Timeline for Ordering Ambiguity in Quantum Hamiltonian
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 26, 2013 at 19:44 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added ref.
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| Oct 1, 2012 at 14:17 | vote | accept | Jaswin | ||
| Sep 30, 2012 at 13:53 | comment | added | Freedom | @Jaswin, see below, but the reordering of terms is not the inversion relationship they are referencing. | |
| Sep 30, 2012 at 13:17 | answer | added | Freedom | timeline score: 0 | |
| Sep 30, 2012 at 4:30 | answer | added | Alex Nelson | timeline score: 4 | |
| Sep 28, 2012 at 5:24 | comment | added | Jaswin | @Qmechanic : I was reading from $Mirror$ $Symmetry$, chapter 10, equation 10.68. It is available freely in Claymath.org library. claymath.org/library/monographs/cmim01.pdf | |
| Sep 28, 2012 at 5:23 | comment | added | Jaswin | @LubošMotl : Yes, now I could figure it out, probably he meant that $\dfrac{1}{2} g^{ij} P_iP_j \neq \dfrac{1}{2} P_i P_j g^{ij} $ quantum mechanically, but classically it is true. | |
| Sep 28, 2012 at 4:28 | comment | added | Luboš Motl | Dear Jaswin, differently ordered products of operators (those that exist classically) always differ by terms proportional to $\hbar$ or its positive powers, so in the classical $\hbar\to 0$ limit, they're the same. I would have to go to higher, 5th order polynomials for a good example. | |
| Sep 27, 2012 at 17:17 | comment | added | Jaswin | @LubošMotl : Ok I get the "related" part now. Thanks for that interpretation, but is there an example of showing that two different definitions of Hamiltonian leads to same classical Hamiltonian. May be considering different definitions of momenta leading to same classical Hamiltonian. Earlier in the text, conjugate momentum was defined as $P_i = \dfrac{\delta S}{\delta \dot{X}^i} = g_{ij} \partial_t X^j $ | |
| Sep 27, 2012 at 15:44 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
Tried to make title more informative. Dear Jaswin, if u don't like my changes please roll back or use the parts u like.
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| Sep 27, 2012 at 15:33 | comment | added | Luboš Motl | The operator is clearly "equal" to a Laplacian only if the metric $g$ is flat and positively definite. Otherwise it's just similar, that's why they say it's "related". Also, there are ordering ambiguities because $g^{ij}$ are functions of $X$ which don't commute with $P$. Does it answer all your questions? | |
| Sep 27, 2012 at 15:21 | history | asked | Jaswin | CC BY-SA 3.0 |