= Solution
Remove the commutator gauge by setting $\Psi_j=\mu_je^{i\theta\sigma_3}$. These matrices all satisfy the ordinary first-order systems
$$
\Psi_x=(ik\sigma_3+Q)\Psi,\qquad
\Psi_t=(-4ik^3\sigma_3+\widetilde Q)\Psi.
$$
Both coefficient matrices are traceless, so the normalization implies $\det\Psi_j=\det\mu_j=1$. For any two solutions, differentiating $\Psi_2^{-1}\Psi_j$ shows that it is independent of both $x$ and $t$. Calling these constant matrices $s$ and $S$ gives
$$
\boxed{\mu_3=\mu_2e^{i\theta\widehat\sigma_3}s,\qquad
\mu_1=\mu_2e^{i\theta\widehat\sigma_3}S.}
$$
At $(0,0)$, $\mu_2=I$, so $s=\mu_3(0,0,k)$. More generally,
$$
s=e^{-ikx\widehat\sigma_3}[\mu_2(x,0,k)^{-1}\mu_3(x,0,k)].
$$
At $(0,T)$ the other normalization is $\mu_1=I$, so
$$
\boxed{S=e^{4ik^3T\widehat\sigma_3}\mu_2(0,T,k)^{-1}.}
$$
Thus $s$ is determined by the initial spatial <Volterra integral equation>, while $S$ is determined by the boundary time equation evaluated at $T$. The latter uses all boundary values occurring in $\widetilde Q$, including the initially unknown second-derivative trace. It is the <global relation>, rather than a claim of independence of those traces, that supplies their constraint.
Back to article page
