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\newcommand{\ie}{\emph{i.e.}}
\newcommand{\dkkl}[1]{$\spadesuit${\sl #1}$\spadesuit$}
\newcommand{\jtc}[1]{$\clubsuit${\sl #1}$\clubsuit$}

\newcommand{\bS}{{\bf S}}
\newcommand{\bp}{{\bf p}}
\newcommand{\bpperp}{{\bf p}_\perp}
\newcommand{\ppara}{p_\parallel}
\newcommand{\bm}{{\bf m}}
\newcommand{\br}{{\bf r}}
\newcommand{\bC}{{\bf C}}
\newcommand{\bH}{{\bf H}_{\rm eff}}
\newcommand{\be}{{\bf e}}
\newcommand{\sR}{{\sf R}}
\newcommand{\sP}{{\sf P}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\lap}{\partial^2}
\newcommand{\rhoinv}{\frac{1}{\rho}}
\newcommand{\arho}[1]{\frac{1}{\rho}(\partial_{#1}a)\,}
\newcommand{\di}{\partial_i}
\newcommand{\dj}{\partial_j}
\newcommand{\dk}{\partial_k}
\newcommand{\dl}{\partial_l}
\newcommand{\dx}{\partial_x}
\newcommand{\dy}{\partial_y}
\newcommand{\bnab}{{\bf\nabla}}
\newcommand{\cE}{{\cal E}}
\newcommand{\hex}{H_J}
\newcommand{\hexperp}{H_{J\perp}}
\newcommand{\hfluc}{H_{\rm fluc}}
\newcommand{\Vh}{V_{\rm H}}
\newcommand{\Vb}{V_{\rm bare}}
\newcommand{\sth}{\sin\theta}
\newcommand{\cth}{\cos\theta}
\newcommand{\pa}{p_\alpha}
\newcommand{\pb}{p_\beta}
\newcommand{\bea}{\be_a}
\newcommand{\beb}{\be_b}
\newcommand{\epij}{\epsilon_{ij}}
\newcommand{\epab}{\epsilon_{ab}}
\newcommand{\bA}{{\bf A}}
\newcommand{\cEt}{\tilde{\cal E}}

\newcommand{\Bi}{B_i}
\newcommand{\Bj}{B_j}
\newcommand{\bi}{b_i}
\newcommand{\bj}{b_j}
\newcommand{\bic}{b^*_i}
\newcommand{\bjc}{b^*_j}
\newcommand{\Ai}{A_i}
\newcommand{\Aj}{A_j}
\newcommand{\Di}{D_i}
\newcommand{\Dj}{D_j}
\newcommand{\qi}{q_i}
\newcommand{\qj}{q_j}

\noindent{\bf NOTES ON SPIN WAVES}\bigskip

Consider a small rotation $\sR(\bp)$ of the spins around $\bp$, so
that the spins rotate by $\bS \to \sR\bS$:
%
\begin{equation}
 \sR = e^{\sP}; \qquad [\sP]^{ab} = \epsilon^{abc}p^c
\end{equation}
%
So, $\delta\bS \simeq  -\bp\times\bS +\bp\times(\bp\times\bS)/2$.

This parametrisation is related to Ginzburg (1978) by $\bp=2\varphi$.


The {\bf charge density}, given by:
$\rho = \epij \bS\cdot(\di\bS\times\dj\bS)/2$,
becomes:
\[
\rho + \delta\rho = \half \epij \sR\bS\cdot
\left[\di(\sR\bS)\times\dj(\sR\bS)\right]
\]
so that
\begin{eqnarray}
\delta\rho &=& \half \epij \bS\cdot
        \left[2(\sR^{-1}\di\sR) \bS \times \dj\bS +
         (\sR^{-1}\di\sR)\bS \times (\sR^{-1}\dj\sR)\bS\right]
        \nonumber\\
&=& -\epij \di\bp\cdot\dj\bS +
        \half\epij \di\bp\cdot\dj(\bp\times\bS) + O(p^3)
\end{eqnarray}
using $(\sR^{-1}\di\sR)\bS \simeq (\di\sP - [\sP,\di\sP]/2)\bS = 
\bS\times \di\bp + (\di\bp\times\bp)\times\bS/2$. We can drop
higher-order terms for $\delta\rho$ if $|\bp|\ll 1$ and $|\nabla\bp|
\ll |\nabla\bS|$ (so that $\delta\rho / \rho \ll 1$.)\\


