Extract spectral invariants

What does the Toda Lax equation tell us about conserved quantities? You will turn $\dot L=[B,L]$ into a proof that traces of powers are constant, then compute the three-particle characteristic polynomial from three such traces. The result identifies conserved spectral data; independence and Poisson commutativity remain separate questions.

Required background. Construct and check the matrices in Build the Toda Lax pair. You should be able to multiply small matrices. The entry repair explains the trace identity used below.

Helpful background. The Lax-equation reference summarizes the finite-matrix argument and its scope.

Conserved traces of the Toda matrix

Use the dimensionless finite open chain with real canonical $(q,p)$, $a_i=e^{(q_i-q_{i+1})/2}\gt0$, and $b_i=p_i$. The symmetric matrix $L$ has diagonal $b_i$ and nearest off-diagonal entries $a_i$. Its skew partner has $B_{i,i+1}=-a_i/2$ and $B_{i+1,i}=a_i/2$. The previous lesson established $\dot L=BL-LB$ with the missing bonds $a_0=a_N=0$.

For every positive integer $k$, define

$$I_k=\frac1k\operatorname{tr}(L^k).$$

Here the trace is the sum of the diagonal entries. For $N=3$, we will use the names

$$P=I_1,\qquad H=I_2,\qquad J=I_3.$$

The factor $1/k$ is part of the definition, so $J$ means $\operatorname{tr}(L^3)/3$, not $\operatorname{tr}(L^3)$. The finite Toda characteristic polynomial as a source of constants of motion appears in Moser 1975, § 2, p. 473, equation (2.8), PDF; the normalization above is the one used in this course.

Entry check and repair

For

$$X=\begin{pmatrix}0&1\\0&0\end{pmatrix}, \qquad Y=\begin{pmatrix}0&0\\1&0\end{pmatrix},$$

compute $XY$, $YX$, and their traces. Are equal traces sufficient to conclude that the two matrices are equal?

Repair. $XY=\operatorname{diag}(1,0)$ and $YX=\operatorname{diag}(0,1)$. Each has trace $1$, but the matrices differ. More generally, for finite square matrices,

$$\operatorname{tr}(XY)=\sum_{i,j}X_{ij}Y_{ji} =\sum_{j,i}Y_{ji}X_{ij}=\operatorname{tr}(YX).$$

Repeated use gives the cyclic rule $\operatorname{tr}(XYZ)=\operatorname{tr}(ZXY)$. This permits cyclic rotations inside a trace, not arbitrary rearrangements of factors.

Why the traces are conserved

The derivative of a matrix power must initially retain every factor order:

$$\frac{d}{dt}L^k =\sum_{r=0}^{k-1}L^r\dot L L^{k-1-r}.$$

After taking the trace, cyclicity makes all $k$ terms equal. Therefore

$$\frac{dI_k}{dt}=\operatorname{tr}(L^{k-1}\dot L).$$

Insert the Lax equation:

$$\begin{aligned} \frac{dI_k}{dt} &=\operatorname{tr}(L^{k-1}BL-L^kB)\\ &=\operatorname{tr}(BL^k-L^kB)=0. \end{aligned}$$

We used neither $L\dot L=\dot L L$ nor $LB=BL$. Those equalities generally fail. The argument applies to differentiable finite matrices satisfying the stated Lax equation; infinite-dimensional operator traces need additional hypotheses and are outside this calculation.

