Consider the differential operator acting on $L^2(\mathbb{R})$, $$L=-\frac{d^2}{dx^2}, 0\leq x<\infty$$ with self adjoint boundary conditions $$\psi(0)/\psi'(0)=\tan \theta$$ for some fixed angle $\theta$.
a) Show that when $\tan\theta<0 $ there is a single negative eigenvalue with a normalizeable eigenfunction $\psi_0(x)$ localized near the origin, but none when $\tan\theta>0$.
If anyone could be kind enough to help me with any of these problem parts I would really appreciate it.They are way over my head right now and I need to somehow learn how to do them. :(
For $\lambda\in\mathbb{C}$ there is a unique classical solution of $$ -f'' = \lambda f, \;\; f(0)=\sin\theta,\; f'(0)=\cos\theta. $$ That solution can be expressed as $$ f_{\lambda}(x)=\cos(\sqrt{\lambda}x)\sin\theta + \frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}\cos\theta. \tag{*} $$ Assume the branch cut of $\sqrt{\lambda}$ to be along the positive real axis.
(a) The only issue of whether or not $f$ is an eigenfunction of the operator $L$ defined by $Lf=-f''$ is whether or not $f$ is square integrable on $[0,\infty)$. In terms of exponentials, $$ f_{\lambda}(x) = \frac{1}{2}\left(\sin\theta+\frac{1}{i\sqrt{\lambda}}\cos\theta\right)e^{i\sqrt{\lambda}x} + \frac{1}{2}\left(\sin\theta -\frac{1}{i\sqrt{\lambda}}\cos\theta\right)e^{-i\sqrt{\lambda}x}. \tag{**} $$ Eigenvalues have to be real for this selfadjoint operator. You can use. For real and positive $\lambda$, $f \notin L^2[0,\infty)$. If $\lambda < 0$, then $\sqrt{\lambda}=i|\lambda|^{1/2}$, and $e^{i\sqrt{\lambda}x}=e^{-|\lambda|^{1/2}x}\in L^2$, which gives $f\in L^2$ iff $$ \sin\theta - \frac{1}{i\sqrt{\lambda}}\cos\theta=0 \\ \sin\theta +\frac{1}{|\lambda|^{1/2}}\cos\theta = 0 \\ \tan\theta = -|\lambda|^{1/2}. $$ So there is one solution $f \in L^2$ when $\tan\theta < 0$ and none otherwise.
(b) For real and positive $\lambda$, the function $f_{\lambda}(x)$ is bounded for $x\in[0,\infty)$, and may be written as $$ f_{\lambda}(x)=A(\lambda)\sin(\sqrt{\lambda}x+\phi_{\lambda}). $$