An operator on the other side of the Schrödinger equation

Question

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 1

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.

Answer 2

Yes, the operator you are referring to is the energy operator,

$$\hat{E}=i\hbar \frac{\partial}{\partial t}$$

Answer 3

I would like to add to JamalS's answer that the way I see this is that the energy operator is the generator of time evolution. A generator, in this context, means that we exponentiate it to get the transformation, $\hat{U}=e^{-\frac{i}{\hbar} \hat{E} t}$.

The Schrodinger equation becomes an operator equation between the time-evolution generator and generators of other transformations, so that $\hat{E}=\hat{H}$ is just writing $\hat{E}$ in terms of other operators (classically, we also use $H$ to write the energy in terms of $p$ and $x$).

So, if we say that $\hat{H}=\hat{p}$ for example, we then plug it into Schrodinger's equation to find that the generator of time evolution is $\hat{p}$, which is the generator of translations in space, so as time evolves everything would just shift with the same speed of 1. The usual $\hat{p}^{2}$ will just case dispersion. Having an $\hat{x}$ means that now momenta will shift because $\hat{x}$ is the generator of translations in $p$ just like $\hat{p}$ is for $x$.

Another interesting example is when we have an spin in a magnetic field $\hat{H}=-m \hat{\vec{\sigma}} \cdot \vec{B}$, where $\hat{\vec{\sigma}}$ are the Pauli spin matrices, and $m$ is the magnetic moment. Here then well have $\hat{E}$ being equal to something proportional to $\hat{\vec{\sigma}}$, which is the generator of spin rotations. We can immediately predict that in a magnetic moment an spin will rotate (precess) in some uniform fashion!

By the way, this only works if $\hat{E}$ is explicitly independent of time. If not, one cannot directly make the connection that it is the generator of time evolution and one has to generalize this (involves using Dyson formula and other things).