin and one physical spin, let $X=|\uparrow\rangle\langle\uparrow|_a\otimes\sigma_x$ and $Y=|\uparrow\rangle\langle\uparrow|_a\otimes\sigma_z$. Compute $\operatorname{tr}_a(XY)$ and $\operatorname{tr}_a(YX)$. Why does this not invalidate the auxiliary R-matrix step in the proof?

Hint

The auxiliary projector has trace one. Use $\sigma_x\sigma_z=-i\sigma_y$ and $\sigma_z\sigma_x=i\sigma_y$.

Solution

The partial traces are $-i\sigma_y$ and $i\sigma_y$, so they differ. Both $X$ and $Y$ contain physical operators whose order matters. In the valid similarity step, the moved matrix $R$ has only numerical auxiliary entries and acts as the identity on the physical space. Index relabeling moves those numbers without exchanging physical operator products. Full trace cyclicity must not be applied indiscriminately to a partial trace.

Independent practice: recover the closing bond

For $N=3$, omit $P_{a1}$ from the regular monodromy. Show that its traced derivative contribution is $-P_{23}$ and that multiplication by $\tau_0^{-1}$ gives $-iP_{31}$. Repeat for the other two omissions. What Hamiltonian term would be lost by treating the chain as open?

Hint

Here $U=P_{12}P_{23}$, $\tau_0=-iU$, and $\tau_0^{-1}=iU^{-1}$. Use the two-swap trace from the preceding lesson.

Solution

The remaining scalar is $i^2=-1$, and $\operatorname{tr}_a(P_{a3}P_{a2})=P_{23}$. Since $U^{-1}P_{23}=P_{31}$, the contribution to $\tau_0^{-1}\tau'_0$ is $-iP_{31}$. Omitting factors $2$ and $3$ similarly gives $-iP_{12}$ and $-iP_{23}$. Their sum is $-iS$, as required.

The open-chain operator would omit $J(I-P_{31})/2$. This is not merely an additive constant: it acts with eigenvalue $J$ on a singlet of sites $3,1$ and zero on their triplet. Dropping that bond changes the physical dynamics as well as the constant in the energy convention.

Transfer: change the scalar normalization

Replace every local factor by $\widetilde L_{an}(\lambda)=f(\lambda)L_{an}(\lambda)$, with scalar $f$ analytic and nonzero near $\lambda_0$. Find $\widetilde\tau_0^{-1}\widetilde\tau'_0$. If the original Hamiltonian extraction formula is used unchanged, how does the resulting operator differ from $H$?

Hint

There are $N$ local factors, so $\widetilde\tau=f^N\tau$. Differentiate before substituting $\lambda_0$.

Solution

One obtains

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

Using the same displayed extraction formula would therefore give

$$\widetilde H =H-\frac{iJN}{2}\frac{f'(\lambda_0)}{f(\lambda_0)}I.$$

The eigenvectors and commutators are unchanged, but the energy zero shifts. For arbitrary complex $f$, even that scalar shift need not be real, so recovering the chosen Hermitian Hamiltonian requires explicitly subtracting it. Nonvanishing $f(\lambda_0)$ is essential for this inverse and logarithmic derivative. Scalar freedom in the Yang–Baxter equation is not freedom to ignore Hamiltonian normalization.

Connect the algebra to the spectrum

You have proved a commuting transfer family and identified the periodic XXX Hamiltonian inside it. Next, construct regular Bethe vectors: act with one or two creation blocks, derive the unwanted terms and cancel them with Bethe equations. The coordinate course gives the wave interpretation, while the Library article summarizes the operator proof and its hypotheses. General completeness and thermodynamic limits require further arguments.

After the regular construction, investigate the singular-state reproduction: recover a known four-site eigenstate by regularization, and use the transfer matrix at its regular point to reject the same formal root pair on five sites.

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.