# 5 Interatomic exchange coupling

From the very outset of this course, we established the quantum origin of atomic magnetic moments. In the present chapter, we would like to move forward and consider how elementary magnetic moments couples in solids. It is straightforward to understand that this issue is actually a quantum many-body problem. Though well-defined, this problem is practically not tractable on a large scale from first principles, i.e., from *ab initio* approaches. Still nowadays it is, therefore, common to rely on effective spin Hamiltonians for the theoretical rationalization of the magnetic materials, at least as an intermediate step between *ab initio* approaches and experiments.

When we evaluated the possible transitions from the ground state of a He atom to its excited states, we introduced the *effective* Hamiltonian:

\[\begin{equation}
\tag{5.1}
\mathcal{H}_{\rm exch}= -2J\,\hat{\mathbf s}_1\cdot\hat{\mathbf s}_2 \,.
\end{equation}\]
The constant \(J\) was dubbed exchange integral. For the specific case of the excited levels of the He atom \(J\) is positive, namely it favors the spin triplet state.
The 1st Hund’s rule prescribes the maximization of the total spin of an atom and, thus, can be summarized with a similar expression as Eq.(5.1), with \(J>0\).

Since Hund’s rules act at the atomic level, we defined this interaction *intra-atomic* exchange interaction. Dictating how single-electron levels shall be filled, the intra-atomic exchange interaction determines the number of unpaired electrons in the outern shell (i.e. the total atomic spin \(S\)) and their total angular momentum \(L\); therefore, the intra-atomic exchange interaction indirectly determines the Landè factoor and the number of levels associated with the atomic magnetic moment. On the other side, *inter-atomic* exchange interaction determines magnetic ordering on a large spatial scale. This second interaction is also written in the same way as Eq.(5.1). But, while for the intra-atomic exchange interaction the subscripts \(\hat{\mathbf s}_1\) and \(\hat{\mathbf s}_2\) indicate the spin of different electrons belonging to the same atom, for the inter-atomic exchange interaction \(\hat{\mathbf s}_1\) and \(\hat{\mathbf s}_2\) indicate the total *effective* spins of atoms sitting on different lattice sites.

The purpose of this chapter is to provide some insight into the origin of the inter-atomic exchange interaction and briefly review the most important mechanisms that produce it.

## 5.1 H\(_2\) and O\(_2\) molecules

As a first step toward the understanding of inter-atomic exchange interaction we need to clarify the role of spin in the chemical bond. To this aim, we reproduce the main passages of the Heitler-London calculation for the H\(_2\) molecule. The Hamiltonian consists of the following contributions \[\begin{equation} \tag{5.2} \mathcal{H}_{\rm H_2} = \mathcal{H}_{a} +\mathcal{H}_{b} + \mathcal{H}_{\rm ee} + \mathcal{H}_{\rm NN} + \mathcal{H}_{1b} + \mathcal{H}_{2a} \end{equation}\] The first term on the right-hand side is the Hamiltonian of the independent H atom with nucleus \(a\):\[\begin{equation} \tag{5.3} \mathcal{H}_{a} = \frac{\hat{\mathbf p}^2_1}{2m_{\rm e}} -\frac{e^2}{4\pi \epsilon_0} \frac{1}{r_{1a}} \,. \end{equation}\] The Hamiltonian \(\mathcal{H}_{b}\) is analogous but contains the linear momentum of the second electron \(\hat{\mathbf p}^2_2\) and the distance \(r_{2b}\).

\[\begin{equation} \tag{5.4} \mathcal{H}_{2a} = -\frac{e^2}{4\pi \epsilon_0} \frac{1}{r_{2a}} \qquad\text{and}\qquad \mathcal{H}_{1b} = -\frac{e^2}{4\pi \epsilon_0} \frac{1}{r_{1b}} \end{equation}\] represent the Coulomb interaction of the electron 2 with the nucleus \(a\) and the interaction of the electron 1 with the nucleus \(b\), respectively. Namely, both vectors \(\underline{r}_{1a}\) and \(\underline{r}_{1b}\) give the position of the electron 1, but in the former the nucleus \(a\) is assumed as the origin of coordinates while in the latter nucleus \(b\) is assumed as the origin (see Eq.(5.7)); the same convention applies to the electron 2. \[\begin{equation} \tag{5.5} \mathcal{H}_{\rm ee}=\frac{e^2}{4\pi \epsilon_0}\frac{1}{r_{12}} \end{equation}\]

