Is there any other difference between self-adjoint operators, and Hermitian operators, than that mathematicians seem to talk mostly about self-adjoint operators, and physicists seem to talk mostly about Hermitian operators?
I had difficulty understanding this because I knew only one definition for A* which was for bounded A only (through the Riesz's representation theorem). I've just learned that for unbounded operators A, A* can still be defined, but differently. I think I got this now.IIRC correctly:
-An operator is Hermitian if <u,Av>=<Au,v> for all u,v, in the domain of A.
-An operator is self-adjoint if A*=A.
The subtle difference being that the domains of A and A* may not coincide in general.
These definitions coincide for bounded operators.
I've seen some places where a hermitian operator is bounded by definition, others relax that condition. I`m not too sure of the exact definitions so don't quote me here.
Correct. But it's more than that. A Hermitian operater is a self adjoint operator on a complex inner product space with a particular inner product. So, even if the field is complex, if the inner product is is the typical one, the operater would be merely selfadjoint.And it is allowable to call T Hermitian only if U is a vector space over complex field?
too much, while I didn't know what was relevant. The rest of the post was already containing the answer, which was the same answer as given by Galileo... But I'm not convinced that everything was fine with the sets U and V hereThe Hermitian (more correctly, the Hermitian adjoint) of an operator only apply to operators on vector spaces over the complex numbers.
With arbitrary norm spaces the adjoint would be between the duals, T*:V*->U*. With Hilbert spaces, I assume, it goes like I showed now.If U and V are any inner product spaces and T is a linear transformation from U to V, the "adjoint" of T, T*, is a linear transformation from V to U such that, for any u in U and v in V, <Tu,v>= <u,T*v>. The two inner products are taken in V and U respectively.