Spin-half.tex

\chapter{Spin one-half: Derivation of nonrelativistic Hamiltonian}

\section{Structure of the spin one-half theory}

The goal is to obtain a nonrelativistic theory from the relativistic.  To that end it'll help to have a clear understanding of the structure of the two theories.

\subsection{Relativistic framework for spin one-half}
In the relativistic theory, the electrons are part of a fermion field that also includes the positron anti-particle.  Since both are spin-1/2, each has two spin orientations.  There are, then, a total of four degrees of freedom. 

The relativistic Lagrangian for the fermion fields is
\beqa
\mathcal{L} &=&	
	\Psigbar(i \partial \cdot \gamma - e A \cdot \gamma - m)\Psig  - \frac{1}{4} F_{\mu\nu} F^{\mu\nu}.
\eeqa

The fermion fields are $\Psig$ ($\Psi$ being reserved for the fermion fields in the nonrelativistic theory), the photon field is $A$.   $m$ is the particle's mass, and $e$ the electron's charge.  The $\gamma$ matrices mix the different components.  %TODO better explanation of gamma matrices

It will be convenient to work in a representation which already suggests the nonrelativistic behavior.  At low momenta, it should be expected that the free electron and positron fields act approximately as independent fields.  This is exactly the case for the Dirac representation.   In the rest frame, a free particle can be said to be definitively an electron or positron, and in the Dirac representation these correspond to the upper and lower parts of the bispinor.  

In this representation, the gamma matrices are written
\beq
	\gamma^0 = \Mblock{1}{0}{0}{-1},
\eeq

\beq
	\gamma^i = \Mblock{0}{\sigma_i}{-\sigma_i}{0}.
\eeq	




The gamma matrices by themselves do not form a complete basis for this space.  To find such a basis products of the matrices can be considered.  It will make sense to consider combinations such that bilinears are Hermitian.  
 

There is the identity --- such bilinears transform as a scalar.

Symmetric combinations of gamma matrices need not be considered because $\{ \gamma^\mu, \gamma^\nu \}= g^{\mu\nu}$.  The antisymmetric combinations are explicitly
 \beqa
	  {[\gamma^0, \gamma^i]}
		&=&  2 \gamma^0 \gamma^i = \begin{pmatrix}	0 & 2\sigma_i \\ 2\sigma_i & 0\end{pmatrix},	\\
	  {[\gamma^i, \gamma^j ]}
		&=&	 \begin{pmatrix}	-2i\epsilon_{ijk}\sigma_k & 0 \\ 0 & -2i\epsilon_{ijk}\sigma_k\end{pmatrix}.
\eeqa
Using these a tensor like structure arises:
\beq
	\sigma^{\mu \nu} = i \frac{1}{2} [ \gamma^\mu, \gamma^\nu].	%defined as in Peskin p.50
\eeq

The specific form in the Dirac representation is
\beq
	\sigma^{ij} = \frac{i}{2} [ \gamma^i, \gamma^j] 
		= \epsilon_{ijk} \Mblock{\sigma_k}{0}{0}{\sigma_k},
\eeq
\beq
	\sigma^{0i} = i \gamma^0 \gamma^i 
			= \Mblock{0}{i \sigma_i}{i \sigma_i}{0}.
\eeq

The product of each gamma matrix in turn gives a pseudo-scalar:
\beq
	\gamma^5 = i \gamma^0 \gamma^1 \gamma^2 \gamma^3 = \Mblock{0}{1}{1}{0}.
\eeq
 
 And the product of $\gamma^5$ and $\gamma^\mu$ gives a pseudo-vector
 \beq
 	\gamma^5 \gamma^\mu .
 \eeq
 
 
 
Solutions of the Dirac equation in momentum space are the spinors $\sr$.  They have four degrees of freedom, and can be treated as a bispinor consisting of an upper and lower spinor.  In the chiral basis the upper and lower components have opposite helicity; in the Dirac basis, opposite charge:
\beq
	\sr = \begin{pmatrix} \eta \\ \chi \end{pmatrix}.
\eeq



 
\subsection{Nonrelativistic framework for spin one-half}
The nonrelativistic theory describes a single particle, without its oppositely charged antiparticle.  The nonrelativistic fields will be denoted by $\fnr$, which had two components.  The wave function that appears in amplitudes will be the Schrodinger-like $\phis$.  The relation between the relativistic and nonrelativistic amplitudes is obtained through the relations
\beqa
	\eta &=& \left( 1  - \frac{\v{p}^2}{8m^2} \right ) \phis,	\\
	\chi &=& 	\frac{ \gv{\sigma} \cdot \v{p}}{2m} \left(1 - \frac{3\v{p}^2}{8m^2} \right ) \phis.	\\
\eeqa



 \section{Foldy-Wouthyusen approach}
 
The goal is to derive a nonrelativistic Hamiltonian or Lagrangian starting from relativistic theory.  (Having obtained one, we can easily obtain the other, of course.)  One method is to take the relativistic equations of motion and use them to obtain a Schrodinger like equation.

