- #1

spiffing_abhijit

- 9

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter spiffing_abhijit
- Start date

- #1

spiffing_abhijit

- 9

- 0

- #2

malawi_glenn

Science Advisor

Homework Helper

Gold Member

2022 Award

- 6,197

- 1,687

http://www.falstad.com/qmatom/

free electrons propagate as a "wave" in that sense that it has a deBroigle wavelength: (plane wave solution to SE, also called deBroigle wavefunction)

[tex]\Psi (\vec{x},t) = N e^{i(\vec{p}\cdot\vec{x}-Et)} [/tex]

Remember that the 'wave nature' of particles has to do with its wave function, it is not like a water wave or a standing wave on a string.

So the concept in QM is that we can't really say how things move etc, we can only work out the wave function and what observables that it contains. I hope you are familiar with Heisenbergs uncertainty relation: [tex] \Delta x \Delta p > \hbar [/tex], so if you know where the particle are, then you have no idea of what its momentum is.

Spin is an intrinsic degree of freedom for subatomic particles. Angular momentum we can derive from rotation symmetry in 3D, and we will obtain commutator relations for angular momentum operators. Then we see what happens if we move to 2D, and then we get spin.

Spin is manifested in how particles react on magnetic and electric fields. Compare with classical magnetic dipoles.

-> Stern-Gerlach experiment

- #3

spiffing_abhijit

- 9

- 0

Thanks

Share:

- Replies
- 2

- Views
- 775

- Last Post

- Replies
- 3

- Views
- 572

- Replies
- 2

- Views
- 300

- Last Post

- Replies
- 11

- Views
- 740

- Last Post

- Replies
- 0

- Views
- 253

- Last Post

- Replies
- 2

- Views
- 778

- Replies
- 3

- Views
- 502

- Replies
- 1

- Views
- 525

- Replies
- 3

- Views
- 645

- Replies
- 14

- Views
- 2K