is the Coulomb repulsion between the two electrons while

\[\begin{equation} \tag{5.6} \mathcal{H}_{\rm NN}=\frac{e^2}{4\pi \epsilon_0} \frac{1}{R} \end{equation}\]

is the Coulomb repulsion between the two nuclei. With respect to a common reference frame, the coordinates appearing in the equations above are given by \[\begin{equation} \tag{5.7} \begin{split} &\underline{r}_{1a}=\underline{r}_{1}-\underline{R}_{a} \qquad \qquad \underline{r}_{2b}=\underline{r}_{2}-\underline{R}_{b} \\ &\underline{r}_{12}=\underline{r}_{1}-\underline{r}_{2} \qquad \qquad \underline{R}=\underline{R}_a -\underline{R}_b \\ &\underline{r}_{1b}=\underline{r}_{1}-\underline{R}_{b} \qquad \qquad \underline{r}_{2a}=\underline{r}_{2}-\underline{R}_{a} \end{split} \end{equation}\] where \(\underline{r}_{1}\) and \(\underline{r}_{2}\) are the electron coordinates, \(\underline{R}_{a}\) and \(\underline{R}_{b}\) those of the two nuclei. In the Heitler-London calculation the two nuclei are kept at a fixed distance \(R\). This distance enters the problem only as a

*parameter*and, for this reason, the kinetic energy of the nuclei has not been included in the Hamiltonian Eq.(5.2). The basic idea is to evaluate the expectation value of this Hamiltonian on a pair of trial wave functions, compatible with the triplet and the singlet spin state. The spatial parts of these trial wave functions are built from the ground-state wave function of an isolated H atom, that is the \(\psi_{1s}(\underline{r})\) orbital. The last one was obtained in the first chapter taking the nucleus of an H atom as the origin of the reference frame. However, because they are not distinguishable, either electron 1 or 2 can be assigned to one nucleus, e.g., \(a\). Therefore, an educated guess for the spatial wave functions associated with the triplet or singlet spin state is \[\begin{equation} \tag{5.8} \begin{split} &\Psi^{\rm T}(\underline{r}_1,\underline{r}_2)=%-\Psi^{\rm T}(\underline{r}_2,\underline{r}_1)= \frac{1}{\sqrt{2(1-S^2)}}\left[\psi_{1s}(\underline{r}_{1a})\psi_{1s}(\underline{r}_{2b})-\psi_{1s}(\underline{r}_{1b})\psi_{1s}(\underline{r}_{2a})\right] \\ &\Psi^{\rm S}(\underline{r}_1,\underline{r}_2)=%\Psi^{\rm S}(\underline{ r}_2,\underline{r}_1)= %\hline \frac{1}{\sqrt{2(1+S^2)}}\left[\psi_{1s}(\underline{r}_{1a})\psi_{1s}(\underline{r}_{2b})+\psi_{1s}(\underline{r}_{1b})\psi_{1s}(\underline{r}_{2a})\right] \end{split} \end{equation}\] where the superscripts “T” and “S” indicate the spatial wave function associated with the triplet and singlet spin state, respectively, and \[\begin{equation} S=\langle \psi_{1s}(\underline{r}_{1a})|\psi_{1s}(\underline{r}_{1b})\rangle = \langle \psi_{1s}(\underline{r}_{2b})|\psi_{1s}(\underline{r}_{2a})\rangle = \int\psi_{1s}(\underline{r}_{1a})\psi_{1s}(\underline{r}_{1b}) \,d^3r_1 \end{equation}\] is the

