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Copyright, 1995-2017, all rights reserved, Mark Jarrell (Dept.of
Physics and Atronomy, Louisiana State University, LA 70803). This
material may not be reproduced for profit, modified or published in any
form (this includes electronic redistribution) without the prior written
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\title{Chapter 8: Magnetism}
\author{Holstein \& Primakoff}
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%\Large
\section{Introduction}
Magnetism is one of the most interesting subjects in condensed matter
physics. Magnetic effects are responsible for heavy fermion behavior,
ferromagnetism, antiferromagnetism, ferrimagnetism and probably high
temperature superconductivity.
Unlike our previous studies, most magnetic systems are not well
described by simple models which ignore intersite correlations. The
reason is simple: magnetism is inherently due to electronic correlations \
of moments on different sites.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{moment.pdf}}
\caption[]{\em{Both moment formulation and the correlation between these
moments ($J$) are due to Coulombic effects}}
\end{figure}
As we will see, systems without these inter-spin correlations (or those
without well defined moments to be correlated) have uninteresting and
unimportant (energetically) magnetic properties.
\subsection{The Relevance of Magnetostatics}
Perhaps the term magnetism is a misnomer, or rather describes only the
external probe ($B$) which we use to study magnetic behavior. Magnetic
effects are mainly due to electronic correlations, those mediated or due
to Coulombic effects, and not due to magnetic correlations between moments
(these are smaller by orders of $v/c$, ie., they are relativistic
corrections). For example, consider the magnetic correlation
between two moments separated by a couple of Angstroms.
\begin{eqnarray}
U = \frac 1{r^3} [\mdm_1 \cdot \mdm_2 &-&3(\mdm_1 \cdot \nn )
(\mdm_2 \cdot \nn)]\\
\noalign{\hbox{then,}}
U_{dipole-dipole} & \approx & \frac{m_1m_2}{r^3} \\
\noalign{\hbox{If we let,}}
m_1 \sim m_2 & \sim & g \mu _B \sim \frac{e\hbar}{m} \nonumber \\
r & \approx & 2 \AA \nonumber \\
\noalign{\hbox{then,}}
U \sim \frac{(g \mu _{B})^2}{r^3} & \sim & \left( \frac{e^2}{ \hbar
c}\right) ^2
\left( \frac{a_0}{r}\right)^3 \frac{e^2}{a_0} \sim 10^{-4} eV
\end{eqnarray}
Or roughly one degree Kelvin! Magnetic correlations due to magnetic
interactions would be destroyed by thermal fluctuations at very low
temperatures.
\subsection{Non-interacting Magnetic Systems}
We will define non-interacting magnetic systems as those for which the
independent moments do not interact with each other, and only interact with
the probing field. For the moment let's consider only the magnetic moments due
to electrons (as we will see, since they can interact with each other, they are
by far the most important moments in the system). They have a moment
\beq
\approx \mu_B({\bf{L}}+2\S)
\eeq
The system energy will change by an amount
\beq
\Delta E\sim g\mu _B B(L+2S)\sim \mu _B B, \qquad \mu_B =
\frac{e\hbar}{2m}\sim 5.8\times10^{-5} \frac{eV}{Te}
\eeq
in an external field. The largest field which can regularly be produced
in a lab is $\sim \ 10Te$ ($100 Te$ or more can be produced at LANL by
blowing things up), thus
\beq
\Delta E ^< \! \! \! \! \: \! _{\sim} \ 10^{-3} eV \quad or \ \sim \
k_B(10^oK)
\eeq
This is a very small energy. Thus magnetic effects are wiped out
by thermal fluctuation for
\beq
k_BT > \mu _BB
\eeq
at about 10 K! Thus experiments which measure the magnetism of
non-interacting systems must be carried out at low temperatures.
These experiments typically measure the susceptibility of
the system with a Faraday balance or a magnetometer (SQUID).
For a collection of isolated moments (spin 1/2), the susceptibility may be
calculated from the moment
\begin{eqnarray}
\uparrow s=\frac 12 \qquad & \left< m\right> \ & = g\mu _{B}\frac 12
\frac{\left( e^{-\beta \frac 12 g\mu _BB} - e^{\beta \frac
12g\mu _BB}\right) }{\left( e^{-\beta \frac 12 g\mu _BB} +
e^{\beta \frac 12g\mu _BB}\right) } \\
& & \simeq g\frac{\mu _B}{2} \tanh (\beta \mu _BB) \approx
\ \mu _B \frac{\mu_BB}{T} \qquad (g \sim \ 2) \nonumber \\
& \chi & = \frac{\partial \left< m\right> }{\partial B} \approx \frac{\mu
_B^2}{k_BT}
\end{eqnarray}
Once again, the energy of the moment-field interaction is roughly
\beq
E \sim \ \frac{\mu_B^2}{T}B^2 \sim \ \frac{10^{-8}\left( \frac{eV}{Te}
\right) ^2}{T\left( \frac{10^{-4}eV}{^{\circ }K}\right) ^2}B^2,
\qquad (k=1)
\eeq
When $E \sim \ T$, thermal fluctuations destroy the orientation of the
moments with the external field. If $B \sim \ 10Te$
\beq
E \sim \ \frac{100}{T} ^{\circ }K
\eeq
or $E \sim \ T$ at $ 10^{\circ }K$!
\begin{figure}[htb]
\centerline{\includegraphics[width=0.35\textwidth,clip=true]{surface.pdf}}
\caption[]{\em{ In a metal, only the electrons near the Fermi
surface, which are not paired into singlets, contribute to the bulk
susceptibility $\chi \ \sim \ \frac{kT}{E_F} \frac{\mu _B^2}{k_BT}
\sim \ \mu_B^2D(E_F)$}}
\label{fig:surface}
\end{figure}
However, we know that systems such as iron exist for which a small field can
induce a relatively large moment at room temperature. This is surprising
since for a metal, or a free electron gas, the susceptibility is much
smaller than the free electron result, since only the spins near the
Fermi surface can participate, $\chi=\frac{\mu _B^2}{E_F}$.
Note that this is {\it even smaller} than the free electron result by a
factor of $\frac{k_BT}{E_F} \ll \ 1$!
\section{Coulombic Correlation Effects}
\subsection{Moment Formation}
Of course real materials are not composed of free isolated electrons.
Nevertheless some insulators act almost as if they are composed of
non-interacting atoms (ions) with moments given by Hunds rules: maximum
$S$ maximum $L$ which leads to large atomic moments. Hunds rules reflect the
atom's attempt to lower its Coulombic energy, see Fig.~\ref{fig:centrifuge}.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{centrifuge.pdf}}
\caption[]{{\em Hunds rules, Maximize $S$ and $L$, both result
from minimizing the Coulomb energy.}}
\label{fig:centrifuge}
\end{figure}
By maximizing the total spin $S$,
the spin part of the wavefunction becomes symmetric under electron
exchange (i.e.\ for two electrons with $s=\frac 12 \uparrow \quad \downarrow $
the maximum value of total spin is $S=1$ with a wavefunction
$ \left| \uparrow \downarrow \right> + \left| \downarrow \uparrow \right> $ ).
Then, since the total wavefunction must be antisymmetric under exchange,
the spatial part must be antisymmetric requiring it to have a node.
The node keeps the electrons apart, minimizing their Coulomb energy.
The second Hunds rule is also due to Coulombic interactions. Maximizing
$L$ tends to keep the electrons apart, much like a centrifuge.
(alternatively, the radial Schroedinger equation obtains an
angular momentum barrier $ L(L+1)/r^2$).
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{tightbind.pdf}}
\caption[]{{\em A simple tight-binding model with a local Coulomb
repulsion $U$. If $U=0$, the rate that electrons hop on and off
any site may be approximated using Fermi's golden rule $\sim \ \pi |B|^2D(E_F) \sim \frac{1}{\Delta t}$. Then by the uncertainty principal
$\Delta E \Delta t \sim \ \hbar$ so each site energy acquires an uncertainty
or width $\Delta E \sim \ \frac{\hbar }{\Delta t} \sim \ \pi |B|^2D(E_F)
\equiv \Gamma $. The sites will form moments (see Fig.~\ref{fig:mforms})
if $\Gamma \gg U,\; |\ep|$}}
\label{fig:tightbind}
\end{figure}
To illustrate how band formation modifies this scenerio, lets
consider a simple tight binding model (See Fig.~\ref{fig:tightbind}).
By Fermi's goldon rule, each level acquires a width (uncertainty in its
energy) $\Gamma=\pi B^2 D(E_F)$ and each level can be in one of the four
states shown in Fig.~\ref{fig:mforms}.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{mforms.pdf}}
\caption[]{\em{A moment forms on an orbital provided that
$\Gamma \gg U,\; |\ep|$. $U \sim \ \frac{e^2}{a}e^{-\frac a{r_{TF}}}$
$r_{TF}$ is small for a metal, large for an insulator}}
\label{fig:mforms}
\end{figure}
Clearly, the states $-\! \! \! -\! \! \! -\! \! \! -\! \! \! \! \! \!
\bigcirc \! \! \! \! \! -\! \! \! -\! \! \! - $ and $-\! \! \! -\! \! \!
\uparrow \! \! \! -\! \! \! \! \! -\! \! \! \downarrow \! \! \! \!-\! \! \!
-$ do not have a moment, and the states $-\! \! \! -\! \! \! \! \! \uparrow \! \! \!
-\! \! \!\! \! -$ and $ -\! \! \!\! \! -\! \! \! \downarrow \! \! \! \! \!-\!
\! \! -$ do. If these states mix equally a moment will not form.
The mixing between the states with moments is only through one of the other
two states ($-\! \! \! -\! \! \! -\! \! \! -\! \! \!
\! \! \! \bigcirc \! \! \! \! \! -\! \! \! -\! \! \! - $ or $-\! \! \! -\!
\! \! \uparrow \! \! \! -\! \! \! \! \! -\! \! \! \downarrow \! \! \! \!-\!
\! \! -$) and may be suppressed, as can the occupancy of the moment-less
states, by increasing the energy of the states without moments.
Ie., a moment will form on each site if $-\epsilon \ \gg \ \Gamma $ and
$\epsilon \ + U \gg \ \Gamma $.
