GeneralSpin.tex

\chapter{Arbitrary spin: Derivation of nonrelativistic Hamiltonian}

%name? 
Above, two particular relativistic theories have been considered: those sectors of QED involving spin half and spin one particles.  A nonrelativistic theory for each was derived in two ways, each starting from the relativistic Lagrangian; once by solving the equations of motion, the other by comparing scattering amplitudes so as to determine NRQED coefficients.

Although the two nonrelativistic Lagrangians do differ, it turns out that all terms contributing to the bound $g$-factor coincide, at least at the second order in $\alpha^2$.  This offers the hope that this remains true for all charged particles, no matter their spin.

The obvious obstacle is that in the previous chapters a specific, known relativistic Lagrangian was the starting point.  However, a key point is that, in the NRQED approach, only the one-photon vertex for the charged particle was necessary to fix all relevant NRQED coefficients.  If this holds true for the case of general spin, then only that portion of a relativistic Lagrangian representing the one photon interaction need be considered.  And there are enough constraints on this interaction to render the problem quite tractable.

First, the form of the nonrelativistic Lagrangian and fields will be constructed for general spin.  Then the relativistic fields will be treated similarly.  Finally, by considering physical results from both theories, the coefficients will be fixed for the general spin Lagrangian.



\section{The nonrelativistic Lagrangian}
\subsection{Spinors for nonrelativistic theory}
The first question is how to represent the fields in a nonrelativistic theory.  A single particle of spin $s$ has $2s+1$ degrees of freedom.  A regular spinor of rank $2s$ has the correct transformation properties under rotation, and by considering only such spinors which are symmetric in all indices, the necessary number of degrees of freedom is obtained.


\subsection{Building the NRQED Lagrangian}
%TODO add refs back to them
Previously, NRQED Lagrangians were developed for spin half and spin one.  They included all the terms that could arise at the required order for such theories.  As discussed in developing the Lagrangian for spin one, new terms will arise because the increasing degrees of spin freedom allow a larger set of spin operators.  So to describe the NRQED Lagrangian for general spin, a complete description of spin space operators are needed.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%   Spin space operators input
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\input{SpinSpaceOperators}


\subsection{New terms arising at higher spin}
With the general set of spin operators in hand, what new terms can occur in the Lagrangian involving them?  Although a large number of such operators exist for high spin, only a small number need be considered for the NRQED Lagrangian.  That is because only position space operators up to a certain order are considered.

The different combinations of position space operators were catalogued in describing the spin half Lagrangian.  The first new spin structure to consider is $\Sb_{ijk}$, which has three indices.  The only allowed set of position space operators that can be contracted with this operator is $\v{B}, \v{D}, \v{D}$.  Considering Hermitian combinations, two new terms can arise:
\beq \label{eq:Sg:Lnewterms}
  c_{T_1} \frac{ e\Sb_{ijk} D_i B_j D_k }{8m^3}  + c_{T_2} \frac{ e\Sb_{ijk} (D_i D_j B_k + B_i D_j D_i) }{8m^2}.
\eeq   

Only one combination of position space operators existed at the needed order, $D_i D_j D_k D_\ell$.  This cannot be coupled to $\Sb_{ijk\ell}$ because it would spoil the kinetic term.  So the only new terms are those in  \eqref{eq:Sg:Lnewterms}.  Adding these to the Lagrangian \eqref{eq:S1:Lnr} found for spin one, the result is
\small \beq \label{eq:Sg:nrLFull}
\begin{split}
\mathcal{L}_{NRQED} = & \fnrb \Bigg\{
		iD_0 +  \frac{\v{D}^2}{2m}  + 	\frac{\v{D}^4}{8m^2}
		 + c_F \frac{e}{m} \v{S} \cdot \v{B}
		+ c_D \frac{e (\v{D} \cdot \v{E} - \v{E} \cdot \v{D})}{8m^2} 
		+ c_Q \frac{eQ_{ij}(D_i E_j - E_i D_j)}{8m^2}
\\	& + c_S \frac{ i e \v{S} \cdot(\v{D} \times \v{E} - \v{E} \times \v{D})}{8m^2}
		+ c_{W1} \frac{ e \v{D}^2 \v{S} \cdot \v{B} + \v{S} \cdot \v{B} \v{D}^2 }{8m^3}
		- c_{W2} \frac{e D_i (\v{S} \cdot \v{B}) D_i}{4m^3}
\\	&		+c_{p'p} \frac{ e [ (\v{S}\cdot \v{D})(\v{B} \cdot \v{D}) + (\v{B} \cdot \v{D})(\v{S}\cdot \v{D})]}{8m^3}
 	+ c_{T_1} \frac{ e \bar{S}_{ijk} (D_i D_j B_k + B_k D_j D_i)}{8m^3}
\\&		+ c_{T_2} \frac{ e \bar{S}_{ijk} D_i B_j D_k }{8m^3} 
		\Bigg \} \fnr.
\end{split}
\eeq
\normalsize
\subsection{Scattering off an external field in NRQED}

To calculate scattering off an external field, that part of the Lagrangian involving only a single power of $A$ is needed.  The terms which arise for spin greater than one are the only modifications from \eqref{eq:S1:nrqedScatter}
\small
\beq
\begin{split}
\mathcal{L}_A =& \fnrb (  -eA_0 - ie  \frac{ \{\nabla_i, A_i \} }{2m} -ie \frac{ \{\grad^2, \{\nabla_i, A_i \}  \} }{8m^3} 
		+ c_F e \frac{\v{S} \smalldot \v{B}} {2m}   	
		+ c_D \frac{ e(\v{\grad} \smalldot \v{E} - \v{E} \smalldot \v{\grad} ) }{8m^2}	
	\\&	+ c_Q \frac{e Q_{ij} (\nabla_i E_j - E_i \nabla_j) }{8m^2}	
		+ c^{1}_S \frac{ ie \v{S} \smalldot ( \v{\grad} \times \v{E} - \v{E} \times \v{\grad} )}{8m^2}
		+ c_{W_1} \frac{ e [ \v{\grad}^2 (\v{S} \smalldot \v{B} ) + (\v{S} \smalldot \v{B} ) \v{\grad}^2] }{8m^3}	
	\\&	- c_{W_2} \frac{ e \nabla^i (\v{S} \smalldot \v{B} ) \nabla^i }{4m^3}
		+ c_{p'p} \frac{ e [ (\v{S} \smalldot \v{\grad}) (\v{B} \smalldot \v{\grad}) + (\v{B} \smalldot \v{\grad})(\v{S} \smalldot \v{\grad}) }{8m^3} 
	\\&	+ c_{T_1} \frac{ e \bar{S}_{ijk} (D_i D_j B_k + B_k D_j D_i)}{8m^3}
		+ c_{T_2} \frac{ e \bar{S}_{ijk} D_i B_j D_k }{8m^3} 
		\big )\fnr.
\end{split}
\eeq
\normalsize

The calculation of the scattering amplitude goes just as before, with a couple of extra terms.  The result is
\beq 
\begin{split} \label{eq:Sg:nrqedScatter}
	iM_{\text{NR}} =
		ie\wnrb \Bigg(  -A_0 +  \frac{ \v{A} \cdot \v{p} }{m} - \frac{  (\v{A} \cdot \v{p}) \v{p}^2   }{2m^3} 
		+ c_F  \frac{\v{S} \smalldot \v{B}} {2m}   	
		+ c_D \frac{ ( \partial_i E_i ) }{8m^2}	
		+ c_Q \frac{ Q_{ij} ( \partial_i E_j ) }{8m^2}	
		+ c^{1}_S \frac{  \v{E} \times \v{p} }{4m^2}
	\\	- (c_{W_1} -c_{W_2}) \frac{   (\v{S} \smalldot \v{B} ) \v{p}^2  }{4m^3}	
		-  c_{p'p} \frac{   (\v{S} \smalldot \v{p}) (\v{B} \smalldot \v{p})  }{4m^3} 
		+ (c_{T_1} + c_{T_2} )\frac{ e \bar{S}_{ijk} D_i B_j D_k }{8m^3}
		\Bigg )\wnr.
\end{split}
\eeq

%Such a representation is equivalent to just the sum of several spin half representations.  If $\eta$ is a spinor of rank $2s$, then, the spin operator is
%\beq
%	\v{S}  = \frac{1}{2}\displaystyle \sum_{n=1}^{2s} \gv{\sigma}_n 	
%\eeq
%where $\gv{\sigma}_n$ is a Pauli spin matrix, acting on the $n$th index of $\eta$.