The starting point is the relativistic equations of motion, which can come from the Lagrangian of the relativistic theory.  Those equations can then be written in terms of the noncovariant quantities that appear in the nonrelativistic theory.  In doing so, the energy of the particle will now explicitly appear.

The relativistic theory will contain not only the particle of interest (the electron) but also its anti-particle (the positron.)  The nonrelativistic theory should contain only the electron.  Before obtaining an expression for the nonrelativistic Hamiltonian it will be necessary to somehow disentangle the two fields.  This is impossible in the general case, but as long as the energy and momenta in question are nonrelativistic, can be accomplished to any desired order.

Formally this is accomplished by the Foldy-Wouthyusen transformation, the result of which is that all operators are diagonal, the coupling between the particle and anti-particle suppressed to which ever order is desired.  However, practically the same result can be obtained by examining the normalisation of the two theory's particles.  By demanding that the Schrodinger like wave functions are appropriately normalized, the relationship between relativistic and nonrelativistic spinors can be established.

The result of this procedure will be an equation for the energy of the electron, accurate at some order in the nonrelativistic expansion.  However, it will not perfectly replicate the predictions of the high energy theory.  Unlike the process of NRQED, it does not truly incorporate the high energy sector of the theory.



\subsection{Equations of motion}

To reestablish the problem considered, the system to be examined is an electron placed in a loosely bound system with another charged particle, subject to an infinitesimal and constant magnetic field.  There will be, because of the bound system, an electric field acting on the electron as well as the external magnetic field.  When recoil effects are ignored, the electric field can just be taken as given.

The corrections to the gyromagnetic ratio of the electron are to be established.  There are two small scales that appear in the problem, the velocity $v$ of the electron and the infinitesimally small magnetic field $eB$.  The precision desired requires terms of up to order $mv^4$ and $ (e/m)Bv^2$.

The starting point will be the relativistic Lagrangian of Dirac.  However, remember that the technique to be used simply ignores behavior introduced by the high energy sector of the theory, even if it might effect the low energy behavior.  One such effect is corrections to the free gyromagnetic ratio of the electron, which first arise when one loop diagrams are considered.  Without such corrections the $g$-factor will be exactly $g=2$.

Knowing that there will actually be bound-state corrections proportional to $g-2$, it is necessary to somehow include this anomalous term.  The way to do so is to introduce a new local interaction into the Lagrangian, coming from the high energy radiative corrections which dress the electron vertex.  The Lagrangian to be used is, then
\beqa
\mathcal{L} &=&	
	\Psigbar(\cancel{D} - m)\Psig + \frac{1}{2} \mu' \Psigbar \sigma^{\mu\nu}F_{\mu\nu} \Psig.
\eeqa
%TODO check \mu_0
$\mu'$ is the correction to the classical moment $\mu_0 = \frac{e}{2m}$, and is equal to $(g-2)/2 \mu_0$.  $D$ is the long derivative $\partial + ieA$.


From this Lagrangian the equations of motion of the particle may be obtained from the Euler-Lagrange method.  The Euler-Lagrange equation is
\beq
\pd{ \mathcal{L}}{{ \Psigbar } } - \partial_\mu \frac{\partial \mathcal{L}} {\partial (\partial_\mu { \Psigbar )} } 
	= 0	.
\eeq

In the Lagrangian above, we can consider that all differential operators act only on the right field $\Psi$.  (This freedom of choice comes from being able to rewrite the Lagrangian through integration by parts, without changing its physical meaning.)  So the second term in the Euler-Lagrange equation can be ignored, and after differentiating with respect to $\bar{\Psi}$ the following equation is obtained:
\beq
	(i\cancel{D} - m + \frac{1}{2} \mu'  \sigma^{\mu\nu}F_{\mu\nu} )\Psig = 0	.
\eeq
Writing explicitly in terms of the $\gamma$ matrices, this is
\beq \label{eq:Sh:eom}
	\left( (i\partial_\mu- eA_\mu)\gamma^\mu -m + i\mu' \frac{1}{4}[ \gamma^\mu, \gamma^\nu]F_{\mu\nu} \right) \Psig
		=0 .
\eeq
This equation of motion is invariant under Lorentz transformations.  It is written in terms of the Dirac bispinor $\Psi$, external fields $A_\mu$ and $F_{\mu\nu}$, and the gamma matrices.  To apply in to a nonrelativistic problem, the very first step will be to rewrite it in terms of the sorts of quantities that appear in that domain: three-vectors and scalars.

