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appendix.tex
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\appendix
\chapter{Appendices}
\numberwithin{equation}{section}
\input{bilinear_transforms}
\section{Spin one identities}
Simplify $ \v{W} \times \v{B}$:
\begin{eqnarray*}
(\v{W} \times \v{B})_i
&=& \epsilon_{ijk} W_j B_k \\
&=& i(S_k)_{ij} W_j B_k\\
&=& i(\v{S} \cdot \v{B})_{ij} W_j \\
&=& i([\v{S} \cdot \v{B}] \v{W})_i
\end{eqnarray*}
Simplify $\v{D} \times ( \v{D} \times \v{W} ) $
\begin{eqnarray*}
(\v{D} \times [ \v{D} \times \v{W} ] )_i
&=& \epsilon_{ijk} D_j (\v{D} \times \v{W})_k \\
&=& \epsilon_{ijk} \epsilon_{k\ell m} D_j D_\ell W_m \\
&=& -(S_j)_{ki} (S_\ell)_{mk} D_j D_\ell W_m \\
&=& -(\v{S} \cdot \v{D})_{ki} (\v{S} \cdot \v{D})_{mk} W_m \\
&=& -\left( [\v{S} \cdot \v{D}]^2 \right)_{im} W_m \\
&=& -\left( [\v{S} \cdot \v{D}]^2 \v{W} \right)_i \\
\end{eqnarray*}
The spin matrices are represented as:
$${(S_k)}_{ij}=-i \epsilon_{ijk}$$
They have the commutator
$$ [S_i, S_j] = i \epsilon_{ijk} S_k $$
The product of two such spin matrices is given by:
\begin{eqnarray*}
{(S_k S_\ell)}_{ij}
& = & {(S_k)}_{ia} {(S_\ell)}_{aj} \\
& = & -\epsilon_{iak} \epsilon_{aj\ell} \\
& = & (\delta_{k\ell} \delta_{ij} - \delta_{kj} \delta_{\ell i} )
\end{eqnarray*}
This implies that:
\begin{eqnarray*}
(S_i S_j A_j B_i \v{v} )_l
&=& {(S_i)}_{lm} {(S_j)}_{mn} A_j B_i v_n \\
&=& {(S_iS_j)}_{ln} A_j B_i v_n \\
&=& (\delta_{ij} \delta_{ln} - \delta_{in} \delta_{lj} ) A_j B_i v_n \\
&=& (\v{A} \cdot \v{B}) v_l - A_l ( \v{B} \cdot \v{v}) \\
\end{eqnarray*}
Or
$$ \v{A} (\v{B} \cdot \v{v}) = (\v{A} \cdot \v{B} - S_i S_j A_j B_i) \v{v} $$
This can be used this to establish some identities:
\begin{eqnarray*}
E^i \v{D} \cdot \v{\eta}
&=& \left( \left[\v{E} \cdot \v{D} - S_j S_k E_k D_j\right] \v{\eta} \right)^i \\
&=& \left( \left[\v{E} \cdot \v{D} - (S_k S_j - [S_k, S_j] ) E_k D_j \right] \v{\eta} \right)^i \\
&=& \left( \left[\v{E} \cdot \v{D} - (S_k S_j - i \epsilon_{kjl} S_l ) E_k D_j \right] \v{\eta} \right)^i \\
&=& \left( \left[\v{E} \cdot \v{D} - (\v{S} \cdot \v{E})( \v{S} \cdot \v{D}) + i S_l (\v{E} \times \v{D})_l \right] \v{\eta} \right)^i \\
&=& \left( \left[\v{E} \cdot \v{D} - (\v{S} \cdot \v{E})( \v{S} \cdot \v{D}) + i \v{S} \cdot (\v{E} \times \v{D})\right] \v{\eta} \right)^i
\end{eqnarray*}
Since E and W commute:
\begin{eqnarray*}
E^i \v{W} \cdot \v{E}
&=& \left( \left[\v{E}^2 - S_j S_k E_k E_j\right] \v{W} \right)^i \\
&=& \left( \left[\v{E}^2 - (\v{S} \cdot \v{E} )^2 \right] \v{W} \right)^i \\