\section{Relativistic theory for general spin particles}


\subsection{Relativistic bispinors}
To work out a relativistic theory that describes particles of arbitrary spin, the first step will be to consider how the fields should be represented.  The tactic here will be to represent the spin state by an object that looks like a generalization of the Dirac bispinor.


It is easiest to start with the chiral basis, where the upper and lower components of the bispinor are objects of opposite helicity, each transforming as an object of spin $s$.

To that end define an object
\beq \label{eq:PsiDef}
\Psig  = \frac{1}{\sqrt{2}} \begin{pmatrix} \xi \\ \eta \end{pmatrix},
\eeq
the components of which will have the desired properties.  Each component should transform as a particle of spin $s$, but with opposite helicity.  Under reflection the upper and lower components transform into each other.

Spinors which are separately symmetric in dotted and undotted indices are representations of the proper Lorentz group.  If $\xi$ is an object with $p$ undotted and $q$ dotted indices
\beq
	\xi = \{ \xi^{\alpha_1 \ldots \alpha_p}_{\dot\beta_1 \ldots \dot\beta_q} \},
\eeq
then this can be a representation of a particle of spin $s = (p+q)/2$.

There is some free choice in how to partition the dotted/undotted indices --- the same scheme won't work for all spin as long as both types of indices are present.  However, separately consistent choices can be made for integral and half-integral spin.  For integral spin let $p=q=s$, while for the half-integral case choose $p=s+\frac{1}{2}$, $q=s-\frac{1}{2}$.

$\xi$ and $\eta$ should transform as objects of opposite helicity.  Under reflection they will transform into each other.  So the choices made for $\xi$ then dictate that
\beq
	\eta = \{ \eta_{\dot \alpha_1 \ldots \dot \alpha_p}^{\beta_1 \ldots \beta_q} \}.
\eeq




In the rest frame of the particle, there is no helicity.  Both objects will have clearly defined and identical properties under rotation.    The rest frame spinors are equivalent to rank $2s$ nonrelativistic spinors.  So the bispinor in the rest frame looks like
\beq \label{eq:PsiRest}
	\Psig = \frac{1}{\sqrt{2}} \begin{pmatrix} \xi_0 \\ \xi_0 \end{pmatrix},
\eeq
where
\beq \label{eq:xi0def}
	\xi_0 = \{ (\xi_0)_{\alpha_1 \ldots \alpha_p \beta_1 \ldots \beta_q}  \},
\eeq
and is symmetric in all indices.

We can obtain the spinors in an arbitrary frame by boosting from the rest frame.  The upper and lower components we have defined to have opposite helicity, and so will act in opposite ways under boost by some rapidity $\rapidity$:
\beq \label{eq:xi0boosted}
	\xi = \exp{ \left (\frac{\v{\Sigma} \cdot \gv{\rapidity}}{2} \right ) } \xi_0,  
	\hspace{3em} 
	\eta = \exp{ \left(-\frac{\v{\Sigma} \cdot \gv{\rapidity}}{2} \right) } \xi_0.
\eeq

%TODO check dotted/undotted transformations are correct
What form should the operator $\v{\Sigma}$ have?  Under an infinitesimal boost by a rapidity $\phi$, a spinor with a single undotted index is transformed as
\beqB
	\xi_\alpha \to \xi'_\alpha = \left(\delta_{\alpha \beta} + \frac{\gv{\rapidity}\cdot \gv{\sigma}_{\alpha \beta} }{2} \right) \xi_\beta, 
\eeqB
while one with a dotted index will transform as
\beqB
\xi_{\dot\alpha} \to \xi'_{\dot\alpha} = \left(\delta_{\dot \alpha \dot \beta} - \frac{\gv{\rapidity}\cdot \gv{\sigma}_{\dot \alpha \dot\beta} }{2} \right) \xi_{\dot \beta}.
\eeqB
The infinitesimal transformation of a higher spin object with the first $p$ indices undotted and the last $q$ dotted would then be
\beqB
	\xi \to \xi' = \left(1 
		+  \sum\limits_{a=0}^p \frac{\gv{\sigma}_a \cdot \gv{\rapidity} }{2}
		- \sum\limits_{a=p+1}^{p+q} \frac{\gv{\sigma}_a \cdot \gv{\rapidity} }{2}
	\right ) \xi,
\eeqB
where $a$ denotes which spinor index of $\xi$ is operated on.


So if $\gv{\Sigma}$ is defined as
\beq \label{eq:SigDef}
	\v{\Sigma} = \sum\limits_{a=0}^p \gv{\sigma}_a - \sum\limits_{a=p+1}^{p+q} \gv{\sigma}_a,
\eeq
then the infinitesimal transformations would be
\beqB
	\xi \to \xi' = \left( 1 + \frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2} \right) \xi,
\eeqB
\beqB
	\eta \to \eta' = \left( 1 - \frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2} \right) \eta.
\eeqB
That is satisfied if the exact transformations are
\beqB
		\xi \to \xi' = \exp\left( \frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2} \right) \xi,
\eeqB
\beqB
	\eta \to \eta' = \exp \left( -\frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2} \right) \eta.
\eeqB
  
So the bispinor of some particle boosted by $\gv{\phi}$ from rest will be
% TODO check passive v. active boost
\beq \label{eq:PsiByXi0}
\Psig = \frac{1}{\sqrt{2}} \begin{pmatrix} 
		\exp\left( \frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2} \right)\xi_0 \\ 
		\exp \left( \frac{-\gv{\Sigma} \cdot \gv{\rapidity} }{2} \right) \xi_0 
	\end{pmatrix}.
\eeq

%TODO fix sqrt(2) factors in the right place
The helical basis was a sensible one in which to examine the transformation properties of the bispinors.  But in descending to the nonrelativistic theory, a basis that separates the particle and antiparticle parts of the field will be more convenient.  If the upper component is supposed to be the particle, then in the rest frame the lower component will vanish, and for low momentum will be small compared to the upper component.