*overlap integral*. As for the He atom, the spin triplet is associated with an

*antisymmetric*spatial wave function, while the spin singlet with a

*symmetric*one. The expectation value of the Hamiltonian \(\mathcal{H}_{\rm H_2}\) in Eq.(5.2) evaluated on the spatial wave functions defined in Eq.(5.8) is given by

\[\begin{equation} \tag{5.9} \begin{split} &E^{\rm T} = \langle \Psi^{\rm T} | \mathcal{H}_{\rm H_2} | \Psi^{\rm T} \rangle = 2E_{1s} + \frac{Q-X}{1-S^2}\\ &E^{\rm S} = \langle \Psi^{\rm S} | \mathcal{H}_{\rm H_2} | \Psi^{\rm S} \rangle = 2E_{1s} + \frac{Q+X}{1+S^2} \,. \end{split} \end{equation}\] Similarly to the case of the He atom, \(Q\) is called

*Coulomb*integral and \(X\) is the

*exchange*integral (the reason why here we do not use the letter \(J\) will become clear later on). The former reads \[\begin{equation} \begin{split} Q &= \langle \psi_{1s}(\underline{r}_{1a})\psi_{1s}(\underline{r}_{2b}) | \left[\mathcal{H}_{\rm NN} + \mathcal{H}_{\rm ee} + \mathcal{H}_{1b} + \mathcal{H}_{2a} \right] |\psi_{1s}(\underline{r}_{1a})\psi_{1s}(\underline{r}_{2b}) \rangle \\ &= \frac{e^2}{4\pi \epsilon_0} \left\{ \frac{1}{R} + \langle \psi_{1s}(\underline{r}_{1a})\psi_{1s}(\underline{r}_{2b}) | \frac{1}{r_{12}} |\psi_{1s}(\underline{r}_{1a})\psi_{1s}(\underline{r}_{2b}) \rangle \right. \\ &\left. - \langle \psi_{1s}(\underline{r}_{1a})|\frac{1}{r_{1b}}|\psi_{1s}(\underline{r}_{1a})\rangle - \langle \psi_{1s}(\underline{r}_{2b})|\frac{1}{r_{2a}}|\psi_{1s}(\underline{r}_{2b})\rangle \right\} \,. \end{split} \end{equation}\] The first term on the right-hand side is the Coulomb repulsion between the two nuclei; the second term is the electron-electron repulsion; the third term is the Coulomb interaction between the electron 1 and the nucleus \(b\), while the forth term is the interaction between the electron 2 and the nucleus \(a\). Practically, the last three terms should be evaluated performing integrals over the positions of the two electrons, namely integrating over \(d^3r_1\) and/or \(d^3r_2\). As the evaluation of those integrals is not our scope, we chose to use the Dirac notation because it helps focus on the labels of electrons (1,2) and nuclei (\(a\), \(b\)). The

*exchange*integral reads \[\begin{equation} \tag{5.10} \begin{split} X &= \langle \psi_{1s}(\underline{r}_{1a})\psi_{1s}(\underline{r}_{2b}) | \left[\mathcal{H}_{\rm NN} +\mathcal{H}_{\rm ee} + \mathcal{H}_{1b} + \mathcal{H}_{2a} \right] |\psi_{1s}(\underline{r}_{1b})\psi_{1s}(\underline{r}_{2a}) \rangle \\ &= \frac{e^2}{4\pi \epsilon_0} \left\{ \frac{S^2}{R} + \langle \psi_{1s}(\underline{r}_{1a})\psi_{1s}(\underline{r}_{2b}) | \frac{1}{r_{12}} |\psi_{1s}(\underline{r}_{1b})\psi_{1s}(\underline{r}_{2a}) \rangle \right. \\ &\left. - S\, \langle \psi_{1s}(\underline{r}_{1a})|\frac{1}{r_{1b}}|\psi_{1s}(\underline{r}_{1b})\rangle - S\, \langle \psi_{1s}(\underline{r}_{2b})|\frac{1}{r_{2a}}|\psi_{1s}(\underline{r}_{2a})\rangle \right\}\,. \end{split} \end{equation}\] All the contributions appearing in the \(X\) integral arise from the requirement of symmetrization of the total spatial wave functions \(\Psi^{\rm T}(\underline{r}_1,\underline{r}_2)\) and \(\Psi^{\rm S}(\underline{r}_1,\underline{r}_2)\). Both \(Q\) and \(X\) depend

*parametrically*on the distance \(R\) (see Eq.(5.7)).