\end{eqnarray*}
And
\begin{eqnarray*}
D^i \v{D} \cdot \v{\eta}
&=& \left( \left[\v{D}^2 - S_j S_k D_k D_j\right] \v{\eta} \right)^i \\
&=& \left( \left[\v{D}^2 - (S_k S_j + [S_j, S_k]) D_k D_j\right] \v{\eta} \right)^i \\
&=& \left( \left[\v{D}^2 - (\v{S} \cdot \v{D})^2 + i\epsilon_{jkl} S_l D_k D_j )\right] \v{\eta} \right)^i \\
&=& \left( \left[\v{D}^2 - (\v{S} \cdot \v{D})^2 + i\v{S} \cdot (\v{D} \times \v{D}) \right] \v{\eta} \right)^i \\
&=& \left( \left[\v{D}^2 - (\v{S} \cdot \v{D})^2 + e \v{S} \cdot \v{B}) \right] \v{\eta} \right)^i \\
\end{eqnarray*}
Similarly:
\begin{eqnarray*}
D^i \v{E} \cdot \v{W}
&=& \left( \left[\v{D} \cdot \v{E} - S_j S_k D_k E_j\right] \v{\eta} \right)^i \\
&=& \left( \left[\v{D} \cdot \v{E} - (S_k S_j + [S_j, S_k]) D_k E_j \right] \v{\eta} \right)^i \\
&=& \left( \left[\v{D} \cdot \v{E} - (S_k S_j +i\epsilon_{jkl} S_l) D_k E_j \right] \v{\eta} \right)^i \\
&=& \left( \left[\v{D} \cdot \v{E} - (\v{S} \cdot \v{D})(\v{S} \cdot \v{E}) + i \v{S} \cdot (\v{D} \times \v{E})\right] \v{\eta} \right)^i \\
\end{eqnarray*}
\subsection*{Product of $H_{12}H_{21}$}
It is necessary to calculate $\left( \frac {\gv{\pi}^2} {2} - (\v{S} \cdot \gv{\pi})^2 + (g-2)\frac{2}{m} \v{S} \cdot \v{B} \right )^2 $ to first order in magnetic field strength. As a first step of simplification
\[
\left( \frac {\gv{\pi}^2} {2} - (\v{S} \cdot \gv{\pi})^2 + \frac{g-2}{2}\frac{e}{m} \v{S} \cdot \v{B} \right )^2
= \left( \frac {\gv{\pi}^2} {2} - (\v{S} \cdot \gv{\pi})^2 \right )^2
+\frac{g-2}{2}\frac{e}{m} \left \{ \frac{p^2}{2} - (\v{S} \cdot \v{p})^2, \v{S} \cdot \v{B} \right \}
\]
\subsubsection*{First term}
To simplify the first term, consider one element of this matrix operator:
\small
\begin{eqnarray*}
\left\{ \left( \frac {\gv{\pi}^2} {2} - (\v{S} \cdot \gv{\pi})^2 \right )^2 \right \} _{ac}
&=& \left ( \frac {\gv{\pi}^2} {2} - S_i S_j \pi_i \pi_j \right)_{ab}
\left ( \frac {\gv{\pi}^2} {2} - S_l S_m \pi_l \pi_m \right)_{bc}\\
&=& \left ( \frac {\gv{\pi}^2} {2}\delta_{ab} - [S_i S_j]_{ab} \pi_i \pi_j \right)
\left ( \frac {\gv{\pi}^2} {2}\delta_{bc} - [S_l S_m]_{bc} \pi_l \pi_m \right)\\
&=& \left ( \frac {\gv{\pi}^2} {2}\delta_{ab} - [\delta_{ab}\delta_{ij} - \delta_{aj}\delta_{bi}] \pi_i \pi_j \right)
\left ( \frac {\gv{\pi}^2} {2}\delta_{bc} - [\delta_{bc}\delta_{lm} - \delta_{bm}\delta_{cl}] \pi_l \pi_m \right)\\
&=& \left ( -\frac {\gv{\pi}^2} {2}\delta_{ab} + \pi_b \pi_a \right)
\left ( -\frac {\gv{\pi}^2} {2}\delta_{bc} + \pi_c \pi_b \right) \\
&=& \frac{ \gv{\pi}^4 } {4} \delta_{ac}
