[proofplan] We perform a perturbative expansion of the scattering solution $\phi_k$ for a small potential $u = \varepsilon q$. Writing $\phi_k = e^{-ikx}(1 + \varepsilon y_k + O(\varepsilon^2))$ and substituting into the eigenvalue equation yields a first-order ODE for $y_k'$ involving the integrating factor $e^{-2ikx}$. Integrating from $-\infty$ and taking the limit $x \to +\infty$, we read off the leading-order contributions to $a(k)$ and $b(k)$ by comparing with the standard scattering asymptotics, yielding $R(k) = b/a$ to leading order in $\varepsilon$. [/proofplan] [step:Set up the perturbation expansion $\phi_k = e^{-ikx}(1 + \varepsilon y_k + O(\varepsilon^2))$] The scattering solution satisfies \begin{align*} -\phi_k'' + \varepsilon q(x)\, \phi_k = k^2 \phi_k, \end{align*} with the boundary condition $\phi_k \sim e^{-ikx}$ as $x \to -\infty$. For $\varepsilon = 0$, the solution is simply $\phi_k^{(0)} = e^{-ikx}$. We seek a correction of the form \begin{align*} \phi_k(x) = e^{-ikx}\bigl(1 + \varepsilon\, y_k(x) + O(\varepsilon^2)\bigr), \end{align*} where the function $y_k: \mathbb{R} \to \mathbb{C}$ satisfies $y_k(x) \to 0$ as $x \to -\infty$ (so that the boundary condition at $-\infty$ is preserved at all orders). [guided] The idea behind the perturbation expansion is natural: for a weak potential $\varepsilon q$, the scattering solution should be close to the free solution $e^{-ikx}$. We factor out the leading behaviour $e^{-ikx}$ and expand the correction in powers of $\varepsilon$. The function $y_k$ captures the first-order deviation from free propagation. The boundary condition $\phi_k \sim e^{-ikx}$ as $x \to -\infty$ means that the wave is incident from the left with unit amplitude. For $\varepsilon = 0$ there is no potential, so the wave passes through unchanged. For small $\varepsilon$, the potential induces a small correction $\varepsilon y_k$ that builds up as the wave propagates through the potential region. [/guided] [/step] [step:Substitute the expansion into the Schrodinger equation and collect terms at order $\varepsilon$] We compute the derivatives of $\phi_k = e^{-ikx}(1 + \varepsilon y_k)$: \begin{align*} \phi_k' &= -ik\, e^{-ikx}(1 + \varepsilon y_k) + \varepsilon\, e^{-ikx} y_k', \\ \phi_k'' &= -k^2 e^{-ikx}(1 + \varepsilon y_k) - 2ik\,\varepsilon\, e^{-ikx} y_k' + \varepsilon\, e^{-ikx} y_k''. \end{align*} Substituting into $-\phi_k'' + \varepsilon q\, \phi_k = k^2\phi_k$: \begin{align*} k^2 e^{-ikx}(1 + \varepsilon y_k) + 2ik\,\varepsilon\, e^{-ikx} y_k' - \varepsilon\, e^{-ikx} y_k'' + \varepsilon\, q\, e^{-ikx}(1 + \varepsilon y_k) = k^2 e^{-ikx}(1 + \varepsilon y_k). \end{align*} The $k^2 e^{-ikx}(1 + \varepsilon y_k)$ terms cancel on both sides. Dividing by $\varepsilon\, e^{-ikx}$ and dropping the $O(\varepsilon)$ correction: \begin{align*} 2ik\, y_k' - y_k'' + q = O(\varepsilon). \end{align*} At leading order in $\varepsilon$, the equation for $y_k$ is \begin{align*} y_k'' - 2ik\, y_k' = q. \end{align*} [/step] [step:Reduce to a first-order equation using the integrating factor $e^{-2ikx}$] Define $w_k := y_k'$. The equation $y_k'' - 2ik\, y_k' = q$ becomes \begin{align*} w_k' - 2ik\, w_k = q. \end{align*} Multiply both sides by the integrating factor $e^{-2ikx}$: \begin{align*} \frac{d}{dx}\bigl(e^{-2ikx} w_k\bigr) = e^{-2ikx} q(x). \end{align*} This can be verified directly: $\frac{d}{dx}(e^{-2ikx} w_k) = e^{-2ikx}w_k' - 2ik\, e^{-2ikx} w_k = e^{-2ikx}(w_k' - 2ik\, w_k) = e^{-2ikx} q$. Integrating from $-\infty$ to $x$, with the boundary condition $w_k(-\infty) = y_k'(-\infty) = 0$ (since $y_k \to 0$ as $x \to -\infty$): \begin{align*} e^{-2ikx} w_k(x) = \int_{-\infty}^{x} e^{-2ikz}\, q(z)\, dz. \end{align*} Solving for $w_k = y_k'$: \begin{align*} y_k'(x) = e^{2ikx} \int_{-\infty}^{x} e^{-2ikz}\, q(z)\, dz. \end{align*} [/step] [step:Integrate