Discrete and Essential spectrum of Laplacian in $\mathbb R_{+}$ (with weird boundary conditions) I am given 
on Hilbert Space $\mathcal H=L^2(\mathbb R_{+})$
$$
Af(x)=-f''(x)
$$
 and 
Domain of A is 
$$
D(A)=\{f\in H_2(\mathbb R_{+})\;\;| \;\;f'(0)+\alpha f(0)=0\}
$$
for some $\alpha \in \mathbb R$
I need to find discrete and essential spectrum of this unbounded operator. I only know one way of finding the spectrum of Laplacian, that is by Fourier transform (and its variant for half line). I do not know how to handle this situation.
 A: Consider the operator 
$$
A_0f(x)=-f''(x)
$$
$$
D(A_0)=\{f\in H_2(\mathbb R_{+})\;\;| \;\;f'(0)=f(0)=0\}
$$
It is a symmetric operator with (finite and equal) defect indices $(1,1)$. All self-adjoint extensions (including operator $A$ depending on $\alpha\in\mathbb R$, and the one with '$\alpha=\infty$', which corresponds to the boundary condition $f(0)=0$) have the same essential spectrum. In this question it is shown that this spectrum is $[0,\infty)$.
Assume that $A$ has a point spectrum, i.e. $Af=-f''=\lambda f,\ \lambda\in\mathbb R$. The fundamental solution of $f''+\lambda f=0$ is either $$\{\sin{\sqrt{\lambda}x},\ \cos{\sqrt{\lambda}x}\},\ \lambda>0$$ or $$\{e^{\sqrt{-\lambda}x},e^{-\sqrt{-\lambda}x}\},\ \lambda<0$$ or $$\{1,x\},\ \lambda=0.$$
Since $f\in H_2(\mathbb R_{+})$, the only possible eigenvector of $A$ is $e^{-\sqrt{-\lambda}x}$ for $\lambda <0$. The boundary condition $f'(0)+\alpha f(0)=0$ implies 
$$-\sqrt{-\lambda}e^{0}+\alpha e^{0}=0$$ i.e. $\alpha=-\sqrt{-\lambda},\ \lambda<0$.
Hence, $A$ has a point spectrum iff $\alpha<0$. In this case $\sigma_p(A)=\{-\alpha^2\}.$
