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 $$D(T)=\left\{f\,:\, \iint (x^2+y^2)|f(x,y)|^2 \,dx\,dy<\infty\right \}$$
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.
