EquationsOfMotion.tex

\chapter{Spin-1 equations of motion}

\input{ExactEquationsSpin1}

\subsection{Derivation of relativistic Hamiltonian}

We want to derive a "Schrodinger-like" equation governing the nonrelativistic motion of the vector particle; an equation of the form $i\partial_t \Psi = \hat{H} \Psi$.  Since it is a vector particle, $\Psi$ is a spinor with three components.  We will first find such an equation for a bispinor in the relativistic theory, having the form:
\[
i\partial_t 
\begin{pmatrix}
	\Psi_u	\\	\Psi_l
\end{pmatrix} 
=
\hat{H}
\begin{pmatrix}
	\Psi_u	\\	\Psi_l
\end{pmatrix}
\]


If we have a Lagrangian describing the interaction, such an equation can be obtained from the Euler-Lagrange equations.  


We have such a Lagrangian:
\begin{eqnarray*}
\mathcal{L} 
	&=&	-\frac{1}{2} (D \times W)^\dagger \cdot (D \times W) 
				+ m_w^2 W^\dagger W 
				- i e {W^\dagger}^\mu W^\nu F_{\mu \nu}	\\
\end{eqnarray*}

Where
\begin{eqnarray*}
		D^\mu	=	\partial^\mu - i e A^\mu ,
	&&
		D \times W = D^\mu W^\nu - D^\nu W^\mu
\end{eqnarray*}
The last term in the Lagrangian, $-ie{W^\dagger}^\mu W^\nu F_{\mu\nu}$, is needed to produce the "natural" magnetic moment of g=2.  We can introduce an anomalous magnetic moment by changing the coefficient of that term, from 1 to $1+(g-2)$.  For now call this coefficient $\lambda$.  Now
\begin{eqnarray*}
\mathcal{L} 
	&=&	-\frac{1}{2} (D \times W)^\dagger \cdot (D \times W) 
				+ m_w^2 W^\dagger W 
				- \lambda i e  {W^\dagger}^\mu W^\nu F_{\mu \nu}	\\
\end{eqnarray*}



We can find the equations of motion with the Euler-Lagrange method.
\begin{eqnarray*} \pd{ \mathcal{L}}{{W^\dagger}^\alpha} - \frac{d}{dt} \frac{\partial \mathcal{L}} {\partial (\partial_t {W^\dagger}^ \alpha)} 
	&=& 0	\\
\frac{ \partial}{\partial  {W^\dagger}^\alpha} (D \times W)^\dagger_{\mu \nu} (D \times W)^{\mu \nu}
	&=&	\frac{ \partial}{\partial  {W^\dagger}^\alpha}
				(D_\mu W_\nu - D_\nu W_\mu)^\dagger(D^\mu W^\nu - D^\nu W^\mu) \\
	&=&	\frac{ \partial}{\partial  {W^\dagger}^\alpha}
				2(D_\mu W_\nu)^\dagger(D^\mu W^\nu - D^\nu W^\mu) \\
	&=& 	\frac{ \partial}{\partial  {W^\dagger}^\alpha}
				2W^\dagger_\mu D_\nu (D^\mu W^\nu - D^\nu W^\mu) \\
	&=& 	2g_\mu^\alpha D_\nu(D^\mu W^\nu - D^\nu W^\mu) \\
	&=& 	2 D_\nu(D^\alpha W^\nu - D^\nu W^\alpha) \\ 
\frac{ \partial}{\partial  {W^\dagger}^\alpha}  m_W^2 {W^\dagger}^\mu W_\mu
	&=&	m_W^2 W_\alpha	\\
\frac{ \partial}{\partial  {W^\dagger}^\alpha} \lambda i e {W^\dagger}^\mu W^\nu F_{\mu \nu}s
	&=&	\lambda i e g^{\alpha \mu} W^\nu F_{\mu \nu}	\\
	&=&	- \lambda i e W_\nu F^{\nu \mu}
\end {eqnarray*}


