Complex conjugate of the Schrödinger equation? This might be a very simple question but I don't understand how to compute the complex conjugate of the Schrödinger equation:
$$
i\partial_t \psi = H\psi
$$
where $H$ is an hermitian operator. How to proceed now? All answers I found via the search function were not satisfactory for me.
First idea: Complex conjugate both sides:
$$
(i\partial_t \psi)^* = (H\psi)^*
$$
The problem is that I am not sure how to proceed here. What is the conjugate of an operator $A$ acting on a function $f$? So what is $(Af)^*$? Does the following hold: $(Af)^* = A^\dagger f^*$ ? If so, I get
$$
-i\partial_t^\dagger \psi^* = H\psi^*
$$
The problem is now, that I don't know what $\partial_t^\dagger$ is? If $\partial_t^\dagger = \partial_t$ holds then I have the common result!
Second idea: Say $C$ is the operator that conjugates a function: $Cf=f^*$. Then I can write
$$
H\psi^* = HC\psi = (HC + CH - CH)\psi = [H,C]\psi + CH\psi = [H,C]\psi + Ci\partial_t\psi 
$$
where $[\cdot,\cdot]$ is the commutator. Moreover I find
$$
Ci\partial_t\psi = Ci\partial_t\psi - i\partial_tC\psi + i\partial_tC\psi = \{i\partial_t,C\}\psi - i\partial_t\psi^* 
$$
where $\{\cdot,\cdot\}$ is the anticommutator. Thus I get
$$
H\psi^* = - i\partial_t\psi^* + [H,C]\psi + \{i\partial_t,C\}\psi
$$
If I could prove that $[H,C]= \{i\partial_t,C\} = 0$ then I would also get the common result.  
But unfortunately I get stuck here... can someone resolve my problems?
 A: Generally speaking the complex conjugate of an operator is not a standard notion of operator theory, though it can be defined after having introduced some general notions.
Definition. A conjugation $C$ in a Hilbert space $\cal H$ is an antilinear map $C : \cal H \to \cal H$ such that is isometric ($||Cx||=||x||$ if $x\in \cal H$) and  involutive ($CC=I$). 
There are infinitely many such maps, at least one for every Hilbert basis in $\cal H$ (the map which conjugates the components of any vector with respect to that basis). 
On $L^2$ spaces there is a standard conjugation $$C : L^2(\mathbb R^n, dx) \ni \psi \mapsto \overline{\psi}\in L^2(\mathbb R^n, dx)\:,  $$ where $\overline{\psi}(x) := \overline{\psi(x)}$ for every $x \in \mathbb R^n$ and where $\overline{a+ib}:= a-ib$ for $a,b \in \mathbb R$.
Definition. An operator $H : D(H) \to \cal H$ (where henceforth $D(H)\subset \cal H$) is said to be real with respect to a conjugation $C$ if $$CHx=HCx \quad \forall x \in D(H)$$ (which implies $C(D(H)) \subset  D(H)$ and thus $C(D(H)) =  D(H)$ in view of $CC=I$, so that the written condition can equivalently be re-phrased $CH=HC$).
The complex conjugate $H_C$ of an operator $H$ with respect to a conjugation $C$, can be defined as $$H_C :=CHC\:.$$ This operator with domain $C(D(H))$ is symmetric, essentially self-adjoint, self-adjoint if $H$ respectively if is symmetric, essentially self-adjoint, self-adjoint. Obviously it coincides with $H$ if and only if  $H$ is real with respect to $C$.
Let us come to your issue. Let us start form Schroedinger equation
$$-i \frac{d}{dt} \psi_t = H\psi_t\:.$$
Here we have a vector valued map $$\mathbb R \ni t \mapsto \psi_t \in \cal H\:,$$
such that $\psi_t \in D(H)$ for every $t \in \mathbb R$ and the derivative is computed with respect to the topology of the Hilbert space whose norm is $||\cdot|| = \sqrt{\langle \cdot| \cdot \rangle}$:
$$\frac{d}{dt} \psi_t = \dot{\psi}_t \in \cal H$$
means 
$$ \lim_{h\to 0} \left|\left| \frac{1}{h} (\psi_{t+h} -\psi_t) - \dot{\psi}_t  \right|\right|=0\:.$$
(See the final remark)
If $C : \cal H \to \cal H$ is a conjugation, as it is isometric and involutive, in view of the definition above of derivative, we have 
$$C\frac{d}{dt} \psi_t = \frac{d}{dt} C\psi_t\tag{1}$$ 
where both sides exist or do not simultaneously. 
Summing up, given a conjugation $C$, and the (self-adjoint) Hamiltonian operator $H$, both in the Hilbert space $\cal H$, the complex conjugate of the Schroedinger equation 
$$-i \frac{d}{dt} \psi_t = H\psi_t\:.\tag{2}$$
is a related equation satisfied by $C\psi_t$ and just obtained by applying $C$ to both sides of (2) and taking (1) and $CC=I$  into account, obtaining
$$i \frac{d}{dt} C\psi_t = H_C\: C\psi_t\:.$$
If $H$ is real with respect to $C$ (this is the case for a particle without spin described in $L^2(\mathbb R^3)$, assuming the Hamiltonian of the form $P^2/2m + V$ and $C$ is the standard complex conjugation of wavefunctions), the equation reduces to
$$i \frac{d}{dt} C\psi_t = H C\psi_t\:.$$

REMARK. It is worth stressing that $-i \frac{d}{dt}$ is not an operator in the Hilbert space $\cal H$ as, for instance, $H$ is. To compute $H\psi$, it is enough to know the vector $\psi \in D(H)$. To compute 
$\frac{d}{dt}\psi_t$ we must know a curve of vectors $$\gamma :\mathbb R \ni t \mapsto \psi_t \in \cal H\:.$$
$\frac{d}{dt}$ computes the derivative of such curve defining another curve
$$\dot{\gamma} :\mathbb R \ni t \mapsto \frac{d}{dt}\psi_t \in \cal H\:.$$
More weakly one may view $\frac{d}{dt}|_{t_0}$ as a map associating vector-valued curves defined in aneighborhood of $t_0$ to vectors $\frac{d}{dt}|_{t_0}\psi_t$. In both cases it does not make sense to apply the derivative to a single vector $\psi$, contrarily to $H\psi$ is well defined.
A: You actually want the transpose-conjugate, which applies to everything: $$
(i\partial_t |\psi\rangle )^\dagger = (H\|psi\rangle)^\dagger 
$$
This gives:
$$
-i\partial_t |\psi\rangle^\dagger = |\psi\rangle^\dagger H^\dagger 
$$
which reduces to:
$$
-i\partial_t \langle\psi| = \langle\psi|H
$$
The differential operator remains unchanged.
Note if your units don't have $\hbar=1$, it belongs on the LHS.