Define the new bispinor as
\[
	\Psig' = \begin{pmatrix} \phi \\ \chi \end{pmatrix}.
\]
The components of this bispinor are, in terms of the old components $\xi$ and $\eta$
\[
	\phi = \frac{1}{\sqrt{2}}(\xi + \eta),
\]
\[
	\chi = \frac{1}{\sqrt{2}}( \eta - \xi).
\]


And $\Psig'$ can obtained by a unitary transformation:
\[
	\Psig' = \frac{1}{\sqrt{2}} \begin{pmatrix}1 & 1 \\ -1 & 1 \end{pmatrix} \Psig.
\]

In terms of the original rest frame spinors, the components of $\Psig'$ are
\beq \label{eq:phiDef}
	\phi =  \cosh \left( \frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2} \right ) \xi_0,
\eeq

%TODO consider: Sign confusion compared to original equation again?
\beq \label{eq:chiDef}
	\chi =  \sinh \left( \frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2} \right ) \xi_0.
\eeq


\subsection{Properties of $\gv{\Sigma}$ matrices}

The matrices $\Sigma_i$ act on wave functions with $p+q$ indices.  If we treat the two sets of indices as two separate spaces, with spins $\frac{p}{2}, \frac{q}{2}$ respectively, we can write it as the sum of those spaces spin operators: $\Sigma_i = 2(S^P_i - S^Q_i)$.  We can then write the total spin operator as $S^I_i = S^P_i + S^Q_i$.  (The case of spin one-half is degenerate: $\Sigma_i =  2 S_i = \sigma_i $.)

Because the two spaces are orthogonal, $[S^P_i, S^Q_j]=0$.  Using that and the above definitions, the algebra of these matrices is found.  
\beq \label{eq:Sg:comSSig}
	[S^I_i, \Sigma_j] = i\epsilon_{ijk}\Sigma_k
\eeq
\beq \label{eq:Sg:comSigSig}
	[\Sigma_i, \Sigma_j] = 4i\epsilon_{ijk}S^I_k
\eeq

Wave functions of definite spin are also eigenfunctions of $\Sigma^2$ and $\Sigma \cdot S^I$.  This can be shown by calculating various scalar quantities:
\beqa
	{S^P}^2 &=& \frac{p}{2} ( \frac{p}{2} + 1)	\\
	{S^Q}^2 &=& \frac{q}{2} ( \frac{q}{2} + 1)	\\
	(S^I)^2 	&=& \frac{p+q}{2} ( \frac{p+q}{2} + 1 )	\\
		&=& \frac{p}{2} (\frac{p}{2} + 1) + \frac{q}{2}(\frac{q}{2} + 1 ) + 2 \frac{pq}{4}	\\
		&=& {S^P}^2 + 2\frac{pq}{4} + {S^Q}^2	\\
	 {S^P} \cdot {S^Q} &=& \frac{pq}{4}	\\
	\Sigma ^2 &=& 4(S^P - S^Q)^2	\\
		&=&	4({S^P}^2 + {S^Q}^2 - 2 S^P \cdot S^Q)	\\
		&=& 	4\left(
				\frac{p}{2} ( \frac{p}{2} + 1)
				+\frac{q}{2} ( \frac{q}{2} + 1)
				- \frac{pq}{2}
			\right )	\\
		&=&	p^2 + q^2 - 2pq + 2(p + q)	\\
		&=& 	(p-q)^2 + 2(p+q)	\\
	\Sigma \cdot S &=& 2 (S^P - S_Q) \cdot (S^P + S_Q)	\\
		&=& 2 (S^P)^2 - 2 (S^Q)^2	\\
		&=&	\frac{1}{2} ( p^2 - q^2 + p - q )	
\eeqa

Expressing these in terms of $I= \frac{p+q}{2}$ and $\Delta = p-q$ we can write
\begin{eqnarray} 
	S^2 		&=&	I(I+1)	\\
	\Sigma^2	&=&	4I + \Delta	 \label{eq:Sg:SigScalars}  \\
	\Sigma \cdot S	&=&	\frac{\Delta}{2} ( 2I + 1)	
\end{eqnarray}

%TODO write properly in terms of matrix elements and so forth
It will eventually be necessary to take the matrix element of terms containing $\Sigma_i \Sigma_j$ in the nonrelativistic theory.  These must be expressible in terms of the spin operator's matrix element, because that operator spans that space.  Considering the traceless and symmetric structure, it must be proportional to the similar structure composed of spin matrices.
\beq \label{eq:Sg:SigTens}
	\avg{\Sigma_i \Sigma_j + \Sigma_j \Sigma_i - \frac{2}{3} \delta_{ij} \gv{\Sigma}^2}
		=	\lambda \avg{ \left \{ I_i I_j + I_j I_i - \frac{2}{3} \delta_{ij} \v{I}^2 \right\} }	
\eeq
To determine the constant $\lambda$, consider the $zz$ component of the tensor structure acting on a wave function with all spin projected in the $z$ direction.  Then $I_3 \to \frac{p+q}{2} = I$, and $\Sigma_3 \to p-q = \Delta$, giving
\beqa
	2 \Delta^2 - \frac{2}{3} (4I + \Delta) 
		&=& \lambda  \left \{ 2 I^2 -\frac{2}{3} (I^2 + I) \right \}	\\
		&=& \lambda \frac{2}{3}I(2I - 1 )
\eeqa
For integer spin $\Delta = 0$, giving
\beqa
	-\frac{8}{3} I &=& \lambda_1 \frac{2}{3}I (2I -1)	\\
	\lambda_1 &=& -\frac{4}{2I-1}	\\
\eeqa
For half spin we have $\Delta = 1$ 
\beqa
	\frac{4}{3} (1  - 2I) &=& \lambda_{\frac{1}{2}}	 \frac{2}{3}I (2I -1)	\\
	\lambda_{\frac{1}{2}} &=&-\frac{4}{2I}	\\
\eeqa



\subsection{Bilinears in the relativistic theory}
Recall that in the case of spin one-half it was possible to catalogue a complete set of field bilinears with definite Lorentz transformation properties.  The structure of the electron vertex was necessarily expressible in terms of those bilinears.  These bilinears were built out of the $4\times 4$ $\gamma^\mu$ matrices, which were themselves built out of $2 \times 2$ $\sigma$ matrices.

What is the generalisation of these bilinears for the case of general spin?  The fields have been written as bispinors, with upper and lower components spinors with transformation properties defined above.  Whereas in the case of spin one-half $\{ \mathcal{I}, \sigma_i\}$ formed a basis for operators acting on these spinors, here more complicated quantities built out of $S$ and $\Sigma$ are allowed.

It is still possible to write down a set of bilinears with definite transformation properties.  The greater the spin (and thus the degrees of freedom of the spinors) the larger their number and rank.  However, for the purposes of constructing the electromagnetic current and then calculating its nonrelativistic limit, only bilinears of up to rank 2 tensors need be considered.

In the appendix \ref{chap:bilinear} the transformations of various bilinears are worked out.  There are two necessary for constructing the electromagnetic current; the simple scalar $\Psigbar \Psi$ and the antisymmetric tensor $\Psigbar \TensBi_{\mu\nu} \Psig$.

The tensor structure is defined by

\beq \label{eq:Sg:SigDef}
	\Sigma_{ij} = -2 i \epsilon_{ijk} \Mblock{S_k}{0}{0}{S_k}
\eeq
\beq 
	\Sigma_{0i} = \Mblock{0}{\Sigma_i}{\Sigma_i}{0}
\eeq
 
For the case of spin one-half, where $2S_i = \Sigma_i = \sigma_i$, this reduces to the familiar anti-symmetric tensor $\sigma_{\mu\nu} = [\Gamma_\mu, \Gamma_\nu]$.
%Section on bilinear transforms
%\input{bilinear_transforms}

\subsection{Electromagnetic Interaction}
%TODO insert diagram, showing the type of interaction we're talking about, defining momenta of particles in question.
Knowing how the fields themselves behave, what form might the electromagnetic interaction take?  In general, it can be written
\[
	M = e A_\mu j^\mu 
\]
where $j^\mu$ is the electromagnetic current.


