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The Schrödinger equation is given by:

$$ i \hbar \partial_t |\Psi\rangle = \hat{H}|\Psi\rangle $$

The right hand side is just an operator acting on a state vector, so we are free to consider its expectation value

$$ \langle \Psi|\hat{H}|\Psi\rangle $$

which would just be the average of the energies we would measure if we measured identically prepared systems many times. But from the equality this should be equal to:

$$ i\hbar \langle \Psi|\partial_t|\Psi\rangle $$

But what does this even mean? I know that the time derivative is somewhat unique in not being an operator in quantum mechanics, so I could see this term not even making sense. But I haven't done anything questionable here: I just hit both sides of the Schrödinger equation with the bra $\langle \Psi |$, and got something reasonable on the right hand side.

Is there a reasonable interpretation of the term on the left?

gabe
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  • I imagine that if you chose, for instance, the position basis $\psi(x)=\langle x|\psi\rangle$ and evaluated the inner product of the time derivative as an integral you would find that both sides match. It is perfectly fine to evaluate expectation values of derivative operators for instance in the momentum operator (again in the position basis though). – Charlie Oct 16 '20 at 18:16
  • @Charlie But my issue is that the time derivative is not an operator. I agree that the left hand side could be evaluated if you went to a basis, but it is not clear to me what it means. – gabe Oct 16 '20 at 18:17
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    Does this previous question help at all? – Charlie Oct 16 '20 at 18:44
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    Relevant: this and this – jacob1729 Oct 16 '20 at 18:59
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    Technically speaking, the time evolution of a physical state is given by a map $\mathbb R\to\mathcal H,~t\mapsto\vert\Psi(t)\rangle$, where $\mathcal H$ is the Hilbert space. Its derivative is also a map $\mathbb R\to\mathcal H$, this time mapping $t\mapsto\partial_t\vert\Psi(t)\rangle$. So $\partial_t\vert\Psi(t)\rangle$ is a state in the Hilbert space, so you can calculate the scalar product $\langle \Psi(t)\vert\partial_t\vert\Psi(t)\rangle$ in a straightforward manner. – Vercassivelaunos Oct 16 '20 at 21:42

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