Skew-adjoint differential operator $B$ with spectrum $\sigma(B)=i(-\infty,-1]$

Question

Consider the Hilbert space $X=L^{2}\left(\mathbb{R}^n\right)$ and the Schrödinger operator $A=i\Delta$ defined on the domain $D(A)=H^2(\mathbb{R}^n)$. It is known that the spectrum of $A$ is $\sigma(A)=i(-\infty,0]:=\{i\omega,\ \omega\leq0 \}$. I am just asking if there exists a skew-adjoint differential operator $B$ with a spectrum $\sigma(B)=i(-\infty,-1]$.

Answer 1

This is just an intuition for the much simpler case of finite dimension. Let $A$ be any matrix and $v$ an associated eigenvector, i.e. $Av = \lambda v$. Then $(A-I)v = Av-v = (\lambda-1)v$, and thus $\sigma(A-I)= \sigma(A)-1$.

And now an attempt for infinite dimension, suppose that $(A-i\lambda I)$ is not invertible for every $\lambda \in (-\infty, 0]$ but then $(A-i(\lambda+1)I) = ((A-iI)-i\lambda)$ is not invertible for every $\lambda \in (-\infty,-1]$, i.e. $\sigma(A-iI)=i (-\infty,-1]$.