On taking exponential of a general operator on Hilbert space

Question

Can we do functional calculus for nonnormal operators? Like breaking into two self adjoints and applying the fa again

Answer 1

If $A$ is a bounded linear operator, then you can define $e^{A}$ and more generally $f(A)$ for functions $f$ that are holomorphic on a neighborhood of the spectrum of $A$. This is known as holomorphic functional calculus und works as follows. Let $\Gamma$ be a closed curve that encloses $\sigma(A)$ such that $f$ is holomorphic on the "inside" of $\Gamma$. Then one defines $$ f(A)=\frac 1{2\pi i}\int_\Gamma f(\zeta)(\zeta-A)^{-1}\,d\zeta. $$ If $f(z)=\sum_{k=0}^\infty a_k z^k$ on a neighborhood of $\sigma(A)$, then $$ f(A)=\sum_{k=0}^\infty a_k A^k. $$ In particular, $$ e^{A}=\sum_{k=0}^\infty \frac{A^k}{k!}. $$ Decomposing $A$ into real and imaginary part does not work as easily as you might expect. One is tempted to define $e^{A}=e^{(A+A^\ast)/2}e^{(A-A^\ast)/2}$, guided by the formula $e^{z+w}=e^{z}e^{w}$. But the identity $e^{A+B}=e^{A}e^{B}$ does not necessarily hold for normal operators unless they commute, so this is not a good definition. However, there is the Trotter product formula $$ e^{A+B}x=\lim_{n\to\infty}(e^{A/n}e^{B/n})^nx. $$ If the operator $A$ is unbounded, then there is in general no satisfactory way to define $e^{A}$. If you want $(e^{tA})$ to be a $C_0$-semigroup with generator $A$ (which is a reasonable assumption, of course you could give $e^{A}$ just any meaning), then there are certain restriction on the spectrum of $A$ specified by the Hille-Yosida theorem. For example, a self-adjoint operator with spectrum $\mathbb{R}$ does not generate a $C_0$-semigroup.

One can define a functional calculus for classes of non-normal operators even in the unbounded case, for example for sectorial operators. The idea is similar to the holomorphic functional calculus for bounded operators. The spectrum is no longer bounded, but it is contained in a sector of the complex plane. Instead of a curve enclosing the sector, one now takes a curve coming from infinity, going around the sector and then going to infinity again (and one also needs certain decay conditions at infinity).