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This question already has an answer here:

I would like to know if there is some expression for the magnetic field generated by a electron spin.

As far as I know, the spin provides a magnetic moment to the electron and a magnetic dipoles momentum $\vec m$ produces a field:

$H = {1\over4\pi} \left({3\vec r(\vec m\cdot\vec r)\over r^5} - {\vec{m}\over r^5}\right)$

However, the spin can be measured in each at any direction so, is the previous expression valid for the electron spin?

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marked as duplicate by Pieter, Kyle Kanos, Jon Custer, Chris, Community Mar 10 '18 at 6:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The spin of a particle is an intrinsic property of the particle, not a physical classical spinning of charge. As such it doesn't make sense to speak of the magnetic field of an electron $\endgroup$ – Triatticus Mar 8 '18 at 23:50
  • $\begingroup$ That should be an answer @Triatticus.... $\endgroup$ – tom Mar 8 '18 at 23:54
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    $\begingroup$ When computing the fine & hyperfine corrections to the Hydrogen atom, some of the interaction terms that manifest in the Hamiltonian can be read as the interaction between (the magnetic moment produced by) the electron spin and the magnetic field produced by (the magnetic moment produced by) the nucleus spin. For more details you can check chapter XII and the complements therein in Cohen, Tannoudji, Diu and Laloe's book, Quantum Mechanics, vol II. In that sense, I'd say it is legitimate to talk about the magnetic field produced by spin, regardless of the fact it is an intrinsic property. $\endgroup$ – secavara Mar 9 '18 at 0:41
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    $\begingroup$ Possible duplicate of What does the magnetic field of the (quantum-mechanical) electron look like? See the answer by Lubos Motl. $\endgroup$ – Pieter Mar 9 '18 at 10:01
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Separation of up down spin electrons yields a 50 / 50 split. If an additional gradient magnetic field at 90 degrees to all up spin electrons is applied then the up / down spin ratio returns to a 50 / 50 split ? Does the electron have spin or is the a magnetic gradient imposing spin on the electron? How could all spin up electrons end up being 50 / 50 spin up and spin down?

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  • $\begingroup$ This does not appear to be an answer to the OP's question. $\endgroup$ – PM 2Ring Mar 10 '18 at 5:47
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The magnetization of for example an iron magnet is almost completely due to the spin moment of the electrons. I have not checked OP's expression, but the field outside a magnet should be the integral over all those electrons.

At close range, the dipole-dipole interaction between electrons is usually negligible in comparison with the exchange interaction (which is an electrostatic term due to the Pauli principle). But it is there. It also plays a role in the muon-electron interaction. An electron has a dipole magnetic field.

See also Lubos Motl's answer to a previous query and his blog post.

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Rewriting my comment in answer form: The issue is that spin is an intrinsic property of a particle, there isn't a classical analogue to this property as it is purely a quantum mechanical phenomenon. So the electron itself isn't to be viewed as a spinning sphere of charge and as such doesn't create a magnetic field because of it.

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    $\begingroup$ That is not correct. The field of a magnet is usually almost completely due to the spin moments. $\endgroup$ – Pieter Mar 9 '18 at 7:39
  • $\begingroup$ It has a magnetic moment...not a magnetic field which is exactly what op was asking. There are no closed field loops due to magnetism around an electron in the sense of a magnetic field $\endgroup$ – Triatticus Mar 9 '18 at 7:57
  • $\begingroup$ Maybe Wilczek's essay would convince you: frankwilczek.com/electronTakeThreeB.pdf $\endgroup$ – Pieter Mar 11 '18 at 10:25
  • $\begingroup$ @Pieter Your link is dead, but you are correct. Wikipedia's page on magnetic moment should clear things up. $\endgroup$ – thermomagnetic condensed boson Jan 16 at 8:52

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