Build the Toda Lax pair

Can one matrix equation reproduce the open Toda equations? You will build a symmetric matrix $L$ and a skew-symmetric matrix $B$, then verify $\dot L=[B,L]$ entry by entry for three particles. The check includes entries that must remain zero: matching the visible nonzero entries alone is insufficient.

Required background. Use the open-chain equations and exponential variables from Derive the open Toda equations. The entry repair reviews matrix multiplication.

Helpful background. The Lax-equation reference records the signs and normalizations used across the course.

A symmetric matrix for the three-particle chain

The model has $q,p\in\mathbb R^3$ and dimensionless Hamiltonian

$$H=\frac12\sum_{i=1}^{3}p_i^2+e^{q_1-q_2}+e^{q_2-q_3}.$$

Set $a_1=e^{(q_1-q_2)/2}\gt0$, $a_2=e^{(q_2-q_3)/2}\gt0$, and $b_i=p_i$. The equations we must reproduce are

$$\begin{aligned} \dot a_1&=\tfrac12a_1(b_1-b_2),& \dot a_2&=\tfrac12a_2(b_2-b_3),\\ \dot b_1&=-a_1^2,& \dot b_2&=a_1^2-a_2^2,\qquad \dot b_3=a_2^2. \end{aligned}$$

Put the momenta on the diagonal and each bond variable in two symmetric positions:

$$L=\begin{pmatrix} b_1&a_1&0\\ a_1&b_2&a_2\\ 0&a_2&b_3 \end{pmatrix}, \qquad B=\frac12\begin{pmatrix} 0&-a_1&0\\ a_1&0&-a_2\\ 0&a_2&0 \end{pmatrix}.$$

Here $[B,L]=BL-LB$. Our factor $1/2$ and upper-diagonal minus signs are consequential. Moser uses $a_i^{\mathrm M}=a_i/2$ and $b_i^{\mathrm M}=-b_i/2$; see Moser 1975, § 2, p. 470, equation (2.1), and pp. 472–473, equation (2.7), PDF. The reference conversion explains how the complete matrices correspond at the same time variable.

Entry check and repair

Take

$$X=\begin{pmatrix}0&c\\-c&0\end{pmatrix}, \qquad Y=\begin{pmatrix}r&s\\s&t\end{pmatrix}.$$

Compute $(XY)_{12}$, $(YX)_{12}$, and $[X,Y]_{11}$. Which order appears in the commutator?

Repair. The rule is $(XY)_{ij}=\sum_kX_{ik}Y_{kj}$: row $i$ of the first matrix meets column $j$ of the second. Thus $(XY)_{12}=ct$, $(YX)_{12}=rc$, and $[X,Y]_{12}=c(t-r)$. For the diagonal, $(XY)_{11}=cs$ and $(YX)_{11}=-sc$, hence $[X,Y]_{11}=2cs$. The products $XY$ and $YX$ must be calculated in their stated order.

To match the left $2\times2$ block of our Toda matrices, use $c=-a_1/2$, not $c=a_1$.

Work through three different entry types

For the first diagonal entry,

$$\begin{aligned} [B,L]_{11} &=B_{12}L_{21}-L_{12}B_{21}\\ &=-\tfrac12a_1^2-\tfrac12a_1^2=-a_1^2. \end{aligned}$$

This is exactly $\dot b_1$. For the first upper off-diagonal entry,

$$\begin{aligned} [B,L]_{12} &=B_{12}L_{22}-L_{11}B_{12}\\ &=-\tfrac12a_1b_2+\tfrac12b_1a_1 =\tfrac12a_1(b_1-b_2), \end{aligned}$$

which equals $\dot a_1$. Now check the corner. Although $B_{13}=L_{13}=0$, each product has a nonzero contribution through site $2$:

$$\begin{aligned} [B,L]_{13} &=B_{12}L_{23}-L_{12}B_{23}\\ &=-\tfrac12a_1a_2-(-\tfrac12a_1a_2)=0. \end{aligned}$$

The cancellation keeps the matrix tridiagonal. It is a necessary part of the Lax identity, not an optional check.

Because $L^{\mathsf T}=L$ and $B^{\mathsf T}=-B$,

$$[B,L]^{\mathsf T} =L^{\mathsf T}B^{\mathsf T}-B^{\mathsf T}L^{\mathsf T} =-LB+BL=[B,L].$$

Thus the lower off-diagonal entries follow from the upper ones. Completing the remaining three upper-triangle entries gives

$$[B,L]=\begin{pmatrix} -a_1^2&\tfrac12a_1(b_1-b_2)&0\\ \tfrac12a_1(b_1-b_2)&a_1^2-a_2^2&\tfrac12a_2(b_2-b_3)\\ 0&\tfrac12a_2(b_2-b_3)&a_2^2 \end{pmatrix}.$$

This is $\dot L$ obtained by differentiating the entries of $L$ along the Toda flow.

A numerical state checked exactly

For $q=(0,0,0)$ and $p=(1,0,-1)$,

$$L_0=\begin{pmatrix}1&1&0\\1&0&1\\0&1&-1\end{pmatrix}, \qquad B_0=\frac12\begin{pmatrix}0&-1&0\\1&0&-1\\0&1&0\end{pmatrix}.$$

Ordinary matrix multiplication gives

$$[B_0,L_0]= \begin{pmatrix}-1&1/2&0\\1/2&0&1/2\\0&1/2&1\end{pmatrix}.$$

The diagonal is the force vector $(-1,0,1)$, and the upper off-diagonal entries are the bond derivatives $(1/2,1/2)$ calculated in the previous lesson. Its trace is zero, consistently with momentum conservation. This one-state calculation checks arithmetic; the symbolic identity above checks the whole stated three-particle system.

