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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.

Qmechanic
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Dargscisyhp
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    possible duplicate of Time as a Hermitian operator in QM? – Kyle Kanos Aug 11 '15 at 19:29
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    See also http://physics.stackexchange.com/q/34243/, http://physics.stackexchange.com/q/56081, and http://physics.stackexchange.com/q/83701 (all closed as duplicates of the link in my above comment) – Kyle Kanos Aug 11 '15 at 19:30
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    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. – Dargscisyhp Aug 11 '15 at 19:33
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    Essentially a duplicate of http://physics.stackexchange.com/q/17477/2451 and links therein. – Qmechanic Aug 11 '15 at 19:53
  • perhaps rephrase the question? I was already going to answer the traditional "because H is bounded from below" – arivero Aug 11 '15 at 19:53
  • Qmechanic, I had not seen that question before I posted mine. It is a very similar question, certainly. That said, I feel ACuriousMind's answer answers my question more adequately than the answers on that thread. – Dargscisyhp Aug 11 '15 at 19:58
  • Arivero: How should I rephrase it to make my question clearer? – Dargscisyhp Aug 11 '15 at 19:58

1 Answers1

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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.

ACuriousMind
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