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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?

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  • $\begingroup$ 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
    Commented Apr 30, 2018 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
    Commented Apr 30, 2018 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
    Commented Apr 30, 2018 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
    Commented May 1, 2018 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
    Commented May 1, 2018 at 15:10

3 Answers 3

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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.

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  • $\begingroup$ 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
    Commented May 1, 2018 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
    Commented May 1, 2018 at 17:47
  • $\begingroup$ Okay.. @tparker $\endgroup$
    – SRS
    Commented May 1, 2018 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
    Commented May 1, 2018 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
    Commented May 1, 2018 at 20:26
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tparker's answer is complete. I just wanted to add a classical approach as well.

If you have two vectors of norm $1/2$, if their sum has norm $1$, then they are aligned by the triangular inequality. This corresponds to the case $s=1$, which is why you say that it is ferromagnetic order. In general, if you have $N$ spins $s$, then the state of total spin $S = Ns$ is a state of ferromagnetic order since classically all the spins must be aligned (like constructive interference).

Conversely, if their sum has norm zero, then they are anti aligned. This corresponds to the case $s=0$, which is why you say that it is the antiferromagnetic order.

Mathematically, this classical picture works with spin $1/2$ since a spin $1/2$ state is an eigenvector of an appropriately oriented spin component.

Hope this helps.

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I'll try to elaborate more on tparker's answer, which is very good.

First of all, consider a single spin $\frac{1}{2}$, which can be described by a superposition of two basis states: let them be $|\uparrow\rangle$ or $|\downarrow\rangle$ if we are using as a basis the eigenstates of $S_z$, or let them be $|\rightarrow\rangle$ and $|\leftarrow\rangle$ if we are using the eigenstates of $S_x$. The relation between the two basis stes is the following: $$ |\downarrow\rangle = \frac{1}{\sqrt{2}} \left(|\rightarrow\rangle + |\leftarrow\rangle \right) $$ $$ |\uparrow\rangle = \frac{1}{\sqrt{2}} (|\rightarrow\rangle - |\leftarrow\rangle ) $$ (if you want to derive these relations, just take the eigenvectors of the first Pauli matrix...).

Now how can we rewrite the triplet state $|s=1, m=0\rangle$ and the singlet state $|s=0, m=0\rangle$ in this basis? With some simple algebra we get $$ |s=1,m=0\rangle = \frac{1}{\sqrt{2}} \left( |\rightarrow\rightarrow\rangle - |\leftarrow\leftarrow\rangle \right) \,\,\,\,\,\,\, \text{triplet} $$ $$ |s=0,m=0\rangle = \frac{1}{\sqrt{2}} \left( |\rightarrow\leftarrow\rangle - |\leftarrow\rightarrow\rangle \right) \,\,\,\,\,\,\, \text{singlet}. $$ From this we can clarify a first important difference between the two states: the former is a quantum superposition of two oppositely magnetized states in the positive and negative $x$ direction; the latter is a superposition of two non-magnetized states. However, if you compute the expectation value of the total magnetization $\langle S_{1a} + S_{2a} \rangle$ in any direction $a=x,y,z$, you get zero both with the singlet and with the triplet states considered (this is also suggested by the fact that the magnetic quantum number m is 0).

The best classical analogue that comes into my mind to visualize the basis of the triplet is a vector preceding around a $z$-axis at an angle of 45°, 90° and 135° respectively. When the vector preceeds at 45°, its projection (averaged over a precession period) along the $z$-axis is positive; when it preceeds at 135°, the projection is negative, and at 90° the projection is zero. If you look at the projection (averaged over time) of the vector along the $x$ or $y$ axis, it is zero regardless of the angle.

So far (anti)ferromagnetism has not been mentioned, so now let's talk about this. First of all, I would define a ferromagnetic state as a state with non-vanishing magnetization vector, i.e. a state such that $\sum_i\langle \vec{S}_{i}\rangle \neq \vec{0}$. So, I would call "ferromagnetic" states such as $|\uparrow\uparrow\rangle$, $|\downarrow\downarrow\rangle$, $|\rightarrow\rightarrow\rangle$, $|\leftarrow\leftarrow\rangle$, etc., but I wouldn't call ferromagnetic the state $|s=1,m=0\rangle$.

Similarly, I would define an antiferromagnetic state as a state with non vanishing staggered magnetization, i.e. with $\sum_i (-1)^i \langle \vec{S}_i \rangle \neq \vec{0}$, so neither the singlet $|s=0,m=0\rangle$, nor the triplet $|s=1,m=0\rangle$ are antiferromagnetic. Examples of antiferromagnetic states are $|\uparrow\downarrow\rangle$, $|\downarrow\uparrow\rangle$, $|\rightarrow\leftarrow\rangle$, $|\leftarrow\rightarrow\rangle$, etc.

A possible source of confusion is the fact that the ground state of a Heisenberg model on a finite lattice is the singlet if the interaction is antiferromagnetic and (a sort of) spin "triplet" if the interaction is ferromagnetic (see tparker's answer). In the thermodynamic limit, which is the only case where a spontaneous symmetry breaking can happen, the ground state becomes an antiferromagnet in the first case and a ferromagnet in the second, which might bring us to the misleading statement that "triplet = ferromagnet and singlet = antiferromagnet".

Hope this helps!

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