In this limit $U \gg \ B$, the system will act more like a system of free
moments than a free electron gas. Thus, one might expect
\beq
\chi _{insulator} \gg \ \chi _{metal}
\eeq
for noninteracting systems. However, this is not the case. The reason is
that I have only told you half of the story. A real atom, or a system
composed of such atoms, has a diamagnetic response due to the angular
momentum ${\bf{L}}$ of the rotating electrons. This effect is due to
Lenz' Law.
\begin{figure}
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{rotelec.pdf}}
\caption[]{{\em Here $\xi \ = -\frac 1c \frac{\partial \phi }{\partial
t}$, $m = -\mu _BL$ and $\phi \ = B\pi a^2$}}
\label{fig:rotelec}
\end{figure}
So that any introduced magnetic induction will induce an EMF and
hence a current that opposes the electron current which reduces
the moment. $\Rightarrow $ diamagnetism with $\chi \ \propto \ a^2$. In
the free electron limit (see J.M. Ziman, )
\beq
\chi \ = \mu _B^2D(E_F) \left[1 - \frac 13 \left( \frac{m}{m^*}\right)
^2\right]
\eeq
\beq
\Delta E \sim 10^{-8} \left( \frac{eV}{T} \right)
^2a10^{-1}eV^{-1}B^2 \sim 10^{-9}eVB^2(T)
\eeq
For insulators, often with $\frac{m^*}{m} < 1$ the diamagnetism wins;
whereas for metals, generally with $\frac{m^*}{m} \geq 1$, the Pauli
paramagnetism wins.
\subsection{Magnetism and Intersite Correlations }
From both Hund's rules and a simple tight binding picture, we argued that
moment formation in solids results from local Coulomb correlations between
electrons. We also saw that a collection of such isolated moments is
rather boring since all magnetic behavior is washed out by thermal
fluctuations at very low temperatures. Consider once again an isolated
moment of magnitude $m\mu_B$ in an external field.
\begin{eqnarray}
\chi &\approx& \frac{(m\mu_B)^2}{k_BT} \\
E&\approx&\frac{(m\mu_BB)^2}{k_BT}
\end{eqnarray}
For magnetism to be significant at room temperature ($300K$)
we must increase the energy of our system in a field. This may be accomplished
by increasing the effective moment m by correlating adjacent moments. If the
range of this correlation is $\xi $, so that roughly $\frac{4\pi \xi ^3}{3a^3}$
moments are correlated, then let $\frac{\xi }{a} = 3$ so that $\sim \ 10^2$
moments are correlated and ${\rm{m}} \sim \ 10^2$.
\begin{figure}
\centerline{\includegraphics[width=0.35\textwidth,clip=true]{correlating.pdf}}
\caption[]{\em{If the magnetic moments in a small volume
are correlated, then the magnetic susceptibility is strongly enhanced.}}
\label{fig:correlating}
\end{figure}
This increases $E$ by about $10^4$, so that $E \sim \ k_bT$ at $T \sim \
10^3K$. The observed (measured) susceptibility also then increases by
about $10^4$, all by only correlating moments in a range of 3 lattice spacings.
Clearly correlations between adjacent spins can make magnetism in
materials relevant. Such correlations are due to {\it electronic} effects and
are hence usually short ranged due to electronic (Thomas-Fermi) screening. If
we consider two $s = \frac 12$ spins, $\uparrow_1 \ \downarrow _2 $,
then the correlation is usually parameterized by the Heisenberg
exchange Hamiltonian, or
\beq
H = -2J \sigma _1 \cdot \ \sigma _2
\eeq
where $J$ is the exchange splitting between the singlet and triplet energies.
\begin{eqnarray}
& \left\{ \begin{array}{lll}
\left| \uparrow \ \uparrow \right> \\
\left| \uparrow \ \uparrow \right> \ + \left|
\downarrow \ \uparrow \right>
\\
\left| \downarrow \ \downarrow \right>
\end{array} \right\} \ E_t \\
& \left\{ \left| \uparrow \ \downarrow
\right> \ - \left| \downarrow \ \uparrow \right> \right\} E_s
\end{eqnarray}
\beq
E_t - E_s = -J
\eeq
The trick then is to calculate $J$!
\subsubsection{The Exchange Interaction Between Localized Spins}
Imagine that we have two hydrogen atoms A and B which localize
two electrons 1 and 2. As these two electrons approach, their spins
will become correlated.
\begin{figure}
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{interaction.pdf}}
\caption[]{Geometry of two electrons, 1 and 2, bound to two ions
A and B.}
\label{fig:interaction}
\end{figure}
\beq
H = H_1 + H_2 + H_{12}
\eeq
\beq
H_1 = -\frac{\hbar ^2}{2m}\nabla ^2 - \frac{e^2}{r_{1A}} -
\frac{e^2}{r_{1B}}
\eeq
\beq
H_{12} = \frac{e^2}{r_{12}} + \frac{e^2}{R_{AB}}
\eeq
As we did in Chap.~1 to describe binding, we will use the atomic
wave functions to approximate the molecular wavefunction $\psi _{12}$.
\begin{eqnarray}
\psi _{12} & = & \left( \phi _A(1) + \phi _B(1)\right) \left( \phi
_A(2) + \phi _B(2)\right) \otimes \mbox{spin part} \nonumber \\
& = & \lep \phi _A(1)\phi _A(2) + \phi _B(1)\phi _B(2) + \phi _A(1)\phi
_B(2) + \phi _A(2)\phi _B(1)\rip\nonumber\\
&& \otimes \mbox{spin part}
\end{eqnarray}
If $\frac{e^2}{r_{12}}$ is strong (it is) then the first two {\it states with
both electrons on the same ion are suppressed}, especially if the ions
are far apart. Thus we neglect them, and make the Heitler-London
approximation; for example
\beq
\psi _{12} \simeq \ \left( \phi _A(1)\phi _B(2) + \phi _B(1)\phi
_A(2)\right) \otimes \mbox{spin singlet}
\eeq
The spatial wave function is symmetric, and thus appropriate for the
spin singlet state since the total electronic wave function must
be antisymmetric. For the symmetric spin triplet states, the
electronic wave function is
\beq
\psi _{12} = \left( \phi _A(1)\phi _B(2) - \phi _B(1)\phi
_A(2)\right) \otimes \mbox{ spin triplet}
\eeq
or
\beq
\psi _{12} = \phi _A(1)\phi _B(2) \pm \phi _B(1)\phi _A(2)
\otimes \mbox{ spin part}
\eeq
The energy of these states may then be calculated by evaluating $\frac{\left<
\psi _{12}|H|\psi _{12}\right> }{\left< \psi _{12}|\psi _{12}\right> }$.
\beq
E = \frac{\left< \psi _{12}|H|\psi _{12}\right> }{\left< \psi
_{12}|\psi _{12}\right> } = 2E_I + \frac{C \pm \ A}{1 \pm S},
\quad +\mbox{ singlet }, -\mbox{ triplet}
\eeq
where
\beq
E_I = \int d^2r_1 \phi _A^*(1)\left\{ -\frac{\hbar ^2}{2m}\nabla _1^2 -
\frac{e^2}{r_{1A}}\right\}\phi _A(1) < 0
\eeq
the Coulomb integral
\beq
C = e^2\int d^3r_1d^3r_2\left\{ \frac 1{R_{AB}} + \frac 1{r_{12}} -
\frac 1{r_{2A}} - \frac 1{r_{1B}}\right\} \left| \phi
_A(1)\right| ^2\left| \phi _B(2)\right| ^2 < 0
\eeq
the exchange integral
\beq
A = e^2\int d^3r_1d^3r_2\left\{ \frac 1{R_{AB}} + \frac 1{r_{12}} -
\frac 1{r_{2A}} - \frac 1{r_{1B}}\right\} \phi _A^*(1)\phi
_A(2)\phi _B(1)\phi _B^*(2)
\eeq
and finally, the overlap integral is
\beq
S = \int d^3r_1d^3r_2\phi _A^*(1)\phi _A(2)\phi _B(1)\phi _B^*(2)
\;\;\;(0~~ 0 \nonumber\\
& J = & 2\frac{A - SC}{1 - S^2} < 0
\eeqa
where the inequality follows since the last two terms in the $\{\}$
dominate the integral for $A$ and in the Heitler-London approximation
$S\ll 1$. Or, for the effective Hamiltonian.
\beq
H = -2J \, \sigma _1 \cdot \sigma _2, \qquad J < 0
\eeq
Clearly this favors an antiparallel or {\em{antiferromagnetic }} alignment
of the spins (See Fig.~\ref{fig:arrows1}) since then (classically)
$\sigma _1 \cdot \ \sigma _2 < 0$ and $E < 0$, so minimizing the energy.
This type of interaction is clearly appropriate for insulators which may be
approximate as a collection of isolated atoms. Indeed antiferromagnets are
generally insulators for this and other reasons.
\begin{figure}
\centerline{\includegraphics[width=0.6\textwidth,clip=true]{arrows1.pdf}}
\caption[]{}
\label{fig:arrows1}
\end{figure}
\beq
H = -2J \sum _{\left< ij\right> }\sigma _i \cdot \sigma _j, \qquad J
< 0
\eeq
\subsubsection{Exchange Interaction for Delocalized Spins}
Ferromagnetism, where adjacent spins tend to align forming a bulk
magnetic moment, is most often seen in conducting metals such as Fe.
As we will see in this section, the Pauli principle, the Coulomb
interactions, and the itinerancy of free (metallic) electrons favors
a ferromagnetic ($J>0$) exchange interaction.
Consider two like-spin (triplet) free electrons in a volume V (See
Fig.~\ref{fig:2electrons}).
\begin{figure}[htb]
\centerline{\includegraphics[width=0.5\textwidth,clip=true]{2electrons.pdf}}
\caption[]{{\em $\left| \uparrow \ \uparrow \right>$
triplet-symmetric}}
\label{fig:2electrons}
\end{figure}
If we describe the spatial part of their wave function with plane
waves, then
\begin{eqnarray}
\psi_{ij} \ & = & \frac 1{\sqrt{2}V}\left\{ e^{ik_i\cdot r_i}e^{ik_j\cdot
r_j} - e^{ik_i\cdot r_j}e^{ik_j\cdot r_i} \right\}
\nonumber \\
& = & \frac 1{\sqrt{2}V}e^{ik_i\cdot r_i}e^{ik_j \cdot r_j}\left\{ 1 -
e^{i(k_i - k_j)\cdot (r_i - r_j)}\right\}
\qquad
\end{eqnarray}
The probability that the electrons are in volumes $d^3r_i$ and $d^3r_j$ is
\beq
|\psi _{ij}|^2 d^3r_id^3r_j = \frac 1{V^2}\left\{ 1 - \cos \left[ (k_i
- k_j) \cdot (r_i - r_j)\right] \right\} d^3r_id^3r_j
\eeq
As required by the Pauli principle, this probability
vanishes when $r_i = r_j$. This would not be the case for electrons in
the singlet spin state (if the coulomb interaction continues to
be ignored). Thus there is a hole, called the ``exchange hole'', in the
probability density for $r_i \approx r_j$ for triplet spin electrons, but
not singlet spin ones.
