quantum mechanics operators - Hermitian or complex conjugate? Let $f(x)$ be a normalised state in a 1-D system.
Let $g(x) = iA f(x)$, where $A$ is a Hermitian operator.
I want to find the inner product of $g(x)$ with itself. Is it $$\int \left(-i A^\dagger f^*(x)\right)\left(iAf(x)\right)dx$$ or 
$$ \int \left(-iA^*f^*(x)\right)\left(iAf(x)\right)dx$$ Here $*$ = complex conjugate and $\dagger$ = hermitian conjugate
 A: Both versions are wrong! The right version is, evidently,
$$\langle g| g \rangle =\int g(x)^* g(x) dx = \int \left((iA f)(x) \right)^* (iAf)(x) dx = \langle iAf| iAf \rangle = \langle f| (iA)^\dagger iA f \rangle =  \langle f| (-i)A^\dagger iA f \rangle =  \langle f| (-i)i A^\dagger A f \rangle =
\int f(x)^* (A^\dagger Af)(x) dx  = \int f(x)^* (A Af)(x) dx =
\int f(x)^* (A^2f)(x) dx \:,$$
where, I used the  hypothesis that $A=A^\dagger$ only in the penultimate passages. 
Notice that $(Af)(x)$ above means that first the operator $A$ acts on the function $f$ producing the function $Af$, next I evaluate this function at $x$: $(Af)(x)$.
Instead, $Af(x)$ is ambiguous especially in situations like the one you are considering (obviously it can safely be used if one understands well what he/she is doing).
The "complex conjugate" of an operator is a notion which hardly makes sense, though it something appears in textbooks giving rise to confusion and misleading artificial mathematical problems.
A: As a rule the correspondence between bra and kets are thus:
$$
\langle a|A^{\dagger}\leftrightarrow A|a\rangle
$$ 
Thus the correct expectation value for an operator S, in a $A|a>$ state is usually given by:
$$
\langle a|A^{\dagger}SA|a\rangle
$$
If $S\equiv I$ we get:
$$
\langle a|A^{\dagger}A|a\rangle
$$
And if furthermore A is unitary we are in luck!! :)
If $S\equiv I$ we get:
$$
\langle a|A^{\dagger}A|a\rangle=\langle a|a\rangle
$$
So to answer your question... as a rule, dagger is the way to go...
