Suppose $T: C^\infty_c(J) \to L^2(J)$ is the kinetic operator $T\psi = -\frac{\hbar^2}{2m} \partial_x^2\psi$ on the open half-line $J=(0,\infty)$. This operator is symmetric and has infinitely many self-adjoint extensions, because the positive and negative deficiency indices are non-zero and coincide ($d_\pm(T)=1$). Then each self-adjoint extension is parametrized by a phase factor $e^{i\alpha}$, due to the unitary isomorphism $$U_0:\ker(T^*-i1)\simeq\Bbb C\to\Bbb C\simeq\ker(T^*+i1)$$ and the domain of each self-adjoint extension $T\subset T'_\alpha \subset T^*$ is $$\operatorname{dom}(T'_\alpha) = \{\psi \in \operatorname{dom}(T^*)=W^{2,2}(J)\ |\ \exists!\tilde\psi \in \operatorname{dom}(\overline T),\ \psi_0 \in \ker(T^*-i1)\ :\ \psi = \tilde\psi + (1-U_0)\psi_0 \}.$$ We can rewrite the generic element $(1-U_0)\psi_0 \in \ker(T^*-i1)\oplus \ker(T^*+i1)$ as $\psi_0=z(\psi_--e^{i\alpha}\psi_+)$, where $z \in \Bbb C$ and $\psi_\pm$ are normalized elements of $\ker(T^*\pm i1)$. Now, by solving $T^*\psi_\pm \pm i\psi_\pm = 0$ in the strong sense (thanks to the Sobolev embedding theorem) we obtain, remembering the constraint $\psi_\pm \in W^{2,2}(J)$, $$\psi_\pm(x) = a_\pm e^{-i\omega_\pm x}, $$ where $\omega_\pm = \frac{\sqrt m}{\hbar} (\pm 1 - i)$. Normalizing yields $$a_\pm = \sqrt{\frac{2\sqrt m}{\hbar}} =: \nu. $$
To turn the algebraic condition $\psi = \tilde\psi + (1-U_0)\psi_0$ into a boundary condition on $\psi$, I should evaluate the generic $\psi$ at $0^+$, as well as its derivative. What I get is $$\require{cancel}\begin{split} \psi(0^+) &= \cancel{\tilde \psi(0^+)} + \nu z( e^{-i\omega_- 0^+} - e^{i\alpha} e^{-i\omega_+ 0^+}) = \nu z (1-e^{i\alpha}); \\ \partial_x\psi(0^+)&= \cancel{\partial_x\tilde \psi(0^+)} + \nu z (-i\omega_- e^{-i\omega_-0^+} - e^{i\alpha}(-i\omega_+) e^{-i\omega_+0^+}) \\ &= \nu z \left( -i \frac{\nu^2}{2} (-1-i) - e^{i\alpha}(-i) \frac{\nu^2}2 (1-i) \right) \\ &= \frac{\nu^3}2 z \left( -(1-e^{i\alpha}) + i(1+e^{i\alpha}) \right), \end{split} $$ where the $\tilde\psi,\partial_x\tilde \psi$ cancel out because they are (limits of) compactly supported functions on $J$. All in all, I obtain the Robin-type condition $\psi(0^+) + f(\alpha)\partial_x\psi(0^+) = 0$ with $$f(\alpha) = \frac{2}{\nu^2} \frac{1}{\cot(\alpha/2) + 1}, $$ whereas I expected something along the lines of a much simpler $f(\alpha)\propto\tan(\alpha)$ (see the end of this answer). What am I getting wrong?