Can the $s=1, m_s=0$ member of the spin triplet state lead to ferromagnetic alignment? 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?
 A: 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.
