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relation between spin and its magnetic field |
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| Mar16-07, 01:09 PM | #1 |
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relation between spin and its magnetic field
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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| Mar17-07, 09:17 AM | #2 |
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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. |
| Mar17-07, 10:19 AM | #3 |
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Recognitions:
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or how the light was produced. |
| Mar17-07, 01:13 PM | #4 |
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relation between spin and its magnetic fieldI'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. |
| Mar20-07, 11:14 PM | #5 |
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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? |
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