Compute the first three invariants

For three particles,

$$L=\begin{pmatrix} p_1&a_1&0\\ a_1&p_2&a_2\\ 0&a_2&p_3 \end{pmatrix}.$$

The first trace gives

$$P=p_1+p_2+p_3.$$

Each off-diagonal square occurs twice in $\operatorname{tr}L^2$, so

$$H=\frac12(p_1^2+p_2^2+p_3^2)+a_1^2+a_2^2.$$

This is exactly the Hamiltonian because $a_i^2=e^{q_i-q_{i+1}}$. For the cubic trace, set $u=a_1^2$ and $v=a_2^2$. Direct multiplication gives

$$\begin{aligned} (L^3)_{11}&=p_1^3+u(2p_1+p_2),\\ (L^3)_{22}&=p_2^3+u(p_1+2p_2)+v(2p_2+p_3),\\ (L^3)_{33}&=p_3^3+v(p_2+2p_3). \end{aligned}$$

Summing and dividing by three yields

$$J=\frac13(p_1^3+p_2^3+p_3^3) +u(p_1+p_2)+v(p_2+p_3).$$

The mixed terms matter. The sum of momentum cubes alone is not generally conserved by the interacting chain.

A worked spectrum

At $q=(0,0,0)$ and $p=(1,0,-1)$, the initial matrix is

$$L_0=\begin{pmatrix}1&1&0\\1&0&1\\0&1&-1\end{pmatrix}, \qquad L_0^2=\begin{pmatrix}2&1&1\\1&2&-1\\1&-1&2\end{pmatrix}.$$

Thus $P=0$ and $H=3$. The cubic formula gives $J=0$: the momentum-cube contribution vanishes, and the two bond contributions are $1$ and $-1$.

For an arbitrary three-particle state, expansion of the determinant gives

$$\begin{aligned} \chi_L(\lambda) &=\det(\lambda\mathrm{Id}_3-L)\\ &=(\lambda-p_1)(\lambda-p_2)(\lambda-p_3)\\ &\quad-u(\lambda-p_3)-v(\lambda-p_1). \end{aligned}$$

At the initial state this becomes

$$\chi_{L_0}(\lambda)=\lambda^3-3\lambda =\lambda(\lambda-\sqrt3)(\lambda+\sqrt3).$$

The eigenvalues are $-\sqrt3$, $0$, and $\sqrt3$. They remain the spectrum along this exact Toda trajectory even though the matrix entries change. In the guided exercise you will show that $P,H,J$ determine all three characteristic-polynomial coefficients, making that conclusion independent of any numerical diagonalization.

The values $P=J=0$ at this state do not say that $P$ and $J$ are the same function, or that their gradients vanish. Also, a fixed spectrum does not specify all positions and momenta along a trajectory.

Exercises

Guided practice: recover the characteristic polynomial

Let the eigenvalues be $\lambda_1,\lambda_2,\lambda_3$. Write

$$\chi_L(\lambda)=\lambda^3-e_1\lambda^2+e_2\lambda-e_3.$$

Use $\sum_j\lambda_j=P$, $\sum_j\lambda_j^2=2H$, and $\sum_j\lambda_j^3=3J$ to derive

$$e_1=P,\qquad e_2=\frac{P^2}{2}-H, \qquad e_3=\frac{P^3}{6}-PH+J.$$

Begin by expanding $(\lambda_1+\lambda_2+\lambda_3)^2$. For the cubic step, use

$$\sum_j\lambda_j^3=e_1\sum_j\lambda_j^2-e_2\sum_j\lambda_j+3e_3.$$

Explain why conservation of these traces fixes the spectrum as a multiset, including multiplicities.

Hint · Full solution

Independent practice: two routes to one polynomial

Use the unequal-bond state $q=(\log4,0,-\log9)$ and $p=(2,-1,0)$. Calculate $P,H,J$ and obtain $\chi_L$ from the formulas above. Independently expand $\det(\lambda\mathrm{Id}_3-L)$ and compare all coefficients. What is $\det L$? Is the conserved cubic trace equal to that determinant?