once more to find $y_k(x)$ and reconstruct $\phi_k$] Integrating $y_k'(x) = e^{2ikx}\int_{-\infty}^{x} e^{-2ikz} q(z)\, dz$ from $-\infty$ to $x$ (with $y_k(-\infty) = 0$) and using integration by parts, we find the full first-order correction. However, for the purpose of reading off the scattering coefficients, we only need the $x \to +\infty$ behaviour of $\phi_k$. As $x \to +\infty$ (beyond the support of $q$), the integral $\int_{-\infty}^{x} e^{-2ikz} q(z)\, dz$ stabilises to \begin{align*} \hat{q}(2k) := \int_{-\infty}^{\infty} e^{-2ikz}\, q(z)\, dz, \end{align*} which is the Fourier transform of $q$ evaluated at frequency $2k$. Therefore, for $x$ beyond the support of $q$: \begin{align*} y_k'(x) = \hat{q}(2k)\, e^{2ikx}. \end{align*} Integrating from some point $x_0$ (beyond the support of $q$) to $x$: \begin{align*} y_k(x) = y_k(x_0) + \frac{\hat{q}(2k)}{2ik}\, e^{2ikx} - \frac{\hat{q}(2k)}{2ik}\, e^{2ikx_0}. \end{align*} Write $y_k(x) = \alpha_0 + \frac{\hat{q}(2k)}{2ik}\, e^{2ikx}$ for $x$ beyond the support, where $\alpha_0$ is a constant that depends on the detailed structure of $q$. The full scattering solution is then, for large $x$: \begin{align*} \phi_k(x) = e^{-ikx}\Bigl(1 + \varepsilon\alpha_0 + \frac{\varepsilon\,\hat{q}(2k)}{2ik}\, e^{2ikx} + O(\varepsilon^2)\Bigr). \end{align*} Distributing: \begin{align*} \phi_k(x) = (1 + \varepsilon\alpha_0)\, e^{-ikx} + \frac{\varepsilon\,\hat{q}(2k)}{2ik}\, e^{ikx} + O(\varepsilon^2). \end{align*} [/step] [step:Read off $a(k)$ and $b(k)$ and compute $R(k)$ to leading order] Comparing the $x \to +\infty$ asymptotics \begin{align*} \phi_k(x) \sim a(k)\, e^{-ikx} + b(k)\, e^{ikx} \end{align*} with the expression derived above, we identify \begin{align*} a(k) &= 1 + \varepsilon\alpha_0 + O(\varepsilon^2) = 1 + O(\varepsilon), \\ b(k) &= \frac{\varepsilon}{2ik}\int_{-\infty}^{\infty} e^{-2ikz}\, q(z)\, dz + O(\varepsilon^2). \end{align*} The reflection coefficient is \begin{align*} R(k) = \frac{b(k)}{a(k)} = \frac{\varepsilon\,\hat{q}(2k)/(2ik) + O(\varepsilon^2)}{1 + O(\varepsilon)} = \frac{\varepsilon}{2ik}\int_{-\infty}^{\infty} e^{-2ikz}\, q(z)\, dz + O(\varepsilon^2), \end{align*} where the division by $1 + O(\varepsilon)$ does not affect the leading $O(\varepsilon)$ term. This establishes that to leading order in $\varepsilon$, the reflection coefficient $R(k)$ is (up to the factor $1/(2ik)$) the Fourier transform of $q$ evaluated at $2k$: \begin{align*} R(k) = \frac{\varepsilon}{2ik}\,\hat{q}(2k) + O(\varepsilon^2). \end{align*} [guided] The final step is purely algebraic: matching coefficients of the two linearly independent functions $e^{-ikx}$ and $e^{ikx}$ in the asymptotic region. The coefficient of $e^{-ikx}$ gives $a(k)$, and the coefficient of $e^{ikx}$ gives $b(k)$. The $O(\varepsilon)$ correction to $a(k)$ (the constant $\alpha_0$) does not affect $R(k)$ at leading order, because $R = b/a$ and $b$ is already $O(\varepsilon)$, so the correction from $a = 1 + O(\varepsilon)$ in the denominator contributes only at $O(\varepsilon^2)$. The result has a clean physical interpretation: in the Born approximation (first-order perturbation theory), the reflection from a weak potential is determined by its Fourier content. A potential $q$ that has no Fourier component at frequency $2k$ does not reflect waves of wavenumber $k$ to leading order. This is the quantum-mechanical analogue of Bragg diffraction: reflection occurs when the wavelength matches the spatial periodicity of the scattering potential. Note also that $R(k) \to 0$ as $k \to \infty$ (since $\hat{q}(2k) \to 0$ by the Riemann-Lebesgue lemma, provided $q \in L^1$), consistent with the general principle that high-energy waves are transmitted through any bounded potential. [/guided] [/step]