- \pi_c \pi_a \frac{\gv{\pi}^2}{2}
- \frac{\gv{\pi}^2}{2} \pi_c \pi_a
+ \pi_b \pi_a \pi_c \pi_b \\
\end{eqnarray*}
\normalsize
It's very useful to have the following identity:
\begin{eqnarray*}
e(\v{S} \cdot \v{B})_{ab}
&=& e(S_i)_{ab} B_i \\
&=& -i e \epsilon_{iab} B_i \\
&=& -i e \epsilon_{iab} (\epsilon_{ijk}\partial_j A_k) \\
&=& -i e \epsilon_{iab} \epsilon_{ijk} \frac{1}{2} (\partial_j A_k \partial_k A_j) \\
&=& - \epsilon_{iab} \epsilon_{ijk} \frac{1}{2} [\pi_j, \pi_k] \\
&=& -\epsilon_{iab} \epsilon_{ijk} \pi_j \pi_k \\
&=& -(\delta_{aj} \delta_{bk} - \delta_{ak} \delta_{bj} ) \pi_j \pi_k \\
&=& \pi_b \pi_a - \pi_a \pi_b \\
\end{eqnarray*}
Therefore,
$$ e(\v{S} \cdot \v{B})_{ab} = [\pi_b, \pi_a] $$
Using this, and the fact that $\pi$ commutes with S and B:
\begin{eqnarray*}
\pi_b \pi_a \pi_c \pi_b
&=& \pi_b \pi_c \pi_a \pi_b
- \pi_b (e \v{S} \cdot \v{B})_{ac} \pi_b \\
&=& \pi_b \pi_c \pi_a \pi_b
-\gv{\pi}^2 (e \v{S} \cdot \v{B})_{ac} \\
\end{eqnarray*}
Also,
\begin{eqnarray*}
\pi_b \pi_c \pi_a \pi_b
&=& \pi_b \pi_c \pi_b \pi_a
+ \pi_b \pi_c (e \v{S} \cdot \v{B})_{ba} \\
&=& \pi_b \pi_b \pi_c \pi_a
+ \pi_b \pi_a (e \v{S} \cdot \v{B})_{bc}
+ \pi_b \pi_c (e \v{S} \cdot \v{B})_{ba}
\end{eqnarray*}
So now:
\begin{eqnarray*}
\pi_b \pi_a \pi_c \pi_b
- \pi_c \pi_a \frac{\gv{\pi}^2}{2}
- \frac{\gv{\pi}^2}{4} \pi_c \pi_a
&=& \gv{\pi}^2 \pi_c \pi_a
+ \pi_b \pi_a (e \v{S} \cdot \v{B})_{bc}
+ \pi_b \pi_c (e \v{S} \cdot \v{B})_{ba} \\
&& -\gv{\pi}^2 (e \v{S} \cdot \v{B})_{ac}
- \pi_c \pi_a \frac{\gv{\pi}^2}{2}
- \frac{\gv{\pi}^2}{4} \pi_c \pi_a \\
&=& \frac{1}{2} [\gv{\pi}^2, \pi_c \pi_a]
+ \pi_b \pi_a (e \v{S} \cdot \v{B})_{bc}
\\&& + \pi_b \pi_c (e \v{S} \cdot \v{B})_{ba}
-\gv{\pi}^2 (e \v{S} \cdot \v{B})_{ac}
\end{eqnarray*}
Evaluating the commutator:
\begin{eqnarray*}
[\pi_b, \pi_c \pi_a]
&=& [\pi_b, \pi_c]\pi_a - \pi_c [\pi_a, \pi_b] \\
&=& (e \v{S} \cdot \v{B})_{cb} \pi_a
- \pi_c (e \v{S} \cdot \v{B})_{ba}
\end{eqnarray*}
\begin{eqnarray*}
[ \gv{\pi}^2, \pi_c \pi_a]
&=& [\pi_b \pi_b, \pi_c \pi_a] \\
&=& \pi_b [\pi_b, \pi_c \pi_a] + [\pi_b, \pi_c \pi_a] \pi_b \\
&=& (e \v{S} \cdot \v{B})_{cb} ( \pi_b \pi_a + \pi_a \pi_b)
- (e \v{S} \cdot \v{B})_{ba} (\pi_b \pi_c + \pi_c \pi_b)
\end{eqnarray*}
This gives the result:
\small
\begin{eqnarray*}
\pi_b \pi_a \pi_c \pi_b
- \pi_c \pi_a \frac{\gv{\pi}^2}{2}
- \frac{\gv{\pi}^2}{4} \pi_c \pi_a
&=& \frac{1}{2} \left [ (e \v{S} \cdot \v{B})_{cb} ( \pi_b \pi_a + \pi_a \pi_b)
- (e \v{S} \cdot \v{B})_{ba} (\pi_b \pi_c + \pi_c \pi_b) \right ] \\
