# What is a 'magnetic substate'? (Magnetic-Optical Traps)

1. Jul 2, 2009

### Shukie

I'm doing a small paper on Magneto-Optical Traps, trying to explain how they work and I just encountered another paper which talked about ground states and excited states (J = 0, J = 1) and magnetic substates (m = 0 for J = 0 and m = -1, 0 or 1 for J = 1). I understand what ground states and excited states are, but can someone explain magnetic substates?

Also, I'm confused by this:

It's from near the bottom of page 2 from http://physweb.bgu.ac.il/COURSES/LAB_C/Laser%20Cooling%20and%20trapping/colorado%20mot.pdf [Broken]. Could anyone tell me in laymans terms what exactly the bolded sentence means? As far as I understand, the atoms absorb photons, which excites them, then they emit them. What does this have to do with the magnetic substate?

Edit: I see now that I probably should have put this in the Atomic, Solid State, Comp. Physics subforum.

Last edited by a moderator: May 4, 2017
2. Jul 3, 2009

### Redbelly98

Staff Emeritus
Electrons possess angular momentum, due to both their intrinsic spin as well as their motion around the nucleus. We use J to denote the value of the angular momentum, in this case 0 or 1 for the two energy levels.

As for the magnetic substates ... since angular momentum is a vector, we can talk about a component in a particular direction, in this case the propagation direction of the laser beam. We denote this vector component of angular momentum by m.

If J=0 then magnitude of the angular momentum is zero, so obviously the component m can only be 0. But for J=1, the component of angular momentum can be m = -1, 0, or 1.

The beams are circularly polarized, so that each photon has an angular momentum of 1. Moreover, the angular momentum vector points in the direction of propagation: for the beam from the left it points to the right, so m=+1 for photons in the beam from the left. Likewise, m=-1 for photons in the beam from the right.

For a ground state atom, m=0. When it absorbs a photon, by conservation of angular momentum, the excited state atom will have whatever angular momentum the absorbed photon had: +1 for photons absorbed by the beam from the left, or -1 for the other beam.

Hope this helps ... there's a lot here, so feel free to ask followup questions if anything is still confusing you.

EDIT:
Slightly off-topic, but note the appearance of current DOE Secretary Steven Chu in many references at the end of the link you gave,

Last edited by a moderator: May 4, 2017
3. Jul 3, 2009

### Shukie

Thank you very much, that was extremely helpful. I do have some more questions. This is from the same article, from the bottom of page 2 again and the top of page 3:

An atom in groundstate absorbs a photon and will take over whatever angular momentum the photon had. (1) This is what is meant by a $$\Delta$$m = +1 (or -1) transition, as described in the quote?

(2) The magnetic field itself does not exert a force on the atoms, right? The magnetic field is decreasing in -x direction and increasing in the +x direction. The left beam excites transitions to m = +1 and the right beam to m = -1. This somehow changes the resonance frequencies of the atoms. The lasers are set to below these resonance frequencies and due to the Doppler shift, atoms traveling toward a laser beam, will scatter more photons from that laser than the one they are moving away from and so it will experience a force in the direction of the origin. Is this correct?

If so, I don't quite understand the role the magnetic substates play in this. Is it because the magnetic substates makes the resonance frequencies position dependant and so you don't have to change the frequency of the laser due to the decreasing velocity of the atom (since the atom's resonance frequency changes)?

4. Jul 3, 2009

### Shukie

Thank you very much, that was extremely helpful. I do have some more questions. This is from the same article, from the bottom of page 2 again and the top of page 3:

An atom in groundstate absorbs a photon and will take over whatever angular momentum the photon had. (1) This is what is meant by a $$\Delta$$m = +1 (or -1) transition, as described in the quote?

(2) The magnetic field itself does not exert a force on the atoms, right? The magnetic field is decreasing in -x direction and increasing in the +x direction. The left beam excites transitions to m = +1 and the right beam to m = -1. This somehow changes the resonance frequencies of the atoms. The lasers are set to below these resonance frequencies and due to the Doppler shift, atoms traveling toward a laser beam, will scatter more photons from that laser than the one they are moving away from and so it will experience a force back to the origin. Is this correct?

If so, I don't quite understand the role the magnetic substates play in this. Is it because the magnetic substates make the resonance frequencies position dependant and so you don't have to change the frequency of the laser due to the decreasing velocity of the atom (since the atom's resonance frequency changes due to the Doppler shift)?

5. Jul 3, 2009

### Redbelly98

Staff Emeritus
(1) Yes.

(2)
The magnetic field would exert a force on excited state atoms, because (a) those atoms possess a magnetic dipole moment, and (b) the magnetic field has a gradient.

However, that is not the force being discussed here. Nor are they talking about Doppler shifts creating a viscous force for atoms depending on what direction they are moving in.

Are you familiar with the Zeeman effect? That is at the root of this. Since the upper-state energy shifts with magnetic field, and the magnetic field is location-dependent, the upper state energy (and therefore the transition frequency) depends on location. So there is preferential absorption from one laser beam or the other depending on the atom's location, i.e. the location where the Zeeman shift brings the energy-level difference close to resonance with the laser.

6. Jul 3, 2009

### Shukie

I wasn't entirely familiar with the zeeman effect, but I knew it was involved. Once again your explanation cleared things up and I was able to finish my paper. Thanks again!