Unbounded operator whose spectrum is the entire complex plane?

The question is simple: how to find an unbounded operator $$T:H\to H$$ where $$H$$ is a Hilbert space such that $$\text{Sp} T = \mathbb C$$? This seems a very basic thing, but I have not found an example in the literature.

In some proofs, we need to consider this case separately. This example should be quite important.

Yes, such an example can be found on page 254 of Reed & Simon's Methods of Modern Mathematical Physics I.

Let $$AC[0,1]$$ be the family of all absolutely continuous functions on $$[0,1]$$ whose derivatives are in $$L^2[0,1]$$. Let $$T:L^2[0,1] \to L^2[0,1]$$ be the densely defined operator $$i \frac{d}{dx}$$ whose domain is the set $$D(T) = \{ f \in L^2[0,1] : f \in AC[0,1] \}.$$ $$T$$ is then a closed operator whose spectrum is the entire complex plane $$\Bbb C$$. Indeed, observe that $$(\lambda I - T)e^{-i\lambda x} = 0$$ and the function $$f(x) = e^{-i\lambda x}$$ belongs to $$D(T)$$ for each $$\lambda \in \Bbb C$$.

Here is an elementary example.

Let $$a_n$$ be a sequence dense in the complex plane. For example $$a_n$$ is an enumeration of $$\mathbb{Q}\oplus i\mathbb{Q}.$$ Consider the operator $$T$$ on $$\ell^2(\mathbb{N})$$ given by $$T\{x_n\}=\{a_nx_n\}$$ with domain $$D(T)=\left\{\{x_n\}\,:\, \sum |a_nx_n|^2<\infty\right \}$$ Then $$T$$ is closed and the spectrum is equal to the entire complex plane.

Another example: let $$(Tf)(x,y)=(x+iy)f(x,y)$$ act on $$L^2(\mathbb{R}^2)$$ with domain $$D(T)=\left\{f\,:\, \iint (x^2+y^2)|f(x,y)|^2 \,dx\,dy<\infty\right \}$$

• Let the record show I did not see your second example when I wrote my answer ;). In any event, to the OP, the first of these examples is more instructive, it's how you typically build operators with particular properties. Commented Mar 24 at 18:13
• @operatorerror It seems we got the same example at the same time :) There was an error concerning the domain, which I have just corrected. Commented Mar 24 at 18:56

Consider $$L^2(\mathbb C,dA(z))$$ with $$dA(z)$$ the area measure and take $$Tf(z)=zf(z).$$ Recall that the spectrum is the (essential) range of the function defining a multiplication operator.