Now consider the effects of the electron-ion and the
electron-electron coulomb interactions (See Figure ~\ref{fig:coulomb}).
\begin{figure}[htb]
\centerline{\includegraphics[width=0.3\textwidth,clip=true]{coulomb.pdf}}
\caption[]{{\em Electron a screens the potential seen by electron b,
raising its energy. Anything which keeps pairs of electrons apart,
but costs no energy like the exchange hole for the electronic
triplet, will lower the energy of the system. Thus, triplet
formation is favored thermodynamically.}}
\label{fig:coulomb}
\end{figure}
If one electron comes near an ion, it will screen the potential
of that ion seen by other electrons; thereby raising their
energy. Thus the effect of allowing electrons to approach each
other, is to increase the electron-ion coulomb energy, and of
course the electron-electron Coulomb energy. Thus, anything
which keeps them apart without an energy cost, like the exchange
hole for triplet spin electrons, will reduce their energy. As a
result, like-spin electrons have lower energy and are
thermodynamically favored $\Rightarrow $ Ferromagnetism.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.35\textwidth,clip=true]{FMexchange.eps}}
\caption[]{Geometry to calculate the exchange interaction.}
\label{fig:FMexchange}
\end{figure}
To determine the range of this FM exchange interaction, we must average
the effect over the Fermi sea. If one of the electrons is fixed at the
origin (See Figure \ref{fig:FMexchange}), then the probability that a second
is located a distance $\r$ away, in a volume element $d^3r$ is
\begin{eqnarray}
P_{\uparrow\uparrow}(r)d^3r = n_{\uparrow}d^3r &
\overline{\underbrace {\left( 1 -
\cos \left[ (k_i - k_j) \cdot r\right] \right) }} \\
& \mbox{Fermi sea average} \nonumber
\end{eqnarray}
\beq
n_\uparrow = \frac 12n = \frac 12 \ \frac{\mbox{\# electrons}}{\mbox{volume}}
\eeq
In terms of an electronic charge density, this is
\begin{eqnarray}
\rho _{ex}(r) & = & \frac{en}2 \overline{{\left( 1 - \cos \left[ (k_i -
k_j) \cdot r\right] \right) }} \nonumber \\
& = & \frac{en}2\left\{ 1 - \frac 1{(\frac 43 k_F^3)^2}\int
_o^{k_F}d^3k_id^3k_j \frac 12\left( e^{\imath (k_i - k_j) \cdot
r} + e^{-\imath (k_i - k_j) \cdot r}\right) \right\}
\nonumber\\
& = & \frac{en}2\left\{ 1 - (\frac 43 k_F^3)^{-2}\int _0^{k_F}d^3k_i
e^{\imath k_i \cdot r}\int _0^{k_F}d^3k_j e^{\imath k_j \cdot
r}\right\} \nonumber \\
& = & \frac{en}2\left\{ 1 - 9\frac{(\sin k_Fr - k_Fr\cos
k_Fr)^2}{(k_Fr)^6} \right\}
\end{eqnarray}
Note that both of the exponential terms in the second line are the same,
since we integrate over all $k_i\rightarrow -k_i \ \& \ k_j\rightarrow -k_j$.
Since we have only been considering Pauli-principle effects, the electronic
density of spin down electrons remains unchanged. Thus, the total
charge density around the up spin electron fixed at the origin is
\begin{figure}
\centerline{\includegraphics[width=0.5\textwidth,clip=true]{exchole.pdf}}
\caption[]{{\em Electron density near an electron fixed at
the origin. Coulomb effects would reduce the density for small $r$
further, but would not significantly effect the size of the exchange
hole or the range of the corresponding potential, both $\sim 1/k_F$.}}
\label{fig:exchole}
\end{figure}
\beq
\rho _{eff}(r) = en\left\{ 1 - \frac 92 \frac{ (\sin k_Fr - k_Fr\cos
k_Fr)^2}{(k_Fr)^6} \right\}
\eeq
The size of the exchange hole, and the range of the corresponding
ferromagnetic exchange potential, is $\sim \ \frac 1{k_F} \sim \
a$ which is rather short.
\section{Band Model of Ferromagnetism}
Due to the short range of this potential, its Fourier transform is essentially
flat in $k$. This fact may be used to construct a band theory of FM where the
mean effect of a spin-up electron is to lower the energy of all other band
states of spin up electrons by a small amount, independent of $k$.
\beq
E_{\uparrow }(k) = E(k) - \frac{IN_{\uparrow}}{N}; \qquad I \alt \ 1eV
\eeq
Likewise for spin down
\beq
E_{\downarrow }(k) = E(k) - \frac{IN_{\downarrow}}{N}.
\eeq
Where I, the stoner parameter, quantifies the exchange hole
energy. The relative spin occupation R is related to the bulk
moment
\beq
R = \frac{(N_{\uparrow } - N_{\downarrow })}{N}, \qquad M = \mu
_B\left( \frac NV\right) R
\eeq
Then
\begin{eqnarray}
E_{\sigma}(k) & = E(k) - \frac{I(N_{\uparrow} + N_{\downarrow })}{2N} -
\frac{\sigma IR}{2}, \qquad (\sigma \ = \pm ) \\
& \equiv \tilde{E}(k) - \frac{\sigma IR}{2}.
\end{eqnarray}
If $R$ is finite and real, then we have ferromagnetism.
\beqa
R &=& \frac 1N \sum _k \frac 1{\exp \left\{ (\tilde{E}(k) - IR/2 -
E_F)/k_BT\right\} + 1} \nonumber \\
&&- \frac 1{ \exp \left\{ (\tilde{E}(k) +
IR/2 - E_F)/k_BT \right\} + 1}
\eeqa
For small $R$, we may expand around $\tilde{E}(k) = E_F$.
\beq
f(x-a) - f(x+a) = -2af^{\prime } - \frac 2{3!}a^3f^{\prime \prime
\prime }
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{deriv.eps}}
\caption[]{}
\label{fig:deriv}
\end{figure}
All derivatives will be evaluated at $\tilde{E}(k) = E_F$, so $f^{\prime } < 0$
and $f^{\prime \prime \prime } > 0$. Thus,
\beq
R = \left. -2 \frac{IR}2\frac 1N \sum _k \frac{\partial f}{\partial
\tilde{E}(k)} \right| _{E_F} - \left. \frac 26\left(
\frac{IR}2\right) ^3\frac 1N \sum _k \frac{\partial
^3f}{\partial \tilde{E}^3(k)} \right| _{E_F}
\eeq
This is a quadratic equation in $R$
\beq
-1 - \left. \frac IN\sum _k \frac{\partial f}{\partial E(k)} \right|
_{E_F} = \left. \frac 1{24}I^3R^2\frac 1N\sum _k \frac{\partial
^3f}{\partial E^3(k)} \right| _{E_F}
\eeq
which has a real solution iff
\beq
-1 - \left. \frac IN\sum _k \frac{\partial f}{\partial E(k)} \right|
_{E_F} > 0
\eeq
Or, the derivative of the Fermi function summed over the BZ must
be enough to overcome the -1 and produce a positive result.
Clearly this is most likely to happen at $T=0$, where $\left.\frac {\partial
f}{\partial E(k)}\right|_{E_F} \rightarrow \ -\delta (\tilde{E} - E_F)$
\beq
T=0, \quad -\frac 1N\sum _k \frac{\partial f}{\partial E_k} = \int
d\tilde{E} \frac V{2N} D(\tilde{E}) \delta (\tilde{E} - E_F) =
\frac V{2N} D(E_F) = \tilde{D}(E_F)
\eeq
So, the condition for FM at $T=0$ is $I\tilde{D}(E_F) > 1$. This is known as
the Stoner criterion.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{Stonercrit.pdf}}
\caption[]{}
\label{fig:Scrit}
\end{figure}
I is essentially flat as a function of the atomic number, thus materials
such as Fe, Co, \& Ni with a large $\tilde{D}(E_F)$ are favored to be FM.
\subsection{Enhancement of $\chi$}
Even those systems without a FM ground state have their
susceptibility strongly enhanced by this mechanism. Let us
reconsider the effect of an external field ($gS = 1$) on the band
energies.
\beq
E_{\sigma }(k) = E(k) - \frac{In_{\sigma}}N - \mu _B\sigma B
\eeq
Then
\begin{eqnarray}
R & = -\frac 1N \sum _k \frac{\partial f}{\partial \tilde{E}_k} (IR +
2\mu _BB) \nonumber \\
& = \tilde{D}(E_F)(IR + 2\mu _BB)
\end{eqnarray}
or as $M = \mu _B \frac NV R$, we get
\beq
M = 2\mu _B^2 \frac NV \frac{\tilde{D}(E_F)}{1 - I\tilde{D}(E_F)}B
\eeq
or
\beq
\chi \ = \frac{\partial M}{\partial B} = \frac{\chi _0}{1 -
I\tilde{D}(E_F)}
\eeq
\begin{eqnarray}
\chi _0 & = 2\mu _B^2 \frac NV \tilde{D}(E_F) \\
& = \mu _B^2 D(E_F)
\end{eqnarray}
Thus, when $I\tilde{D}(E_F) \alt \ 1$, the susceptibility can be considerably
enhanced over the non-interacting result $\chi _0$. However, this
approximation usually overestimates $\chi $ since it neglects diamagnetic
contributions, and spin
fluctuations (at $T \neq \ 0$). As we will see, the latter especially are
important for estimating $T_c$.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.5\textwidth,clip=true]{correlating2.pdf}}
\caption[]{\em{Spin fluctuations can reduce the total moment within
the correlated region, and even reduce $\xi$ itself. Both effects lead to
a reduction in the bulk susceptibility $\chi\sim \mbox{moment}^2$}}
\label{fig:correlating2}
\end{figure}
\subsection{Finite T Behavior of a Band Ferromagnet }
In principle, one could start from an ab-initio calculation of
the electronic band structure of $E(k)$ and $I$, such as Ni,
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{bandstrut.pdf}}
\caption[]{\em{In metallic Ni, the d-orbitals are compact and
hybridize weakly due to low overlap with the s-orbitals (due to symmetry)
and with each other (due to low overlap). Thus, moments tend to form
on the d-orbitals and they contribute narrow features in the electronic
density of states. The s-orbitals hybridize strongly and form a broad
metallic band.}}
\label{fig:bandstrut}
\end{figure}
and calculate the temperature dependence of $R$ (and hence the
magnetization) using
\beq
R = \frac{1}{N} \sum _k f(\tilde{E}_k - \frac{IR}2 - \mu _BB_0 - E_F) -
f(\tilde{E}_k + \frac{IR}2 + \mu _BB_0 - E_F)
\eeq
with $f(x) = \frac 1{e^{\beta x} + 1}$.
However, this would be pointless since all of the approximations
made to this point have destroyed the {\it quantitative} validity of
the calculation. However, it still retains a {\it qualitative} use.