To obtain a coupled set of first order equations, we introduce the fields $W^{\mu \nu} = D^\mu W^\nu - D^\nu W^\mu$.
\begin{eqnarray}
	D_\mu W^{\mu \nu} + m_W^2 W^\nu + \lambda i e W_\mu F^{\mu \nu} = 0	\\
	W^{\mu \nu} = D^\mu W^\nu - D^\nu W^\mu	
\end{eqnarray}
$W^{\mu\nu}$ is antisymmetric and so has six degrees of freedom, corresponding to six independent fields.  Together with $W^\mu$ this represents a total of ten fields.  However, upon examination only some of these fields are dynamic.  The  fields $W^{0i}$ and $W^{i}$ appear in the equations with time derivatives, while the fields $W^{ij}$ and $W^0$ never do.  So it is only necessary to consider the former six fields.  So that these six fields all have the same dimension, we will define $\frac{W^{i0}}{m} = i \eta^i$.

We will now eliminate the extraneous fields and solve for $iD_0 W^i$, $iD_0 \eta^i$.

Consider (1) with $\nu=0$:

\begin{equation*}
-D^i W ^{i0} + m^2 W^0 - \lambda i e W^{i} F^{i0} = 0
\end{equation*}
or equivalently, using $F^{i0} = -E^i$:
\begin{eqnarray*}
W^0	&=&	\frac{1}{m} \left [ D^i \frac{W^{i0}}{m} + \lambda ie \frac{\v{W}}{m} \cdot \v{E}  \right ]	\\
	&=&	\frac{i}{m} \left [ \v{D} \cdot \gv{\eta} + \lambda ie \frac{\v{W}}{m} \cdot \v{E}  \right ]
\end{eqnarray*}


Now consider (1) with $\nu=i$:
\begin{equation*}
D^0 W^{0i} - D^j W^{ji} + m^2 W^{i} + \lambda ieW^0 F^{0i} - \lambda ieW^{j} F^{ji} = 0
\end{equation*}
Using the definition of $W^{ij}$, the previously defined $\eta^i$, and that $F^{ij} = \epsilon_{ijk} B_k$
\begin{equation*}
-i D^0 m \eta^i - D^j (D^j W^i - D^i W^j) + m^2 W^i - \lambda ieW^0 E^i  - \lambda ieW^{j} \epsilon_{jik} B_k = 0
\end{equation*}

This gives
\begin{equation} \label{eq:eta1}
iD^0 \eta^i = \frac{1}{m} \left (\v{D} \times (\v{D} \times \v {W}) \right)^i + m W^i + \lambda\frac{e}{m^2} E^i \left( \v{D} \cdot \gv{\eta} + \lambda \frac{e}{m} \v{W} \cdot \v{E} \right) + \lambda \frac{ie}{m} (\v{W} \times \v{B} )^i
\end{equation}

Now, look at (2) with $\nu=0, \mu=i$:

\begin{eqnarray*}
W^{i0} 
	&=& 	D^i W^0 - D^0 W^i		\\
	&=&	 \frac{i}{m} D^i \left [ \v{D} \cdot \gv{\eta} + \lambda ie \frac{\v{W}}{m} \cdot \v{E}  \right ] - D^0 W^i
\end{eqnarray*}

This yields an equation for $iD^0 W^i$:
\begin{equation}\label{eq:W1}
i D^0 W^i = 
	m \eta^i - \frac{1}{m} D^i \left [ \v{D} \cdot \gv{\eta} + \lambda ie \frac{\v{W}}{m} \cdot \v{E}  \right ]
\end{equation}

We have two equations for $i D_0 W^i$ and $i D_0 \eta^i$.  But we need these equations written so that each term should be written as an operator acting on either $\v{W}$ or $\gv{\eta}$. For this purpose, we introduce into the equations the spin matrices $S_i$.  Then, using the identities shown in the appendix, we can rewrite \eqref{eq:eta1} it as:

\begin{equation}
\begin{split}
iD^0 \gv{\eta} =&	-\frac{1}{m} (\v{S} \cdot \v{D})^2 \v{W} 
								+ m\v{W} 
								+ \lambda \frac{e}{m^2} \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 ] \gv{\eta} 	\\
							&	- \lambda \frac{e}{m} \v{S} \cdot \v{B} \v{W}
								+ \lambda^2 \frac{e^2}{m^3} \left [ \v{E}^2 - (\v{S} \cdot \v{E})^2 \right ] \v{W}
\end{split}
\end{equation}

