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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.

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    $\begingroup$ possible duplicate of Time as a Hermitian operator in QM? $\endgroup$
    – Kyle Kanos
    Aug 11, 2015 at 19:29
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    $\begingroup$ 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) $\endgroup$
    – Kyle Kanos
    Aug 11, 2015 at 19:30
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    $\begingroup$ 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. $\endgroup$
    – Dargscisyhp
    Aug 11, 2015 at 19:33
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    $\begingroup$ Essentially a duplicate of physics.stackexchange.com/q/17477/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Aug 11, 2015 at 19:53
  • $\begingroup$ perhaps rephrase the question? I was already going to answer the traditional "because H is bounded from below" $\endgroup$
    – arivero
    Aug 11, 2015 at 19:53

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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.

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