Ferromagnetism corresponds to the parallel alignment of spins, and it is said that for ferromagnetic alignment one requires the spin part of the wavefunction must be in one of the three symmetric triplet states: $|\uparrow\uparrow\rangle$, $\frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle)$ or $|\downarrow\downarrow\rangle$ with $m=1,0,-1$ respectively. I understand that the first and the third states, the spins are aligned which is not the case for the second state. Hence, I have the following question.

Question How does the second member of the triplet i.e., the state $\frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle)$ with $s=1, m=0$ lead to ferromagnetic alignment while the singlet state $\frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)$ with $s=0, m=0$ lead to antiferromagnetic alignment? In both states, I find that the spins are oppostitely aligned, and therefore, must lead to anti-ferromagnetic order. Where am I wrong?

  • $\begingroup$ It is not the spin part of the wavefunction (which results in the negligible contribution of electron magnetic dipole moment) that results in magnetism. Because the spin part is symmetric, the spatial part has to be antisymmetric. The electrons will therefore lie further apart. When the spatial wavefunction is symmetric, the need to minimise Pauli repulsion (and hence raising of energy) meant they need to be antiparallel, hence when the spatial wavefunction is antisymmetric, the electrons tend to be parallel to each other. This contributes most to magnetism. $\endgroup$ – Melvin Apr 30 '18 at 19:44
  • $\begingroup$ All three triplet states have parallel spins. They differ in their orientation to the so-called quantisation axis, usually the direction of the magnetic field. $\endgroup$ – my2cts Apr 30 '18 at 21:27
  • $\begingroup$ As Melvin said, it is the orbital part of the wave function that is responsible for ferromagnetic coupling between different atoms. It is the spins that generate the magnetic field. $\endgroup$ – my2cts Apr 30 '18 at 21:30
  • $\begingroup$ @Melvin Sorry I did not get the point. My understanding is that if the spin state is $|\uparrow\uparrow\rangle$ or $|\downarrow\downarrow\rangle$, the spins are aligned i.e., ferromagnetic ordering. $\endgroup$ – SRS May 1 '18 at 15:07
  • $\begingroup$ @my2cts "All three triplet states have parallel spins. They differ in their orientation to the so-called quantisation axis, usually the direction of the magnetic field." Can you explain? $\endgroup$ – SRS May 1 '18 at 15:10

Remember that in quantum mechanics you can always take superpositions of degenerate ground states to get another ground state. For a Heisenberg ferromagnet, any state with all the spins aligned in the same direction is a ground state, but by taking superpositions of such states aligned along different directions, you can get other ground states that are not of this simple form.

The $m=0$ triplet state is a superposition of states $|\rightarrow\rangle$ aligned along the $+x$ axis and states $|\uparrow\rangle$ or $|\downarrow\rangle$ aligned along the $z$-axis: $$|s=1;m=0\rangle = \sqrt{2}\, |\rightarrow\rangle|\rightarrow\rangle - \frac{1}{\sqrt{2}}|\uparrow\rangle|\uparrow\rangle - \frac{1}{\sqrt{2}}|\downarrow\rangle|\downarrow\rangle.$$ (This might not look correctly normalized, but it actually is (unless I messed up), because the states in the superposition aren't orthogonal.) It can also be expressed as other superpositions of states with the spins aligned along other axes.

It turns out that if you have a Heisenberg ferromagnet with $N$ spin-1/2's and you take the span of all the all-aligned states $\bigotimes\limits_{n=1}^N (\alpha |\uparrow\rangle + \beta |\downarrow\rangle)$, then only $N+1$ of them are linearly independent, so the ground state manifold is $(N+1)$-fold degenerate. In the $S^z$-basis, this corresponds to the $N+1$ different states $$\left|s = \frac{N}{2}; m \in \left\{-\frac{N}{2}, -\frac{N}{2}+1, \dots, \frac{N}{2}-1, \frac{N}{2} \right\}\right\rangle,$$ but these states are not so easy to interpret physically, because all of them except for the two $|m = \pm N/2\rangle$ states are nontrivial superpositions over different directions of states in which all the spins are aligned in that direction.

  • $\begingroup$ I need some time to digest the answer. Does your answer explain why $s=0,m=0$ not ferromagnetic order? Did you fix the typo in the first eq.? @tparker $\endgroup$ – SRS May 1 '18 at 15:45
  • $\begingroup$ @SRS Yes, my answer does explain why $|s=0, m=0\rangle$ is not ferromagnetic: it cannot be expressed as any linear combination of states with aligned spins, because all such states lie in the $s = 1$ eigenspace of the $\hat{S}_\text{tot}$ operator. And you still haven't told me what the alleged typo in the first equation is; when I said "There is..." I didn't mean "There is a typo", I meant "There is (already) an equality in the equation." $\endgroup$ – tparker May 1 '18 at 17:47
  • $\begingroup$ Okay.. @tparker $\endgroup$ – SRS May 1 '18 at 17:49
  • $\begingroup$ I'm having a problem understanding how is the first equation equivalent to ferromagnetic ordering? In a comment addressed at Melvin I explain what I understand by ferromagnetic ordering. @tparker $\endgroup$ – SRS May 1 '18 at 17:50
  • $\begingroup$ @SRS Ferromagnetic ordering does not mean "there is a single direction along which all the spins point," as you seem to believe. Technically, it is defined to mean that $S_\text{tot}$ is maximal; more intuitively, it means that "in every term in the superposition, every spin is pointing in the same direction." But that direction is allowed to be different across different terms in the superposition, as in my first equation. The key point is that within each term, the spins are aligned. A state is ferromagnetic if and only if it can be written as such a superposition. $\endgroup$ – tparker May 1 '18 at 20:26

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