The electromagnetic current must be built out of the particle's momenta and bilinears of the charged particle fields in such a way that they have the correct Lorentz properties.  It must also obey current conservation: the equation $q_\mu j^\mu = 0$ must hold.  

%TODO address concern about p_0 terms
As higher spins are considered, higher order bilinears will appear.  But the electromagnetic current must ultimately be related to the physically relevant vectors of the problem: the momentum and the spin four-vectors.  This means that only two independent terms will enter.

 And on such physical grounds, there will necessarily be some relation (like the Gordon identity in the case of spin one-half) that relates any seemingly independent vector-like term to the two appearing in the electromagnetic current.  

So if we can come up with two terms that transform as Lorentz vectors, that will be sufficient to capture the essence of the current.  One such is a scalar bilinear coupled with a single power of external momenta.  In order to fulfill the current conservation requirement, the exact combination will be
\[
	\frac{p^\mu + p'^\mu}{2m} \Psigbar^\dagger \Psig
\]
This obeys current conservation because $q = p' -p$, and $ (p+p')\cdot(p'-p) = p^2-p'^2=0$

The other type of term will be a tensor term contracted with a power of momenta.  To fulfill current conservation, the tensor must be antisymmetric, and contracted with $q$:
\[
	\frac{q_\nu}{2m} \Psigbar^\dagger \TensBi^{\mu\nu} \Psig
\]

So the most general current would look like
\beq \label{eq:khr_current}
	j^\mu = F_e \frac{p^\mu + p'^\mu}{2m} \Psigbar^\dagger \Psig + F_m 	\frac{q_\nu}{2m} \Psigbar^\dagger \TensBi^{\mu\nu} \Psig	
\eeq
The form factors might have quite complicated dependence on $q$, but these corrections will be too small.  They will be suppressed beyond the order of the calculation.  At leading order $F_e$ will just be the electric charge of the particle in question, and $F_m$ will, as will be seen after calculating the nonrelativistic limit, be related to the particle's $g$-factor.  So to the order needed, the current can be written
%TODO check definition of form factors F_e and F_m
\beq 
	j^\mu =  e \frac{p^\mu + p'^\mu}{2m} \Psigbar^\dagger \Psig +   e g \frac{q_\nu}{2m} \Psigbar^\dagger \TensBi^{\mu\nu} \Psig
\eeq

%The tensor bilinear $\Psigbar^\dagger T^{\mu\nu} \Psig$ itself has some free parameters.  However, it'll turn out that, when computing actual nonrelativistic amplitudes, the two types of anti-symmetric tensors collapse into the same general form.

This captures the essence of the interaction between a charged particle of general-spin and a single photon.


\section{Comparison of scattering amplitudes}

\subsection{Connection between the spinors of the two theories}
Having sufficient knowledge of the relativistic theory, the approach of NRQED can be applied.  In NRQED, the amplitude for scattering off an external field has been calculated.  To compare the same process in this relativistic theory, first a way to write the nonrelativistic spinors in terms of the relativistic ones is needed.

In the rest frame, there are two independent bispinors which represent particle and antiparticle states.  The particle state is represented by
\beq
	\Psig = \begin{pmatrix} \xi_0 \\ 0 \end{pmatrix}
\eeq
the antiparticle by
\beq
	\Psig = \begin{pmatrix} 0 \\ \xi_0 \end{pmatrix}
\eeq

However, when considering a particle with non-zero momentum it is not the case that the upper component of the bispinor can be directly associated with the Schrodinger like wave-function of the particle --- probability would not be properly conserved, for there is some mixing with the lower component, and thus a nonzero chance of the particle being in such a state.

To obtain a relation between $\xi_0$ and the Schrodinger amplitude $\phi_s$,  consider the current density at zero momentum transfer.  For $\phis$ it will be $j_0 =  \phis^\dagger \phis$.  For the relativistic theory, as calculated above:
\beq
	j^0 = F_e \frac{p^0 + p'^0}{2m} \Psigbar^\dagger \Psig + F_m 	\frac{q_\nu}{2m} \Psigbar^\dagger T^{0\nu} \Psig	
\eeq
At $q=0$ the expression simplifies

\beq
	j^0(q=0) = F_e \frac{p_0}{m} \Psigbar^\dagger \Psig
	= F_e  \frac{p_0}{m}( \phi^\dagger \phi - \chi^\dagger \chi )
\eeq

$\phi$ and $\chi$ are both related to the rest frame spinor $\xi_0$.  In such terms, $j^0$ is given by
\beq
	j^0 = F_e \frac{p_0}{m} \xi_0^\dagger \left \{ 
		\cosh^2( \frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2})
		- \sinh^2( \frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2})
	\right \} \xi_0  
		=	F_e \frac{p_0}{m} \xi_0^\dagger \xi_0
\eeq
where the last equality follows from the hyperbolic trig identity.


Demanding that the two current densities be equal to each other, the result is that
\beq
	\frac{p_0}{m} \xi_0^\dagger \xi_0 = \phi_s^\dagger \phi_s
\eeq
As throughout the calculation, only corrections of order $1/m^2$ are needed.  So appoximating
\beq
	\left( 1 + \frac{\v{p}^2}{2m} \right) \xi_0^\dagger \xi_0 = \phi_s^\dagger \phi_s
\eeq

This will hold to the necessary order if
\beq
	\xi_0 = \left( 1 - \frac{\v{p}^2}{4m} \right) \phi_s
\eeq


%TODO fix cosh expansion

That relates the the Schrodinger like wave functions to the quantities $\xi_0$.  To write the relativistic bispinors in terms of $\phis$, approximations for $\cosh( \frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2})$ and $\sinh( \frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2})$ are also needed.  The rapidity is needed only to the leading order: $\gv{\rapidity} \approx \v{v} \approx \frac{\v{p} }{m}$. 

\beq
	\cosh( \frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2}) 
		\approx 1 + \frac{1}{2}\left( \frac{\gv{\Sigma} \cdot \v{p} }{2m} \right)^2
\eeq
\beq
	\sinh( \frac{\gv{\Sigma} \cdot \gv{\rapidity} }{2}) 
		\approx   \frac{\gv{\Sigma} \cdot \v{p} }{2m}
\eeq
Using this, the two bispinor components are
%Non relativistic expression for \phi and \chi
\begin{align}
\phi 
	&\approx  \left(  1 + \left[ \frac{1}{2}\frac{\gv{\Sigma} \cdot \v{p} }{2m} \right]^2 \right) \xi_0 \notag \\
	&\approx  \left(  1 + \frac{(\gv{\Sigma} \cdot \v{p})^2 }{8m^2} - \frac{\v{p}^2}{4m} \right ) \phis	 \label{eq:nrPhi} \\
 \chi
 	&\approx	\frac{\gv{\Sigma} \cdot \v{p} }{2m} \xi_0 \notag \\
 	&\approx	\frac{\gv{\Sigma} \cdot \v{p} }{2m} \phis  \label{eq:nrChi}
\end{align}




%TODO remember to hunt out all the \phi where I mean \rapidity
\subsection{Bilinears in terms of nonrelativistic theory}
The next step is to express the relativistic bilinears, built out of the bispinors $\Psig$, in terms of the Schrodinger like wave functions.