In Fig.5.1a the \(R\) dependence of \(E^{\rm S}-2E_{1s}\) and \(E^{\rm T}- 2E_{1s}\) is plotted. Coming from large values of \(R\), the energy of the singlet state decreases with decreasing separation between nuclei up to \(R_0\), where it starts increasing because of the Coulomb repulsion between nuclei. On the contrary, the energy associated with the triplet spin state increases monotonically as the two nuclei approach each other. This means that the bound state, in which the H\(_2\) molecule is actually stable, is only compatible with a singlet spin wave function. Within the Heitler-London treatment one thus interprets \(R_0\simeq 1.51 a_0\) (\(a_0=0.529\) \(\,\) is the Bohr radius) as the equilibrium separation between nuclei of the H\(_2\)-molecule. The experimental value is 0.7395 . The exchange constant \(J\), deduced from the Heitler-London calculation, can be defined in the same way as for the excited states of the He atom, i.e., such that the singlet-triplet splitting equals \(2J\):
\[\begin{equation}
\tag{5.11}
E^{\rm S} - E^{\rm T} = 2J =2\frac{X -QS^2}{1-S^4} \,.
\end{equation}\]
For zero overlap between \(| \Psi^{\rm T} \rangle\) and \(| \Psi^{\rm S} \rangle\), e.g. realized when \(R\rightarrow \infty\), one has \(J = X=0\)^{15}. For \(R=R_0\), instead, \(X\) (and thus \(J\)) has a negative sign and this is essential to have a stable chemical bond in the H\(_2\) molecule. Since the triplet state corresponds to the unstable (antibonding) state, the singlet-triplet energy separation is actually the dissociation energy of the H\(_2\) molecule, i.e. 4.476 eV, which implies \(J=2.24\) eV. This energy is of the same order of magnitude as the intra-atomic exchange energy. Later on in this course we will see that magnetic properties of solids are usually determined by the competition between inter-atomic exchange interaction – which favors magnetic ordering – and thermal fluctuations – which favor a disordered phase. However, for laboratory temperatures, thermal fluctuations cannot compete with energies in the eV range. Therefore, for this specific case, thermal fluctuations do not manage to populate the excited triplet state significantly and the H\(_2\) molecule remains diamagnetic also at finite temperature.

Before discussing other mechanisms that produce an inter-atomic exchange interaction with strength comparable to the thermal energy, let us make a quick digression on a different way to look at the role of spin in the chemical bond. In an alternative and more general approach than the Heitler-London method, molecular orbitals are deduced as Linear Combination of Atomic Orbitals (LCAO) and successively filled following the same principles underlying Hund’s rules: Pauli principle and Coulomb repulsion. The diagram of empty molecular orbitals of the H\(_2\) molecule obtained as LCAO is shown in Fig.5.1b. The occupation of molecular orbitals corresponding to the triplet and the singlet spin state of the H\(_2\) molecule are given in Fig.5.1c and d, respectively. We remind that in this representation two antiparrallel arrows indicate the singlet spin state, while two parrallel arrows indicate the triplet spin state:
\[\begin{equation}
\tag{5.12}
\begin{split}
& \uparrow\, \downarrow \quad \quad \text{stands for} \quad |S=0\rangle \quad \text{spin singlet} \\
& \uparrow\, \uparrow \quad \quad \text{stands for} \quad |S=1\rangle \quad \text{spin triplet} \,.
\end{split}
\end{equation}\]
Even if it is customary to speak of “parallel” and “antiparallel” spins in this context, the reader should keep in mind the true meaning of this diagrammatic representation that is summarized in Eq.(5.12). Once clarified this nomenclature issue, the hand-waving expressions “parallel” and “antiparallel” spins can be useful. In the next section, for instance, we will adopt them in the description of exchange mechanisms in metal oxides. For completeness, the situation in which two H atoms are far apart is schematically represented in Fig.5.1e putting the two electrons in the atomic orbitals rather than in the molecular orbitals. Both the Heitler-London and the LCAO approach produce a ground state for the H\(_2\) molecule in which spins are in the singlet state. Therefore, the Pauli principle is indeed essential for the chemical bond but at *the same time* seems to work against “parallel” spin alignment.

