= Solution
Use the half-line <Fourier transform> and finite-time boundary transforms
$$
\widehat q(k,t)=\int_0^\infty e^{-ikx}q(x,t)\,dx,\qquad G_j(k,t)=\int_0^t e^{ik^2s}g_j(s)\,ds,\quad g_j(t)=\partial_x^jq(0,t),\quad j=0,1.
$$
The half-line <Fourier transform> is analytic for $\operatorname{Im}k<0$ under spatial decay. All spectral integrals below have their usual oscillatory, or vanishing Gaussian damping, interpretation until absolute convergence is established.
The <free Schrodinger equation> has the local divergence identity
$$
\partial_t(e^{-ikx+ik^2t}q)=i\partial_x\left[e^{-ikx+ik^2t}(q_x+ikq)\right].
$$
Integrating in $x,t$ gives the <global relation for the half-line free Schrodinger equation>
$$
\boxed{e^{ik^2t}\widehat q(k,t)=\widehat q_0(k)+kG_0(k,t)-iG_1(k,t),\qquad\operatorname{Im}k\leq0.}
$$
Let $D_+=\{\operatorname{Re}k>0,\operatorname{Im}k>0\}$. Orient its boundary from $i\infty$ down to $0$, and then from $0$ to $+\infty$, so that $D_+$ lies on the left. <Fourier inversion> followed by <contour integration> in the second quadrant yields
$$
\boxed{q(x,t)=\frac1{2\pi}\int_{\mathbb R}e^{ikx-ik^2t}\widehat q_0(k)\,dk+\frac1{2\pi}\int_{\partial D_+}e^{ikx-ik^2t}\left[kG_0(k,t)-iG_1(k,t)\right]dk.}
$$
This is a complex spectral representation involving the initial trace and both boundary traces. The contour deformation works because each boundary-time integrand contains $e^{-ik^2(t-s)}$ with $s\leq t$, which decays in the second quadrant, as well as $e^{ikx}$ for $x>0$. This fixes both the quadrant and the orientation; changing either without changing the signs would produce an incorrect representation.
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