From 69fa3eb16d32efc948d72951dbe6e39ba98d1be8 Mon Sep 17 00:00:00 2001
From: Sander Otte <a.r.otte@tudelft.nl>
Date: Wed, 11 Apr 2018 21:39:38 +0200
Subject: [PATCH] lecture week 8, Thursday version

---
 docs/figures/RKKY.svg          | 438 +++++++++++++++++++++++++++++++++
 docs/figures/crystalfield.svg  | 129 ++++++++++
 docs/figures/superexchange.svg |  71 ++++++
 docs/lecture_8.md              |  86 ++++++-
 4 files changed, 718 insertions(+), 6 deletions(-)
 create mode 100644 docs/figures/RKKY.svg
 create mode 100644 docs/figures/crystalfield.svg
 create mode 100644 docs/figures/superexchange.svg

diff --git a/docs/figures/RKKY.svg b/docs/figures/RKKY.svg
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diff --git a/docs/figures/crystalfield.svg b/docs/figures/crystalfield.svg
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diff --git a/docs/figures/superexchange.svg b/docs/figures/superexchange.svg
new file mode 100644
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+++ b/docs/figures/superexchange.svg
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diff --git a/docs/lecture_8.md b/docs/lecture_8.md
index 792c37b4..bcb02cb2 100644
--- a/docs/lecture_8.md
+++ b/docs/lecture_8.md
@@ -101,11 +101,7 @@ $$
 \tilde{g}\ =\frac{1}{2}(g+1)+\frac{1}{2}(g-1)\left[\frac{S(S+1)-L(L+1)}{J(J+1)}\right]=\frac{3}{2}-\frac{L(L+1)-S(S+1)}{2J(J+1)}
 $$
 
-The result is a ladder of equally spaced levels. The bigger the contribution of $S$ inside $J$, the bigger te level spacing.
-
-![](figures/lande.svg)
-
-Now, the magnetisation for $n$ atoms becomes:
+The result is a ladder of equally spaced levels. The bigger the contribution of $S$ inside $J$, the bigger te level spacing. Now, the magnetisation for $n$ atoms becomes:
 
 $$
 M=n\frac{\sum_{m_{J}=-J}^{J}\tilde{g}\ \mu_{\rm B}m_{J}{\rm e}^{-\beta\mu_{\rm B}m_J B}}{\sum_{m_{J}=-J}^{J}{\rm e}^{-\beta\mu_{\rm B}m_J B}}
@@ -115,4 +111,82 @@ The resulting susceptibility is:
 
