# Why isn't the time-derivative considered an operator in quantum mechanics? [duplicate]

Based on my understanding when doing quantum mechanics we deal with a small set of mathematical objects: namely scalars, kets, bras, and operators. But then in the Schrodinger equation we have this time-derivative thing which looks an awful lot like an operator, but from what I've been told it's not considered an operator. My question, then, is what exactly is a time-derivative in quantum mechanics and why is it not considered an operator.

Edit: The issue that this is a possible duplicate of threads linked in the comments below has been brought up. I saw the threads linked below before I posted this question, and I thought this question was sufficiently different because I'm not interested in a time-operator per se, but rather what $\partial_t$ is mathematically. I've been told it's not an operator in the same way the momentum operator is, but it does look like an operator that scales the state vector proportional to its energy.

• possible duplicate of Time as a Hermitian operator in QM? Aug 11, 2015 at 19:29
• See also physics.stackexchange.com/q/34243, physics.stackexchange.com/q/56081, and physics.stackexchange.com/q/83701 (all closed as duplicates of the link in my above comment) Aug 11, 2015 at 19:30
• I did see those before I posted. I thought this question was sufficiently different because I'm not interested in a time-operator per se, but rather what $\partial_t$ is mathematically. I've been told it's not an operator in the same way the momentum operator is, but it does look like an operator that scales the state vector proportional to its energy. Aug 11, 2015 at 19:33
• Essentially a duplicate of physics.stackexchange.com/q/17477/2451 and links therein. Aug 11, 2015 at 19:53
• perhaps rephrase the question? I was already going to answer the traditional "because H is bounded from below" Aug 11, 2015 at 19:53

Let $\mathcal{H}$ be the space of states of our theory. Then, time evolution is given by a unitary operator $U(t_2,t_1) : \mathcal{H}\to\mathcal{H}$ that evolves "stuff" from time $t_1$ to time $t_2$. For time-independent Hamiltonians it is just $\mathrm{e}^{\mathrm{i}H(t_2 - t_1)}$.
If we are in the Schrödinger picture, we say states "carry the time evolution" in the sense that a Schrödinger state is given by a map $$\psi : \mathbb{R}\to\mathcal{H}, t \mapsto \psi(t)$$ so that $\psi(t)$ is a state for every choice of $t\in\mathbb{R}$ and $\psi(t_1) = U(t_1,t_2)\psi(t_2)$. The time derivative then acts on $\psi$ and hence on the space $C^1(\mathbb{R},\mathcal{H})$, so the time derivative is not an operator on $\mathcal{H}$ itself, but on the differentiable functions into it.
If we are in the Heisenberg picture, the operators carry the time evolution in the sense that a Heisenberg operator is given by a map $$A : \mathbb{R}\to\mathcal{O}(\mathcal{H}), t \mapsto A(t)$$ where $\mathcal{O}(\mathcal{H})$ is the algebra of quantum mechanical operators on $\mathcal{H}$ and $A(t_1) = U(t_1,t_2)A(t_2)U(t_2,t_1)$. The time derivative acts thus on differentiable functions $C^1(\mathbb{R},\mathcal{O}(\mathcal{H}))$.
In neither case is the time derivative an operator on $\mathcal{H}$ itself.