Above the bispinors were written terms of $\phi_s$, so those identities can be used to express the bilinears in the same manner.  


\subsubsection{Scalar bilinear}

Calculating the scalar bilinear is just straightforward substitution and expansion.  To express the commutator of $\Sigma$ operators, the identity \eqref{eq:Sg:comSigSig} is needed.
\small \beqa
\Psigbar^\dagger(p') \Psig(p)
	&=&	\phi^\dagger \phi - \chi^\dagger \chi	\\
	&=&	\phi_s^\dagger \left[1 + \frac{(\gv{\Sigma} \cdot \v{p'})^2 }{8m^2}  - \frac{\v{p'}^2}{4m^2} \right ]
			 \left[1 + \frac{(\gv{\Sigma} \cdot \v{p})^2 }{8m^2}  - \frac{\v{p}^2}{4m^2} \right ] \phi_s
		- \phi_s^\dagger \left[
			\frac{ ( \gv{\Sigma} \cdot \v{p'}) (\gv{\Sigma} \cdot \v{p}) }{4m^2}
		\right ] \phi_s	\\
	&=&	\phi_s^\dagger \left (
			1 - \frac{ \v{p}^2 + \v{p'}^2 }{4m^2}
			+ \frac{1}{8m^2} \left \{
				( \gv{\Sigma} \cdot \v{p'})^2 +  (\gv{\Sigma} \cdot \v{p})^2 
				 - 2 ( \gv{\Sigma} \cdot \v{p'}) (\gv{\Sigma} \cdot \v{p})
			\right \}
	\right ) \phi_s	\\
	&=& \phi_s^\dagger \left (
			1 - \frac{ \v{p}^2 + \v{p'}^2 }{4m^2}
			+ \frac{1}{8m^2} \left \{
				[ \gv{\Sigma} \cdot \v{p},  \gv{\Sigma} \cdot \v{q}]  + ( \gv{\Sigma} \cdot \v{q})^2 
			\right \}
	\right ) \phi_s	\\
	&=& \phi_s^\dagger \left (
			1 - \frac{ \v{p}^2 + \v{p'}^2 }{4m^2}
			+ \frac{1}{8m^2} \left \{
				[ 4 i \epsilon_{ijk} p_i q_j S_k  + ( \gv{\Sigma} \cdot \v{q})^2 
			\right \}
	\right ) \phi_s
\eeqa \normalsize
%TODO at some point need to translate \Sigma functions into spin functions in NR theory, but where?

\subsubsection{Tensor $ij$ component}



In calculating the nonrelativistic limit of the antisymmetric tensor bilinear, the $0i$ and the $ij$ components will be treated separately.  First consider $\Psigbar \Sigma_{ij} \Psig$.  The value of $\TensBi_{ij}$ itself was written in \eqref{eq:Sg:SigDef}
\scriptsize
\beqa
	\Psigbar \TensBi_{ij} \Psig 
		&=& \Psigbar (-2\epsilon_{ijk} S_k) \Psig	\\
		&=&	-2i\epsilon_{ijk} ( \phi^\dagger S_k \phi - \chi^\dagger S_k \chi)	\\
		&=&	-2i\epsilon_{ijk} \Big( \phis^\dagger \left[ 1 + \frac{( \gv{\Sigma} \cdot \v{p'})^2}{8m^2}  -\frac{\v{p'}^2}{4m^2} \right] S_k \left[ 1 + \frac{( \gv{\Sigma} \cdot \v{p})^2}{8m^2} -\frac{\v{p}^2}{4m^2}\right ] \phis - \phis^\dagger \frac{ ( \gv{\Sigma} \cdot \v{p'})S_k ( \gv{\Sigma} \cdot \v{p })}{4m^2} \phis \Big )	\\
		&=&	-2i\epsilon_{ijk} \phis^\dagger \left \{
				S_k \left( 1 - \frac{ \v{p}^2 + \v{p'}^2}{4m^2}  \right )
				+ \frac{1}{8m^2} \underbrace{\Big[ ( \gv{\Sigma} \cdot \v{p'})^2 S_k + S_k ( \gv{\Sigma} \cdot \v{p})^2 - 2 ( \gv{\Sigma} \cdot \v{p'})S_k ( \gv{\Sigma} \cdot \v{p}) \Big ]}_{T_k}
			\right \} \phis
\eeqa
\normalsize
Call the term in square brackets $T_k$.  It should be written explicitly in terms of $\v{p}$ and $\v{q}$.
\beqa
T_k	&=& ( \gv{\Sigma} \cdot \v{p})^2 S_k + S_k ( \gv{\Sigma} \cdot \v{p}) -2 ( \gv{\Sigma} \cdot \v{p}) S_k ( \gv{\Sigma} \cdot \v{p})
		+ \{ ( \gv{\Sigma} \cdot \v{p}) ( \gv{\Sigma} \cdot \v{q}) 
	\\ &&	+ ( \gv{\Sigma} \cdot \v{q}) ( \gv{\Sigma} \cdot \v{p}) \}S_k
		- 2 ( \gv{\Sigma} \cdot \v{q}) S_k ( \gv{\Sigma} \cdot \v{p})
		+ ( \gv{\Sigma} \cdot \v{q})^2 S_k
\eeqa

Many of these terms may be expressed as commutators, and then these commutators simplified.
\beqa
T_k	 &=&	\gv{\Sigma} \cdot \v{p} [ \gv{\Sigma} \cdot \v{p}, S_k] + [S_k, \gv{\Sigma} \cdot \v{p}] \gv{\Sigma} \cdot \v{p}
		+ 2 \gv{\Sigma} \cdot \v{q} [ \gv{\Sigma} \cdot \v{p}, S_k] - [\gv{\Sigma} \cdot \v{q}, S_k] \gv{\Sigma} \cdot \v{p}
		+ (\gv{\Sigma} \cdot \v{q})^2 S_k
	\\&=& i\epsilon_{ijk} p_j \{ (\gv{\Sigma} \cdot \v{p})\Sigma_i - \Sigma_i (\gv{\Sigma} \cdot \v{p}) \}
		+ 2 i\epsilon_{ijk} \{ (\gv{\Sigma} \cdot \v{q}) \Sigma_i p_j -  \Sigma_i (\gv{\Sigma} \cdot \v{p}) q_j ) \}
		+ (\gv{\Sigma} \cdot \v{q})^2 S_k
	\\&=& i\epsilon_{ijk} p_j [ (\gv{\Sigma} \cdot \v{p}), \Sigma_i  ]
		+ 2 i\epsilon_{ijk} \{ (\gv{\Sigma} \cdot \v{q}) \Sigma_i p_j -  \Sigma_i (\gv{\Sigma} \cdot \v{p}) q_j ) \}
		+ (\gv{\Sigma} \cdot \v{q})^2 S_k
	\\&=& 4( \v{p}^2 S_k - (\v{S} \cdot \v{p}) p_k ) 
		+ 2 i\epsilon_{ijk} \{ (\gv{\Sigma} \cdot \v{q}) \Sigma_i p_j -  \Sigma_i (\gv{\Sigma} \cdot \v{p}) q_j ) \}
		+ (\gv{\Sigma} \cdot \v{q})^2 S_k
\eeqa
Thus the whole bilinear is
\beq \begin{split}
\Psigbar \TensBi_{ij} \Psig = 
	-2i\epsilon_{ijk} \phis^\dagger \Bigg \{
				S_k \left( 1 - \frac{ \v{p}^2 + \v{p'}^2}{4m^2}  \right )
				+ \frac{1}{8m^2} \Big[ 
				4( \v{p}^2 S_k - (\v{S} \cdot \v{p}) p_k ) 
			\\	+ 2 i\epsilon_{\ell m k} \{ (\gv{\Sigma} \cdot \v{q}) \Sigma_\ell p_m -  \Sigma_\ell (\gv{\Sigma} \cdot \v{p}) q_m ) \}
				+ (\gv{\Sigma} \cdot \v{q})^2 S_k
			 \Big ]
			\Bigg \} \phis
\end{split}
\eeq

\subsubsection{Tensor $\Sigma_{0i}$ component}

Now calculate the $0i$ component, $\Psigbar \Sigma_{0i} \Psig$.

