Relation between spin and its magnetic field

In summary, the magnetic field due to spin is due to the spin of an electron and the angle of rotation is constant * distance * magnetic field.
  • #1
BeauGeste
49
0
I understand that the magnetic moment of an electron is associated with its intrinsic spin. But what is the magnetic field due to spin?
I'm asking this while thinking about the Faraday Effect where the magnetic field in the direction of the light's propagation is due to the spin of an electron.
The angle of rotation is constant * distance * magnetic field. What do I put in for magnetic field? The only thing I see is the magnetic moment but that does not have the right units.
Thanks.
 
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  • #2
I've have also noted, that I haven't seen answer to this. I would guess the reason for this is, that you cannot give classical magnetic field that would be produced by a quantum mechanical particle.

Perhaps you should deal with the electromagnetic field quantum mechanically also? I have no clue of how to do this. QED stuff then. Or semiclassical approximations alternatively.
 
  • #3
BeauGeste said:
I understand that the magnetic moment of an electron is associated with its intrinsic spin. But what is the magnetic field due to spin?
I'm asking this while thinking about the Faraday Effect where the magnetic field in the direction of the light's propagation is due to the spin of an electron.
The angle of rotation is constant * distance * magnetic field. What do I put in for magnetic field? The only thing I see is the magnetic moment but that does not have the right units.
Thanks.
If by "Faraday Effect" you mean rotation of the polarization of light by a magnetic field that has nothing to do with QM or the spin of the electron,
or how the light was produced.
 
  • #4
jostpuur said:
I've have also noted, that I haven't seen answer to this. I would guess the reason for this is, that you cannot give classical magnetic field that would be produced by a quantum mechanical particle.

Hmmm, maybe that's the case. I didn't think of it like that. I guess this is just another reason why spin angular momentum is so different than orbital angular momentum. For the electron's orbital angular momentum you should be able to calculate the B-field (considering the electron as a current). But for spin angular momentum, there is no classical analog to calculate B (because electron is not actually spinning, i.e. no moving charge). Does that make sense?
I'm wondering if I've been thinking of the magnetic force incorrectly; I've always thought of it as B being the fundamental quantity that determines it but maybe I should think of it being the magnetic moment instead and quantity B being a 'luxury'.

Also, in regards to the Faraday Rotation question: the situation I'm considering is a system of polarized electron spins. A polarized laser is incident up on the spin system. Due to the Faraday effect, the polarized spins will rotate the laser beams polarization through an angle. I've only seen the Faraday effect described with the quantity B, so I was wondering how it was different for the case of polarized spins where I seem to not have an equation for B. I think the name for the technique is Time Resolved Faraday Rotation.

Thanks for the comments so far. Any more would be greatly appreciated.
 
  • #5
Ok, i was thinking more about this and I thought that a spin is essentially a magnetic dipole with a magnetic moment so it's B-field will be the B-field from a magnetic dipole (formula in griffiths):
[tex] \vec{B} = \frac{\mu_0}{4 \pi r^3} (3 (\vec{\mu} \cdot \hat{r})\hat{r} - \vec{\mu})[/tex]
where [tex] \vec{\mu} [/tex] is the magnetic moment of a spin 1/2 particle which is proportional to S. So I know this is a classical formula so is it at all valid here?
When you work out the x-component you get the field depending on a quantity something like 2Sx + 3Sy + 3Sz. Since these are non-commuting operators, what does this even mean?
 
Last edited:

1. What is spin and how does it relate to magnetic field?

Spin is an intrinsic property of particles, such as electrons, that describes their angular momentum. This spin creates a magnetic moment, which interacts with external magnetic fields. The strength of this interaction is determined by the spin of the particle.

2. How does the direction of spin affect the magnetic field?

The direction of spin determines the orientation of the magnetic moment. A particle with spin pointing in the same direction as an external magnetic field will experience a lower energy state, while a particle with spin pointing in the opposite direction will experience a higher energy state.

3. Is there a relationship between the strength of spin and the strength of the magnetic field?

Yes, there is a direct relationship between the strength of spin and the strength of the magnetic field. As the spin of a particle increases, so does the strength of its magnetic moment, resulting in a stronger interaction with external magnetic fields.

4. How does the spin-magnetic field interaction affect the behavior of particles?

The spin-magnetic field interaction affects the behavior of particles in several ways. It can cause particles to align with an external magnetic field, creating a magnetization effect. It can also lead to precession, where the spin axis of a particle rotates around the magnetic field direction.

5. Can the spin-magnetic field interaction be manipulated?

Yes, the spin-magnetic field interaction can be manipulated using various techniques, such as applying an external magnetic field or using spin-polarized currents. This manipulation is key in many technologies, such as MRI machines and magnetic storage devices.

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