Likewise, write \eqref{eq:W1} as:
\begin{equation}
iD^0 \v{W} =	m\gv{\eta}
		- \frac{1}{m}\left [ \v{D}^2 - (\v{S} \cdot \v{D})^2 + e \v{S} \cdot \v{B} \right ] \v{\eta}
		- \lambda \frac{e}{m^2} \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{W}
\end{equation}

Now, these two equations can be written together in matrix form, from which the Hamiltonian can be easily obtained.
\begin{equation*}
iD_0
\begin{pmatrix}
\eta	\\	W
\end{pmatrix} 
=
\begin{pmatrix}
	\lambda \frac{e}{m^2} \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 ]
&
	m
	-\frac{1}{m} (\v{S} \cdot \v{D})^2 
	- \lambda \frac{e}{m} \v{S} \cdot \v{B}
	+ \lambda^2 \frac{e^2}{m^3} \left [ \v{E}^2 - (\v{S} \cdot \v{E})^2 \right ] 
\\
	m
	- \frac{1}{m}\left [ \v{D}^2 - (\v{S} \cdot \v{D})^2 + e \v{S} \cdot \v{B} \right ] 
&
	- \lambda \frac{e}{m^2} \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 ] 
\end{pmatrix}
\begin{pmatrix}
\eta	\\	W
\end{pmatrix}
\end{equation*}

\subsubsection{Current}

We can also derive the conserved current from the Lagrangian:
\begin{eqnarray*}
\mathcal{L} 
	&=&	-\frac{1}{2} (D \times W)^\dagger \cdot (D \times W) 
				+ m_w^2 W^\dagger W 
				-\lambda i e {W^\dagger}^\mu W^\nu F_{\mu \nu}	\\
\end{eqnarray*}

Where
\begin{eqnarray*}
		D^\mu	=	\partial^\mu - i e A^\mu ,
	&&
		D \times W = D^\mu W^\nu - D^\nu W^\mu
\end{eqnarray*}


We want the conserved current corresponding to the transformation $ W_i \to e^{i \alpha}W_i $, which in infinitesimal form is:
\begin{equation*}
	W_\mu \to W_\mu + i \alpha W_\mu, \;
	W_\mu^\dagger \to W_\mu^\dagger - i \alpha W_\mu^{\dagger}
\end{equation*}

The 4-current density will be:
\begin{equation*}
j^{\sigma} = -i \pd {\mathcal{L} }{W_{\mu, \sigma}} W_\mu  +  i\pd {\mathcal{L} }{W^\dagger_{\mu, \sigma}} W^\dagger_\mu
\end{equation*}

Only one term contains derivatives of the field:

\begin{eqnarray*}
 \pd {\mathcal{L} }{W_{\alpha, \sigma}} 
		&=& \pd{}{W_{\alpha, \sigma}} \left \{ - \frac{1}{2} (D_\mu W_\nu - D_\nu W_\mu)^\dagger(D^\mu W^\nu - D^\nu W^\mu) \right \}\\
		&=& - \frac{1}{2} (D_\mu W_\nu - D_\nu W_\mu)^\dagger (g_{\sigma \mu} g_{\alpha \nu} - g_{\sigma \nu}g_{\alpha \mu})\\
		&=& - (D_\alpha W_\sigma - D_\sigma W_\alpha)^\dagger
\end{eqnarray*}
Likewise:
\begin{eqnarray*}
	\pd {\mathcal{L} }{W_{\alpha, \sigma}^\dagger} 
		&=& - (D_\alpha W_\sigma - D_\sigma W_\alpha)
\end{eqnarray*}

If  we define $W_{\mu \nu} =  D^\mu W^\nu - D^\nu W^\mu$ then the 4-current and charge density are:

\begin{eqnarray*}
	j_\sigma &=& i W_{\sigma \mu}^\dagger W^{\mu} - i W_{\sigma \mu} {W^{\dagger}}^\mu \\
	j_0 	&=& i W_{0 \mu}^\dagger W^{\mu} - i W_{0 \mu} {W^{\dagger}}^\mu \\
		&=& i W_{0 i}^\dagger W^i - i W_{0 i} {W^{\dagger}}^i \\
\end{eqnarray*}
Where the last equality follows from the antisymmetry of $W_{\mu \nu}$.