The scalars that appear will be $\partial_0= \partial_t$ and $A_0 = \Phi$.  The external fields $\v{E}$ and $\v{B}$ will appear explicitly, while the vector field $\v{A}$ will appear in the gauge-invariant operator $\gv{\pi} = \v{p} - e\v{A}$.  The gamma matrices can be written in terms of the Pauli spin matrices $\gv{\sigma}$.  Finally, the bispinor will be written in terms of its upper and lower components.

Of the terms that appear in \eqref{eq:Sh:eom}, all except the last are trivial to write in this manner.  To deal with that last term, the antisymmetric tensor $\sigma^{\mu\nu} =  \frac{1}{2}[\gamma^\mu, \gamma^\nu]$ needs to be written explicitly.

Using the antisymmetry of $\sigma^{\mu\nu}$, and that we deal with time-independent fields: 
\beqa
	F_{\mu\nu} \sigma^{\mu\nu} &=& F_{i}\sigma^{ij} - F_{0i}\sigma^{0i} 	- F_{i0}\sigma^{i0} +F_{00}\sigma^{00}	\\
		&=&	F_{ij} \sigma^{ij} -2F_{0i} \sigma^{0i}	\\
		&=&	2 \partial_i A_j \sigma^{ij} - 2\partial_i \Phi \sigma^{0i}	\\
		&=&	-2i \begin{pmatrix} \sigdot{B} & 0 \\ 0 & \sigdot{B}\end{pmatrix}	
			-2 \begin{pmatrix} 0 & \sigdot{E} \\ \sigdot{E} & 0 \end{pmatrix}.	
\eeqa

So far the discussion has been in position space.  To work out the nonrelativistic form it will be easier to talk about the equations of motion directly in terms of energy and momentum.  So replace $i\partial_t$ with $p_0$, and $i\partial_i$ with $p_i$.  Likewise, replace the gauge invariant derivative  $iD_i = \pi_i = p_i - eA_i$.

A solution of definite momentum $p$ to the equation is written in terms of upper and lower components
\beq
	u = \begin{pmatrix} \eta \\ \chi \end{pmatrix}.
\eeq


With these considerations, the \eqref{eq:Sh:eom} can be rewritten acting explicitly on the bispinor.
\scriptsize
\beq \label{eq:Sh:matrixEOM}
	\left\{
		\begin{pmatrix}
			p_0 - e\Phi	- m &	0	\\
			0	&	-p_0 + e\Phi - m	\\
		\end{pmatrix}
		+
		\begin{pmatrix}	0 & -\sigdotg{\pi} \\  \sigdotg{\pi} & 0 \end{pmatrix} 
		+\mu'\left [
			\begin{pmatrix}
				\sigdot{B} & 0 \\ 0 & \sigdot{B}
			\end{pmatrix}
			-i \begin{pmatrix}
				0 & \sigdot{E} \\ \sigdot{E} & 0
			\end{pmatrix}
		\right ] 
	\right\} \begin{pmatrix} \eta \\ \chi \end{pmatrix}
		= 0.
\eeq
\normalsize
This gives rise to exact coupled equations for $\eta$ and $\chi$.  So far this is in principle the same as the relativistic equation, only the form in which it is written is non covariant.

\subsection{Nonrelativistic limit}

The particle under consideration is a nonrelativistic electron.  Roughly, the expectation is that $\eta$ corresponds to the electron field and $\chi$ to that of the positron.  The off-diagonal terms in the equation above represent some sort of mixing between the electron and positron: the electron wave function still has some small positron component, that decreases as momentum is decreased.    The off-diagonal component that \emph{does} vanish at 0 momentum is proportional to $\mu'$, the term introduced to account for a high-energy process.

The upshot is that although the equation above is really a set of coupled equations for $\eta$ and $\chi$, $\chi$ will be small compared to $\eta$ --- the very leading order diagonal term will indicate that $\chi \sim \sigdotg{\pi} \eta$.