For Ni, we can do this by approximating the very narrow d-electron feature
in $D(E)$ as a $\delta$ function and performing the integral. {\it However},
only the d-electrons have a strong exchange splitting $I$ and hence
only they will tend to contribute to the magnetization. {\it Thus our
$\tilde{D}(E)$ should reflect only the d-electron contribution}, we will
accommodate this by setting
\beq
\tilde{D}(E) \approx C\delta (E -E_F), \qquad (C<1)
\eeq
$C$, an unknown constant, will be determined by the $T=0$ behavior. Then
\beq
R = C\left\{ f(-\frac{IR}2 - \mu _BB_0) - f(\frac{IR}2 + \mu
_BB_0)\right\}
\eeq
Let $\frac RC \equiv \tilde{R}$ and $T_c = \frac {IC}{4k_B}$,
{\it then if $B_0 = 0$}
\beq
\tilde{R} = \frac 1{\exp \left( \frac{-2\tilde{R}T_c}{T}\right) + 1} -
\frac 1{\exp \left( \frac{2\tilde{R}T_c}{T}\right) + 1} = \tanh
\frac{\tilde{R}T_c}{T}
\eeq
If $T = 0$, then $\tilde{R} = 1 = \frac RC = \frac 1C \frac {n_{\uparrow } -
n_\downarrow }{N}$. For Ni, the measured ground state magnetization per Ni
atom is $\frac{\mu{eff}}{\mu_B} = 0.54 = \frac{n_{\uparrow } -
n_{\downarrow }}{N}$. Therefore, $C = 0.54 = \frac{\mu_{eff}}{\mu_B}$.
For small $x$, $\tanh x \simeq x - \frac 13x^3$, and for large $x$
\begin{eqnarray}
\tanh x & = & \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x +
e^{-x}} = \frac{1 - e^{-2x}}{1 + e^{-2x}} \nonumber \\
& = & (1 - e^{-2x})(1 - e^{-2x}) \simeq \ 1 - 2e^{-2x}
\end{eqnarray}
Thus,
\begin{eqnarray}
\tilde{R} & = 1 - 2e^{-\frac{2T_c}{T}}, \qquad & \mbox{for } T \ll
T_c \label{eq:TllTc} \\
\tilde{R} & = \sqrt{3}(1 - \frac T{T_C})^{\frac 12}, \qquad &
\mbox{for small $\tilde{R}$ or } T \alt T_c \label{eq:TaltTc}
\end{eqnarray}
\begin{figure}[htb]
\centerline{\includegraphics[width=0.45\textwidth,clip=true]{Magnetization.eps}}
\caption[]{}
\label{fig:Magnetization}
\end{figure}
However, neither of the formulas is verified by experiment. The critical
exponent $\beta \ = \frac 12$ in Eq.~\ref{eq:TaltTc} is found to be reduced to
$\approx \ \frac 13$, and Eq.~\ref{eq:TllTc} loses its exponential form, in
real systems. Using more realistic $\tilde{D}(E)$ or values of $I$ will
not correct these problems. Clearly something fundamental is missing from
this model (spin waves).
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{spinwave.pdf}}
\caption[]{\em{Local spin-flip excitations, left, due to thermal
fluctuations are properly treated by mean-field like theories such
as the one discussed in Secs.~3 and 4. However, non-local spin fluctuations
due to intersite correlations between the spins are neglected in mean-field
theories. These low-energy excitations can fundamentally change the nature
of the transition.}}
\label{fig:spinwave}
\end{figure}
\subsubsection{Effect of B}
If there is an external field $B_0 \neq 0$, then
\beq
\tilde{R} = \tanh \left\{ \frac{\tilde{R}T_c + {\mu _BB_0}/{2k_B}}{T}
\right\}
\eeq
Or for small $R$ and $B_0$, (or rather, large $T \gg T_c$.)
\beq
\tilde{R} = \frac{\mu _B}{2kT}B_0 + \frac{T_c}{T}\tilde{R} \Rightarrow
\ \tilde{R} = \frac{\mu _B}{2k} \frac 1{T - T_c} B_0
\eeq
Thus since $M = \frac{\mu _BN}VR = \frac{C\mu _BN}V\tilde{R} \Rightarrow \ \chi
= \frac{\partial M}{\partial B_0} = \frac{C\mu _B^2}{2kV} \frac{N}{T - T_c}$.
This form for $\chi$
\beq
\chi = \frac{\mbox{Const}}{T - T_c}
\eeq
is called the Curie-Weiss form which is qualitatively satisfied
for $T \gg T_c$; however, the values of Const and $T_c$ predicted by
band structure are inaccurate. Again, this is due to the neglect of
low-energy excitations.
\section{Mean-Field Theory of Magnetism}
\subsection{Ferromagnetism for localized electrons (MFT)}
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{spinfield.pdf}}
\caption[]{\em{Terms in the Heisenberg Hamiltonian
$H = -\sum _{i\delta} J_{i\delta }S_i \cdot S_{i\delta} - g\mu_BB_0\sum _i S_i$
Here $i$ refers to the sites and $\delta $ refers to the neighbors of site
$i$.}}
\label{fig:spinfield}
\end{figure}
Some of the rare earth metals or ionic materials with valence d
or f electrons are both ferromagnetic and have largely localized
electrons for which the band theory of FM is inappropriate (A good example
is CeSi$_{2-x}$, with $x>0.2$). As we have seen, systems with localized
spins are described by the Heisenberg Hamiltonian.
\beq
H = -\sum _{i\delta } J_{i\delta }S_i \cdot S_{i\delta } - g\mu
_BB_0\sum _i S_i
\eeq
In general, this Hamiltonian has no solution, and we must
resort to (further) approximation. In this case, we will
approximate the field (exchange plus external magnetic) felt by
each spin as the average field due to the neighbors of that spin
and the external field. (See Fig. ~\ref{fig:spinfield2}.)
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{spinfield2.pdf}}
\caption[]{{\em The mean or average field felt by a spin $S_i$ at site $i$,
due to both its neighbors and the external magnetic field, is
$\frac 1{g\mu _B}\left< \sum _{\delta} J_{i\delta }S_{i\delta }\right> \ + B_0 = B_{i eff}$.
Where $\left< \sum _{\delta} J_{i\delta }S_{i\delta }\right> $ is the
internal field, due to the neighbors of site $i$.}}
\label{fig:spinfield2}
\end{figure}
Then
\beq
H \approx \ -\sum _i g\mu _BB_{ieff} \cdot \ S_i = -\sum _i S_i \cdot
\left\{ \sum _{\delta} J_{i\delta }\left< S_{i\delta
}\right> \ + g\mu _BB_0\right\}
\eeq
If $J_{i\delta }=J$ is a constant (independent of $i$ and $\delta $)
describing the exchange between the spin at site $i$ and its $\nu$ nearest
neighbors, then
\beq
B_{ieff} = \frac{J \sum _{i\delta } \left< S_{i\delta
}\right> \ + g\mu _BB_0}{g\mu _B} = \frac{J\nu }{g\mu
_B}\left< S\right> \ + B_0
\eeq
\beq
M = g\mu _B\frac NV \left< S\right> ; \qquad \nu = \# nn\,.
\eeq
For a homogeneous, ordered system,
\beq
B_{eff} = \frac V{Ng^2\mu _B^2} \nu JM + B_0 = B_{MF} + B_0
\eeq
and
\beq
H \approx -g\mu _BB_{eff} \cdot \sum _i S_i
\eeq
ie., a system of independent spins in a field $B_{eff}$. The
probability that a particular spin is up, is then
\beq
P_{\uparrow } \propto \ e^{-\beta \left(-g\mu _BB_{eff} \frac
12\right)}
\eeq
and
\beq
P_{\downarrow } \propto \ e^{-\beta \left(+g\mu _BB_{eff} \frac
12\right)}
\eeq
so, on average
\beq
\frac{N_{\downarrow }}{N_{\uparrow }} = e^{-\beta g\mu _BB_{eff}}
\eeq
and, since $N_{\uparrow } + N_{\downarrow } = N$
\beq
M = \frac 12g\mu _B \frac{N_{\uparrow } - N_{\downarrow }}{V} = \frac
12g\mu _B \frac NV\tanh \left( \frac \beta 2g\mu _BB_{eff}\right)
\eeq
Since $\tanh $ is odd and $B_{eff} \propto M$, this will only have
nontrivial solutions if $J > 0$ (if $B_0 = 0$). If we identify
\beq
M_s = \frac NV \frac{g\mu _B}2; \qquad T_c = \frac 14 \nu \frac Jk
\eeq
\beq
\frac M{M_s} = \tanh \left( \frac{T_c}{T}\frac{M}{M_s}\right)
\label{eq:transmoment}
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[width=0.5\textwidth,clip=true]{tanh.eps}}
\caption[]{\em{Equations of the form $a=\tanh(ba)$, i.e.\
Eq.~\ref{eq:transmoment}, have nontrivial solutions ($a\neq 0$) solutions
for all $b>1$.}}
\end{figure}
and again for $T=0, M(T=0) = M_s$, and for $T \alt T_c$
\beq
\frac M{M_s} \simeq \ \sqrt{3}\left( 1 - \frac{T}{T_c}\right) ^{\frac
12}
\eeq
Again, we get the same (wrong) exponent $\beta = \frac 12$.
When is this approximation good? When each spin really
feels an "average" field. Suppose we have an ordered solid, so
that
\beq
B_{MF} = \frac{J\nu }{2g\mu _B} = B_{real}
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[width=0.3\textwidth,clip=true]{spinfield3.pdf}}
\caption[]{\em{The flip of a single spin adjacent to site $i$
makes a significant change in the effective exchange field, felt by
spin $S_i$, if the site has few nearest neighbors.}}
\label{fig:spinfield3}
\end{figure}
Now, consider one spin flip excitation adjacent to site $i$ only,
Fig.~\ref{fig:spinfield3}.
If there are an infinite \# of spins then $B_{MF}$ remains unchanged
but for $\nu \ < \infty $
\beq
B_{real} = \frac{J\nu }{2g\mu _B} \frac{\nu \ -2}{\nu } \neq \ B_{MF}
= \frac{J\nu }{2g\mu _B}\,.
\eeq
Clearly, for this approximation to remain valid, we need $B_{real} = B_{MF}$,
which will only happen if $\frac{\nu \ - 2}{\nu } = 1$ or $\nu \ \gg \ 2$. The
more nearest neighbors to each spin, the better MFT is! (This remains true
even when we consider other lower energy excitations, other than a local
spin flip, such as spin waves).
\subsection{Mean-Field Theory of Antiferromagnets}
Oxides of Fe Co Ni and of course Cu often display antiferromagnetic
coupling between the transition-metal d orbitals. Lets assume we have
such a magnetic system on a bipartite lattice composed of two
inter-penetrating sublattices, like bcc.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{lattice.pdf}}
\caption[]{\em{Antiferromagnetism (the Neel state) on a bcc lattice
is composed of two interpenetrating sc sublattices lattices.}}
\label{fig:bcc_neel}
\end{figure}
We consider the magnetization of each lattice separately: For
example, the central site shown in Fig.~\ref{fig:bcc_neel} feels a mean field
from the $\nu \ = 8$ near-neighbor spins on the ``down'' sublattice.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{lattice2.eps}}
\caption[]{}
\label{fig:bcc_neel_BMF}
\end{figure}
so
\begin{eqnarray}
M^+ & = & \frac 12g\mu _B\frac{N^+}{V}\tanh \left\{ \frac{g\mu
_B}{2kT} \frac{V}{N^-g^2\mu _B^2}\nu JM^-\right\} \\
M^- & = & (+ \leftrightarrow \ -) \ldots
\end{eqnarray}
where $M^+$ is the magnetization of the up sublattice.