Now, we defined the fields $\eta_i = -i \frac{W_{i0}}{m}$.  In terms of these fields,
$j_0 =  m (\eta_i^\dagger  W^i + \eta_i {W^\dagger}^i )$. 

We can do the same to find the vector part of the current.
\begin{eqnarray*}
	j_i &=& i W_{i \mu}^\dagger W^{\mu} - i W_{i \mu} {W^{\dagger}}^\mu 	\\
	&=&	i W_j^\dagger W_{ij}  + i W_{i0}^\dagger W_0 + c.c.
\end{eqnarray*}

We have $W_{ij} = D_i W_j - D_j W_i$.  Using the identities developed in the appendix, we can obtain
\[ D_j W_i = D_i W_j - D_k(S_i S_k)_{ja} W_a 	\]
Then
\[ W_{ij} = D_k (S_i S_k)_{ja} W_a 	\]

In the absence of an electric field E, $W_{0} = \frac{i}{m} D_j \eta_j$, with $W_{i0} = i m \eta_i$.
\[	W_{i0}^\dagger W_0 = - \eta_i^\dagger D_j \eta_j 	\]
Again we introduce spin matrices to get the equation in the desired form, and obtain
\[	W_{i0}^\dagger W_0 = - \eta_j^\dagger D_k (\delta_{ik} - S_k S_i) \eta_j 	\]

This leads to
\[ j_i = i W_j^\dagger D_k (S_i S_k W)_j - i \eta_j^\dagger D_k ([\delta_{ik} - S_k S_i]\eta)_j + c.c. \]

Writing this in terms of the bispinor $\begin{pmatrix}\eta \\ W\end{pmatrix}$, the expression for the current is

\begin{equation}	j_i	=
		\frac{i}{2} \begin{pmatrix}\eta^\dagger && W^\dagger \end{pmatrix} \left [
		(\{S_i, S_j\} - \delta_{ij})  
		\begin{pmatrix} 
			1 & 0 \\
			0 & 1 \\ 
		\end{pmatrix}
		- ([S_i, S_j] +\delta_{ij})	\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}
		\right ]
		D_j \begin{pmatrix}\eta \\ W\end{pmatrix} + c.c.
\end{equation}


\subsubsection{Hermiticity of Hamiltonian}


It's clear from inspection that the above Hamiltonian is not Hermitian in the standard sense.  (Of the operators in use, the only one which is not self adjoint is $D_i^\dagger = - D_i$.)  Noticing that 
\[
	\left [ \v{E} \cdot \v{D} - (\v{S} \cdot \v{E})( \v{S} \cdot \v{D}) + i \v{S} \cdot \v{E} \times \v{D} \right]^\dagger 
		= - \left [ \v{D} \cdot \v{E} - (\v{S} \cdot \v{D}) (\v{S} \cdot \v{E}) + i \v{S} \cdot \v{D} \times \v{E} \right ] 
\]
we can see it has the general form 
\[
	H = 
\begin{pmatrix}
	A	&	B	\\
	C	&	A^\dagger
\end{pmatrix}
\]
where the off diagonal blocks are Hermitian in the normal sense: $B^\dagger = B$ and $C^\dagger=C$.

At this point we should consider that an operator is defined as Hermitian with respect to a particular inner product.  In quantum mechanics this inner product is normally defined as:
\[	<\Psi, \phi> = \int d^3x \Psi^\dagger \phi	\]
This definition, however, does not produce sensible results for the states in question.  We need the inner product of a state with itself to be conserved; in other words, it should play the role of a conserved charge.  From the considerations above we already have one such quantity: the conserved charge $\int d^3x j_0$, where 

\[	
	j_0 = m [\eta^\dagger W + W^\dagger \eta] 
		= 	m \begin{pmatrix} \eta^\dagger & W^\dagger \end{pmatrix}
			\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
			\begin{pmatrix} \eta \\ W \end{pmatrix}
\]

A more general definition of the inner product includes some weight M: 
\[	<\Psi, \phi> = \int d^3x \Psi^\dagger M \phi	\]
In normal quantum mechanics M would be the identity matrix, but here, as implied by the charge density, we want $M=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.  Such a definition will lead to the inner product $<\Psi, \Psi>$ being conserved.