To first order in $p$, the component of $\bp$ parallel to $\bS$ does
not affect the charge density, so we can choose $\bp\cdot\bS=0$. At
this order, $\delta\rho$ can be written as:
\begin{equation}
  \delta\rho = \di\cE_i\;, \qquad \cE_i = \epij(-\bp\cdot\dj\bS+\dj a)
\end{equation}
for some ``gauge field'' $a$. From continuity $\delta\dot\rho +
\di J_i = 0$, the {\bf current density} is:
\begin{equation}
J_i = \epij\dot\bp\cdot\dj\bS \quad\textrm{(+ divergence-free part)}
\label{current}
\end{equation}
Note that we have only identified the \emph{transport current}, and
not any circulating current in the bulk. Therefore, this would not give the
correct $\sigma_{xy}$.\dkkl{Need Faraday's law for motion of CS flux?}\\

For each disorder realisation, the $T=0$
current correlation function is related to the spin dynamics by:
\begin{eqnarray}
\langle J_i(\br,t)J_j(\br',t') \rangle &=& 
  \left[\delta_{ij}\dk S^a(\br)\dk S^b(\br') - 
        \dj S^a(\br)\di S^b(\br')\right]
     \langle \dot{p}^a(\br,t)\dot{p}^b(\br',t')\rangle\nonumber\\
&=&
  S^a(\br) S^b(\br')
  \left[\delta_{ij}\langle \dk\dot{p}^a(\br,t)\dk\dot{p}^b(\br',t')\rangle - 
         \langle\dj\dot{p}^a(\br,t)\di\dot{p}^b(\br',t')\rangle\right]
\end{eqnarray}
The second form is obtained using
$J_i=-\epij\bS\cdot\dj\dot\bp$.  The off-diagonal part is
symmetrical and should average to zero (ignoring UCF). The trace is
given by:
\begin{eqnarray}
\langle J_i(\br,t)J_i(\br',t') \rangle &=& 
  \dk S^a(\br)\dk S^b(\br')\langle \dot{p}^a(\br,t)\dot{p}^b(\br',t')\rangle
\nonumber\\
&=& S^a(\br) S^b(\br')\langle \dk\dot{p}^a(\br,t)\dk\dot{p}^b(\br',t')\rangle 
\end{eqnarray}

\goodbreak
To second order in $p$, we can write $\delta\rho$ in terms of the
components of $\bp$ normal and parallel to $\bS$: 
$\bp = \bpperp + \ppara \bS$. 
Then, $\delta\rho = \delta\rho_\perp + \delta\rho'$ where:
\begin{eqnarray}
\delta\rho_\perp &=& \di\cE_i +
        \half\epij\left\{(\di\bpperp\times\dj\bpperp)\cdot\bS + 
        \di\bpperp\cdot(\bpperp\times\dj\bS)\right\}\\
\delta\rho' &=&
 \half\epij \di(\ppara\bpperp)\cdot(\dj\bS\times\bS)
\end{eqnarray}
Note that the linear-$p$ term does not depend on $\ppara$. Also,
$\delta\rho'$ vanishes if we pick $\bp\cdot\bS=0$. (In fact, it is a
divergence: $\di(\ppara p_{i\perp}\rho)/2$, if we write
$\bpperp=p_{i\perp}\di\bS$.)
\\

\goodbreak
{\bf Exchange energy}
\begin{eqnarray}
\hex &=& \half |\nabla\bS|^2 \to H + \delta \hex =
    \half |\nabla\bS + (\sR^{-1}\nabla\sR)\bS|^2\\
\Rightarrow \delta \hex &=&  (\sR^{-1}\di\sR)\bS \cdot \di\bS + 
        \half |(\sR^{-1}\di\sR)\bS|^2\nonumber\\
&=& \di\bp\cdot (\di\bS \times \bS)\quad +\nonumber\\
&&  \half|\nabla\bp|^2 - (\nabla\bp\cdot\bS)^2  +
        \half(\di\bp\cdot\bS)\di(\bp\cdot\bS) -
        \half(\bp\cdot\bS)(\di\bp\cdot\di\bS)
\end{eqnarray}

This can be separated into a term involving the
component of $\bp$ normal to $\bS$ and other terms. Writing 
$\bp = \bpperp + \ppara \bS$, the terms of $O(p^2)$ are $\delta \hex =
\delta \hexperp + \delta \hex'$ with:
\begin{eqnarray}
 \delta \hexperp &=& \di\bpperp\cdot (\di\bS \times \bS) +
\half|\nabla\bpperp|^2 - |\nabla\bpperp\cdot\bS|^2
\label{hamperp}\\
 \delta \hex' &=& \half \di(\ppara \bpperp)\cdot\di\bS
\end{eqnarray}
The exchange energy does depend on $\ppara$ at order $O(p^2)$.
If we integrate by parts, then $\delta \hex'$ vanishes in the regions where
$\lap\bS$ and $\bS$ are collinear.
\\

