The exponent of self-adjoint operator If $X$ is a Hilbert space and $A$ is an unbounded self-adjoint operator on $X$, is it necessarily that $A^2$ is self-adjoint as well?(admittedly, $A^2$ is densely defined)
 A: Yes.  More generally whenever $A$ is closed and densely defined operator on Hilbert space, the operator $A^* A$ is also closed and densely defined, and in fact is self-adjoint.  (Wikipedia calls this von Neumann's theorem but I do not think this is a standard name for it.)
As plausible as this sounds, it is highly nontrivial.  (If you think a little bit about what is actually being asserted, it is far from obvious why the domain of $A^* A$ must include even one nonzero vector, let alone be dense.)
For the details of this and a great deal more on unbounded operators, consult your local library.  Any textbook that treats unbounded operators on Hilbert space from a mathematically coherent point of view (and not simply to motivate applications to physics or some non-analysis-oriented field) should include a proof.  It is key, for example, in constructing the polar decomposition of a closed operator.  
There is also a functional calculus for unbounded self-adjoint operators of which this discussion is a very special case (if $f$ is any real-valued function on $\mathbb{R}$ one can, with a great deal of work, give meaning to a self-adjoint operator $f(A)$ in a way that respects the usual algebra of functions, and in particular, in a way that assigns to $f(x) = x^2$ the operator $A^2$).
A standard reference for this kind of thing is Volume 1 of Reed and Simon's "Methods of modern mathematical physics" but many other books cover it too.  The Wikipedia entry for unbounded operator is also worth a look (although much of the mathematical internet's treatment of unbounded operators is nonrigorous, it is an OK start).