These equations have the same form as that for the ferromagnetic case! We can
make a closer analogy by realizing that $N^+ = N^-$ and $M^+ = -M^-$, so that
\begin{eqnarray}
M^+ & = & \frac 12g\mu _B\frac{N^+}{V}\tanh \left\{ -\frac{V\nu
JM^+}{2kTN^+g\mu _B}\right\}, \qquad \mbox{$J<0$} \\
M^- & = & -M^+
\end{eqnarray}
Again, these equations will saturate at
\beq
M_s^+ = -M_s^- = \frac 12g\mu _B\frac{N^+}V
\eeq
so
\beq
\frac{M^+}{M_s^+} = \tanh \left\{ \frac{T_N}{T}
\frac{M^+}{M_s}\right\}
\eeq
where $T_N = -\frac 14 \frac{\nu J}{k_B}$
Now consider the effect of a {\it small external field $B_0$}. This will
yield a small increase or decrease in each sublattice's magnetization $\Delta
M^{\pm }$.
\begin{eqnarray}
M^+ + \Delta M^+ & = & \frac 12g\mu _B \frac{N^+}V \tanh \left\{
\frac{g\mu _B}{2kT}\left[ B_0 + \frac{V\nu J}{N^-g^2\mu
_B^2}\left(M^- + \Delta M^-\right) \right] \right\} \nonumber\\
M^- + \Delta M^- & = & (+ \leftrightarrow \ -) \ldots
\end{eqnarray}
Or, since $\frac d{dx} \tanh x = \frac 1{\cosh ^2x}$, then $\left\{ \Delta M =
\frac{\partial M}{\partial B_{eff}}\Delta B_{eff}\right\} $.
\beq
\Delta M = \Delta M^+ + \Delta M^- = \frac 12g\mu _B \frac{N^+}V
\frac{g\mu _B}{2kT} \frac 1{\cosh ^2x}\left[ B_0 + \frac{V\nu
J}{2N^-g^2\mu _B^2}\Delta M\right]
\label{eq:AF_deltaM}
\eeq
where $x = \frac{T_N}T \frac{M^+}{M_s^+}$. For $T > T_N, M^+ = 0$ and so $x =
0$, and
\beq
\Delta M = 2\frac{g^2\mu _B^2N}{8Vk_BT}\left[ B_0 -
\frac{4k_BT_NV}{Ng^2\mu _B^2}\Delta M\right]
\eeq
\beq
T\Delta M = \frac{g^2\mu _B^2N}{4Vk_B}B_0 - \Delta MT_N
\eeq
\beq
\Delta M = \frac{g^2\mu _B^2N}{4Vk_B(T + T_N)}B_0
\eeq
\beq
\chi \ = \frac{g^2\mu _B^2N}{4Vk_B(T + T_N)}
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{suscept.eps}}
\caption[]{\em{Sketch of $\chi={\rm{Const}}/(T+T_N)$. Unlike
the ferromagnetic case, the bulk susceptibility $\chi$ does not diverge
at the transition. However, as we will see, this equation only applies
for the paramagnetic state ($T>T_N$), and even here, there are important
corrections.}}
\end{figure}
Below the transition, $T < T_N$, the susceptibility displays different
behaviors depending upon the orientation of the applied field. For
$T \ll \ T_N$ and a small $B_0$ {\bf{parallel to the axis}} of the sublattice
magnetization, we can approximate $M^+(T) \approx \ M_s^+$ and $x \approx
\frac{T_N}T$ in Eq.~\ref{eq:AF_deltaM}
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{lattice3.pdf}}
\caption[]{\em{When $T\ll T_N$, a weak field applied parallel to the
sublattice magnetization axis only weakly perturbs the spins.
Here $M^+(T) \approx \ M_s^+$ and $x \approx \frac{T_N}T$}}
\label{fig:Neel_B0par}
\end{figure}
\beq
\chi _{\-} \simeq \ \frac{g^2\mu _B^2N}{4Vk_B} \frac 1{T\cosh ^2\left(
\frac{T_N}T\right) + T_N}
\eeq
\beq
\chi _{\-} \simeq \ \frac{g^2\mu _B^2N}{4Vk_B} e^{-2\frac{T_N}T}
\eeq
Now consider the case where $B_0$ is perpendicular to the magnetic
axis.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{lattice4.pdf}}
\caption[]{\em{When $T\ll T_N$, a weak field $B_0$ applied
perpendicular to the sublattice magnetization, can still cause a rotation
of each spin by an angle proportional to $B_0/B_{MF}$.}}
\label{fig:Neel_B0perp}
\end{figure}
The external field will cause each spin to rotate a small angle $\alpha $. (See
Fig.~\ref{fig:Neel_B0perp})
The energy of each spin in this external field and the mean field $\propto \nu
\frac J{g\mu _B}$ to the first order in $B_0$ is
\beq
E = -\frac 12g\mu _BB_0\sin \propto + \frac 12\nu J\cos \alpha
\eeq
Equilibrium is obtained when $\frac{\partial E}{\partial \alpha } = 0$. Since
$B_0$ is taken as small, $\alpha \ \ll \ 1$.
\beq
E \sim \ -\frac 12 g\mu _BB_0\alpha + \frac 12\nu J\left( 1 - \frac
12\alpha ^2\right)
\eeq
or
\beq
\frac{\partial E}{\partial \alpha} = 0 = \frac 12 g\mu _BB_0 + \frac
12\nu \alpha \ \Rightarrow \alpha \ = -\frac{g\mu _BB_0}{\nu
J}
\eeq
The induced magnetization is then
\beq
\Delta M = \frac 12 \frac{g\mu _BB_0}{V}\alpha \ = -\frac{g^2\mu
_B^2NB_0}{2\nu JV}
\eeq
so
\beq
\chi _{\bot} = \frac{g^2\mu _B^2N}{2\nu \left| J\right| V} =
\mbox{constant}
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{suscept_neel.pdf}}
\caption[]{\em{Below the Neel transition, the lattice responds
very differently to a field applied parallel or perpendicular to the
sublattice magnetization. However, in a powdered sample, or for a
field applied in an arbitrary direction, the susceptibility looks
something like the sketch on the right.}}
\label{fig:suscept_neel}
\end{figure}
Of course, in general, in a powdered sample, the susceptibility will
reflect an average of the two forms, see for example Fig.~\ref{fig:bill_sus}.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{Y123.pdf}}
\caption[]{{\em High temperature superconductor Y123 with 25\% Fe
substituted on Cu, courtesy W. Joiner, data from a SQUID magnetometer.}}
\label{fig:bill_sus}
\end{figure}
\section{Spin Waves}
We have discussed the failings of our mean-field approaches to
magnetism in terms of their inability to account for low-energy processes, such
as the flipping of spins. ($S^{\alpha } \rightarrow \ -S^{\alpha }$) However,
we have yet to discuss the lowest energy spin flip processes which are spin
waves.
We will approach spin-waves two ways. First following Ibach and
Luth we will determine a spin wave in a ferromagnet. Second, we
will argue that they should be quantized and then introduce a
(canonical) transformation to a Boson representation.
Consider a ferromagnetic Heisenberg Model
\beq
H = -J\sum _{i\delta } S_i \cdot \ S_{i+\delta}
\eeq
where $S_i = {\bf \hat{x}}S_i^x + {\bf \hat{y}}S_i^y + {\bf \hat{z}}S_i^z$.
If we define $\left| \al \right> \ =
\begin{pmatrix} 1 \cr 0 \cr \end{pmatrix}
$ (i.e. $\left| \uparrow \right>$),
$\beta \ = \begin{pmatrix} 0 \cr 1 \cr \end{pmatrix}$ (i.e. $\left| \downarrow \right>$)
so that
\beq
S^z \begin{pmatrix} 0 \cr 1 \cr \end{pmatrix} =
-\frac 12 \begin{pmatrix} 0 \cr 1 \cr \end{pmatrix} \cdots
\eeq
and
\beq
S^x = \frac 12 \begin{pmatrix} 0 & -i \cr
i & 0 \cr \end{pmatrix} \qquad
S^y = \frac 12 \begin{pmatrix} 0 & 1 \cr
1 & 0 \cr \end{pmatrix} \qquad
S^z = \frac 12 \begin{pmatrix} 1 & 0 \cr
0 & -1 \cr \end{pmatrix}
\eeq
\beq
\left[ S^{\alpha },S^{\beta }\right] \ = i \epsilon _{\alpha \beta
\gamma}S^\gamma
\eeq
It is often convenient to introduce spin lowering and raising
operators $S^-$ and $S^+$.
\begin{eqnarray}
S^+ = S^x + iS^y & = \begin{pmatrix} 0 & 1 \cr 0 & 0 \cr \end{pmatrix} \qquad & \left[
S^z, S^{\pm }\right] \ = \pm S^{+/-} \\
S^- = S^x - iS^y & = \begin{pmatrix} 0 & 0 \cr 1 & 0 \cr \end{pmatrix} \qquad & \left[
S^2, S^{\pm }\right] \ = 0
\end{eqnarray}
\beq
S^+ \begin{pmatrix} 0 \cr 1 \cr \end{pmatrix} =
\begin{pmatrix} 1 \cr 0 \cr \end{pmatrix}, S^- \begin{pmatrix} 0
\cr 1 \cr \end{pmatrix} = 0 \ldots
\eeq
They allow us to rewrite $H$ as
\beq
H = -J\sum _{i\delta } S_i^zS_{i+\delta }^z + \frac 12\left(
S_i^+S_{i+\delta }^- + S_-S_{i+\delta }^+\right)
\eeq
Since $J > 0$, the ground state is composed of all spins oriented,
for example
\beq
\left| 0\right> = \Pi _i \left| \al \right> _i \qquad \mbox{(i.e. all
up)}
\eeq
This is an eigenstate of H, since
\beq
S_i^+S_{i+\delta }^- \left| 0\right> \ = 0
\eeq
and
\beq
S_i^zS_{i+\delta }^z \left| 0\right> \ = \frac 14\left| 0\right>
\eeq
so
\beq
H\left| 0\right> \ = -\frac 14J\nu N\left| 0\right> \ \equiv E_0\left|
0\right>
\eeq
where $N$ is the number of spins each with $\nu $ nearest neighbors.
