QM tells us that only one component of angular momentum is measurable which conventionally taken to be Lz= 1/2. The other two components have an uncertain magnitude and direction and this is usually depicted as a Cone around measurement axis.

In a system of two or more electrons, how does one electron see the spin of others? Does it feel the magnetic moment of another electron in only parallel to it or antiparallel to it with a value of 1/2? The confusion which I face comes from the ferromagnetic materials in which all the spins tend to align with each other. What is meant by this alignment? Since two components of any individual spin cannot be determined, I think there will always be some uncertainty. Or just the z components of all spins align?

  • $\begingroup$ +1 This is a great question because there is the "easy" answer using mean-field theory, but deeper answer using entanglement that extends to the anyiferromagnet case. I hope someone who is familiar with these can give a great answer. $\endgroup$ – KF Gauss Jan 28 at 7:23
  • $\begingroup$ @KFGauss Don't you think that mean-field is somewhat a circular explanation? $\endgroup$ – Syed Hashim Shah Hashmi Jan 28 at 7:47
  • 1
    $\begingroup$ A long time ago, I remember my physics course which showed that the Curie temperature was too high to explain spin interaction in ferromagnet as a dipolar magnetic interaction (typical energy ${{k}_{B}}T$ a lot too high). It is necessary to involve the interaction of exchange which sometimes leads to a alignment of the spins. (sometimes to more complicated situations) You can take a look at en.wikipedia.org/wiki/Exchange_interaction $\endgroup$ – Vincent Fraticelli Jan 28 at 8:14
  • $\begingroup$ The alignment should be about the expectation value. $\endgroup$ – K_inverse Jan 28 at 9:27
  • $\begingroup$ @VincentFraticelli is right, and the exchange interaction is a Coulomb interaction. It is easiest to understand for atoms, where it gives rise to Hund's rule: hyperphysics.phy-astr.gsu.edu/hbase/Atomic/Hund.html#c2 $\endgroup$ – Pieter Jan 28 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.