I asked a related question (so far unanswered) on mathoverflow (here: https://mathoverflow.net/questions/269076/exponentially-weighted-spaces-effect-on-spectrum ). To clarify some concepts related to it, I am asking this question.
Consider an unbounded non-self adjoint operator $L$ on two different spaces , $H_a$=$L^2[\mathbb{R},e^{-x^2}dx]$, and $H_b=L^2[\mathbb{R},dx]$.
I am interested in numerically computing the spectrum.
CASE 1: $H_b$
So if I work in $H_b$, the usual way would be to either
a). Pick a finite domain $[-L,L]\in \mathbb{R}$, and use finite-difference to numerically estimate the operator. Then compute its eigenvalues/vectors.
or
b). Pick a spectral cut-off frequency, and discretize the fourier-transform of the operator, and again find eigenvalues.
Note that doing either of the above can result in artificial "eigenvalues", i.e. numbers which are not really eigenvalues (e.g. consider $L=\nabla^2$, which has no eigenvalues on real line).
CASE 2: $H_a$
Now on the other hand, let us work on $H_a$. In this case, one can use Hermite functions, which are known to be orthonormal-basis of $H_a$. Hence, the operator $L$ can then be written down in the eigenbasis given by the Hermite functions. The "cut-off" is now $N$, i.e. we take the basis given by first $N$ Hermite functions. My question is in two parts:
1). What kind of "spurious" eigenvalues can we expect in CASE 2 ?
2). What is "really" going under the hood here: why do we get a countable basis by using Hermite functions (CASE 2), while an uncountable when using finite different/ fourier transform cut-off (CASE 1)