\goodbreak
{\bf Potential energy}
\begin{eqnarray}
  \delta H_\rho &=& \Vh(\br) \delta\rho(\br) 
  +\half\int \delta\rho(\br) U(\br-\br')\delta\rho(\br')d^2\br'\nonumber\\ 
\Vh(\br) &=& \Vb(\br) + \int U(\br-\br')\rho(\br')d^2\br'.
\end{eqnarray}
where $\Vh$ is the Hartree potential, $\Vb$ is the bare
potential measured from the chemical potential $\mu$, and $U$ is the
interaction. (The unit of energy is the exchange coupling.)\\

Note that there are two contributions to the energy density involving
$\ppara$ : $\Vh(\br)\delta\rho'(\br)$ and $\hex'(\br)$. These
cancel (after integrating over all space) if $\bS$ is
the ground-state configuration (\ref{localfield}).

\goodbreak\newpage
{\bf Ground state and Fluctuations}\\

In the ground state, $\delta\hex + \delta H_\rho$ should not
have a linear-$p$ term:
\begin{eqnarray}
&&\int \di\bp\cdot\left\{ \di\bS \times \bS 
        - \epij\Vh(\br)\dj\bS \right\}\, d^2\br = 0 \quad\forall\bp
\nonumber\\  
&\Rightarrow& 
  \di\left[ \di\bS \times \bS - \epij\Vh(\br)\dj\bS \right] = 0
\end{eqnarray}
This means that 
\begin{equation}
\epij\dj\bS\times\bS + \Vh\di\bS = \di\bC
\label{gndstate}
\end{equation}
for some $\bC(\br)$. Belavin-Polyakov gives $\bC=\pm\bS$ for $\Vh=0$, but
I cannot deduce this from here. 

It also means that:
\begin{eqnarray}
&&  \lap\bS \times \bS = \epij(\di\Vh)\dj\bS
\label{ground}\\ 
&\Rightarrow& 
  \lap\bS + \epij (\di \Vh) \dj\bS\times\bS
                = - |\nabla\bS|^2 \bS
\label{localfield}
\end{eqnarray}
(I have used
$\di(\bS\cdot\di\bS)=\bS\cdot\lap\bS+\di\bS\cdot\di\bS=0$.)  This is
the condition for a stationary configuration that the spin should be
collinear with a local effective field which has contributions from
the exchange term and the Hartree term. (See eqn (\ref{lifshitz}).)
This condition does not depend on $\mu$, because we are only considering 
variations at fixed global charge.
\\

A Belavin-Polyakov skyrmion of charge $N$ can be written in the form:
\begin{equation}
\tan(\theta/2)=(R/r)^{|N|}\qquad\phi = N\varphi
\end{equation}
where $\theta$ and
$\phi$ are the polar angles in spin space and $\varphi$ is the
azimuthal angle around the origin in real space.
\\

We can use Polyakov's parametrization of the spin configuration:
\begin{eqnarray}
\di \bS &=& \Bi^a \bea\nonumber\\
\di \bea &=& \Ai^{ab}\beb - \Bi^a \bS
\end{eqnarray}
where we choose the local orthonormal basis in spin space to be
($\be_1$,$\be_2$,$\bS$) such that
$\bea\times\be_b=\epab\bS$. Orthonormality requires that
$\Ai^{ab}=-\Ai^{ba}=\epab \Ai$. 
Note that:
\begin{eqnarray}
 \epij\di\dj\bS = 0\quad &\Rightarrow&\quad \epij D_i^{ab} \Bj^{b} = 0 
        \qquad\qquad D_i^{ab}\equiv \delta_{ab}\di - \Ai^{ab}
\label{constraint1}
\\  
 \epij\di\dj\bea = 0\quad &\Rightarrow&\quad \epij (\Di^{bc}\Aj^{ac} + \Bi^a\Bj^b) = 0 
\label{constraint2}
\end{eqnarray}

The topological charge density (quantized in units of $2\pi$)
can be written in terms of $F_{ij} \equiv \di\Aj - \dj\Ai$:
\begin{equation}
   \epij \bS\cdot(\di\bS\times\dj\bS)/4 
        = (B_1^1 B_2^2 - B_1^2 B_2^1)/2 = F_{21}/2
\end{equation}
\goodbreak