Hint · Full solution

Changed setting: a moving scalar shift

Let $L(t)$ satisfy the original Lax equation and let $c(t)$ be a differentiable real function. Define a new matrix

$$\widetilde L(t)=L(t)+c(t)\mathrm{Id}_3.$$

Find $\dot{\widetilde L}$ and determine whether it still equals $[B,\widetilde L]$. Compute $\operatorname{tr}\widetilde L$ and $\operatorname{tr}(\widetilde L^2)/2$. Explain what happens to individual eigenvalues and their pairwise differences when $c$ changes in time. How does the answer change if $c$ is constant?

Hint · Full solution

Hints

Polynomial coefficients. The cross terms in $P^2$ are twice $e_2$. Substitute the resulting $e_2$ into the supplied cubic identity and solve for $e_3$.

Unequal bonds. The two mixed contributions to $J$ are $4(2-1)$ and $9(-1+0)$. The constant coefficient of $\chi_L$ is $-\det L$ for a $3\times3$ matrix.

Scalar shift. The identity matrix commutes with $B$. If $Lw=\lambda w$, compute $\widetilde Lw$ for the same nonzero vector $w$.

Solutions and checks

Polynomial coefficients

Expansion gives $P^2=2H+2e_2$, hence $e_2=P^2/2-H$. In the cubic identity,

$$\begin{aligned} 3J&=2PH-P\left(\frac{P^2}{2}-H\right)+3e_3\\ &=3PH-\frac{P^3}{2}+3e_3. \end{aligned}$$

Consequently $e_3=J-PH+P^3/6$, and

$$\chi_L(\lambda)=\lambda^3-P\lambda^2 +\left(\frac{P^2}{2}-H\right)\lambda -\left(\frac{P^3}{6}-PH+J\right).$$

Every coefficient is a function of conserved quantities. The polynomial is therefore constant in time, so its roots, counted with their multiplicities, are unchanged. This is isospectrality. It is not yet a calculation of Poisson brackets between the conserved quantities.

Unequal bonds

The invariant values are

$$P=1,\qquad H=\frac{31}{2},\qquad J=\frac{8-1}{3}+4-9=-\frac83.$$

Thus $e_2=-15$ and $e_3=-18$. The resulting polynomial is

$$\chi_L(\lambda)=\lambda^3-\lambda^2-15\lambda+18.$$

Direct expansion of the matrix

$$L=\begin{pmatrix}2&2&0\\2&-1&3\\0&3&0\end{pmatrix}$$

gives

$$\begin{aligned} \chi_L(\lambda) &=(\lambda-2)(\lambda+1)\lambda -4\lambda-9(\lambda-2)\\ &=\lambda^3-\lambda^2-15\lambda+18. \end{aligned}$$

The determinant is $-18$, which differs from $J=-8/3$. Both are conserved functions, related by $\det L=P^3/6-PH+J$. They agree when $P=0$, but that special equality must not be used at a general state.

Scalar shift

Differentiating the new matrix gives

$$\dot{\widetilde L}=[B,\widetilde L]+\dot c\,\mathrm{Id}_3.$$

The extra term vanishes identically only when $c$ is constant. The shifted traces are

$$\widetilde P=P+3c,\qquad \widetilde H=H+cP+\frac32c^2,$$

and their derivatives are $3\dot c$ and $\dot c(P+3c)$, respectively. They are not generally constant. For example, at the course’s initial invariant values and with $c(t)=t$, they become $3t$ and $3+3t^2/2$.

Each eigenvalue becomes $\widetilde\lambda_j(t)=\lambda_j+c(t)$, so its differences from the other eigenvalues stay fixed. Conserved spectral gaps alone do not imply an unchanged spectrum. If $c$ is constant, the shifted matrix again obeys the homogeneous Lax equation and all its traces are conserved.

Check your understanding

You should now be able to prove trace conservation, identify the Hamiltonian among the traces, and reconcile trace invariants with characteristic-polynomial coefficients. Cyclicity is the key algebraic step; it does not let you commute matrices outside a trace.

Continue to test independence and Poisson commutativity. That lesson supplies the missing checks between isospectral evolution and Liouville integrability.

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.