\beq
	\Psigbar \Sigma_{0i} \Psig = \Psigbar \begin{pmatrix} 0 & \Sigma_i \\ \Sigma_i & 0 \end{pmatrix} \Psig
\eeq

\beq
	=	\phi^\dagger \Sigma_i \chi - \chi^\dagger \Sigma_i \phi
\eeq

Because this tensor structure is coupled to $q_i/m$, $\phi$ and $\chi$ are needed only to first order here.
\beq
	\Psigbar \Sigma_{0i} \Psig = \phis^\dagger \left( \frac{\Sigma_i \Sigma_j p_j - \Sigma_j \Sigma_i p'_j}{2m} \right ) \phis
\eeq
Using $p'=p+q$ the terms involving only $p$ can be simplified using the commutator of $\Sigma$ matrices.
\beq
	\Psigbar \Sigma_{0i} \Psig =\phis^\dagger \left( \frac{ 4i\epsilon_{ijk} p_j S_k - \Sigma_j \Sigma_i q_j}{2m} \right )\phis
\eeq



%%%%%%%  Calculate the Current
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Current in terms of nonrelativistic wave functions}

%add ref to equation
The four-current \eqref{eq:khr_current} was derived above; in Galilean form it is 
\beq
	j_0 =  F_e \frac{p_0 + p'_0}{2m} \Psigbar^\dagger \Psig -  F_m \frac{q_j}{2m} \Psigbar^\dagger \TensBi^{0j} \Psig
\eeq

\beq
	j_i =  F_e \frac{p_i + p'_i}{2m} \Psigbar^\dagger \Psig -   F_m \frac{q_j}{2m} \Psigbar^\dagger \TensBi^{ij} \Psig 
			+F_m \frac{q_0}{2m} \Psigbar^\dagger \TensBi^{i0} \Psig
\eeq

Expressions for the bilinears in terms of the nonrelativistic wave functions $\phis$ have been calculated, so it is fairly straight forward to apply them here.  The calculation of $j_0$ goes:
\small \beqa
F_e \frac{p_0 + p'_0}{2m} \Psigbar^\dagger \Psig 
	  &=& F_e \left(1 + \frac{\v{p}^2 + \v{p'}^2}{4m^2} \right)  \phis^\dagger \left (
			1 - \frac{ \v{p}^2 + \v{p'}^2 }{4m^2}
			+ \frac{1}{8m^2} \left \{
				 4 i \epsilon_{ijk} p_i q_j S_k  + ( \gv{\Sigma} \cdot \v{q})^2 
			\right \}
	\right ) \phis	\\
	&\approx& 	F_e   \phis^\dagger \left (
					1 + \frac{1}{8m^2} \left \{ 4i \v{S} \cdot \v{p} \times \v{q}  + ( \gv{\Sigma} \cdot \v{q})^2 \right \}
				\right ) \phis	\\
F_m \frac{q_j}{2m} \Psigbar^\dagger \TensBi^{0j} \Psig
	&=& F_m \frac{q_i}{2m}\phis^\dagger \left( \frac{ 4i\epsilon_{ijk} p_j S_k - \Sigma_j \Sigma_i q_j}{2m} \right )\phis	\\
	&=&  F_m \phis^\dagger \left( \frac{ 4i \v{S} \cdot \v{q} \times \v{p} -  (\gv{\Sigma} \cdot \v{q})^2 }{4m^2} \right )\phis	\\
\eeqa \normalsize

It turns out that both terms here have the same form, so combining them, 
\beq \label{eq:nrJ0}
j_0 =  	 \phis^\dagger \left (
			F_e + \frac{F_e + 2F_m}{8m^2} \left \{ 4i \v{S} \cdot \v{p} \times \v{q}  + ( \gv{\Sigma} \cdot \v{q})^2  \right \}
		\right ) \phis	\\
\eeq


%Justify dropping derivatives of magnetic field a bit better
To calculate $j_i$ it helps to first simplify things by considering the constraints of this particular problem.  The term with $\Sigma_ij$ can be simplified by dropping terms with more than one power of $q$; these will turn into derivatives of the magnetic field, and this problem concerns only a constant field.  Further, only elastic scattering is considered, and so $q_0=0$.  With those simplifications
\beq
\Psigbar \Sigma_{ij} \Psig \approx
		-2i\epsilon_{ijk} \phis^\dagger \left \{
			S_k \left( 1 - \frac{ \v{p}^2 + \v{p'}^2}{4m^2}  \right )
			+ \frac{\v{p}^2 S_k - (\v{S} \cdot \v{p}) p_k}{2m^2}  
		\right \} \phis
\eeq


\beqa
F_e \frac{p_i + p'_i}{2m} \Psigbar^\dagger \Psig
	&=&			F_e \frac{p_i + p'_i}{2m}  \phis^\dagger \left (
					1 - \frac{ \v{p}^2 + \v{p'}^2 }{4m^2}
					+ \frac{1}{8m^2} \left \{  4 i \epsilon_{\ell jk} p_\ell q_j S_k  + ( \gv{\Sigma} \cdot \v{q})^2	\right \}
				\right ) \phis	\\
	&\approx& 	F_e \frac{p_i + p'_i}{2m}  \phis^\dagger \left (
					1 + \frac{1}{8m^2} \left \{ 4 i \epsilon_{\ell jk} p_\ell q_j S_k  \right \}
				\right ) \phis	\\
F_m \frac{q_j}{2m} \Psigbar^\dagger \TensBi^{ij} \Psig
	&=& 		- F_m\frac{ i\epsilon_{ijk} q_j}{m} \phis^\dagger \left \{
					S_k \left( 1 - \frac{ \v{p}^2 + \v{p'}^2}{4m^2}  \right )
					+ \frac{\v{p}^2 S_k - (\v{S} \cdot \v{p}) p_k}{2m^2}  
				\right \} \phis
\eeqa

So the full spatial part of the current is
\small \beq \label{eq:nrJi}
j_i	=	\phis^\dagger \Bigg \{
			F_e \frac{p_i + p'_i}{2m} \left (
				1 + \frac{ i \epsilon_{\ell jk} p_\ell q_j S_k   }{2m^2}  \right)
			+ F_m   \frac{i\epsilon_{ijk} q_j}{m} \left( 
				S_k \left \{ 1 - \frac{ \v{p}^2 + \v{p'}^2}{4m^2}  \right \}
				+ \frac{\v{p}^2 S_k - (\v{S} \cdot \v{p}) p_k}{2m^2} \right)	
		\Bigg \} \phis
\eeq \normalsize