An operator H is hermitian with respect to this inner product if
\[ <H\Psi, \phi> = <\Psi, H\phi> \to
		\int d^3x \Psi^\dagger H^\dagger M \phi	=	\int d^3x \Psi^\dagger M H \phi
\]
For this equality to hold, it is sufficient for $H^\dagger M = M H$.  With $M=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ and $H=\begin{pmatrix} A & B \\ C & D \end{pmatrix}$ this condition reduces to 
\[
	\begin{pmatrix} A^\dagger & C^\dagger \\ B^\dagger & D^\dagger \end{pmatrix}
	=\begin{pmatrix} D & C \\ B & A \end{pmatrix}
\]
Our Hamiltonian fulfills exactly this requirement, and so is Hermitian with respect to this particular inner product.




\subsection{Non-relativistic Hamiltonian}
Now we will consider the non-relativistic limit of the above Hamiltonian.  To work in this regime constrains the order of both the momentum and the electromagnetic field strength.
\begin{eqnarray*}
	D	&\sim&	mv	\\
	\Phi	&\sim&	mv^2	\\
	E	&\sim&	m^2v^3	\\
	B	&\sim&	m^2v^2
\end{eqnarray*}

We can write the Hamiltonian matrix in terms of the basis of 2x2 Hermitian matrices: $({\bf I}, \rho_i)$:

\begin{eqnarray*}
H &=&	a_0 {\bf I} + a_i \rho_i \\
a_0  	&=& 
 \frac{1}{2}(H_{11}+H_{22}) =
			e\Phi + 
				\lambda \frac{e}{2m^2} 
				\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}
					- \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 ]\\
a_3 	&=&
 \frac{1}{2}(H_{11}-H_{22}) =
				\lambda s\frac{e}{2m^2} 	\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}
					+ \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 ]\\
ia_2	&=& 
 \frac{1}{2}(H_{21}-H_{12}) =
				-\left [	
					\frac{\v{D}^2}{2m} - \frac{1}{m} (\v{S} \cdot \v{D})^2 
					-\frac{\lambda-1}{2} \frac{e}{m} \v{S} \cdot \v{B}
					+ \frac{e^2}{2m^3}(\v{E}^2 -(\v{S}\cdot\v{E})^2)
				\right ]	\\
a_1		&=&	
 \frac{1}{2}(H_{12}-H_{21}) =
				\left [
					m - \frac{\v{D}^2}{2m} - \frac{1+\lambda}{2}\frac{e}{m}\v{S} \cdot \v{B}
					+ \frac{e^2}{2m^3}(\v{E}^2 -(\v{S}\cdot\v{E})^2)
				\right ]
\end{eqnarray*}

We can see that to leading order, the Hamiltonian is
\begin{equation*}
H = m \rho_1 = 
\begin{pmatrix}
0	&	m \\
m	&	0	\\
\end{pmatrix}
\end{equation*}
Since we wish to separate positive and negative energy states, this poses a problem.  So we first switch to a basis where at least the rest energies are separate.  Then, remaining off-diagonal elements can be treated as perturbations.

An appropriate transformation which meets our requirements is a "rotation" in the space spanned by $\rho_i$ matrices, about the $\rho_1 + \rho_3$ axis.  This has the explicit form:
\begin{equation}
U =\frac{1}{\sqrt {2}}  (\rho_1 + \rho_3)	
	= \frac{1}{\sqrt {2}}
\begin{pmatrix}
1	&	1	\\
1	&	-1	\\
\end{pmatrix}
\end{equation}
Any transformation U can be described by it's action on the basis of 4x4 matrices.   This takes:
\begin{eqnarray*}
{\bf I} & \to &	{\bf I}	\\
\rho_1	& \to &	\rho_3	\\
\rho_3	& \to &	\rho_1	\\
\rho_2	& \to &	-\rho_2	\\
\begin{pmatrix}
	\eta	\\	W
\end{pmatrix}
&	\to	&
\begin{pmatrix}
\Psi_u \\	\Psi_\ell
\end{pmatrix}
=	\frac{1}{\sqrt{2}}
\begin{pmatrix}
\eta	+ W \\	\eta - W
\end{pmatrix}
\end{eqnarray*}