Because the off diagonal terms are small, the set of coupled equations may be solved perturbatively.  The particular quantity of interest is the nonrelativistic energy of the particle $\epsilon = p_0 - m$.  For a free particle, this would be
\beq
	\epsilon = p_0 - m \approx \frac{\v{p}^2}{2m} + \frac{\v{p}^4}{8m^3} + \mathcal{O}\left (\frac{\v{p}^6}{m^5} \right ).
\eeq  

In order to perform this perturbative analysis the order of various terms needs to be established.  It's evident that at leading order $\epsilon \sim mv^2$.  From earlier analysis,   $\Phi\sim mv^2$, $\v{\pi} \sim mv$, and $\v{E} \sim m^2v^3$.

First, find an expression for $\chi$ in terms of $\eta$.  The second of the set of equations represented by \eqref{eq:Sh:matrixEOM} is
\beq
	(-p_0 + e\Phi - m) \chi + \sigdotg{\pi} \eta + \mu'( \sigdot{B} \chi - i \sigdot{E}  \eta).
\eeq
Writing $p_0 = \epsilon + m$, and grouping terms, the result is that
\beq
	\left( \epsilon + 2m - \mu' \sigdot{B} \right ) \chi = \left( \sigdotg{\pi} - i\sigdot{E} \right) \eta	.
\eeq
It is necessary now to approximate $\chi$ in terms of $\eta$.  Because $\epsilon$ and $\abs{B}$ are smaller than $m$, and only second order terms are needed for the final result:
\beq
	\chi \approx	\frac{1}{2m} \left ( 1- \frac{\epsilon - e\Phi - \mu' \sigdot{B}}{2m} \right ) (\sigdotg{\pi} - i\mu' \sigdot{E} )\phi.
\eeq

With this expression $\chi$ may be eliminated from the first of the set of equations (at least at the necessary order).  The resulting equation will only involve $\eta$, and so may be used to solve for energy $\epsilon$ of  $\eta$.

The original equation is
\beq
	(p_0 - e\Phi - m) \eta - \sigdotg{\pi} \chi + \mu'( \sigdot{B}\eta  - i \sigdot{E} \chi ) .
\eeq
So again using $p_0 - m = \epsilon$
\beq
	\epsilon \eta 	= (e\Phi - \mu' \sigdot{B} )\eta + (\sigdot{E} + \sigdotg{\pi}) \chi	.
\eeq
Now the expression for $\chi$ in terms of $\eta$ may be used.
\beq
 	\epsilon \eta \approx (e\Phi - \mu' \sigdot{B} )\eta + (\sigdot{E} + \sigdotg{\pi})  \frac{1}{2m} \left ( 1- \frac{\epsilon - e\Phi - \mu' \sigdot{B}}{2m} \right ) (\sigdotg{\pi} - i\mu' \sigdot{E} )\eta .
\eeq				
Writing the $1/m$ and $1/m^2$ terms separately:
\beq
\begin{split}				\epsilon \eta		\approx& \left \{
		e\Phi - \mu' \sigdot{B} + \frac{ \exminus \explus}{2m}	\right. \\
		& \left. +\frac{1}{4m^2} \exminus (\mu' \sigdot{B} - [\epsilon - e\Phi]) \explus 	
	\right \} \eta	.
\end{split}
\eeq

Several of the terms are of too high order to consider.  A term with both $E$ and $B$, for instance, will be of higher order than $(e/m)\abs{B} v^2$.  Likewise, a term of $E \Phi$ or $E \epsilon$ is also too small.  Dropping all such:
\beq
	\epsilon \eta \approx \left \{
				e\Phi - \mu' \sigdot{B} + \frac{(\sigdotg{\pi})^2 - i\mu' [\sigdotg{\pi}, \sigdot{E}]}{2m}
				+\frac{1}{4m^2} \sigdotg{\pi} (\mu' \sigdot{B} - [\epsilon - e\Phi]) \sigdotg{\pi} 
			\right \} \eta.
\eeq
This is an expression for the energy $\epsilon$ of the particle, in terms of operators.  This will yield the nonrelativistic Hamiltonian.  There is still some manipulation required, though, because the right hand side also contains $\epsilon$.  But since the leading order terms don't, it may be perturbatively solved for.  (The above expression could be simplified somewhat, using the properties of $\sigma$ matrices for instance, but for now it is more convenient to write it compactly.)