Figure5.2 shows a qualitative diagram of molecular orbitals for the O\(_2\) molecule. In this case, the progressive filling of levels results in a total spin \(S=1\). The occupied levels with higher energy are the two \(\pi^*\) orbitals filled with a pair of “parallel” spins (forming a triplet state), in accordance with the first Hund’s rule. From the examples of H\(_2\) and O\(_2\) molecules we learn that when atoms are bond together to form a molecule the atomic levels are mixed and split due to the formation of molecular orbitals. Molecular orbitals that “glue” together the molecule (bonding orbitals) coexist with orbitals that, if occupied, would tend to dissociate it (antibonding). The interplay between these counterposed tendencies and the progressive filling of molecular orbitals determine the stability of the bonding. The intuitive picture about the ground state of the H\(_2\) molecule suggests that Pauli principle applies to electrons originally assigned to different atoms that, because of the chemical bond, happen to be delocalized over the same region of space. On the other hand, the application of the 1st Hund’s rule to orthogonal molecular orbitals explains why O\(_2\) is paramagnetic with \(S=1\). The actual energetic ordering of molecular orbitals for a generic molecule obviously play a role but, unfortunately, this cannot be determined quantitatively by simple arguments.

## 5.2 Exchange coupling in transition-metal oxides

With the few ideas summarized at the end of the previous section we can understand qualitatively the main mechanisms producing ferromagnetic or antiferromagnetic coupling between neighboring paramagnetic ions in transition-metal oxides. It is important to remember that the chemical bond involves other orbitals besides the magnetic orbitals (those containing unpaired electrons) or it may even have a dominant ionic character, like in MnO. Said that, we can proceed looking at few examples in which transition-metal ions are packed in a crystal in such a way that an oxygen bridge exists between pairs of neighboring metal ions. The oxygen atoms thus act as the ligands in the crystal-field description given the previously. In particular, they are doubly ionized in the valence O\(^{2-}\). The corresponding electronic configuration is [O\(^{2-}\)]=He\(^2\)(2p)\(^6\), meaning that all the three 2p orbitals are doubly occupied by electrons forming a spin singlet in each orbital.

Referring to Fig.5.3, we start considering the group Cu-O-Cu where the ion Cu\(^{2+}\) has a (3d)\(^9\) electronic configuration. We assume that the only magnetic orbital in the two copper ions is the \(d_{x^2-y^2}\) orbital, which lies on the equatorial plane of the Jahn-Teller-distorted octahedron of coordination. Let us assume that the spin of the electron occupying this orbital in the Cu on the left-hand side points upwards and that one p orbital of the oxygen overlaps significantly with the \(d_{x^2-y^2}\) of the considered Cu ion.
Then an electron with spin pointing “downwards” in the p orbital of the oxygen shall preferentially delocalize on the \(d_{x^2-y^2}\) orbital of the Cu ion on the left (Pauli principle). The remaining electron occupying the same orbital of the oxygen will be delocalized on the \(d_{x^2-y^2}\) of the Cu ion on the right-hand side. The two electrons in the same oxygen orbital are forced by Pauli principle to form a spin singlet, i.e. they must be “antiparallel” in the meaning of Eq.(5.12). Thus, since the spin of the electron preferentially delocalized on the left-hand side of the p orbital points “downwards”, the spin of the electron delocalized on the right-hand side *must* point “upwards”. As a consequence, the unpaired electron on the Cu ion on the right shall preferentially align dwonwards. In this way an effective antiferromagnetic coupling between the two Cu\(^{2+}\) ions is created.