(This transformation will transform the current to
\[	j_0 =  m (\Psi_u^\dagger \Psi_u  - \Psi_\ell^\dagger \Psi_\ell)		\]
and the weight M, used in the inner product, to
\[	M \to M' = U^\dagger M U = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}	\]
The transformed Hamiltonian will of course be Hermitian with respect to the transformed inner product.)

Our equation is now of the following form:
\begin{eqnarray*}
i\partial_0 \begin{pmatrix} \Psi_u \\	\Psi_\ell \end{pmatrix}  
	& =&
	H' \begin{pmatrix} \Psi_u \\	\Psi_\ell  \end{pmatrix}
\end{eqnarray*}

We can see that the Hamiltonian still contains off-diagonal elements, so this represents a pair of coupled equations for the upper and lower components of $\gv{\Psi}$.  But the off-diagonal terms are small, so we can consider the case where $\Psi_\ell$ is small compared to $\Psi_u$. Solving for $\Psi_\ell$ in terms of $\Psi_u$: 

\begin{eqnarray*}
	E \Psi_\ell &=& 	H'_{21} \Psi_u + H'_{22} \Psi_\ell \\
	\Psi_\ell 	&=& 	(E - H'_{22})^{-1} H'_{21} \Psi_u
\end{eqnarray*}

This gives the exact formula:
\begin{eqnarray*}
	E \Psi_u 	&=&		\left( H'_{11} + H'_{12}[E-H'_{22}]^{-1} H'_{21} \right) \Psi_u
\end{eqnarray*}

However, we only need corrections to the magnetic moment of order $v^2$.  With $\frac{e}{m}\v{S} \cdot \v{B} \sim mv^2$, this means we only need the Hamiltonian to at most order $mv^4$.  Examining the leading order terms of the matrix H', the diagonal elements are order m while the off-diagonal elements are order $mv^2$.  To leading order the term $[E-H'_{22}]^{-1}=\frac{1}{2m}$. So we'll need $H'_{11}$ to $\mathcal{O}(v^4)$, and $H'_{12}$, $H'_{21}$, and $[E-H'_{22}]^{-1}$ each to only leading order.


\[	E \Psi_u 	=		\left( H'_{11} + \frac{1}{2m}H'_{12} H'_{21} + \mathcal{O}(mv^6)\right) \Psi_u \]

The needed terms of H are, using $\lambda=g-1$
\begin{eqnarray*}
	H'_{11} 	= a_0+ a_3
			&=& m + e\Phi - \frac{D^2}{2m} - \frac{g}{2}\frac{e}{m} \v{S} \cdot \v{B}	\\
			&& +(g-1)\frac{e}{2m^2} 
				\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}
					- \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 ]
			+\mathcal{O}(mv^6)	\\
	H'_{12} 	= a_1 - ia_2
			&=& \frac{D^2}{2m} - \frac{1}{m}(\v{S} \cdot \v{D})^2 - \frac{g-2}{2}\frac{e}{m} \v{S} \cdot \v{B}
			+\mathcal{O}(mv^4)	\\
	H'_{21}  = a_1 + ia_2
			&=&  -\frac{D^2}{2m} + \frac{1}{m}(\v{S} \cdot \v{D})^2 + \frac{g-2}{2}\frac{e}{m} \v{S} \cdot \v{B}
			+\mathcal{O}(mv^4)
\end{eqnarray*}

The product $H'_{12}H'_{21}$ is calculated in the appendix.  To first order in the magnetic field strength it is:
\begin{eqnarray*}
\frac{1}{2m}H'_{12}H'_{21}
	&=&	-\frac{1}{2m^3}\left(  \frac {\v{D}^2} {2} -  (\v{S} \cdot \v{D})^2  -  \frac{g-2}{2}  e\v{S} \cdot \v{B} \right )^2	\\	
	&=& 	-\frac{1}{2m^3}\left( 
				\frac{ \gv{\pi}^4 } {4}  -  e \v{p}^2  \v{S} \cdot \v{B}   
				-\frac{g-2}{2}e (\v{S} \cdot \v{p}) (\v{B} \cdot \v{p})
			\right)
\end{eqnarray*}