 $$
 \chi=\frac{n\mu_0(\tilde{g}\ \mu_{\rm B})^2}{3}\frac{J(J+1)}{k_{\rm B}T}
-$$
\ No newline at end of file
+$$
+
+### Atoms in solids
+Until now we have considered magnetic atoms in free space. When embedded inside a solid, many things change. Below, we will discuss two effects:
+- Interaction of magnetic atoms with other magnetic atoms $\rightarrow$ _Heisenberg model_
+- Interaction of spins with non-magnetic atoms $\rightarrow$ _Crystal field_
+
+### Heisenberg model
+Consider a one-dimensional chain of spins $\frac{1}{2}$ that are nearest-neighbor coupled with equal coupling strength $J$:
+
+$$
+{\mathcal H}=-\frac{1}{2}J\sum_{\langle i,j \rangle} {\bf S}_i\cdot{\bf S}_j+\sum_i g\mu_{\rm B}{\bf B}\cdot{\bf S}_i=\sum_i \left( -J{\bf S}_i\cdot{\bf S}_{i+1}+ g\mu_{\rm B}{\bf B}\cdot{\bf S}_i \right)
+$$
+
+#### Ferromagnetism
+Let's assume ${\bf B}=0$ for now. When $J>0$, the spins tend to align and form a _ferromagnet_. In this case, $\left|\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\right\rangle$ and $\left|\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\right\rangle$ are both valid eigenstates. The order parameter is the total magnetisation $M=\sum_i S_z^i$.
+
+#### Antiferromagnetism
+When $J<0$, the term favors anti-alignment, leading to _antiferromagnetism_. Antiferromagnets typically have zero magnetization, so we need a different order parameter. We can define the _staggered magnetization_ $\tilde{M} = \sum_i (-1)^i S_z^i$, which will become finite if there is alternating order.
+
+Surprisingly, the so-called _Néel states_ $\left|\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\right\rangle$ and $\left|\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\right\rangle$ are _not_ eigenstates of the Heisenberg Hamiltonian. This can be seen as follows:
+
+$$
+{\mathcal H}=-J\sum_i{\bf S}_i\cdot{\bf S}_{i+1}=-J\sum_i \left( S_x^i S_x^{i+1}+S_y^i S_y^{i+1}+S_z^i S_z^{i+1} \right)
+$$
+
+Using $S_x=\frac{S_{+}+S_{-}}{2}$ and $S_y=\frac{S_{+}-S_{-}}{2i}$, we can rewrite this as:
+
+$$
+{\mathcal H}=-J\sum_i\left(\frac{S_+S_+ + S_+S_- + S_-S_+ + S_-S_-}{4}-\frac{S_+S_+ - S_+S_- - S_-S_+ + S_-S_-}{4}+S_zS_z\right)\\
+=-J\sum_i\left(\frac{S_+S_-}{2}+\frac{S_-S_+}{2}+S_zS_z\right)
+$$
+
+#### Frustration
+Finding the correct behavior of the $J<0$ Heisenberg chain is a challenge, both theoretically and experimentally. As particularly fascinating situation arises if there is _frustration_, where no configuration of the spins can satisfy all bonds (Fig. 20.2). For just three spins the ground state can still be worked out (see Problem 20.2), but nobody knows what happens in large frustrated spin systems.
+
+#### Forms of coupling
+Why do spins sometimes prefer to align and sometimes prefer to anti-align? The most obvious explanation is dipolar interaction. But if we estimate the dipolar interaction between two moments $\mu_{\rm B}$ at a distance of 3 Å, we find:
+
+$$
+\Delta E\sim \mu_B B_{\rm dipolar} \sim \frac{\mu_0\mu_{\rm B}^2}{4\pi r^3} \sim 2\ \mu{\rm eV}
+$$
+
+This corresponds to a thermal energy of only ~20 mK! Therefore, other mechanism must be responsible for magnetic ordering:
+
+- _Exchange interaction_ – The total wavefunction of two electrons needs to be antisymmetric upon exchange:
+
+$$
+\psi({\bf r}_1,{\bf r}_2,{\bf s}_1,{\bf s}_2)=-\psi({\bf r}_2,{\bf r}_1,{\bf s}_2,{\bf s}_1)
+$$
+
+Coulomb interaction favors symmetric spatial wavefunction, resulting in a preferred antisymmetric spin wavefunction $\rightarrow J>0$.
+
+- _Superexchange interaction_ – When magnetic atoms are connected via one non-magnetic mutual neighbor, simultaneous exchange of electrons with the neighbor can favor anti-alignment $\rightarrow J<0$.
+
+![](figures/superexchange.svg)
+
+- _Ruderman-Kittel-Kasuya-Yosida (RKKY) interaction_ - Coupling of magnetic atoms via the itinerant electrons inside a metal can lead to an oscillating behavior of $J$ as a function of separation.
+
+![](figures/RKKY.svg)
+
+### Crystal field
+For a free atom, the orbitals are spherically symmetric (_spherical harmonics_). Inside a crystal, it can happen that, due to the Coulomb interaction with neighboring non-magnetic atoms, the degeneracy between orbitals is broken. As a result, electrons can no longer complete a full circular orbit around an atom, causing the orbital angular momentum to be _quenched_: ${\bf L}\rightarrow 0$.
+
+![](figures/crystalfield.svg)
+
+The remaining spin ${\bf S}$ can favor certain directions for magnetisation (due to spin-orbit coupling). This can be expressed in the Heisenberg Hamiltonian as:
+
+$$
+{\mathcal H}=\sum_i \left( -J{\bf S}_i\cdot{\bf S}_{i+1}+ g\mu_{\rm B}{\bf B}\cdot{\bf S}_i -\kappa(S_i^z)^2 \right)
+$$
+
+If $S>\frac{1}{2}$, e.g. $S=2$, in this case the $S_z = \pm 2$ states will be favored. This is called _crystal field anisotropy_. If the anisotropy becomes very large, it makes sense to reduce the system to just the $S_z = \pm S$ subspace. This is called the _Ising model_:
+
+$$
+{\mathcal H}=\sum_i \left( -J \sigma_i \sigma_{i+1}+ g\mu_{\rm B}B\sigma_i\right)
+$$
+
+where $\sigma_i=\pm S$.
\ No newline at end of file
-- 
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