The five numbers $(a_1,a_2,b_1,b_2,b_3)$ omit the center coordinate $Q=(q_1+q_2+q_3)/3$. Given $L(t)$ and the initial $Q$, recover the gaps from $q_i-q_{i+1}=2\log a_i$ and the center from $\dot Q=P/3$. A Lax matrix alone therefore does not specify the absolute positions.

Exercises

Guided practice: complete the commutator

Calculate $[B,L]_{22}$, $[B,L]_{23}$, and $[B,L]_{33}$ directly from the matrices. For the first calculation, begin with

$$(BL)_{22}=B_{21}L_{12}+B_{23}L_{32}.$$

Do the same for $(LB)_{22}$ before subtracting. Explain why checking these entries, together with the three already worked and symmetry, accounts for every entry of a $3\times3$ matrix.

Hint · Full solution

Independent practice: test unequal bonds

Use $q=(\log4,0,-\log9)$ and $p=(2,-1,0)$. Construct $L$ and $B$, multiply them, and compare $[B,L]$ with $\dot L$ obtained from Hamilton’s equations. Check the corner as well as the diagonal. State what would happen if you replaced $B$ by $-B$ while retaining the convention $[B,L]=BL-LB$ and the same physical time.

Hint · Full solution

Changed setting: add a fourth particle

For the finite open chain with $N=4$, form $L$ with diagonal $b_i$ and nearest off-diagonal entries $a_i$, $1\leq i\leq3$. Form $B$ with upper entries $-a_i/2$ and lower entries $a_i/2$.

Derive the diagonal and nearest off-diagonal commutator entries. Check $[B,L]_{13}$, $[B,L]_{24}$, and $[B,L]_{14}$. Which missing-bond values encode the two endpoints? Does this calculation by itself establish Liouville integrability for four particles?

Hint · Full solution

Hints

Remaining entries. $(BL)_{22}=(a_1^2-a_2^2)/2$, while $(LB)_{22}$ has the opposite sign. For entry $(2,3)$, only the adjacent diagonal values survive.

Unequal bonds. The off-diagonal entries of $L$ are $2$ and $3$. Your commutator should have zero trace even though its middle diagonal entry is nonzero.

Four particles. Two tridiagonal matrices can produce entries two steps from the diagonal. Calculate those entries before assuming they vanish. Entries three steps away have no connecting index in either product.

Solutions and checks

Remaining entries

Subtracting the two products gives

$$\begin{aligned} [B,L]_{22}&=\tfrac12(a_1^2-a_2^2) -\tfrac12(-a_1^2+a_2^2)=a_1^2-a_2^2,\\ [B,L]_{23}&=-\tfrac12a_2b_3+\tfrac12b_2a_2 =\tfrac12a_2(b_2-b_3),\\ [B,L]_{33}&=\tfrac12a_2^2-(-\tfrac12a_2^2)=a_2^2. \end{aligned}$$

There are three diagonal and three strictly upper-triangle entries. Their six checks, including the zero corner, determine the three lower-triangle entries by symmetry. All nine entries agree with $\dot L$.

Unequal bonds

The matrices are

$$L=\begin{pmatrix}2&2&0\\2&-1&3\\0&3&0\end{pmatrix}, \qquad B=\begin{pmatrix}0&-1&0\\1&0&-3/2\\0&3/2&0\end{pmatrix}.$$

Their commutator is

$$[B,L]=\begin{pmatrix}-4&3&0\\3&-5&-3/2\\0&-3/2&9\end{pmatrix}.$$

Hamilton’s equations give $\dot b=(-4,-5,9)$ and $\dot a=(3,-3/2)$, exactly matching this matrix. In particular, $[B,L]_{13}=(-1)3-2(-3/2)=0$.

Replacing only $B$ by $-B$ gives $[-B,L]=-[B,L]$, reversing every nonzero derivative. It does not represent the original forward-time equations. A reversed commutator convention or reversed time would require an explicit corresponding change.

Four particles

With $a_0=a_4=0$, direct multiplication gives

$$\begin{aligned} [B,L]_{ii}&=a_{i-1}^2-a_i^2,&&1\leq i\leq4,\\ [B,L]_{i,i+1}&=\tfrac12a_i(b_i-b_{i+1}),&&1\leq i\leq3. \end{aligned}$$

For $i=1,2$, the entries two steps away satisfy

$$[B,L]_{i,i+2}=-\tfrac12a_ia_{i+1} -(-\tfrac12a_ia_{i+1})=0.$$

Both products have zero $(1,4)$ entry because no index is adjacent to both $1$ and $4$. Symmetry supplies the lower entries. Thus this pair again reproduces the open-chain equations, including $\dot b_1=-a_1^2$ and $\dot b_4=a_3^2$.

The result is a verified Lax representation. A claim of Liouville integrability still needs four independent commuting integrals on the eight-dimensional canonical phase space. The Library proof supplies those further steps for general finite $N$.

Check your understanding

You can now check a proposed Lax pair by testing diagonal entries, nearest neighbors, and entries that must remain zero. Remember that a symmetric $L$ and skew-symmetric $B$ produce a symmetric commutator, not a skew-symmetric one.

Next, use the verified matrix equation to extract spectral invariants. The important gain is that one matrix identity organizes many conservation laws.

References

• Moser, Jürgen. “Finitely many mass points on the line under the influence of an exponential potential—an integrable system.” In Dynamical Systems, Theory and Applications, edited by Jürgen Moser, Lecture Notes in Physics 38, pp. 467–497. Springer, 1975. DOI. Open PDF.