0

I know there is no spin in orbital model. And it is always said there is no visualization for the spin.

But why not just let the oribtals rotate with 4D quaternions in some 3D dynamic model?

Qmechanic
  • 201,751
Randy Welt
  • 552
  • 1
    More power to you, if you can visualize quaternions (I still can't). Having said that, there is a difference between physics and physics visualization. One has to be very careful not to mistake the two for each other. This is especially problematic when we visualize for the purpose of teaching physics to laymen who don't have the time to access the correct level of description that lies behind the visualization. This is already a problem in basic physics education, long before we ever get to spin. Many students get stuck with "forces" being some sort of arrows on a black board. Bad idea! – CuriousOne Oct 13 '14 at 21:30
  • Related: http://physics.stackexchange.com/q/822/2451 and links therein. – Qmechanic Oct 13 '14 at 21:40
  • Spin Matrices can be expressed by quaternions no?! – Randy Welt Oct 13 '14 at 21:44
  • "Rotating with 4D quaternions" is the same as letting them rotate normally, since the unit quaternions are isomorphic to $\mathrm{SU}(2)$, and "rotating with them" means just "thinking of them as a rep of $\mathrm{SO}(3)$". There's just no gain in that idea. – ACuriousMind Oct 13 '14 at 21:46
  • Ok, but it would visualize the spin from a pedagogical perspective. Better then no visualisation?? Same as usually everyone gets to know Bohr modell (while actually more wrong) before learning about orbitals. – Randy Welt Oct 13 '14 at 21:52
  • or another example: string theory they reduce 6dim calabi yau spaces to 2d surfaces. I make a prediction: in 10 years we will see some dynamic orbitals on youtube. cheers :) – Randy Welt Oct 13 '14 at 22:03
  • The reason spin is not orbital motion is the $\mathbf{r} \times \mathbf{p}$ angular momentum gets the wrong quantization to agree with spin. Any kind of "rotating" is going to have that problem. Spin is angular momentum that has nothing to do with $\mathbf{r} \times \mathbf{p}$. That's really all there is to it. – dmckee --- ex-moderator kitten Oct 14 '14 at 03:10

0 Answers0