&& + \pi_b \pi_a (e \v{S} \cdot \v{B})_{bc}
+ \pi_b \pi_c (e \v{S} \cdot \v{B})_{ba}
-\gv{\pi}^2 (e \v{S} \cdot \v{B})_{ac} \\
&=& \frac{1}{2} [(e \v{S} \cdot \v{B})_{cb} ( \pi_a \pi_b - \pi_b \pi_a)
+ (e \v{S} \cdot \v{B})_{ba} (\pi_b \pi_c - \pi_c \pi_b)]
\\&& -\gv{\pi}^2 (e \v{S} \cdot \v{B})_{ac} \\
&=& \frac{1}{2} [(e \v{S} \cdot \v{B})_{cb} (e \v{S} \cdot \v{B})_{ba}
+ (e \v{S} \cdot \v{B})_{ba} (e \v{S} \cdot \v{B})_{cb}]
-\gv{\pi}^2 (e \v{S} \cdot \v{B})_{ac} \\
&=& (e \v{S} \cdot \v{B})_{ab}(e \v{S} \cdot \v{B})_{bc})
-\gv{\pi}^2 (e \v{S} \cdot \v{B})_{ac} \\
&=& \left[ (e \v{S} \cdot \v{B})^2 \right ]_{ac}
-\gv{\pi}^2 (e \v{S} \cdot \v{B})_{ac} \\
\end{eqnarray*}
\normalsize
Since terms of order $B^2$ can be thrown away, the final result is that, to first order in B:
\begin{eqnarray*}
\left( \frac {\gv{\pi}^2} {2} - (\v{S} \cdot \gv{\pi})^2 \right )^2
&=& \frac{ \gv{\pi}^4 } {4} - \gv{\pi}^2 (e \v{S} \cdot \v{B}) \\
&=& \frac{ \gv{\pi}^4 } {4} - e \v{p}^2 \v{S} \cdot \v{B} \\
\end{eqnarray*}
\subsubsection*{Second term}
To simplify the second term, use the following:
\begin{eqnarray*}
\left \{ \frac{\v{p}^2}{2} - (\v{S} \cdot \v{p})^2, \v{S} \cdot \v{B} \right \}
&=& \v{p}^2 \v{S} \cdot \v{B} - [(\v{S} \cdot \v{p})^2 \v{S} \cdot \v{B} + \v{S} \cdot \v{B} (\v{S} \cdot \v{p})^2 ] \\
&=& \v{p}^2 \v{S} \cdot \v{B} - (S_i S_j S_k + S_k S_j S_i) p_i p_j B_k
\end{eqnarray*}
That triple product of spin matrices can be simplified by using their explicit form:
\begin{eqnarray*}
(S_i S_j S_k )_{ab}
&=& i\epsilon_{aci}\epsilon_{cdj}\epsilon_{dbk} \\
&=& i(\delta_{id} \delta_{aj} - \delta_{ij} \delta_{ad})\epsilon_{dbk} \\
&=& i(\delta_{aj} \epsilon_{ibk} - \delta_{ij} \epsilon_{abk}) \\
(S_i S_j S_k + S_k S_j S_i)_{ab}
&=& i(\delta_aj \epsilon_{ibk} + \delta_{aj} \epsilon_{kbi} -\delta_{ij} \epsilon_{abk} -\delta{kj}\epsilon_{abi}) \\
&=& -i(\delta_{ij} \epsilon_{abk} + \delta_{kj} \epsilon_{abi} \\
&=& \delta_{ij} {(S_k)}_ab + \delta_{kj} {(S_i)}_{ab}
\end{eqnarray*}
Now
\begin{eqnarray*}
\v{p}^2 \v{S} \cdot \v{B} - (S_i S_j S_k + S_k S_j S_i) p_i p_j B_k
&=& \v{p}^2 \v{S} \cdot \v{B} - (\delta_{ij} S_k + \delta_{kj} S_i ) p_i p_j B_k \\
&=& - (\v{S} \cdot \v{p}) (\v{B} \cdot \v{p})
\end{eqnarray*}
So
\beq \label{eq:A:SBanticom}
\left \{ \frac{\v{p}^2}{2} - (\v{S} \cdot \v{p})^2, \v{S} \cdot \v{B} \right \}
=
- (\v{S} \cdot \v{p}) (\v{B} \cdot \v{p})
\eeq
\subsubsection*{Result}
At last, the final result is that, to the order desired,
\beq \label{eq:A:crossterm}