Now consider a spin-flip excitation. (See Fig.~\ref{fig:excitation})
\begin{figure}[htb]
\centerline{\includegraphics[width=0.45\textwidth,clip=true]{excitation.pdf}}
\caption[]{\em{A single local spin-flip excitation of a ferromagnetic
system. The resulting state is not an eigenstate of the Heisenberg
Hamiltonian.}}
\label{fig:excitation}
\end{figure}
\beq
\left| \downarrow _j\right> \ = S_j^-\Pi _n\left| \al \right> _n
\eeq
This is {\it not} an eigenstate since the Hamiltonian operator
$S_j^+S_{j+\delta }^-$ will move the flipped spin to an adjacent site, and
hence create another state.
However, if we delocalize this spin-flip excitation,
then we can create a lower energy excitation (due to the non-linear nature
of the inter-spin potential) which is an eigenstate.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{excitation_spread.pdf}}
\caption[]{\em{If we spread out the spin-flip over a wider region
then we can create a lower energy excitation. A spin-wave is the completely
delocalized analog of this with one net spin flip.}}
\label{fig:excitation_spread}
\end{figure}
Consider the state
\beq
\left| {\bf k}\right> \ = \frac 1{\sqrt{N}}\sum _j e^{i\k\cdot \r_j}
\left| \downarrow _j\right>\,.
\eeq
It {\it is} an eigenstate. Consider:
\begin{eqnarray}
H\left| {\bf k}\right> \ = \frac 1{\sqrt{N}}\sum _j e^{i\k\cdot \r_j}
\left\{ -\frac 14\nu J(N - 2)\left| \downarrow _j\right>
\right. \nonumber \\
+ \left. \frac 12\nu J\left| \downarrow _j\right> \ -\frac 12J\sum
_{\delta }\left( \left| \downarrow _{j+\delta }\right> +
\left| \downarrow _{j-\delta }\right> \right) \right\}
\end{eqnarray}
where the sum in the last two terms on the right is over the near-neighbors
$\delta$ to site $j$. The last two terms may be rewritten:
\beq
\frac 1{\sqrt{N}}\sum _j e^{i\k\cdot \r_j}\left| \downarrow _{j+\delta
}\right> \ = 1{\sqrt{N}}\sum _m e^{i\k\cdot(\r_m - \r_{\delta })}\left|
\downarrow _m\right>
\eeq
where
\beq
\r_{j+\delta } = \r_j + \r_{\delta } = \r_m
\eeq
so
\beq
H\left| {\bf k}\right> \ = \left\{ -\frac 14\nu JN + \nu J - \frac
12J\sum _{\delta } \left( e^{i\k \cdot \r_{\delta }} + e^{-i\k
\cdot \r_{\delta }}\right) \right\} \frac 1{\sqrt{N}}\sum _j
e^{i\k \cdot \r_j}\left| \downarrow _j\right>
\eeq
Thus $\left| {\bf k}\right> $ is an eigenstate with an eigenvalue
\beq
E = E_0 + J\nu \left\{ 1 - \frac 1{\nu }\sum _{\delta } \cos \k \cdot
\r_{\delta }\right\}
\eeq
Apparently the energy of the excitation described by $\left| {\bf k}\right> $
vanishes as ${\bf k} \rightarrow \ 0$.
What is $\left| {\bf k}\right> $? First, consider
\begin{eqnarray}
S^z\left| {\bf k}\right> \ = \sum _i S_i^z\left| {\bf k}\right> = \sum
_i S_i^z\frac 1{\sqrt{N}}\sum _j e^{i\k \cdot \ \r_j}\left|
\downarrow _j\right> \nonumber \\
= \frac 1{\sqrt{N}}\sum _j e^{i\k \cdot \r_j}\sum _i S_i^z\left|
\downarrow _j\right> \ = (SN - 1)\left| {\bf k}\right>
\end{eqnarray}
I.e., it is an excitation of the ground state with one spin
flipped. Apparently, since $E_{k=0} = E_0$, the energy to flip a spin in
this way vanishes as $k \rightarrow \ 0$.
\subsection{Second Quantization of Ferromagnetic Spin Waves}
In the ground state all of the spins are up. If we flip a spin,
using a spin-wave excitation, then
\beq
S^z\left| {\bf k}\right> \ = (SN - 1)\left| {\bf k}\right> ,\qquad
S^z\left| 0\right> \ = SN\left| 0\right>
\eeq
If we add another spin wave, then $\left~~~~ = (SN - 2)$. For
spin $\frac 12$, $\left~~~~ = \frac N2 - n$, where $n$ is the
number of the spin waves. Since $S^z$ is quantized, so must be the
number of spin waves in each mode. Thus, we may describe spin waves
using Boson creation and annihilation operators $a^{\dag }$ and $a$.
By specifying the number $n_k$ excitations in each mode $k$, the
corresponding excited spin state can be described by a Boson
state vector $\left| n_1 n_2 \ldots n_N\right> $
We can introduce creation and anhialation operators to
describe the spin excitations {\it on each site}. Suppose, in the ground
state, the spin is saturated in the state $S^z = S$, then $n = 0$. If
$S^z=S-1$, then $n=1$, and so on.
\begin{table}
\centering
\begin{tabular}{|r|c|} \hline
$S^z$ & $n$ \\ \hline
$\frac 32$ & 0 \\
$\frac 12$ & 1 \\
$-\frac 12$ & 2 \\
$-\frac 32$ & 3 \\ \hline
\end{tabular}
\caption[]{\em{The correspondence between $S^z$ and the number of
spin-wave excitations on a site with $S=3/2$.}}
\label{tab:boseexcite}
\end{table}
Apparently
\beqa
S_i^z &=& S - a_i^{\dag }a_i \nonumber \\
S_i^+ & \propto & a_i \nonumber \\
S_i^- & \propto & a_i^{\dag }
\eeqa
If these excitations are Boselike, then
\beq
\left[ a_i,a_i^{\dag }\right] \ = 1
\eeq
\begin{eqnarray}
a_i\left| n\right> \ & = \sqrt{n_i}\left| n - 1\right> \\
a_i^{\dag }\left| n\right> \ & = \sqrt{n_i + 1}\left| n + 1\right>
\end{eqnarray}
This transformation is faithful (canonical) and will maintain the dynamical
properties of the system (given by
$\frac{\partial }{\partial t}\theta = i\hbar \left[ H, \theta \right]$)
if it preserves the commutator algebra
\beq
\left[ S_i^+, S_i^-\right] = 2S_i^z, \;\;
\left[ S_i^-, S_i^z\right] = 2S_i^-, \;\;
\left[ S_i^+, S_i^z\right] = -2S_i^+
\eeq
consider
\beq
\left\{S^+S^- - S^-S^+\right\} \left| n\right> \ = 2S^z \left|
n\right> \ = 2(S - n)\left| n\right>
\eeq
If $S^+ = a$, and $S^- = a^{\dag}$, then the left-hand side of the above
equation would be
$\left\{ (n + 1) - n\right\} \left| n\right> = \left| n\right> \ \neq \ 2(S -
n)\left| n\right>$. In order to maintain the commutators, we need
\beq
S^+ = \sqrt{2S - n}\,a \qquad S^- = a^{\dag }\sqrt{2S - n}
\eeq
Then
\begin{eqnarray}
[S^+,S^-] \left| n\right> \ & =& S^+S^-\left|
n\right> - S^-S^+\left| n\right> \nonumber \\
& =& \sqrt{2S - a^{\dag }a} aa^{\dag } \sqrt{2S - a^{\dag }a}\left|
n\right> - a^{\dag }(2S - a^{\dag }a)a\left| n\right> \nonumber\\
& =& (2S - n)(n + 1)\left| n\right> \ - n(2S - (n - 1))\left| n\right>
\nonumber\\
& =& (2Sn + 2S - n^2 - n - 2Sn + n^2 -n)\left| n\right> \nonumber\\
& =& 2(S - n)\left| n\right>
\end{eqnarray}
You can check that this transformation preserves the other
commutators. Of course, we need one other constraint, since $-S \leq \ S^z
\leq \ S$, we also must have
\beq
n \leq \ 2S
\eeq
This transformation
\beq
S_i^+ = \sqrt{2S - a_i^{\dag }a_i}a_i \qquad S_i^- = a_i^{\dag }\sqrt{2S -
a_i^{\dag }a_i} \qquad S_i^z = S - a_i^{\dag }a_i
\eeq
is called the {\it Holstein Primakoff transformation}.
If we Fourier transform these operators,
\beq
a_i^{\dag } = \frac 1{\sqrt{N}}\sum_k e^{i\k \cdot \R_i}a_k^{\dag };
\qquad a_i = \frac 1{\sqrt{N}}\sum_k e^{-i\k \cdot \R_i}a_k
\eeq
then (since the Fourier transform is {\it unitary}) these new
operators satisfy the same commutation relations
\beq
\left[ a_k, a_{k^{\prime }}^{\dag }\right] \ = \delta _{kk^{\prime }}
\qquad \left[ a_k^{\dag }, a_{k^{\prime }}^{\dag }\right] \ =
\left[ a_k, a_{k^{\prime }}\right] \ = 0
\eeq
To convert the Hamiltonian into this form, assume the number of magnons in each
mode is small and expand
\begin{eqnarray}
S_i^+ & = &\sqrt{2S - n_i}a_i \simeq \ \sqrt{2S}(1 - \frac{n_i}{4S})a_i
\\
& \approx & \left\{ \frac 1{\sqrt{N}}\sum_\k e^{i\k \cdot \R_i}a_k -
\frac 1{4SN^{\frac 32}}\sum_{\k\p\q} e^{i(\p+\q-\k) \cdot
\R_i}a_k^{\dag }a_pa_q\right\}\nonumber
\end{eqnarray}
Of course this is only exact for $n_i \ll \ 2S$, i.e. for low $T$ where
there are few spin excitations, and large $S$ (the classical spin limit).
In this limit
\begin{eqnarray}
S_i^+ & \simeq & \sqrt{\frac{2S}N}\sum_k e^{i\k \cdot \R_i}a_\k \\
S_i^- & \simeq & \sqrt{\frac{2S}N}\sum_k e^{-i\k \cdot \R_i}a_\k^{\dag
} \\
S_i^z & = &S - \frac 1N\sum_{kk^{\prime }} e^{i(\k-\k^{\prime }) \cdot
\R_i}a_k^{\dag }a_{k^{\prime }}\,,
\end{eqnarray}
the Hamiltonian
\beq
H = -J\sum _{i\delta }\left\{ S_i^zS_{i+\delta }^z + \frac
12(S_i^+S_{i+\delta }^- + S_i^-S_{i+\delta }^+)\right\}
\eeq
may be approximated as
\begin{eqnarray}
H \simeq \ -NJ\nu S^2 + 2J\nu S\sum _ka_k^{\dag }a_k \nonumber \\
- 2J\nu S\sum _k\left( \frac 1{\nu }\sum _{\delta }e^{ik \cdot
\ R_{\delta }}\right) a_k^{\dag }a_k + O(a_k^4)
\end{eqnarray}
\beq
H \simeq E_0 + \sum _k 2J\nu S(1 - \gamma_k) a_k^{\dag }a_k + O(a_k^4)
\label{eq:H_Fspinwave}
\eeq
where $\gamma_k=\frac{1}{\nu}\sum_\delta e^{i\k\R_\delta}$.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{interactions.pdf}}
\caption[]{\em{The fourth order correction to Eq.~\ref{eq:H_Fspinwave}
corresponds to interactions between the spin waves, giving them a finite
lifetime}}
\label{fig:interactions}
\end{figure}
This is the Hamiltonian of a collection of harmonic oscillators plus some
other term of order $O(a_k^4)$ which corresponds to interactions between
the spin waves.
