An hermitian operator problem It is possible to have two hermitian operators $A$ et $B$, with :
$B^2 = \mathbb{I}d$
$[A,B] = i * \mathbb{I}d$
where $i$ is the usual (complex) square root of $(-1)$, and $\mathbb{I}d$ is the identity operator ?
(I think that A is necessarily not bounded, due to the last condition)
If it is possible, may we exhibit a explicit representation of these operators $A$ and $B$ ?
 A: The commutator $[A,B]$ is proportional to $\mathbb{I}$ and therefore commutes with everything. So
$$A-BAB=AB^2-BAB=[A,B]B=B[A,B]=BAB-B^2A=BAB-A$$
or
$$A=BAB.$$
But then
$$[A,B]=AB-BA=(BAB)B-BA=BA-BA=0.$$
Btw. from this it follows that the result generalizes to all commutators with $$[A,B]=f(B).$$
A: This is impossible. In fact, I will prove that if $A,B$ are operators in a real or complex algebra such that $B$ has finite multiplicative order ( that is, $B^{k} = I$ for some positive integer $k),$ then we can't have $AB -BA = cI$ for any non-zero scalar $c.$
For otherwise, an induction argument (due to Wielandt, and to be found in the link given in Jonas's comment) and which does not require any boundedness assumption, shows that we
have $A(c^{-1}B)^{n} - (c^{-1}B)^{n}A = n(c^{-1}B)^{n-1}$ for every positive integer $n$. Taking $n =k$ yields a contradiction, since then the left side is $0,$ but the right side is non-zero.
An alternative approach is to note that if $AB - BA = I,$ then $B$ is not an algebraic element (when $B$ is Hermitian, this implies that the spectrum of $B$ must be infinite). By an algebraic operator, we mean an operator $T$ such that $f(T) =0$ for some monic polynomial $f(z) \in \mathbb{C}[z].$ Note that an algebraic operator has a unique minimum polyomial, that is a unique monic polynomial $p(z) \in \mathbb{C}[z]$ of least degree subject to $p(T) =0.$ For suppose that $B$ is algebraic, and let $q(z)$ be the minimum polynomial for $B,$ say of degree $r.$ Then $r >1$ since $AB \neq BA.$ Then we have $0 = Aq(B) - q(B)A = (r-1)B^{r-1} +$ (some linear combination of lower powers of $B$), so that $h(B) =0$ for some monic monic polynomial $h(z) \in \mathbb{C}[z]$ of degree $r-1,$ contrary to the fact that the minimum polynomial of $B$ has degree $r.$
