= Solution
Undo the commutator gauge by putting $\psi_j=\mu_je^{-i\theta\sigma_3}$. Then $\psi_{j,x}=(-ik\sigma_3+Q)\psi_j$ and $\psi_{j,t}=(-2ik^2\sigma_3+V)\psi_j$, with $V=2kQ-iQ_x\sigma_3-i|q|^2\sigma_3$. The <trace> of each coefficient matrix is zero, so all normalized fundamental matrices have determinant one. Their ratios $\psi_2^{-1}\psi_j$ are independent of both $x$ and $t$, because differentiation cancels the common left-hand coefficient.
Call these constant matrices $s$ for $j=3$ and $S$ for $j=1$. Returning to the commutator gauge gives
$$
\boxed{\mu_3=\mu_2e^{-i\theta\widehat\sigma_3}s,\qquad\mu_1=\mu_2e^{-i\theta\widehat\sigma_3}S.}
$$
Here $e^{\alpha\widehat\sigma_3}C=e^{\alpha\sigma_3}Ce^{-\alpha\sigma_3}$. Evaluation at $(0,0)$ gives $s=\mu_3(0,0,k)$, computed solely by the spatial <Volterra integral equation> for the initial potential. At $(0,T)$ the normalization of $\mu_1$ gives
$$
\boxed{S(k)=e^{2ik^2T\widehat\sigma_3}\mu_2(0,T,k)^{-1}.}
$$
Thus $S$ is computed by integrating the time <Lax pair> for $\mu_2(0,t,k)$ from zero to $T$. It uses both boundary traces $q(0,t)$ and $q_x(0,t)$ before their compatibility is enforced.
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