These interactions are a result of our definition of a spin-wave
as an itinerant spin flip in an otherwise perfect ferromagnet.
Once we have one magnon, another cannot be created in a ``perfect''
ferromagnetic background.
Clearly if the number of such excitations is small ($T$ small)
and $S$ is large, then our approximation should be valid.
Furthermore, since these are the lowest energy excitations of our
spin system, they should {\it dominate} the low-$T$ thermodynamic
properties of the system such as the specific heat and the
magnetization. Consider
\beq
\left< E\right> \ = \sum _k \frac{\hbar \omega _k}{e^{\beta \hbar
\omega _k}- 1}\,.
\eeq
For small $k$, $\hbar \omega _k = 2J\nu S(1 - \gamma_k) = 2J\nu Sk^2$ on
a cubic lattice. Then let's assume that the $k$-space is isotropic, so that
$d^3k \sim k^2dk$, then
\beq
\gamma_k = \frac 2{\nu }(\cos k_x + \cos k_y + \cdots ) = 1 -
\frac{k^2}{\nu }
\eeq
and
\beq
\left< E\right> \approx \ \sum _k \frac{2J\nu Sk^2}{e^{\beta
2J\nu Sk^2} - 1}
\propto \ \int _0^\infty \frac{k^4 dk}{e^{\beta \al k^2} - 1}
\eeq
\beq
x = \beta \al k^2 \qquad k = \left( \frac x{\beta \al }\right) ^{\frac
12} \qquad dk = \frac 12\left( \frac 1{\beta \al}\right)
^{\frac 12}x^{-\frac 12} dx
\eeq
so that
\begin{equation}
\left< E\right>
\propto \beta^{-2} \beta^{-1/2} \int_0^\infty dx \frac{x^{3/2}}{e^x-1}\propto T^{5/2}
\end{equation}
Thus, the specific heat at constant volume $C_V\propto T^{3/2}$, which is
in agreement with experiment.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{Mag.pdf}}
\caption[]{\em{The magnetization in a ferromagnet versus
temperature. At low temperatures, the spin waves reduce the magnetization
by a factor proportional to $T^{3/2}$, which dominates the reduction
due to local spin fluctuations, derived from our mean-field theory.
This result is also consistent with experiment.}}
\end{figure}
If we increase the temperature from zero, then the change in the magnetization
is proportional to the number of magnons generated
\beq
M(0) - M(T) = \left< \sum _kn_k\right> \frac{g\mu _B}V
\eeq
since each magnon corresponds to spin flip. Thus
\beq
M(T) - M(0) \sim \ -\int \frac{k^2 dk}{e^{\beta \al k^2} - 1} \sim \
T^{\frac 32}
\eeq
which clearly dominates the exponential form found in MFT ($1 -
2e^{\frac{2T_c}{T}}$). This is also consistent with experiment!
\subsection{Antiferromagnetic Spin Waves}
Since the antiferromagnetic ground state is unknown, the spin wave theory
will perturb around the Neel mean-field state in which there are
both a spin up and down sublattices.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{sublattices.pdf}}
\caption[]{\em{To formulate an antiferromagnetic spin-wave
theory, we once again consider a bipartite lattice, which may be decomposed
into interpenetrating spin up and spin down sublattices.}}
\end{figure}
Spin operators can then be written in terms of the Boson
creation and annihilation operators as before
\begin{eqnarray}
& \mbox{``up'' sublattice} \qquad & \mbox{``down'' sublattice} \nonumber
\\
& S_i^z = S - n_i \qquad & S_i^z = -S + n_i \\
& S_i^+ = \left( S_i^-\right) ^+ = \sqrt{2S}f_i(S)a_i \qquad
& S_i^+ = \left( S_i^-\right) ^+ = \sqrt{2S}a_i^{\dag }f_i(S)\nonumber
\end{eqnarray}
where
\beq
f_i(S) = \sqrt{1 - \frac{n_i}{2S}} \quad \mbox{and} \quad n_i =
a_i^{\dag }a_i
\eeq
Again this transformation is exact (canonical) within the
manifold of allowed states
\beq
0 \leq \ n_i \leq \ 2S \Leftrightarrow \ -S \leq \ S_z \leq \ S\,.
\eeq
The Hamiltonian
\beq
H = -J\sum _{i\delta } S_i^zS_{i+\delta }^z = \frac 12\left(
S_i^+S_{i+\delta }^- + S_i^-S_{i+\delta }^+\right)
\eeq
may be rewritten in terms of Boson operators as
\begin{eqnarray}
H &=& +JS^2N\nu + J\sum _{i\delta } a_i^{\dag }a_ia_{i+\delta }^{\dag}
a_{i+\delta } \nonumber \\
&-& JS\sum _{i\delta }\left\{ a_i^{\dag }a_i +
a_{i+\delta }^{\dag}a_{i+\delta } \right. \nonumber \\
&+& \left. f_i(S)a_i f_{i+\delta }(S)a_{i+\delta } + a_i^{\dag}
f_i(S)a_{i+\delta }^{\dag }f_i(S)\right\}
\end{eqnarray}
Once again, we will expand
\beq
f_i(S) = \sqrt{1 - \frac{n_i}{2S}} = 1 - \frac{n_i}{4S} -
\frac{n_i^2}{32S^2} - \cdots
\eeq
and include terms in $H$ only to ${\cal{O}} (a^2)$
\beq
H \simeq \ JS^2N\nu \ - JS\sum _{i\delta }\left\{ a_i^{\dag }a_i +
a_{i+\delta }^{\dag }a_{i+\delta } + a_ia_{i+\delta } +
a_i^{\dag }a_{i+\delta }^{\dag }\right\}
\eeq
This Hamiltonian may be diagonalized using a Fourier transform
\beq
a_i = \frac 1{\sqrt{N}} \sum _k e^{-i\k \cdot \R_i}a_k
\eeq
and the Bogoliubov transform
\begin{eqnarray}
a_k & = \al _k \cosh u_k - \al _{-k}^{\dag }\sinh u_k \\
a_k^{\dag } & = \al _k^{\dag }\cosh u_k - \al _{-k}\sinh u_k
\end{eqnarray}
\beq
\tanh 2u_k = -\gamma_k
\eeq
\beq
\gamma_k = \frac 1{\nu }\sum _{\delta } e^{ik \cdot \ R_{\delta }}
\eeq
Here the $\al _k$ are also Boson operators $\left[ \al _k, \al _k^{\dag
}\right] = 1$. To see if this transform is canonical, we must ensure that the
commutators are preserved.
\begin{eqnarray}
1 = [a_k, a_{k^{\prime }}^{\dag }] & = & [\al _kC_k - \al _{-k}^{\dag
}S_k, \al _{k^{\prime }}^{\dag }C_{k^{\prime }} - \al
_{-k^{\prime }}S_{k^{\prime }}] \nonumber \\
& = & \left\{ C_k^2[\al _k,\al _k^{\dag }] + S_k^2[\al _{-k }^{\dag
},\al _{-k}]\right\} \delta _{kk^{\prime }} \\
& = & \left\{ C_k^2 - S_k^2\right\} \delta _{kk^{\prime }} = \delta
_{kk^{\prime }} \nonumber
\end{eqnarray}
where $C_k$ ($S_k$) is shorthand for $\cosh u_k$ ($\sinh u_k$).
You should check that the other relations, $[a_k, a_{k^{\prime }}] = [a_k^{\dag
}, a_{k^{\prime }}^{\dag }] = 0$, are preserved.
After this transformation,
\beq
H \approx JN\nu S(S + 1) + \sum _k \hbar \omega _k\left( \al _k^{\dag }\al
_k + \frac 12\right) + O(a^4)
\eeq
where $\hbar \omega _k = -2JS\nu \sqrt{1- \gamma_k^2}$.
Notice that for small $k, \hbar \omega _k \sim \sqrt{2}JS\nu k \equiv \ Ck$
($C = -\sqrt{2}JS\nu $ is the spin-wave velocity).
The ground state energy of this system (no magnons), is
\beq
E_0 = JN\nu S(S + 1) - JS\nu \sum _k\sqrt{1 - \gamma_k^2}
\eeq
If $\gamma_k = 0$, then each spin decouples from the fluctuations of
its neighbors and $E_0 = JN\nu S^2$ ($J < 0$) which is the energy of the
Neel state.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.45\textwidth,clip=true]{arrows.pdf}}
\caption[]{\em{The Neel state of an antiferromagnetic lattice.
Due to zero point motion, this is not the ground state of the
Heisenberg Hamiltonian when $J<0$ and $S$ is finite.}}
\end{figure}
However, since $\gamma_k \neq 0$, the ground state energy $E_0 < E_N$. Thus
the ground state is {\it not the Neel state}, and is thus {\it not composed}
of perfectly antiparallel aligned spins. Each sublattice has a small amount
of disorder $\propto \left< n_i\right> +1/2$ in its spin alignment.
The linear dispersion of the antiferromagnet means that its bulk
thermodynamic properties will emulate those of a phonon lattice.
For example
\begin{eqnarray}
\left< E\right> & \simeq \ & \sum _k \frac{\hbar \omega _k}{e^{\beta
\hbar \omega _k} - 1} \simeq \ \sum _k \frac{\al k}{e^{\beta
\al k} - 1} \nonumber \\
& \sim \ & \int _0^{\infty } \frac{\al k^3 dk}{e^{\beta \al k} - 1}
\nonumber \\
\left< E\right> & \sim \ & T^4 \int_0^{\infty } \frac{x^3 dx}{e^x - 1}
\end{eqnarray}
\beq
C = \frac{\partial \left< E \right>}{\partial T} \sim \ T^3 \qquad
\mbox{like phonons!}
\eeq
Which means that a calorimeter experiment cannot distinguish phonon and
magnon excitations of an antiferromagnet.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{scattering.pdf}}
\caption[]{{\em Polarized neutrons are used for two reasons. First
if we look at only spin flip events, then we can discriminate
between phonon and magnon contributions to $S(k,\omega )$.