To that end, the Hamiltonian can be split into leading order and second order terms.  The leading order will be of $mv^2$ and $(e/m)B$, while the next order will be suppressed by an additional factor of $v^2$.  Because the magnetic field is infinitesimally small no $B^2$ terms are needed. 
 Since the leading order term in H is $\mathcal{O}(mv^2)$, this suggests we split it into two parts: $H = H_0 + H_1 +\mathcal{O}(mv^6, (e/m) B v^4)$, where $H_1$ consists of only second order terms.
\beqa
	\hat{H_0} 
			&=&  e\Phi - \mu' \sigdot{B} + \frac{ (\sigdotg{\pi})^2}{2m},	\\
	\hat{H_1} 
			&=& -\frac{i\mu'}{2m}  [\sigdotg{\pi}, \sigdot{E}]
				+\frac{1}{4m^2} \sigdotg{\pi} (\mu' \sigdot{B} - [\epsilon - e\Phi]) \sigdotg{\pi} .
\eeqa
$H_1$ contains $\epsilon$, along with other terms of total order $mv^2$.  So to eliminate $\epsilon$ from $H_1$ it'll only be necessary to find it to leading order.
\beqa
	\epsilon \eta
			&=& \left(\hat{H_0} + \mathcal{O}(mv^4) \right)\eta 										\\
			&\approx& \left(e\Phi - \mu' \sigdot{B} + \frac{ (\sigdotg{\pi})^2}{2m}	\right) \eta.			
\eeqa
The operators on the right hand side, operating on $\eta$, produce $\epsilon$.  The combination actually needed is $\sigdotg{\pi} \epsilon \sigdotg{\pi}$.  To that end, start with $\sigdotg{\pi}^2 \epsilon$ and use commutation relations.
\beqa
	(\sigdotg{\pi})^2 (\epsilon - e\Phi) \eta		
			&=&	(\sigdotg{\pi})^2 \left ( \frac{(\sigdotg{\pi})^2}{2m}- \mu' \sigdot{B} \right )	\eta	\\
	\sigdotg{\pi} (\epsilon - e\Phi) \sigdotg{\pi}\eta
			&=&	\left( \frac{(\sigdotg{\pi})^4}{2m} 
				- \mu' (\sigdotg{\pi})^2 \sigdot{B} 
				- \sigdotg{\pi}[e\Phi, \sigdotg{\pi}]\right ) \eta
\eeqa
With this $\epsilon$ is eliminated, leaving:
\beq
	\hat{H}_1	=
		-\frac{i\mu'}{2m}  [\sigdotg{\pi}, \sigdot{E}]
		+\frac{1}{4m^2}\left(
			\mu' \sigdotg{\pi} \sigdot{B} \sigdotg{\pi}
			- \frac{(\sigdotg{\pi})^4}{2m} 
			+ \mu' (\sigdotg{\pi})^2 \sigdot{B} 
			+ \sigdotg{\pi}[e\Phi, \sigdotg{\pi}]
		\right).
\eeq
Some terms couple $A$ and $B$; they can be dropped.  Some simplification of the structures involving $\sigma$ matrices can be done.  To start with, simplify $(\sigdotg{\pi})^2$.
\beq
	(\sigdotg{\pi})^2 = \sigma_i \sigma_j \pi_i \pi_j = \pi^2 - i\epsilon_{ijk} \pi_i \pi_j \sigma_k.
\eeq
Since $\v{p} \times \v{p} = \v{A} \times \v{A} = 0$, from $\gv{\pi} \times \gv{\pi}$ only the cross terms survive:
\beq
	(\sigdotg{\pi})^2 = \pi^2 - i e \epsilon_{ijk}(p_i A_j - A_i p_j) = \pi^2 - e \sigdot{B}.
\eeq
Looking at the terms $\mu' \sigdotg{\pi} \sigdot{B} \sigdotg{\pi} + \mu' (\sigdotg{\pi})^2 \sigdot{B}$, they contain an anticommutator involving $B$ and $p$.  Because the magnetic field is assumed to be constant, $p$ and $B$ commute, so: 
\beqa
\{ \sigdot{B}, \sigdot{p} \}	&=&		B_i p_j \{\sigma_i, \sigma_j\}	\\
		&=&	2\v{B}\cdot \v{p}.
\eeqa
The commutator of $\Phi$ and a derivative operator should give the electric field $E$:
\beq
	[ \Phi, \sigdotg{\pi} ]	= [ \Phi, p_i] \sigma_i = -iE_i \sigma_i = -i\sigdot{E} .
\eeq
Using these identities, the Hamiltonian can be expressed as:
\beq
	\hat{H_0} 
			=  e\Phi - \mu' \sigdot{B} + \frac{ \pi^2}{2m} - \frac{e}{2m} \sigdot{B},	\\
\eeq
\beq
	\hat{H}_1	=
		- \frac{\pi^4}{8m^3}  
		+ e\frac{p^2}{4m^3} \sigdot{B}
		-i \frac{\sigdotg{\pi} \sigdot{E}}{4m^2} 
		+ \mu' \left(	
			\frac{ \sigdot{p} \v{B}\cdot \v{p} }{2m^2}
			-\frac{i[\sigdotg{\pi}, \sigdot{E}]}{2m} 
		\right ).
\eeq
There are still some simplifications that can be made to terms quadratic in $\sigma$, but it'll be more convenient for now to keep $H$ written as is.

