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 2×2 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!