\left( \frac {\gv{\pi}^2} {2} - (\v{S} \cdot \gv{\pi})^2 + (g-2)\frac{e}{m} \v{S} \cdot \v{B} \right )^2
= \frac{ \gv{\pi}^4 } {4} - e \v{p}^2 \v{S} \cdot \v{B} \\
-\frac{g-2}{2}\frac{e}{m} (\v{S} \cdot \v{p}) (\v{B} \cdot \v{p})
\eeq
\section{Spin identity}
The spin matrices are represented as:
$${(S_k)}_{ij}=-i \epsilon_{ijk}$$
The product of two such spin matrices is given by:
\begin{eqnarray*}
{(S_k S_\ell)}_{ij}
& = & {(S_k)}_{ia} {(S_\ell)}_{aj} \\
& = & -\epsilon_{iak} \epsilon_{aj\ell} \\
& = & (\delta_{k\ell} \delta_{ij} - \delta_{kj} \delta_{\ell i} )
\end{eqnarray*}
This can be used to write some pair of vector components $u_i v_j$ as a matrix structure. Contract both sides of the above identity with $v_k$ and $u_\ell$. The result is
\[
[(\v{S} \cdot \v{v} )(\v{S} \cdot \v{u})]_{ij}
= \v{u} \cdot \v{v} \delta_{ij} - u_i v_j
\]
or equivalently
\[
u_i v_j = \v{u} \cdot \v{v} \delta_{ij} - [(\v{S} \cdot \v{v} )(\v{S} \cdot \v{u})]_{ij}
\]
The indices on the second term on the right refer to the indices of the product of spin matrices. This identity will be used frequently in the following form:
\[
(\v{W^\dagger} \cdot \v{v} )( \v{u} \cdot \v{W} )
= \v{W^\dagger} \left[ \v{u} \cdot \v{v} - (\v{S} \cdot \v{u} )(\v{S} \cdot \v{v}) \right ] \v{W}
\]
(Note that the vector dotted into $W^\dagger$ appears dotted with right-most spin matrix.)
% \section{Anticommutator identity}
% \begin{eqnarray*}
% \left \{ \frac{\v{p}^2}{2} - (\v{S} \cdot \v{p})^2, \v{S} \cdot \v{B} \right \}
% &=& \v{p}^2 \v{S} \cdot \v{B} - [(\v{S} \cdot \v{p})^2 \v{S} \cdot \v{B} + \v{S} \cdot \v{B} (\v{S} \cdot \v{p})^2 ] \\
% &=& \v{p}^2 \v{S} \cdot \v{B} - (S_i S_j S_k + S_k S_j S_i) p_i p_j B_k
% \end{eqnarray*}
%
% To simplify that triple product of spin matrices, use their explicit form:
% \begin{eqnarray*}
% (S_i S_j S_k )_{ab}
% &=& i\epsilon_{aci}\epsilon_{cdj}\epsilon_{dbk} \\
% &=& i(\delta_{id} \delta_{aj} - \delta_{ij} \delta_{ad})\epsilon_{dbk} \\
% &=& i(\delta_{aj} \epsilon_{ibk} - \delta_{ij} \epsilon_{abk}) \\
% (S_i S_j S_k + S_k S_j S_i)_{ab}
% &=& i(\delta_aj \epsilon_{ibk} + \delta_{aj} \epsilon_{kbi} -\delta_{ij} \epsilon_{abk} -\delta{kj}\epsilon_{abi}) \\
% &=& -i(\delta_{ij} \epsilon_{abk} + \delta_{kj} \epsilon_{abi}) \\
% &=& \delta_{ij} {(S_k)}_{ab} + \delta_{kj} {(S_i)}_{ab}
% \end{eqnarray*}
% Now
% \begin{eqnarray*}
% \v{p}^2 \v{S} \cdot \v{B} - (S_i S_j S_k + S_k S_j S_i) p_i p_j B_k
% &=& \v{p}^2 \v{S} \cdot \v{B} - (\delta_{ij} S_k + \delta_{kj} S_i ) p_i p_j B_k \\
% &=& - (\v{S} \cdot \v{p}) (\v{B} \cdot \v{p})
% \end{eqnarray*}