\subsection{Scattering off an external field}
To compare to the NRQED Lagrangian, scattering off an external field needs to be calculated for an arbitrary spin particle.  The hardest part was calculating the current; with that in hand the amplitude can be found just as
\[
	M = e j_\mu A^\mu = e j_0 A_0 - e \v{j} \cdot \v{A}
\]
With the expressions for both $j_0$ \eqref{eq:nrJ0} and $\v{j}$ \eqref{eq:nrJi} both parts of the amplitude can be 	 directly.
\small \beq
\begin{split}
	e j_0 A_0 = 
		& eA_0 \phis^\dagger \left (
			F_e + \frac{F_e + 2F_m}{8m^2} \left \{ 4i \v{S} \cdot \v{p} \times \v{q}  + ( \gv{\Sigma} \cdot \v{q})^2  \right \}
		\right ) \phis	\\
   e\v{j} \cdot \v{A} =
		& A_i \phis^\dagger \Bigg \{
			F_e \frac{p_i + p'_i}{2m} \left (
				1 + \frac{ i \epsilon_{\ell jk} p_\ell q_j S_k   }{2m^2}  \right)
	\\ &		\hspace{2em} + F_m   \frac{i\epsilon_{ijk} q_j}{m} \left( 
				S_k \left( 1 - \frac{ \v{p}^2 + \v{p'}^2}{4m^2}  \right )
				+ \frac{\v{p}^2 S_k - (\v{S} \cdot \v{p}) p_k}{2m^2} \right)	
		\Bigg \} \phis
\end{split}
\eeq   \normalsize




To compare to the NRQED result, as much as possible terms should be expressed in gauge invariant quantities $\v{B}$ and $\v{E}$.  The relations between these fields and $A_\mu$ in position space and the equivalent equation in momentum space are:
%FIXME add reference to gauge
\beq
\begin{split}
	\v{B} &= \grad \times \v{A}	\notag \to i\v{q} \times \v{A} \\
	\v{E} &= -\grad A_0	 \to -i\v{q} A_0 		\notag
\end{split}
\eeq

There is one term above that can only be put into gauge-invariant form by considering the kinematic constraints of elastic scattering. As previously derived in \eqref{eq:Sh:pqA}, the identity is
\beq \label{eq:Sg:ppqAid}
	i (p_i + p'_i) q_j A_i = - \epsilon_{ijk}B_k (p_i + p'_i)
\eeq 

Now each term involving $q$ can be written in terms of $E$ or $B$.
%TODO maybe write quad term more explicitly in terms of quad moment and term with trace
\beqa
i \v{S} \cdot \v{p} \times \v{q} A_0 
		&=&	-\v{S} \cdot \v{p} \times \v{E}	\\
( \gv{\Sigma} \cdot \v{q})^2 A_0	
		&=&		\Sigma_i \Sigma_j q_i q_j A_0 		\\
		&=&		 \Sigma_i \Sigma_j \partial_i E_j	\\
i\epsilon_{ijk} A_i q_j	
		&=&	-i (\v{q} \times \v{A})_k 			\\
		&=&= - B_k	\\
A_i (p_i + p'_i)  i \epsilon_{\ell jk} p_\ell q_j S_k  
		&=&	\epsilon_{\ell jk}p_\ell S_k  i(p_i + p'_i) q_j A_i		\\
		&=&	- \epsilon_{\ell jk}p_\ell S_k \{ \epsilon_{ijm}B_m (p_i + p'_i) \}			\\
		&=& -(\delta_{\ell i} \delta_{km} - \delta{\ell m} \delta_{ik})p_\ell S_k \{ \epsilon_{ijm}B_m (p_i + p'_i) \}	\\
		&=& 2\{ (\v{B} \cdot \v{p})  (\v{S} \cdot \v{p}) - (\v{B} \cdot \v{S}) \v{p}^2  \}  
\eeqa

Using these
\beqB
	ej_0 A_0 = e\phis^\dagger \left\{
					A_0 + \frac{1 - 2F_2}{8m^2}\left( 4 \v{S} \cdot \v{E} \times \v{p} + \Sigma_i \Sigma_j \partial_i E_j \right)
				\right \}
\eeqB

\beqB
\begin{split}
	e \v{j} \cdot \v{A}	=& e \phis^\dagger \Bigg \{
			\frac{ \v{p} \cdot \v{A} }{m} + \frac{(\v{B} \cdot \v{p})  (\v{S} \cdot \v{p}) - (\v{B} \cdot \v{S}) \v{p}^2 }{m^2} 
		\\&	- F_m \left ( \frac{ \v{S} \cdot \v{B} }{m} \left\{ 1 - \frac{\v{p}^2 + \v{p'}^2}{4m^2} \right \} + \frac{(\v{B} \cdot \v{p})  (\v{S} \cdot \v{p}) - (\v{B} \cdot \v{S}) \v{p}^2 }{2m^2} \right ) \Bigg \} \phi
\end{split}
\eeqB
\beqB
	=e\phis^\dagger \left\{
		 \frac{ \v{p} \cdot \v{A} }{m} + [1 -2F_m] \frac{ (\v{B} \cdot \v{p} )(\v{S} \cdot \v{p})}{m^2} 
				- \v{S} \cdot \v{B} \frac{ \v{p}^2 }{m^2} - \frac{F_m}{m} \v{S} \cdot \v{B} \right \}
\eeqB
From looking at the leading order coefficient of $\v{S} \cdot \v{B}$ it can be seen that $F_m$ is actually $g/2$, so in such terms

\beqB	
	ej_0 A_0 = e\phis^\dagger \left\{
					A_0 - \frac{g-1}{2m^2}\left( \v{S} \cdot \v{E} \times \v{p} + \frac{1}{4}\Sigma_i \Sigma_j \partial_i E_j \right)
				\right \}
\eeqB

\beqB
	e \v{j} \cdot \v{A} = e\phis^\dagger \left\{
		 \frac{ \v{p} \cdot \v{A} }{m} - [g-1] \frac{ (\v{B} \cdot \v{p} )(\v{S} \cdot \v{p})}{m^2} 
				- \v{S} \cdot \v{B} \frac{ \v{p}^2 }{m^2} - \frac{g}{2m} \v{S} \cdot \v{B} \right \}
\eeqB

The complete expression for the amplitude is
\beq \label{eq:Sg:fullScatterSigma}
  \begin{split} M= 	e \phis^\dagger  \Bigg \{
		A_0 - \frac{g-1}{2m^2}\left( \v{S} \cdot \v{E} \times \v{p} + \frac{1}{4}\Sigma_i \Sigma_j \partial_i E_j \right)
		+ \frac{ \v{p} \cdot \v{A} }{m} 
			\\ - [g-1] \frac{ (\v{B} \cdot \v{p} )(\v{S} \cdot \v{p})}{m^2} 
				- \v{S} \cdot \v{B} \frac{ \v{p}^2 }{m^2} - \frac{g}{2m} \v{S} \cdot \v{B} 
	\Bigg \} \phis
\end{split}
\eeq


There is one last vestige of the relativistic notation: the structures $\Sigma_i$ are not appropriate for comparing to the nonrelativistic theory.  In a previous section, identities were developed for scalars and symmetric, traceless tensors built out of $\Sigma$.  So first, express $\Sigma_i \Sigma_j$ in such terms:

\beq
	\Sigma_i \Sigma_j \partial_i E_j 
		=	\frac{1}{2} \partial_i E_j (\Sigma_i \Sigma_j + \Sigma_j \Sigma_i - \frac{2}{3} \delta_{ij} \gv{\Sigma}^2) 
			+ \frac{1}{3} \grad \cdot \v{E} \Sigma^2
\eeq

Then using  \eqref{eq:Sg:SigTens}, this may be written
\beq
	\Sigma_i \Sigma_j \partial_i E_j 
		= \frac{1}{3} \grad \cdot \v{E} \Sigma^2
			- \frac{\lambda}{2} \partial_i E_j (S_i S_j + S_j S_i - \frac{2}{3} \v{S}^2 ) 
		= \frac{1}{3} \grad \cdot \v{E} \Sigma^2
			- \frac{\lambda}{2} \partial_i E_j Q_{ij}
\eeq
The spin dependence found in \eqref{eq:Sg:SigScalars} and \eqref{eq:Sg:SigTens} is most easily written separately for integer and half-integer cases.

