Because (after long university absence) I recently came across field operators again in my QFT lectures (which are not necessarily Hermitian):
What problem is there with observables represented by non-Hermitian operators (by observables, I obviously don't mean the tautological meaning "Hermitian operators")?
One Problem sure is that there are not real eigenvalues. But If I say I want to "measure" some complex quantity, then that alone shouldn't be a problem, I'd be fine with complex eigenvalues then.
This question is answered stating that operators are Hermitian "if and only if it is diagonalizable in an orthonormal basis with real eigenvalues". This doesn't yet seem like a game-stopper to me, as long as I still get an orthonormal basis.
And splitting an arbitrary operator $\hat{O}$ into a Hermitian and an anti-Hermitian part, that would correspond to real and imaginary part of the observable, I could take expectation values just fine.
But maybe I'm forgetting about something, and there are other good reasons why non-Hermitian operators will lead so serious problems.
Best Answer
At the simplest level it is because Hermitian, or more precisely self adjoint, operators have a complete set of eigenstates. The existence of a complete set is essential for the propability interpretation of QM. That the eigenvalues are real is of less importance. Consider, for exmple, the operators $X,Y$ of the $x$ and $y$ coordinates of a particle. One might want, for some reason the the combination $x+iy=z$, which are the eigenvalues of $X+iY$.
To allow operators such as $X+iY$ to be observables, one can relax the self-adjoint condition to allow normal operators which are defined to be those that commute with their adjoint: $[A,A^\dagger]=0$. In this case we can decompose $$ A= \frac 12 (A+A^\dagger) +\frac 12(A-A^\dagger), $$ and as $$ [(A+A^\dagger), (A-A^\dagger)]=0 $$ we can simultaneously diagonalize the hermitian $(A+A^\dagger)$ which has real eigenvalues and $(A-A^\dagger)$ which, being skew hermitian and therefore $i$ times a hermitian operator, has purely imaginary eigenvalues. Then $A$ can be an observable with complex eigenvalues.