So finally,replacing all $\v{D}$ with $\gv{\pi} \equiv  \v{p} - e \v{A}$, we have a direct expression for $\Psi_u$:
\begin{eqnarray*}
	E \Psi_u 
		&=&\left \{ m + e\Phi + \frac{\pi^2}{2m} - \frac{g}{2}\frac{e}{m} \v{S} \cdot \v{B}
			- \frac{\pi^4 } {8m^3}  
			+ \frac{ e \v{p}^2  (\v{S} \cdot \v{B}) }{2m^3}
			+ (g-2)\frac{e}{4m^3} (\v{S} \cdot \v{p}) (\v{B} \cdot \v{p})
				 \right .	\\
		&&	\left . 
			+ (g-1)\frac{ie}{2m^2}  \left [ 
					\v{E} \cdot \gv{\pi} - (\v{S} \cdot \v{E}) (\v{S} \cdot \gv{\pi}) + i \v{S} \cdot \v{E} \times \gv{\pi}
					- \gv{\pi} \cdot \v{E} + (\v{S} \cdot \gv{\pi}) (\v{S} \cdot \v{E}) - i \v{S} \cdot \gv{\pi} \times \v{E} 
			\right ]			
			\right \}\Psi_u
\end{eqnarray*}
The complicated expression in square brackets can be cleaned up a bit:

\begin{eqnarray*}
\v{E} \cdot \gv{\pi} - \gv{\pi} \cdot \v{E}
	&=&	[E_i, \pi_i]			\\
	&=&	[E_i, -i\partial_i]	\\
	&=&	i(\partial_i E_i)		\\
(\v{S} \cdot \v{E}) (\v{S} \cdot \gv{\pi}) - (\v{S} \cdot \gv{\pi}) (\v{S} \cdot \v{E})
	&=&	S_i S_j E_i \pi_j - S_i S_j \pi_i E_j						\\
	&=&	(S_i S_j) (E_i \pi_j - E_j \pi_i - [\pi_i, E_j])					\\
	&=&	[S_i, S_j](E_i \pi_j) - (S_i S_j)(-i \nabla_i E_j)				\\
	&=&	(i\epsilon_{ijk}S_k)E_j \pi_i -  (S_i S_j)(-i \nabla_i E_j)		\\
	&=&	i \v{S} \cdot \v{E} \times \gv{\pi} + i S_i S_j \nabla_i E_j)	\\
(\v{E} \times \gv{\pi} - \gv{\pi} \times \v{E})_k
	&=&	\epsilon_{ijk}(E_i \pi_j - \pi_i E_j)		\\
	&=&	\epsilon_{ijk}(E_i \pi_j + \pi_j E_i)		\\
	&=&	\epsilon_{ijk}(2 E_i \pi_j + [\pi_i, E_j])	\\
	&=&	2 \epsilon_{ijk} E_i \pi_j				\\
	&=&	2 (\v{E} \times \gv{\pi})_k
\end{eqnarray*}

