Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relation between spin and its magnetic field

  1. Mar 16, 2007 #1
    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.
  2. jcsd
  3. Mar 17, 2007 #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.
  4. Mar 17, 2007 #3

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
  5. Mar 17, 2007 #4
    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.
  6. Mar 20, 2007 #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: Mar 20, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook