Is there any particular reason that the Hamiltonian operator was defined in quantum mechanics to be $$\hat H := \frac{\hat p^2}{2m} + V$$ as opposed to $$\hat H := i\hbar \frac{\partial}{\partial t}?$$

This latter definition would put the momentum and Hamiltonian operators in a nice parallel in position space, paring each conserved quantity with its Noether symmetry as a derivative.

$$\hat H = i\hbar \frac{\partial }{\partial t} \longleftrightarrow \hat p = -i\hbar \frac{\partial}{\partial x}$$

It also would put the Schrödinger equation in the following form, which parallels the classical equation.

$$\hat H \Psi = \frac{\hat p^2}{2m} \Psi + V \Psi \longleftrightarrow H = \frac{p^2}{2m} + V$$

Is there any reason why we have instead chosen to write $\hat H = \hat p^2 / 2m + V$ and the Schrödinger equation as $\hat H \Psi = i\hbar\ \partial \Psi / \partial t$? The alternative view seems a bit more enlightening to me, imho.


marked as duplicate by Qmechanic quantum-mechanics Oct 10 '18 at 8:31

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/17477/2451 , physics.stackexchange.com/q/16812/2451 , physics.stackexchange.com/q/220697/2451 and links therein. $\endgroup$ – Qmechanic Sep 26 '18 at 6:06
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    $\begingroup$ Note that $\hat H := \frac{\hat p^2}{2m} + \hat V$ is a special case for a particular kind of quantum mechanical system, and so is not actually a general definition. $\endgroup$ – J. Murray Sep 26 '18 at 6:21
  • $\begingroup$ @Qmechanic Yeah, that one immediately springs to mind, but this one has enough additional sofistication that I don't think it's a duplicate. $\endgroup$ – Emilio Pisanty Sep 26 '18 at 7:38
  • $\begingroup$ @Qmechanic I agree this question is similar to the ones you linked, but the provided answers don't quite address the question at hand here. $\endgroup$ – Trevor Kafka Sep 26 '18 at 20:08
  • $\begingroup$ @J.Murray Can you elaborate on what you mean? I'm unsure of what sort of a quantum mechanical system wouldn't satisfy $\hat H := \frac{\hat p^2}{2m} + V$. $\endgroup$ – Trevor Kafka Sep 26 '18 at 20:09