Second the dispersion may be anisotropic, so excitations
with orthogonal polarizations may disperse differently.}}
\label{fig:scattering}
\end{figure}
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{scattering2.pdf}}
\caption[]{\em{Sketch of neutron structure factor from scattering
off of a magnetic system. The spin-wave peak is centered on the
magnon dispersion. It has a width due to the finite lifetime of
magnon excitations.}}
\label{fig:scattering2}
\end{figure}
Therefore, perhaps the most distinctive experiment one may perform on
an antiferromagnet is inelastic neutron scattering. If spin-polarized
neutrons are scattered from a sample, then only those with flipped
spins have created a magnon. If the neutron creates a phonon, then
its spin remains unchanged. The time of flight of the neutron allows us to
determine the energy loss or gain of the neutron. Thus, if we plot the
differential cross section of neutrons with flipped spins, we learn about
magnon dispersion and lifetime. Notice that the peak in $S(k, \omega )$ has
a width. This is not just due to the instrumental resolution of the
experiment; rather it also reflects the fact that magnons have a finite
life time $\delta t$,which broadens their neutron signature by $\gamma_k$.
\beq
\gamma_k\delta t \sim \ \hbar \qquad \delta t \sim \ \frac 1{\gamma_k}
\eeq
However in the quadratic spin wave approximation the lifetime of
the modes $\hbar \omega _k$ is infinite. It is the neglected terms in $H$, of
order $O(a^4)$ and higher which give the magnons a finite lifetime.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{finite.pdf}}
\caption[]{}
\end{figure}
\section{Criticality and Exponents}
Before reading this section, please read Ken Wilson's [ Scientific American, 241, 158 (Aug, 1979)]
paper on renormalization group (RG). From the this paper, you have learned
a number of new ideas that serve as the basic paradigms of Fisher
scaling which leads to a better understanding of criticality in terms
of critical exponents.
\begin{enumerate}
\item Renormalization Group (RG) operates in the space of all possible Hamiltonian
parameters $\bar{K}=(K_{1},K_{2},K_{3}\cdots)$ which renormalize
such that
\[
\bar{K'}=R(\bar{K)}
\]
where $R$ depends on the system, the RG transform employed (since
RG is not unique) and the renormalization scale $b$.
\item At the fixed point $\bar{K}=\bar{K}^{*}$
\[
\bar{K}^{*}=R(\bar{K}^{*})
\]
\item Transitions are indicated by an unstable fixed point.
\item At or near a transition, the parameters $\bar{K}=(K_{1},K_{2},K_{3}\cdots)$
change very slowly under repeated RG transformations (critical slowing
down).
\end{enumerate}
If $R$ is analytic at $\bar{K}^{*}$, then for small $|\bar{K}-\bar{K}^{*}|$
we may linearlize, so that
\[
K_{\alpha}'-K_{\alpha}^{*}=\sum_{\beta}T_{\alpha\beta}(K_{\beta}-K_{\beta}^{*})
\]
very close to $\bar{K}*$where the T-matrix
\[
T_{\alpha\beta}=\left.\frac{dK'_{\alpha}}{dK_{\beta}}\right|_{\bar{K}=K^{*}}
\]
has real eigenvalues (symmetric, real). Let $\lambda^{(i)}$and $\phi_{\alpha}^{(i)}$be
the left eigenvalues and eigenvectors of T
\[
\sum_{\alpha}\phi_{\alpha}^{(i)}T_{\alpha\beta}=\lambda^{(i)}\phi_{\beta}^{(i)}\,.
\]
We define the scaling variables $u_{i}=\sum_{\alpha}\phi_{\alpha}^{(i)}\left(K_{\alpha}-K_{\alpha}^{*}\right)$
which transform multiplicatively, so that
\begin{eqnarray*}
u'_{i} & = & \sum_{\alpha}\phi_{\alpha}^{i}\left(K'_{\alpha}-K_{\alpha}^{*}\right)\\
& = & \sum_{\alpha}\phi_{\alpha}^{(i)}\sum_{\beta}T_{\alpha\beta}\left(K_{\beta}-K_{\beta}^{*}\right)\\
& = & \sum_{\beta}\lambda^{(i)}\phi_{\beta}^{(i)}\left(K_{\beta}-K_{\beta}^{*}\right)\\
& = & \lambda^{(i)}u_{i}
\end{eqnarray*}
This means that under repeated RG transforms, $u_{i}$ grows or shrinks
depending on the size of $\lambda^{(i)}$. If
\begin{itemize}
\item $\lambda^{(i)}>1$, then the variable $u\_i$ is relevant since repeated
RG transforms will make it grow movig the system away from the fixed point.
\item $\lambda^{(i)}<1$, then it is irrelevant since RG transforms leave
it at the fixed point.
\item $\lambda^{(i)}=1$, then it is marginal.
\end{itemize}
We can use this to describe the scaling behavior. The RG transformation
is constructed to keep the partition function $Z$ the same, and hence
to also keep the free energy the same. So, if $f=F/N$ is the free
energy per site, then
\[
f(K)=f_0(K)+b^{-D}f_{s}(K')
\]
where $f_0$ is the analytic part (no transitions and boring) and $f_{s}$
is the non-analytic part which describes the phase transformation.
So, the singular part scales like $f_{s}(\bar{K})=b^{-D}f_{s}(\bar{K}')$.
Or, in terms of the scaling variables (fields)
\begin{eqnarray*}
f_{s}\left(u_{1},u_{2},\cdots\right) & = & b^{-D}f_{s}\left(\lambda_{1}u_{1},\lambda_{2}u_{2},\right)\\
& = & b^{-nD}f_{s}\left(\lambda_{1}^{n}u_{1},\lambda_{2}^{n}u_{2},\right)
\end{eqnarray*}
where the second line is after $n$ RG steps and includes relevant
variables only so that $\lambda_{1}>1$, $\lambda_{2}>1$, etc.\,
so the arguments grow with $n$. Since $n$ is arbitrary, we may eliminate
it by defining $A$, so that $\lambda_{1}^{n}u_{1}=A\ll1$ (where
the inequality ensures that the linear approximation still holds).
Then $n\ln(\lambda_{1})+\ln(u_{1})=\ln(A)$, which we use to eliminate
$n$, so that
\[
f_{s}(u_{1},u_{2}\cdots)=b^{-D\frac{\ln A}{\ln\lambda_{1}}}b^{D\frac{\ln u_{1}}{\ln\lambda_{1}}}f_{s}\left(A,,\lambda_{2}^{\frac{\ln A}{\ln\lambda_{1}}-\frac{\ln u_{1}}{\ln\lambda_{1}}},\cdots\right)
\]
Now, to make better contact with conventional notation, we define
$y_{1}=\ln\lambda_{1}/\ln b$, $y_{2}=\ln\lambda_{2}/\ln b$, etc.\ so
that $\lambda_{1}=b^{y_{1}}$ etc. These $y$ are called the scaling
exponents. Generally, we define the relevant variables such that $u_{1}=t=(T-T_{c})/T_{c}$,
$u_{2}=h$, or $u_{3}=1/L$, which are the reduced temperature, field
and inverse system size, which are all obviously relevant. The corresponding
set $\{y_{j}\}$ are related to the critical exponents $\{\nu,\gamma,\beta\cdots\}$.
One of the most important applications of these ideas is in a finite-sized
scaling analysis. Suppose we have a simulation (e.g., Monte Carlo
or Molecular Dynamics) of a system of size $L$ and spatial dimension
$D$. Then $1/L$ must be a relevant field $u_{j}$ since if $L$
is finite there can be no transitions, since they require that the
correlation length $\xi$ diverge. Suppose, for the sake of an example,
that the other relevant fields are $t$ and $h$. Under a scale transformation
of size $b$, $f_{s}$ must be scale invariant so that
\[
f_{s}\left(u_{1},u_{2},L^{-1}\right)=b^{-D}f_{s}\left(b^{y_{1}}u_{1},b^{y_{2}}u_{2},b^{1}L^{-1}\right)
\]
where $L^{-1}$ is not only relevant, it must scale with exponent
$y_{L}=1$ as shown. Let's identify, the other relevant variables
as discussed, so that
\[
f_{s}\left(t,h,L^{-1}\right)=b^{-D}f_{s}\left(b^{y_{1}}t,b^{y_{2}}h,bL^{-L}\right)
\]
This form must be true for any $b$, $L^{-1}$, $h$ or $t$ close
to the transition (in the scaling region). So, let's choose $b=L$
to eliminate a parameter, so that
\[
f_{s}\left(t,h,L^{-1}\right)=b^{-D}f_{s}\left(b^{y_{1}}t,b^{y_{2}}h,1\right)
\]
\begin{wrapfigure}{l}{0.4\textwidth}
\includegraphics[width=0.39\textwidth,clip=true]{FSS_chi.pdf}
\label{fig:FSS_chi}
\protect\caption{\emph{Scaled susceptibility versus $t=\left|T-T_c\right|/T_c$.
By scaling the susceptibility with an appropriate power of the cluster size, the
curves for different cluster sizes $L$ may be made to cross at the transition,
identifying $T_c$ and $\gamma/\nu$. A similar finite size scaling ansatz exists
for the specific heat and yields another estimate for $T_c$ and $\nu$
}}
\end{wrapfigure}
Suppose we want to calculate the susceptibility $\chi$ to the
applied field $h$. It involves two derivatives with respect to $h$
at $h=0$)
\[
\chi=L^{-D+2y_{2}}g\left(L^{y_{1}}t,0,1\right)
\]
where $g$ is the second derivative of $f_{s}$ function. Now, some
of these exponents have well known identities that we may as well
use, such as $\xi=\xi_{0}\left|t\right|^{-1/y_{1}}$ so that $\nu=1/y_{1}$,
$\chi\propto t^{-\gamma}$ with the hyperscaling relation $\gamma=(2y_{2}-D)/y_{1}$,
so that
\[
\chi\propto L^{\gamma/\nu}g(L^{1/\nu}t,0,1)
\]
We may apply a similar analysis to the specific heat, which is proportional
to a second derivative of $f_{s}$ with respect to $t$, again for
$h=0$, so that
\[
C\propto L^{2/\nu-D}g'(tL^{1/\nu},0,1)
\]
Now suppose we calculate $\chi$ an $C$ for several different clusters
of size $L$. $L$ must be large enough so that the system is in linear
scaling region of the phase transition. Then we notice that since
$\chi\propto L^{\gamma/\nu}g(L^{1/\nu}t,0,1)$, so that if we plot
$L^{-\gamma/\nu}\chi$ versus $t$, then the curves for different
$L$ must all have the same value and cross at the transition
where $T=T_{c}$ so that $t=0$, and $g(0,0,1)$ is a constant (Fig.~\ref{fig:FSS_chi}). This
allows us to obtain an estimate for $T_{c}$ and $\gamma/\nu$. Likewise,
if I plot $L^{D-2/\nu}C$ versus $t$, then again the curves for different
$L$ must cross at the transition where $g'(0,0,1)$ is constant.
This produces another estimate for $T_{c}$ and for $\nu$, which
when combined with the scaling for $\chi$ gives $\gamma$.
\edo
~~