For integer spin 
\beqa
	(\Sigma^2)_{1} = 4I		&\hspace{4em}&	\lambda_1 = -\frac{4}{2I-1}	
\eeqa
While for half-integer 
\beqa
	(\Sigma^2)_{\frac{1}{2}} = 4I+1		&\hspace{4em}&	\lambda_{\frac{1}{2}} = -\frac{4}{2I}	
\eeqa

Where $I$ is the magnitude of the particle's spin.


\beq \label{eq:Sg:fullScatter}
  \begin{split} M= 	e \phis^\dagger  \Bigg \{
		A_0 
		+ \frac{ \v{p} \cdot \v{A} }{m}
		- \frac{g-1}{2m^2}\left( \v{S} \cdot \v{E} \times \v{p} + \frac{\Sigma^2}{12} \grad \cdot \v{E} + \frac{\lambda}{8}Q_{ij} \right)
		\\ - [g-1] \frac{ (\v{B} \cdot \v{p} )(\v{S} \cdot \v{p})}{m^2} 
				- \v{S} \cdot \v{B} \frac{ \v{p}^2 }{m^2} - \frac{g}{2m} \v{S} \cdot \v{B} 
	\Bigg \} \phis
\end{split}
\eeq

\input{Comparison}



\section{Interpretation of universality in light of BMT equation}
Many of the coefficients are found to be independent of spin.  In the derivation above it is unclear if there is any underlying reason for this.  Another line of logic, that connects the result to the Bargmann-Michel-Telegdi (BMT) equation, helps explain this universality.  The BMT gives the relativistic equation of motion for a particle's spin.  This result is derived from relativistic constraints on the form of this equation, as well as demanding the appropriate nonrelativistic limit.

The time evolution of a particle's spin is related to the Hamiltonian of a system by the Heisenberg equation
\beq
	\dot{\v{S}} = i[H, \v{S}].
\eeq
In regular quantum mechanics, the only part of the Hamiltonian that doesn't commute with spin is $H_{SB}= - 2 \mu \v{S} \cdot \v{B}$.  So the equation for $\dot{S}$ is 
\beq \label{eq:Sum:dotS}
	\dot{\v{S}}	= g \frac{e}{2m} \v{S} \times \v{B}.
\eeq


The idea is to find the relativistic generalisation of this equation.  First let $a$ be a (pseudo) four-vector that is the relativistic generalisation of $\v{S}$.  In the rest frame it is $(0, \v{S})$, and its form in other frames can be found by boosting.  The proper time evolution of $a$ can be highly constrained if only terms linear in the electromagnetic field tensor $F^{\mu\nu}$ are considered, and derivatives of that field are not considered.  Then the only other quantities involved will be $a_\mu$ and the velocity $u = p/m$.    With such constraints, only two independent terms are possible, and the time evolution must have the form
\beq
	\d{a^\mu}{\tau} = 
		\alpha F^{\mu\nu} a_\nu + \beta u^\nu F^{\mu \lambda} u_\mu a_\lambda.
\eeq

The constants $\alpha$ and $\beta$ can be determined by comparison to the nonrelativistic limits.  In the rest frame, $a = (0, \v{S})$ and $u=(1, 0)$.  Then 
\beq
	\d{a^i}{\tau} = \alpha F^{ij} a_j,
\eeq
or in terms of $\v{S}$
\beq
	\d{\v{S}}{\tau} = \alpha \v{S} \times \v{B}
\eeq
Comparing to the nonrelativistic limit \eqref{eq:Sum:dotS}, it follows that $\alpha = g \frac{e}{2m}$.  

To fix $\beta$, consider the time evolution of the velocity.  According to the classical motion of a particle in a field, 
\beq
	m \d{u^\mu}{\tau} = e F^{\mu\nu} u_\nu.
\eeq

With $a \cdot u=0$, it follows that
\beq
	u \cdot \d{a}{\tau} = -a \cdot \d{u}{\tau}
		= \frac{e}{m}F^{\mu\nu} u_\mu a_\nu.
\eeq

Dotting $u$ into the evolution equation earlier gives
\beq
	u \cdot \d{a}{\tau}  = \alpha F^{\mu\nu} a_\nu u_\mu + \beta u^2 F^{\mu\lambda} u_\mu a_\lambda,
\eeq
and substituting the values of $\alpha$ and $u \cdot \d{a}{\tau}$,
\beq
	\frac{e}{m} F^{\mu\nu} u_\mu a_\nu	=	\left( g \frac{e}{2m} + \beta \right)  F^{\mu \nu} u_\mu a_\nu 
\eeq
So 
\beq
	\beta = - (g-2) \frac{e}{2m}.
\eeq 

So the original equation for spin evolution is
\beq
\d{a^\mu}{\tau} = 
		g \frac{e }{2m} F^{\mu\nu} a_\nu - (g-2) \frac{e }{2m} u^\nu F^{\mu \lambda} u_\mu a_\lambda.
\eeq
Written in terms of nonrelativistic quantities, this becomes
\beq
	\d{\v{S}}{t} =
		\frac{e}{2m} \v{S} \times \left\{ 
			\left( g- 2 + \frac{2}{\gamma}\right ) \v{B}\
			- \frac{(g-2)\gamma}{1+\gamma} (\v{v} \cdot \v{B}) \v{v} 
			+ \left( g - \frac{2\gamma}{1 + \gamma} \right ) \v{E} \times \v{v}
		\right \}.
\eeq

Keeping only first order relativistic corrections, and writing $\v{p}-e\v{A}$ instead of $\v{v}$, this is

\beq \label{eq:Sum:NRevol}
		\d{\v{S}}{t} =
		\frac{e}{2m} \v{S} \times \left\{ 
			\left( g -  \frac{ \v{p}^2}{m^2} \right ) \v{B}
			- \frac{(g-2)}{2m^2} (\v{p} \cdot \v{B}) \v{p} 
			+ \left( g - 1 \right ) \frac{ \v{E} \times (\v{p} - e\v{A})}{m}
		\right \}.
\eeq

This equation must match the condition that $\dot{\v{S}} = i[H, S]$.  This implies that terms linear in spin are
\beq	H_S =
		\frac{e}{2m} \v{S} \cdot \left\{ 
			\left( g -  \frac{ \v{p}^2}{m^2} \right ) \v{B}
			- \frac{(g-2)}{2m^2} (\v{p} \cdot \v{B}) \v{p} 
			+ \left( g - 1 \right ) \frac{ \v{E} \times (\v{p} - e\v{A})}{m}
		\right \}.
\eeq 

The logic that led up to \eqref{eq:Sum:NRevol} depended only on relativistic symmetries and classical limits.  These considerations would hold for particles of any spin.  So the partial Hamiltonian is universal, and requires the equivalent coefficients in \eqref{eq:Sg:nrLFull} to likewise be universal.