Talk:Schrödinger equation: Difference between revisions
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Why not put the laplacian in (I guess more known goniometric form (sin/cos, i) )? Besides the idea of an operator somehow comes out of the blue (where it follows 19th centurian hamiltonian physics). Just a few thoughts. [[User:Robert Tito|Robert Tito]] | [[User talk:Robert Tito|Talk]] 14:06, 18 February 2007 (CST) | Why not put the laplacian in (I guess more known goniometric form (sin/cos, i) )? Besides the idea of an operator somehow comes out of the blue (where it follows 19th centurian hamiltonian physics). Just a few thoughts. [[User:Robert Tito|Robert Tito]] | [[User talk:Robert Tito|Talk]] 14:06, 18 February 2007 (CST) | ||
Robert--I think you're confusing the Laplacian (which is simply the three-dimensional total derivative) with Euler's formula, exp(i*theta) = cos theta + i sin theta. [[User:Michael Evans|Michael Evans]] |
Revision as of 00:22, 21 February 2007
Umlaut
Anyone know how to type the O with a colon over it in Schrodinger...?
- I do not know, but you can always copy-paste. --Alex Halicz (hello) 11:43, 8 February 2007 (CST)
Easy when you are on a MAC, cut and paste when you do not have easy access. Robert Tito | Talk 11:44, 8 February 2007 (CST)
concepts in quantum theory
Can I suggest some discussion of its physical implications (eg quantized energy levels in atoms) in addition to the math. The first mention in The Feynman lectures on Physics, Volume III, Feynman's red book is a good model Cheers David Tribe 15:55, 12 February 2007 (CST) David, would you mind sticking to biology? (Attention: David has a big nose.) Robert Tito | Talk 16:53, 12 February 2007 (CST)
Your dutch phraseology RT has deterred me enuf to make sure I leave the article itself alone. So I now hesitate before leaving this as a possible link
- Why Quantum Mechanics Is Not So Weird after All PAUL QUINCEY. Richard Feynman's "least-action" approach to quantum physics in effect shows that it is just classical physics constrained by a simple mechanism. When the complicated mathematics is left aside, valuable insights are gained. David Tribe 19:18, 12 February 2007 (CST)
Good idea; I'll get right on that. Thanks for the link. Michael Evans
And IMHO the most important missing information (at least for a non-physicist who doesn't even try to understand the equations) is: Does this equation have something to do with the observability properties of quantum states (aka "Schrödingers Cat")?
--Markus Baumeister 15:10, 20 February 2007 (CST)
cat
since quantum mechanics is about statistical chances and wave functions being their advocates, the S.Eq. is the mathematical description of these chances. As such the S.Eq. is a statistical equation (at least when solved) and yes: S's cat is about the chance you will find the cat alive after x days with each previous day a chance of X% to get a lethal piece of food. It also is linked to the quantum inflation induced creation of a pair of electron/positron. Both travelling as wave through cosmos. Upon the moment one observer determines the charge of ONE of these electrons the charge of the other particle is immediately known. (instant communication without loss of time). Remember before measuring you don't know what the particle is. Observation determines sometimes the outcome. Robert Tito | Talk 15:16, 20 February 2007 (CST)
3D Schrödinger in Cartesian coordinates, and other coordinates?
Is the plan to give the Q.E. in 3D-cartesian, cilinder, polar coordinate? Robert Tito | Talk 01:31, 18 February 2007 (CST)
I can do that; I'll probably just put the cylindrical and spherical versions underneath the 3-d Cartesian (which is already there, see the eqn. with the Laplacian?). Michael Evans
Why not put the laplacian in (I guess more known goniometric form (sin/cos, i) )? Besides the idea of an operator somehow comes out of the blue (where it follows 19th centurian hamiltonian physics). Just a few thoughts. Robert Tito | Talk 14:06, 18 February 2007 (CST)
Robert--I think you're confusing the Laplacian (which is simply the three-dimensional total derivative) with Euler's formula, exp(i*theta) = cos theta + i sin theta. Michael Evans