\subsection{Foldy-Wouthyusen Transform}

To find a complete description of a single nonrelativistic particle in normal quantum mechanics, we must work in a basis where the lower component $\chi$ is truly negligible, at least at the desired order.  While there exists a formal technique for finding this Foldy-Wouthyusen transformation, for our purposes we can simply demand that the wave function after transformation $\phis = (1 + \Delta)\eta$ obeys $\langle \phis, \phis \rangle = 1$.  This follows from the necessity of probability conservation.

To find the necessary transformation, we can use the relativistic current density, and demand it equal that of the nonrelativistic Schrodinger-like wave functions.   Using the expression for $\chi$ in terms of $\eta$:
\beqa
	\int d^3x (\eta^\dagger \eta + \chi^\dagger \chi) 	
		&=& \int d^3x 	\left[ \eta^\dagger \phi + \left (\frac{\sigdotg{\pi}}{2m} \phi \right)^\dagger
										   \left (\frac{\sigdotg{\pi}}{2m} \eta \right)
						\right ]	\\
		&=& \int d^3x 	\eta^\dagger \left[ 1 + \frac{ (\sigdotg{\pi})^2 }{4m^2} \right ] \eta.
\eeqa
Since we know that $\langle \phis, \phis \rangle=1$ this shows that if $\eta = \left( 1 - \frac{ (\sigdotg{\pi})^2 }{8m^2} \right )\phis$, the current conservation works out correctly.


We now need to find the form of $\hat{H}$ after this transformation.  For now work with the general form:
\beq
	\epsilon \eta = (\hat{H}_0 + \hat{H}_1) \eta .
\eeq
After changing to the Schrodinger like wave functions, this becomes:
\beq
	\epsilon \left (1 - \frac{(\sigdotg{\pi})^2}{8m^2} \right )\phi_S 
		=  (\hat{H}_0 + \hat{H}_1) \left(1 - \frac{(\sigdotg{\pi})^2}{8m^2}\right)\phi_S .
\eeq
To the order needed the inverse of $1 + \sigdotg{\pi}^2 / 8m^2 $ is just $1 - \sigdotg{\pi}^2 / 8m^2 $, so eliminating that on the left hand side gives:
\beq
	\epsilon \phi_S =  (1 + \frac{(\sigdotg{\pi})^2}{8m^2})(\hat{H}_0 + \hat{H}_1) (1 - \frac{(\sigdotg{\pi})^2}{8m^2})\phi_S .
\eeq
Since $H_1$ is already second order, these further corrections don't involve it directly.  Expressing the result as a commutator:
\beq
	\epsilon \phi_S = \left (  \hat{H}_0 + \frac{1}{8m^2}[(\sigdotg{\pi})^2, \hat{H}_0] + \hat{H}_1 \right )\phi_S .
\eeq
So under the FW transformation, the leading order term is unchanged, and the second order term is:
\beq
	 \hat{H}_1 \to \hat{H}_1' = \hat{H}_1 + \frac{1}{8m^2}[(\sigdotg{\pi})^2, \hat{H}_0] .
\eeq

The final step is to simplify the commutator $ [(\sigdotg{\pi})^2, \hat{H}_0] $.
  