If one substitutes the Cu\(^{2+}\) ion on the right-hand side of Fig.5.3 with a vanadyl cation VO\(^{2+}\), which contains a V\(^{4+}\) with electronic configuration (3d)\(^1\), an effective ferromagnetic coupling between the two transition-metal ions can be realized. Referring to Fig.5.4, the situation is practically the same as before for what concerns the Cu ion on the left-hand side. However, the magnetic orbital of V\(^{4+}\) is the (non-bonding) orbital \(d_{xy}\). Therefore, the p orbital of the oxygen that overlaps with the magnetic orbital of the Cu ion will donate some electron density with spin “up” to an empty orbital of the V ion. This empty orbital is necessarily orthogonal to the singly occupied \(d_{xy}\) orbital. In this situation the 1st Hund’s rule applies, which favors parallel spin alignment between the spin of the oxygen electron delocalized on the V ion and the spin of the electron occupying the \(d_{xy}\) orbital in the V\(^{4+}\). The net result is a ferromagnetic coupling between the Cu\(^{2+}\) and V\(^{4+}\) ions.

Another route to ferromagnetic coupling relies on the fact that the oxygen ligand can bridge two non-equivalent orbitals of equivalent transition-metal ions. In Fig.5.5 this case is exemplified for a group Mn-O-Mn. It corresponds to the case in which a magnetic orbital, say \(d_{z^2}\), in an elongated-octahedral Mn\(^{3+}\) ion (left) has non-zero overlap with the empty \(d_{x^2-y^2}\) orbital of the neighboring Mn\(^{3+}\) (right). The delocalization of electron density of the oxygen electron will proceed according to the Pauli principle for the Mn on the left-hand side and according to the 1st Hund’s rule for the Mn on the right-hand side. This eventually produces an effective ferromagnetic coupling between the two Mn\(^{3+}\) ions.

The type of exchange interaction produced with the three mechanisms described above is called *super exchange* interaction.

*double exchange*mechanism is sketched in Fig.@ref(fig:double_exchange). The additional electron on the Mn\(^{3+}\) (left) occupying the \(d_{z^2}\) orbital the can be delocalized on the Mn\(^{4+}\) (right) by an electron-transfer process. But since the \(d_{z^2}\) orbital in this second Mn is empty the 1st Hund’s rule applies, thus determining an effective ferromagnetic coupling between the two magnetic centers.

### Why is magnetite ferrimagnetic?

We began this course mentioning that magnetite was the first magnet discovered by humans. In spite of this fact, the details of its magnetic structure were understood comparatively recently. In this system, with formula Fe\(_3\)O\(_4\), the magnetic contribution is provided by three inequivalent Fe atoms: One Fe\(^{2+}\) and two Fe\(^{3+}\), which are coordinated with the oxygen atoms in such a way that the Fe\(^{2+}\) and *one* Fe\(^{3+}\) feel a distorted octahedral ligand field, while the other Fe\(^{3+}\) feels a tetragonal ligand field. The Fe\(^{2+}\) ion has six electrons and is found in the high-spin multi-electron configuration with spin \(S=2\). The two Fe\(^{3+}\) contribute to the magnetism with a spin \(S=5/2\). However, these last two ions are coupled antiferromangetically via a super-exchange mechanism – as the one discussed in relation to Fig.5.4 – and their magnetic moments cancel out. Therefore, Fe\(^{3+}\) atoms do not contribute to the macroscopic magnetization of magnetite. By virtue of the different valency of the two atoms, a ferromagnetic double exchange coupling is realized between one Fe\(^{3+}\) and the Fe\(^{2+}\) (analogous to the mechanism sketched in Fig.@ref(fig:double_exchange) for Mn\(^{3+}\) and Mn\(^{4+}\)). The magnetism of magnetite is determined only by the uncompensated spin of Fe\(^{2+}\) atoms that provide a magnetic moment of 4\(\mu_{\rm B}\) per Fe\(_3\)O\(_4\) molecule. Properly speaking, the first discovered magnet is thus a ferrimagnet in which the exchange coupling is strong enough to keep it magnetic well above room temperature (\(T_{\rm c}=858\) K).

No overlap between \(| \Psi^{\rm T} \rangle\) and \(| \Psi^{\rm S} \rangle\) means that the product \(\psi_{1s}(\underline{r}_{1a})\psi_{1s}(\underline{r}_{1b})=0\) \(\forall \, \underline{r}_{1}\) and similarly \(\psi_{1s}(\underline{r}_{2a})\psi_{1s}(\underline{r}_{2b})=0\) \(\forall \, \underline{r}_{2}\), therefore all terms on the l.h.s. of Eq.(5.10) vanish.↩︎