Creating a Hermitian function

Question

Say I have an operator $A$ such that $A^\dagger = B$. I want to construct a Hermitian function, $f$, of these operators, $f(A,B)^\dagger = f(A,B)$. Is it possible to construct a function $f$ such that $f$ is not a function of $A+B$?: $f(A,B)\neq f(A+B)$ but $f(A,B)^\dagger = f(A,B)$?

Answer 1

A trivial answer might be any product of $A$ and $A^{\dagger}=B$ is an hermitian function take $f_1(A,B)=AB=AA^{\dagger}$ and than since also every power of $AB$ is an hermitian function: ${(AB)^n}^{\dagger}=B^{\dagger}A^{\dagger}B^{\dagger}A^{\dagger}....B^{\dagger}A^{\dagger}=ABAB....AB = (AB)^n$ and since you can define $$f(A,B)=\sum_{n}\alpha_n(AB)^n$$ with $\alpha_n \in \mathbb{R}$ there you have a large class of functions of $A$ and $B$ which are $f(A,B)=f(AB)$. Another possibility that comes to my mind is a slight generalization of what you wrote(which takes into account just linear combinations of $A$ and $A^{\dagger}$): take any function $g(A,B)$ and define $f(A,B)=g(A,B)+g(A,B)^{\dagger}$ or $f(A,B)=i(g(A,B)-g(A,B)^{\dagger})$. To my knowledge these are the most common cases without any further specification on the properties of $A$ ,edit: beyond the trivial cases of symmetrization and antisymmetrization $A+A^{\dagger}$ and $i(A-A^{\dagger})$

Answer 2

Observe that

$$(\mathrm{i}(A-B))^\dagger = -\mathrm{i}(A^\dagger - B^\dagger) = \mathrm{i}(A-B)$$

so $f(A) = \mathrm{i}(A-A^\dagger)$ is a Hermitian function of $A$ that is not a function of $A+A^\dagger$.