Update: I have found that the detailed answer to my questions is contained in the book "Solitons: an introduction" by P.G. Drazin and R.S. Johnson. Generally speaking, this seems to be a great book for someone looking for a gentle introduction to inverse scattering techniques.
I am trying to understand how to derive the time evolution for the reflection coefficient in the setting of inverse scattering for the 1d cubic NLS
$$ iq_t + q_{xx} = 2|q|^2 q. $$
Setup. [Mostly following Deift-Zhou.] Given a function $q=q(x)$, one studies the scattering problem $$ \partial_x \psi = Q\psi + iz\sigma\psi $$ where $z\in\mathbb{R}$, $\psi$ is a 2x2 matrix,
$$Q=\left[\begin{array}{cc} 0 & q \\ \bar q & 0 \end{array}\right],\quad\text{and}\quad \sigma=\tfrac12\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right].$$
One then finds two solutions $\psi^\pm$ satisfying $$ \psi^\pm \sim e^{ixz\sigma} \quad\text{as}\quad x\to\pm\infty. $$ Because $\text{tr}(Q+iz\sigma)=0$ one finds $\det\psi^\pm \equiv 1$, and by uniqueness one can write $\psi^+ = \psi^- A$ for some $A=A(z)$. By symmetry considerations one can see that $A$ has the form $$ A(z) = \left[\begin{array}{cc} a(z) & \bar b(z) \\ b(z) & \bar a(z)\end{array}\right]. $$ The ratio $r(z) = -\frac{\bar b(z)}{\bar a(z)}$ is the reflection coefficient. One writes $r=R[q]$.
Problem. If $q(t)$ evolves according to the NLS, then $r(t)=R[q(t)]$ is supposed to have the simple evolution $r(t,z) = e^{-iz^2 t}r(0,z)$ (and vice versa). This is what I have been trying to derive.
Attempts. As I understand it, one imposes time evolution on the spectral problem of the form $\partial_t \psi = M\psi$. Then the compatibility of the two differential equations for $\psi$ reduces to $$\partial_t Q = \partial_x M + [M,Q+iz\sigma],$$ where $[\cdot,\cdot]$ is the commutator. If one seeks $M$ as a quadratic polynomial in $z$ then one can recover the NLS (by working down from the $z^3$ terms and then eventually matching the $z^0$ terms in the equation above). In particular, I computed $$ M=-i(z^2+2|q|^2)\sigma -zQ - i\partial_x Q\sigma. $$
My thought was that imposing this time evolution on $\psi^\pm$ should lead to time evolution for $a$ and $b$, and hence for $r$. To see this, we should find formulas for $a,b$ in terms of $\psi^\pm$. From the relation $\psi^+ = \psi^- A$, one finds $$ \psi^+_1 = a\psi^-_1 + b\psi^-_2, $$ where $\psi^+_1$ is the first column of $\psi^+$ and so on. From this (forming the matrix with $\psi^-_j$ as the second column and using $\det \psi^-=1$) one can get $$ a=\det[ \psi^+_1 \ \psi^-_2]\quad\text{and} \quad b = -\det[\psi^+_1 \ \psi^-_1]. $$
Up to this point, what I have done seems to match various references that I have been following. However, if I now try to use the fact that $\psi^\pm$ solve the prescribed evolution equation, I seem to get $a_t=0$ and $b_t=0$ (and hence $r_t=0$). This is because the matrices $[\psi^+_1 \ \psi^-_2]$ and $[\psi^+_1 \ \psi^-_1]$ define solutions to $\partial_t \psi = M\psi$ (don't they?) and $\text{tr}(M)=0$.
What I was expecting was $a$ to be constant and $b(t)=e^{-iz^2t}b(0)$. So I imagine I have a mistake somewhere, or I am failing to understand something (or many things). Any help or insight (or a reference that would bother with such details) would be much appreciated. Thanks!
