Timeline for Extremizing a Hamilton-Jacobi Equation
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 13, 2013 at 19:20 | comment | added | Qmechanic♦ | Related Phys.SE question by OP: physics.stackexchange.com/q/69982/2451 | |
| S Sep 13, 2013 at 16:06 | history | edited | Emilio Pisanty | CC BY-SA 3.0 |
(v3: found the required six characters to correct, based on http://einstein.drexel.edu/~bob/Quantum_Papers/Schr_1.pdf)
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| S Sep 13, 2013 at 16:06 | history | suggested | user12262 | CC BY-SA 3.0 |
(v3: found the required six characters to correct, based on http://einstein.drexel.edu/~bob/Quantum_Papers/Schr_1.pdf)
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| Sep 13, 2013 at 15:40 | review | Suggested edits | |||
| Sep 13, 2013 at 16:06 | |||||
| Sep 13, 2013 at 9:28 | answer | added | Trimok | timeline score: 4 | |
| Sep 13, 2013 at 9:11 | comment | added | bolbteppa | Ahh, nice, didn't look at it as $\psi^2(H - E)$. Well theoretically it simply must hold for that potential, if this idea is at the core of what's going on, because from page 271 of Weinstock's Calculus of Variations onwards the Hydrogen atom is analyzed using this potential & they start their analysis from this functional. However shouldn't the claim be that it holds for any potential? It seems as though it must or else this can't be what's going on, because this is ultimately being used to derive the Schrodinger equation, so I would imagine that implies it must hold for applicable potentials. | |
| Sep 13, 2013 at 8:29 | comment | added | Michael | The thing under the integral isn't the Hamiltonian, it's $\psi^2 (H - E)$ which is (A) (apparently) positive semidefinite for arbitrary $\psi$ and (B) zero for solutions of the Hamilton-Jacobi equation. So by minimizing the integral you get solutions of $H=E$. The subtlety is whether (A) actually holds or not. It's clear for a $q^2$ potential, but $-1/q$? ... | |
| Sep 13, 2013 at 7:48 | comment | added | bolbteppa | I think you might be right, but how does one say this explicitly? If your Hamiltonian to begin with is what's inside that triple integral, then how are you minimizing the square of that? Maybe this is the meaning behind his quadratic form comment? | |
| Sep 13, 2013 at 7:33 | comment | added | Michael | That's what it looks like. Don't know if there is any deeper meaning to his argument or not. | |
| Sep 13, 2013 at 7:03 | comment | added | bolbteppa | Thanks, fixed. I'm honestly not sure, is it? | |
| Sep 13, 2013 at 7:01 | history | edited | bolbteppa | CC BY-SA 3.0 |
added 3 characters in body
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| Sep 13, 2013 at 6:59 | comment | added | Michael | Think you meant $\partial\psi/\partial q$ in the second equation. Is this anything other than saying you can find the solution of $f(x)=0$ by minimizing $f(x)^2$? | |
| Sep 13, 2013 at 6:41 | history | asked | bolbteppa | CC BY-SA 3.0 |