It is also convenient to use the complex notation: $\bi = \Bi^1 +
i\Bi^2$, and $\Di \equiv \di +i\Ai$. If we pick $\be_2$ to be normal
to the $z$-direction in spin space, then $\bi = (\be_1 +
i\be_2)\cdot\di\bS = \di \theta + i \sin\theta\,\di\phi$ in polar
coordinates.\\

Then, eqns.\ (\ref{constraint1}-\ref{constraint2}) become:
\begin{equation}
 \epij D_i \bj = 0, \qquad F_{ij} = i(\bic\bj - \bjc\bi)/2
\end{equation}

Equation (\ref{gndstate}) for the ground state configuration now becomes:
\begin{equation}
  -i\Di \bi - \epij(\di\Vh) \bj = 0
\label{gndstate2}
\end{equation}

\goodbreak

Note that the choice of basis corresponds to a gauge transformation. A
rotation of the basis vectors through an angle $\chi$ implies:
\begin{equation}
   A_i \to A_i + \di\chi  \qquad b_i \to b_i e^{-i\chi}
\label{gaugesym}
\end{equation}

The Belavin-Polyakov self-dual solutions have the additional property:
\begin{eqnarray}
\di\bS = \pm \epij \bS\times\dj\bS
\quad\Rightarrow\quad \Bi^a = \mp \epij\epab\Bj^b\\
\Rightarrow\quad
\Bi^1\Bi^2 = 0, \quad\Bi^1\Bi^1 = \Bi^2\Bi^2, \quad\bi\bi = 0, \quad
 F_{21} = \mp\bic\bi/2
\end{eqnarray}
\\


{\bf Equation of motion}\\

The semiclassical equation of motion is given by:
\begin{eqnarray}
 \dot\bS &=& -\bS\times \frac{\delta H}{\delta \bS}
 = \bS \times \bH(\bS)\nonumber\\
{\rm where\ }\quad \bH &\equiv& \lap\bS+\epij (\di\Vh [\bS ])\,\dj\bS\times\bS
 \label{lifshitz}
\end{eqnarray}
is the same as that on the lhs of eqn (\ref{localfield}).  
Alternatively, we can use $\dot\bS = i[H,\bS]$ and the commutators:
\begin{equation}
\left[ S^a(\br),S^b(\br')\right]= 
        i\epsilon^{abc} S^c(\br')\delta(\br-\br')\ ,\qquad
\left[\di S^a(\br),S^b(\br')\right]= 
        i\epsilon^{abc} S^c(\br')\di\delta(\br-\br')
\end{equation}
to obtain the quantum equation of motion.
\\

From the equation of motion, we can see that the divergence of the 
{\bf spin current} is:
$\di {\bf J}_i = \di(\di\bS\times\bS) + \epij \di\Vh \dj\bS$.
\\

Linearizing eqn (\ref{lifshitz}) and substituting for $\lap\bS$ using
(\ref{localfield}) gives:
\begin{eqnarray}
 \dot\bp\times\bS &=& 
        (\bp\times\bS)\times\lap\bS + \bS\times\lap(\bp\times\bS) +
        \epij\di\Vh \dj(\bp\times\bS)-
        \epij\di(\delta\Vh[\bp]) \dj\bS
        \nonumber\\
%  &=&  \lap\bp + (\bp\cdot\lap\bS- \bS\cdot\lap\bp)\bS 
%       - 2(\bS\cdot\di\bp)\,\di\bS + 
%       \epij\di\Vh (\dj\bp\times\bS+\bp\times\dj\bS)\nonumber\\
%  &=& [\lap\bp - (\bS\cdot\lap\bp)\bS] 
%       - 2(\bS\cdot\di\bp)\di\bS + 
%       \epij\di\Vh (\dj\bp\times\bS)\nonumber\\
{\rm or\quad} \dot\bpperp &=& \bS\times\lap\bpperp 
- 2 (\bS\cdot\di\bpperp)\,\bS\times\di\bS 
+ \epij\di\Vh [\dj\bpperp]_{\perp}
- \epij \di(\delta\Vh[\bp])\, \bS\times\dj\bS
\label{eom}
\end{eqnarray}
where $[\dots]_{\perp}$ projects out the components parallel to $\bS$,
and $\delta\Vh[\bp]$ is the change in $\Vh$ due to $\bp$. It is in
general a non-local function of $\bp$: 
\begin{equation}
\delta\Vh[\bp] = \int U(\br - \br')\,\delta\rho(\br')\, d\br'
= (\epij/2) \int U(\br - \br')\,\di\bS(\br') \cdot\dj\bp(\br')\, d\br'
\end{equation}

