Energy operator and the Hamiltonian operator: Are they same?

[Physics Footnotes]

That's a clear misconception and in fact contradicts quantum theory. The Hamiltonian is NOT $\mathrm{i} \partial_t$ but a function(al) of some set of fundamental operators (in single-particle non-relativistic QT that's the position, momentum, and spin operators). Sometimes $\hat{H}$ is also explicitly time dependent (e.g., the motion of a charged particle in a time-dependent em. field), i.e.,
$$\hat{H}=\hat{H}(\hat{\vec{x}},\hat{\vec{p}},\hat{\vec{S}},t).$$

Ummm... is this addressed to me? If so, where did I mention the word 'Hamiltonian'?? I was addressing a common confusion over the status of the operator $\mathrm{i}\hbar \partial_t$.

 

[Demystifier]

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The two operators are the same on the space of solutions of the Schrödinger equation. On a more general space of all functions of space and time, those are different operators.

 

[dextercioby]

[...] I was addressing a common confusion over the status of the operator $\mathrm{i}\hbar \partial_t$.

And to be really fair (i.e. to show that I've checked the answers on our competition's website - where both of us already posted on a similar topic), the mathematical sense in which these "quantum worldlines" should be looked at is wonderfully presented in this topic (question properly formulated and the answer to it).

https://physics.stackexchange.com/q...r-as-expressed-in-abstract-hilbert-space-vs-a

 

[vanhees71]

Ok, obviously here's a clash of different languages going on. From a physicists point of view it is utmost important to be very clear about the fact that time in quantum theory is NOT an observable but a parameter for the simple reason that otherwise it would be the canonical conjugate to the Hamiltonian, which then due to the analogy with the position-momentum commutation relation would have entire $\mathbb{R}$ as its spectrum and thus there'd be no stable ground state.

In the abstract (representation-free) Hilbert-space formalism the choice of the time dependence is determined only up to a time-dependent unitary transformation, i.e., you can shuffle the time dependence between the operators representing observables and the statistical operator (or state vectors representing the rays standing for the special case of pure states) quite arbitrarily. Of course, this choice of the picture of time evolution doesn't affect any physical outcomes, i.e., probabilities or probability distributions for the outcome of measurements or transition-matrix elements and the like, but I think that's not the issue here.