Timeline for How to derive or justify the expressions of momentum operator and energy operator?
Current License: CC BY-SA 3.0
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Nov 9, 2013 at 23:39 | comment | added | user12262 | jabirali: "[...] perhaps my answer here ..." and there (final section): "Technically, we say that the Hamiltonian is the generator of time translations, and momentum is the generator of space translations." -- Yes, thanks, that seems relevant to what I'm trying to get at, namely: Is the expression $\mathbf{\dot{p}} = -\nabla\Phi(\mathbf{x})$ useful or even required for figuring out which (if any) generator (of "time translations", or "space translations", or "rotations", or who knows what) to call/associate with "momentum, p" ? | |
| Nov 9, 2013 at 10:33 | comment | added | jabirali | I was implicitly thinking in the Schroedinger picture, where the equation doesn't hold neither for operators ($\mathbf{p}$ is time-independent, so $\mathbf{\dot{p}} = 0$) nor for eigenvalues (not simultaneously well-defined). I stand corrected :) | |
| Nov 9, 2013 at 9:55 | comment | added | Trimok | "This implies that this equation is invalid in quantum mechanics". At least in Heinseberg representation, it is correct, as an equation between operators. For instance for the quantum harmonic oscillator, you have : $ \mathbf {\dot P}(t) = -m \omega^2 \mathbf {X}(t) $. See also this previous answer | |
| Nov 9, 2013 at 9:12 | comment | added | user12262 | jabirali: Thanks for answering. Unfortunately it caught me in the middle of editing my question (which I did on formal grounds only). Also, unfortunately, I cannot at all address the contents of your answer at the moment; but I hope to be able to do that within the next 24 h. | |
| Nov 9, 2013 at 8:21 | history | answered | jabirali | CC BY-SA 3.0 |