Lax pair

It happens that sometimes integrable systems could generally be viewed as those equations that can be obtained from an overdetermined system of linear differential equations, with the original PDE playing the role of the compatibility condition. The overdetermined linear system is the Lax pair associated to the equation
This is by no means a general construction: in practice, there are no general methods of computing a Lax pair for a given system of PDEs. However, when a Lax pair for a system is known, it is enormously useful. One can use Lax pairs to derive infinitely many conserved quantities, describe large classes of solutions, and play a crucial role in the celebrated Inverse Scattering Method.

Note: it seems to me that to say that a PDE system admits a Lax pair we only need that the PDE system implies the Lax equation $[L,M]=0$. I guess this is enough to take advantage of the Lax pair...

Finite degree of freedom

Consider a nonlinear system of ODEs

for example one arising in Hamiltonian mechanics,

A Lax pair for this system is a pair of matrices $L$ and $M$ that satisfy the Lax equation:

where $[L,M]=LM-ML$ is the commutator of $M$ and $L$, and $\frac{dL}{dt}$ is the time derivative of $L$. The entries of $L$ and $M$ are typically expressed in terms of the variables $x_i$, and we require that equation (1) is satisfied if and only if (2) is satisfied.

The exact form of $L$ and $M$ depends on the specific system under consideration, and is not unique. For instance, in the case of the simple harmonic oscillator with Hamiltonian $H=\frac{p^2}{2m}+\frac{1}{2}m{\omega }^2{q}^2$, we can choose:

Then, the Lax equation $\frac{dL}{dt}=[L,M]$ reproduces the equations of motion for the harmonic oscillator:

The key point is that the eigenvalues of the matrix $L$ are conserved quantities, i.e., constants of motion (see isospectral property below), and the eigenvectors provide a "good" transformation of variables (????).

Isospectral property

The eigenvalues of $L$ do not change with time, a property known as isospectrality.
Due to the cyclic property of the trace, we can show that the trace of all the powers of $L$ are conserved quantities.
In effect, the trace of $L$ is conserved, since

To show $\frac{d}{dt}\mathrm{tr}(L^2)=0$, observe

where we used the product rule. Substituting the Lax equation for $\frac{dL}{dt}$ gives:

The same is true for $\frac{d}{dt}\mathrm{tr}(L^3)=0$ and son on.

To see that the eigenvalues of $L$ are conserved, only observe that the eigenvalues are determined by the trace of the powers of the matrix.

In the previous example of the harmonic oscillator we can find the eigenvalues $\lambda$ and eigenvectors $\psi$ of the $L$ matrix. We solve the characteristic equation:

where $I$ is the identity matrix. This gives

which simplifies to

We can solve this equation for $\lambda$ to get

These are the eigenvalues of the $L$ matrix, which correspond to the given Hamiltonian.

Spectral parameter

In many cases, Lax pairs depend on an auxiliary variable, the so-called spectral parameter, which is not directly related to the dynamics of the model. The Lax pair $L(u)$, $M(u)$ then obeys the Lax equation at all values of $u\in \mathbb{C}$:

Such a Lax pair, called nonisospectral, must be constructed, as always, such that this equation is equivalent to the complete set of equations of motion. As a functional equation, it is, in principle, much more constraining than the Lax equation without spectral parameter. This feature is useful for mechanical systems with infinitely many degrees of freedom whose equations of motion could thus be formulated by a finite-dimensional Lax pair.

Even for a finite-dimensional system, Lax pairs with spectral parameter often exist. While the spectral parameter is not essential to encode finitely many equations of motion, it is nevertheless useful in several respects.

Infinite degrees of freedom (fields)

Idea: see also this paragraph.

In this case we have PDEs, for example evolution equations like KdV

the transport equation, etc, we can apply similar ideas, but now $L,M$ are operators on a Hilbert space instead of matrices.
For the KdV example we have:

In this cases, Lax equation $L_t=[L,M]$ can be replaced by $[L,M]=0$ if we replace $M$ by $M+{\partial }_t$ (simple computations).

Isospectral property

In this case, it can then be shown that the eigenvalues and more generally the spectrum of $L$ are independent of $t$ in the following way.

The matrices $L(t)$ are all similar by virtue of the existence of matrices $U(t,s)$ such that

Observe that in this case, the eigenvalues of $L(t)$ are the same as those of $L(0)$.
The existence of $U(t,s)$ is due to the existence of solution for the Cauchy problem

where $I$ denotes the identity matrix. (I guess there is a kind of result for general operators!!??)

Indeed, observe that in that case

This means that the operator $U(t,s{)}^{-1}L(t)U(t,s)$ does not change with time, and therefore, it must be equal to a constant matrix. Because $U(s,s)=I$, we have

and then $L(t)=U(t,s)L(s)U(t,s{)}^{-1}$.

Lax equation as a compatibility condition

The Lax equation,

is equivalent to the two linear compatibility conditions,

and

Here, $\psi =\psi (x,t)$ is a nonzero simultaneous solution of both of these two equations for some $\lambda \in \mathbb{C}$.

Let's see that (2) and (3) implies (1). First, by using the operator analogue of the product rule and then eliminating ${\psi }_t$ from equation (3) gives

Second, by using $L\psi =\lambda \psi$ from equation (2) and the fact that $\lambda \in \mathbb{C}$ is a constant, we get

Comparing equations (4) and (5) gives

so

From here, dividing by $\psi$ (assuming $\psi \ne 0$ and that $\frac{dL}{dt}-[L,M]$ is a multiplication operator, since it is supposed to be representing a given PDE like the KdV) gives back the Lax equation (1).

Now, to see that (1) implies (2) and (3), observe that we already know that Lax equation implies conserved eigenvalues. Differentiating the eigenvalue problem equation $L\psi =\lambda \psi$ in time, we obtain:

Using the Lax equation, this becomes:

Rearranging terms and using $L\psi =\lambda \psi$, we get:

Suppose, for simplicity, that the $\lambda$-eigenspace of $L$ is one-dimensional, then $\frac{d\psi }{dt}+M\psi =\beta \psi$, which can be interpreted as a evolution equation for $\psi$:

where $\beta$ is a possibly time-dependent complex number. I think we can get rid of $\beta$ by modifying $M$ at the beginning.

If we replace $\tilde{L}=L-\lambda$ and $\tilde{M}=M+{\partial }_t$ then equation (1) becomes

since

And equations (2) and (3) become

As a zero-curvature condition

Pending task

Relation to QM

In Quantum Mechanics, observables are operators on a Hilbert space, just as $L$ and $M$.
The Lax equation $\dot{L}=[L,M]$ bears a resemblance to the Heisenberg equation of motion for an operator $Q$ in the Heisenberg picture of quantum mechanics:

where $H$ is the Hamiltonian. If $H$ and $Q$ are finite dimensional matrices, then $\text{tr}([Q,H])=0$ so that $\text{tr}(Q)$ is conserved. But often, operators in quantum mechanics are infinite dimensional and unbounded. The trace of the commutator of such operators may not vanish (or even be finite). In such cases, $\text{tr}(Q)$ may not be a (finite) conserved quantity.

Open questions

Can we start with arbitrary operators $L$, $M$ and then the condition $[L,M]=0$ is automatically an integrable system? This should be related with Inverse Scattering Method...
How do we obtain conserved quantities if we have no sense of "trace"?
What does it have to do with the zero-curvature representation?
What are the elements of the Hilbert space on which do these operator act?
What does it have to do with the inverse scattering transform?