How to find an orthonormal basis for $L^2(\mathbb{R},\mathbb{C})$? Consider the Hilbert space 

$X:=L^2(\mathbb{R},\mathbb{C})$

Now consider the operator that takes the second derivative, i.e.

$A := \partial_{x}^2$, i.e. $A: H^2(\mathbb{R},\mathbb{C}) \subset X \to X$

I need to know if there are $(\alpha_n) \subset \mathbb{C}$ for $n \in \mathbb{N}$ and an orthonormal basis $\lbrace e_n : ~  n \in \mathbb{Z}_+ \rbrace$ of $X$ such that

$ A e_n = \alpha_n e_n.$

and additionally I need to calculate the $\lbrace a_n \rbrace$, so I guess I need the basis in closed form. Any hint? 
 A: The solutions of $y''=\alpha y$ are functions of the form $y=Ce^{\beta x}$, $\beta^2=\alpha$. None of them 
belong to $L^2(\mathbb R)$ except for the zero solution. Hence, the one-dimensional Laplacian $\Delta = \frac{d^2}{dx^2}$ has no eigenvalues. 
The spectrum of $\Delta$ is $(-\infty, 0]$. The quickest way to see this is to apply the Fourier transform, which
converts $\Delta$ into multiplication by $-\xi^2$. The spectrum of a multiplication operator is the essential range 
of its symbol (the function by which it multiplies). 
One can also show directly that for any $\alpha\ge 0$ the operator $\Delta+\alpha I$ is not bounded below. Indeed, 
let $y$ be the cosine wave $\cos \sqrt{\alpha} x $, smoothly cut off around $x=\pm M$ to make $y$ compactly supported. 
The application of $\Delta+\alpha I$ to $y$ destroys the periodic part of $y$, leaving only the effects of our 
modification around $x=\pm M$. Making $M$ large enough, we can ensure that  the $L^2$ norm of 
 $y$ is much larger than the $L^2$ norm of $(\Delta+\alpha I)y$. 
