Does $\{ e^{2\pi i (n+\alpha) x}\}_{n=1}^\infty$ form a basis for $L^2[0,1]$?

Question

Let $\alpha \in (0,1)$. Consider the (densely defined) differential operator $-\Delta$ acting on $L^2$ with the boundary condition $$f(1) = e^{i\alpha} f(0), \quad f'(1) = e^{i\alpha} f'(0).$$ It can be easily shown that $e^{2\pi i (n+\alpha) x}$ is an eigenvector of $-\Delta$ that satisfies the boundary condition. It can also be easily shown that these eigenvectors are orthonormal.

Now, I am curious about the following:

1. Does $-\Delta$ self-adjoint?
2. Does $\{ e^{2\pi i (n+\alpha) x}\}_{n=1}^\infty$ form a basis for $L^2[0,1]$?