\beq [(\sigdotg{\pi})^2, \hat{H}_0] = [(\sigdotg{\pi})^2, e\Phi - \mu' \sigdot{B} + \frac {(\sigdotg{\pi})^2}{2m}] . \eeq

Obviously $(\sigdotg{\pi})^2$ commutes with itself, so that term vanishes.  Since $\sigdot{B}$ is constant, that commutator will also disappear.  Writing $\sigdotg{\pi}$ as shown earlier:
$$[(\sigdotg{\pi})^2, \mu' \sigdot{B}] =[\pi^2 - e \sigdot{B}, \mu'\sigdot{B}] = 0 .$$


The non-trivial part is the commutation of the derivative operators with the electric potential $\Phi$. 
\beqa
[(\sigdotg{\pi})^2, \hat{H}_0] &=& [(\sigdotg{\pi})^2, e\Phi] \\
		&=&	e( \sigdotg{\pi}[\sigdotg{\pi},  \Phi] + [\sigdotg{\pi},  \Phi]\sigdotg{\pi}) \\
		&=&	i e( \sigdotg{\pi} \sigdot{E} + \sigdot{E} \sigdotg{\pi} ). 
\eeqa
So, writing down the new $H_1'$:
\beqa
\hat{H}_1' 
	&=&  \hat{H}_1 + \frac{i e}{8m^2}( \sigdotg{\pi} \sigdot{E} + \sigdot{E} \sigdotg{\pi} )	\\
	&=&	 -\frac{\pi^4}{8m^3} + e \frac{p^2}{4m^3}\sigdot{B} - \frac{ie}{8m^2}[\sigdot{E},\sigdotg{\pi}]
		 + \mu' \left( \frac{ (\sigdot{p}) (B \cdot p) }{2m^2} - \frac{i}{2m}[\sigdot{E},\sigdotg{\pi}] \right).
\eeqa

\subsection*{Nonrelativistic Hamiltonian}
From the relativistic equations of motion, a nonrelativistic Hamiltonian was defined in terms of Schrodinger-like wave functions that conserve probability.  The final result may now be written down.  For comparison with other work, rather than writing in terms of $\mu'$, it will be written in terms of the gyromagnetic ratio $g$.  Because terms proportional to $g-2$ may enter separately, it is convenient to express all $g$ dependent terms as linear combinations of $g$ and $g-2$. 

So the entire Hamiltonian, using $\mu' = \frac{g-2}{2}\mu_0 = \frac{g-2}{2}\frac{e}{2m} $, is:
\beqa
H	&=&	e\Phi  + \frac{ \pi^2}{2m} 
		- \frac{\pi^4}{8m^3} 
		- (1 + \frac{g-2}{2})\frac{e}{2m} \sigdot{B}	
		+ e \frac{p^2}{4m^3}\sigdot{B}
\\&&  	+ \frac{ie}{8m^2}(1 + (g-2))[\sigdot{E},\sigdotg{\pi}]
		+ (g-2) \frac{e}{2m}  \frac{ (\sigdot{p}) (\v{B} \cdot \v{p}) }{4m^2}  	\\
&=&		e\Phi  + \frac{ \pi^2}{2m}
		-\frac{\pi^4}{8m^3} 
		- \frac{e}{2m} \Big\{
			\frac{g}{2}\sigdot{B} - \frac{p^2}{2m^2}\sigdot{B}
\\&&		- (g-2) \frac{ (\sigdot{p}) (\v{B} \cdot \v{p}) }{4m^2} 
			+(g-1) \gv{\sigma} \cdot (\v{E} \times \gv{\pi})
		\Big \}	
\\&=&
		e\Phi  + \frac{ \pi^2}{2m} -\frac{\pi^4}{8m^3} 
		- \frac{e}{2m} \Big\{
			\frac{g}{2} \left( 1 - \frac{p^2}{2m^2} \right) \sigdot{B} + \frac{g-2}{2} \frac{p^2}{2m^2}\sigdot{B}
	\\&&		- \frac{g-2}{2} \frac{ (\sigdot{p}) (\v	{B} \cdot \v{p}) }{2m^2} 
			+\left(\frac{g}{2} + \frac{g-2}{2}\right) \gv{\sigma} \cdot (\v{E} \times \gv{\pi})
		\Big \}	.
\eeqa
			


 
 
 

%skeleton of calculation
\section{Effective Lagrangian approach}
%define spinors and structures
Instead of starting from the relativistic equations of motion and rewriting them in the form of a Schrodinger like equation, the method of NRQED can be employed.  The nonrelativistic Lagrangian can be written in the most general way, but ignoring all possible terms that will have too small a contribution to calculations.  The coefficients before each term will be unknown, but there is a straightforward method of fixing them.