Using these identities and collecting terms, and then writing everything in terms of $g$, $g-2$
\begin{eqnarray*}
	E \Psi_u 
		&=&\left\{ m + e\Phi + \frac{\pi^2}{2m} - \frac{\pi^4 } {8m^3}
			- \frac{e}{m} \v{S} \cdot \v{B} \left ( \frac{g}{2} - \frac{p^2}{2m^2} \right )
			+ (g-2)\frac{e}{4m^3} (\v{S} \cdot \v{p}) (\v{B} \cdot \v{p})	\right. \\
		&&	\left.
			- (g-1)\frac{e}{2m^2} 
				\left [ 
					\v{\nabla} \cdot \v{E} 
					- S_i S_j \nabla_i E_j +\v{S} \cdot \v{E} \times \gv{\pi}
				\right ]
			\right\}\Psi_u	\\
		&=& \left\{ m + e\Phi + \frac{\pi^2}{2m} - \frac{\pi^4 } {8m^3}
			- \frac{g}{2}\frac{e}{m} \v{S} \cdot \v{B} \left ( 1 - \frac{p^2}{2m^2} \right )
			- \frac{g-2}{2} \frac{e}{m} \frac{p^2}{2m^2} \v{S} \cdot \v{B} 
			+ (g-2)\frac{e}{4m^3} (\v{S} \cdot \v{p}) (\v{B} \cdot \v{p})	\right.	\\
		&&	\left.
			- \left ( \frac{g}{2} + \frac{g-2}{2} \right) \frac{e}{2m^2} 
				\left [ 
					\v{\nabla} \cdot \v{E} 
					- S_i S_j \nabla_i E_j +\v{S} \cdot \v{E} \times \gv{\pi}
				\right ]
			\right\}\Psi_u
\end{eqnarray*}
			
We have a Hamiltonian for the upper component of the bispinor, as desired.  But is it truly Schrodinger-like?  In the general case we would need to perform the Foldy-Wouthyusen transformation, to a representation where all the physics up to the desired order is contained in the single spinor equation.  But we can show that, at the order we are working at, the Hamiltonian above is correct.



\subsection{Normalization}
What we want is a single equation for the non-relativistic particle.  Because the lower component of the bispinor is small but nonzero, it is not necessarily true that the above equation accurately captures the physics.  The FW transformation to a basis where the lower component is truly negligible might be necessary.

However, here it can be shown that such a transformation will have no effect at the desired order.  The transformation U would have the form::
 $U = e^{iS} $ where S is Hermitian.  Because the Hamiltonian is diagonal at leading order, S must be small, and the transformation will affect $\Psi_u$ as
\begin{eqnarray*} 
\Psi_u \to \Psi'_u= (1 + \Delta)\Psi_u 
\end{eqnarray*}
where $\Delta$ is some small operator.  The probability density must be unaffected by this change, so:
On the one hand, with $\Psi_\ell = \epsilon \Psi_u$, $ \epsilon \sim \mathcal{O}(v^2) $

\begin{eqnarray*}
\int d^3 x (\Psi_u^\dagger \Psi_u - {\Psi^\dagger}_\ell \Psi_\ell)  
	&=& \int d^3 x (\Psi_u^\dagger \Psi_u - (\epsilon {\Psi_u})^\dagger \epsilon \Psi_u) \\
	&=& \int d^3 x \Psi_u^\dagger (1 + \mathcal{O}(v^4)) \Psi_u
\end{eqnarray*}

And on the other hand:
\begin{eqnarray*}
\int d^3 x {\Psi'}_u^\dagger \Psi'_u 
	&=& \int d^3 x ([1 + \Delta]\Psi_u)^\dagger (1+\Delta)\Psi_u \\
\end{eqnarray*}

Comparing the two, it can be seen that $\Delta$ must be no larger than $\mathcal{O}(v^4)$.  Now considering the new equation for $\Psi'_u$:
\begin{eqnarray*}
(E -m)\Psi_u 	
	&=&	\left(H'_{11} +  \frac{1}{2m}H'_{12} H'_{21} \right) \Psi_u\\
(E-m)(1-\Delta)\Psi'_u 
	&=&	\left(H'_{11} +  \frac{1}{2m}H'_{12} H'_{21} \right) (1-\Delta)\Psi'_u\\
(E-m)\Psi'_u 
	&=&	(1 + \Delta)\left(H'_{11} +  \frac{1}{2m}H'_{12} H'_{21} \right) (1-\Delta)\Psi'_u
\end{eqnarray*}

Since $H' \sim \mathcal{O}(mv^2)$ and $\Delta \sim \mathcal{O}(v^4)$, to $\mathcal{O}(v^4)$, $\Psi'_u$ obeys exactly the same equation as $\Psi_u$:
\begin{eqnarray*}
(E -m)\Psi'_u 	
	&=&  \left ( H'_{11} +  \frac{1}{2m} H'_{12} H'_{21}  \right ) \Psi'_u \\
\end{eqnarray*}


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