A global rotation corresponds to a constant $\bp$, but not a constant
$\bpperp$. A constant $\bpperp$ appears to satisfy eqn (\ref{eom})
because the difference is a function of $\ppara$ which does not enter 
into the equation of motion at $O(p)$. However, a constant $\bpperp$ is not
in general possible for a non-collinear spin configuration.
\\
\goodbreak

Let us write $\bpperp=p_a\bea$ and $\psi = p_1 + ip_2$.
Then, the density fluctuations are:
\begin{equation}
\delta\rho = \epij\, {\rm Re} [\bic \,(\Dj\psi) ]
\end{equation}
It can be shown that the equation of motion becomes:
\begin{equation} 
 i \dot\psi = - \Di\Di \psi +i \epij(\di\Vh) \Dj \psi - 
        ( \bic\bi \psi + \bi\bi \psi^* )/2 - \epij \di(\delta\Vh)\bj
\label{bogeqn}
\end{equation}

The choice of spin basis corresponds to a gauge transformation
(\ref{gaugesym}). The wavefunction transforms as $\psi \to \psi
e^{-i\chi}$, and $\Di\psi$ is its covariant derivative. And so,
$\bic\psi$ and $\bic\Dj\psi$ are gauge-invariant objects.\\

There are 3 zero modes, corresponding to the three possible
constant $\bp$:
\begin{equation}
 \Di\psi = - \ppara \bi, \qquad
 (\psi^* \bi + \psi \bic)/2 = \di\ppara, \qquad \ppara = \bp\cdot\bS
\end{equation}

The corresponding Hamiltonian density for the field $\psi$ is:
\begin{equation} 
 H = |(-i\di + {\tilde A}_i)\psi|^2 - \frac{1}{4}|\nabla \Vh|^2 |\psi|^2 - 
        \frac{1}{4} \Psi^+(\br) M_{ii}(\br,\br) \Psi(\br) + (\delta\Vh {\rm\ term})
\label{bogham}
\end{equation}
where $\Psi = (\psi, \psi^*)$, and the ``magnon'' sees a
gauge field $\tilde{A}_i = \Ai + \epij\dj\Vh/2$, corresponding to a
magnetic field of $F_{21} + \nabla^2\Vh/2$. \dkkl{Physical explanation
of extra term?} The matrix $M(\br,\br')$ is:
\begin{equation}
 M_{ij}(\br,\br') = \left(\begin{array}{ll}
        \bi(\br)\bjc(\br')& \bi(\br)\bj(\br') \\
        \bic(\br)\bjc(\br') & \bic(\br)\bj(\br')
        \end{array}\right)
\end{equation}
Note that the original particle
current is given by $J_i = \epij (\dot\psi^* \bj + \dot\psi \bjc)/2$ so that
the current correlation function is:
\begin{equation} 
 \langle J_i(\br,t)J_i(\br',t') \rangle =
        \frac{1}{4}
        \langle\dot\Psi^+(\br,t) M_{ii}(\br,\br') \dot \Psi(\br',t')\rangle
\end{equation} 


{\bf John's Parametrization}\\

Let us write $\delta\bS = q_i \di\bS$. This is valid except along
lines of zero density. Then
\begin{equation}
\bp = \qi\, \di\bS\times\bS\ , \qquad \delta\rho = \di(\rho\qi)\ ,
\qquad J_i = -\rho \dot\qi
\end{equation}
We can separate $\qi$ into a gradient and a curl:
\begin{equation}
\rho\qi = \di\phi + \epij\dj\chi\quad ,\qquad \delta\rho = \partial^2\phi
\end{equation}
Note that the density change affects only curl-free part of $\qi$.
The divergence-free contribution from $\chi$ does not affect the charge.\\
 
In terms of Polyakov's notation, $\bp\cdot\di\bS = \rho\epij\qj = 
(\psi\bic + \psi^*\bi)/2$. Since $\rho = i\epij \bi^*\bj/2$,
we see that: $\psi = i \qi\bi$. Also,
\begin{equation}
  \frac{1}{4} \Psi^+(\br) M_{ij}(\br,\br') \Psi(\br') = 
  \rho(\br) \rho(\br') 
  \left[ q_k(\br) q_k(\br') \delta_{ij} - \qi(\br) \qj(\br') \right]
\end{equation}



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