quantum mechanics - An operator on the other side of the Schrödinger equation

A form of the Schrödinger equation is

$$ \left[-\frac{\hbar^2}{2m} \nabla^2 + V(\vec{r}, t)\right]\Psi = i\hbar \frac{\partial}{\partial t} \Psi $$

The bracketed term is of course the Hamiltonian, and the whole equation can hence be written as

$$ \hat{H}\Psi = i\hbar \frac{\partial}{\partial t} \Psi $$

My question is, is there a existing/named operator represented by $i\hbar \frac{\partial}{\partial t}$? That is, can we further rewrite the equation as

$$ \hat{H}\Psi = \hat{\Theta}\Psi $$

For some predefined operator $\hat{\Theta} = i\hbar \frac{\partial}{\partial t}$?

Answer

No, it is not an operator, at least with the same status as that of the Hamiltonian operator! The states are described by vectors in a Hilbert space, in this case $L^2(\mathbb R^3)$, i.e. functions $\psi=\psi(x)$ with integrable squared absolute value: $\int_{\mathbb R^3}|\psi(x)|^2 d^3x < +\infty$. Operators representing observables act in that space $A : \psi \mapsto A\psi$ where $A\psi \in L^2(\mathbb R^3)$.

Regarding Schroedinger equation, the situation is different. In that case one considers vector-valued curves $\mathbb R \ni \tau \mapsto \psi_\tau \in L^2(\mathbb R^3)$ representing the temporal evolution of an initially given state $\psi_0$.

The equation says that the time derivative (actually computed in the topology of the Hilbert space) of that curve at a given time $t$: $$\frac{d \psi_t}{dt}\tag{1}$$ equals the action of an operator on $\psi_t$ at that time $$-i \hbar^{-1}H\psi_t\tag{2}$$ To compute (2), we need to know just the vector $\psi_t$. Instead, to compute (1) we need to know the curve $\tau \mapsto \psi_\tau$ in a neighborhood of $t$!

Hence $d/dt$ (with some possible constant factor) cannot be considered an operator in the Hilbert space of the theory, $L^2(\mathbb R^3)$, as it acts on curves valuated in that Hilbert space. Obviously this space of curves enjoys a structure of complex linear space and $d/dt$ is an operator with respect to that vector space structure, but that structure does not play any role in QM.