Within the realm of NRQEDs validity, it should produce the same predictions as QED.  The same physical process can be calculated in both frameworks, and then the results compared.  The calculation in QED will then fix the coefficients in NRQED.  In comparing the two calculations it will be necessary to write the QED results nonrelativistically, in the same manner as was done for the equation of motion.
%TODO mention free vs. nonfree spinors?

For spin one-half, up to the order desired, the form of the NRQED Lagrangian involving two fermion fields has been derived in the previous chapter.  It is
\beq \label{eq:Sh:nrL-2f}
\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_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}
		\Bigg \} \fnr.
\end{split}
\eeq
Each term contains two fermion fields, and zero, one or two powers of the photon field.  Terms with a different number of photon fields, of course, correspond to different physical processes.  But because gauge invariance is a necessary feature of the theory, the coefficients of many terms involving the photon field are constrained to be the same as that of terms with a smaller number of such fields.  For instance, the gauge invariant term $\v{D}^2/(2m)$ is: 
\beq	
	\frac{\v{D}^2}{2m} = \frac{ \grad^2 - i e (\grad \cdot \v{A} + e \v{A} \cdot \grad ) - e^2 \v{A}^2 }{2m}.
\eeq
The first term above, containing $\grad^2$, is a purely kinetic term.  The other two represent interactions with one or two photon fields.  To fix these coefficients, two physical processes could be calculated and compared with the QED results, but in the end they must have the same coefficient as guaranteed by gauge invariance.

Obviously not every term goes this same way.  The term with $\v{S} \cdot \v{B}$, for instance, is in and of itself gauge-invariant, and so there is no option but to calculate it from a process involving a single photon.  For the necessary terms appearing in the above Lagrangian, it would suffice to consider just the one photon processes.  However, since in principle some of the interesting coefficients could be calculated from two photon diagrams, this also will be done. 

To fix these terms, some physical process must be chosen.  For the one photon terms, scattering off an external field will be calculated.  For the two photon terms, Compton scattering will be used.

\section{Electron scattering off an external field in QED}
%First form
To fix all the terms in the NRQED Lagrangian which have a single power of the photon field, it suffices to calculate the scattering of an electron off an external field, as represented (in QED) by the diagram

\begin{center}   \includegraphics[scale=0.8]{eps/blob-wave} \end{center} 
	%TODO add additional diagrams


The leading order contributions (in QED) to this interaction come from the fundamental electron vertex.  There are radiative corrections to this process, starting at the one loop order.  In principle, such calculations (up to some desired order in $\alpha$) should be included in the QED calculation.  

However, it turns out that the actual \emph{form} of the interaction is highly constrained by the symmetries of the theory.  No matter the source of contributions to the vertex, their effects can be incorporated into two coefficients or form factors.  The NRQED coefficients can then be written in terms of these form factors, which can later be calculated to whatever precision is necessary.  

The first symmetry that constrains the interaction is that it must be invariant under Lorentz transformations.  Since every term involves the photon field $A_\mu$ and has external fermion legs, then the interaction must be proportional to the general form:
\beq
	A_\mu \srb(p') \Gamma^\mu(p', p) \sr(p),
\eeq 	
and $\srb(p') \Gamma^\mu(p', p) \sr(p) $ must transform as a Lorentz vector.  If it did not, the whole would not be Lorentz invariant.  To leading order $\Gamma^\mu = \gamma^\mu$, since the fundamental vertex is just that.  The corrections, whatever the exact details of the processes which produce them, can only depend on the momenta $p$ and $p'$, in addition to constants $m$ and $e$, and such structures as may act upon the spinors.

A basis for such structures is known, with well defined Lorentz transformations.  That gives scalar, vector, tensor, pseudo-vector, and pseudoscalar terms.  From the momenta in the problem can be constructed scalar and vector quantities.  What symmetries control the allowed terms?

In addition to proper Lorentz invariance, it is necessary that the interaction as a whole preserve parity.  Since $A$ is a vector, the term $\srb \Gamma^\mu \sr$ must also behave as a vector under parity transformations.  And there is no Lorentz invariant combination of external momenta that can be written that is not even under parity.   Because of this, there can be no contribution from the pseudovector and pseudoscalar bilinears.

The vector bilinear $\srb \gamma^\mu \sr$ already has the correct transformation properties.  From the scalar bilinear $\srb \sr$, a vector can be constructed by adjoining a single power of the momentum.  (As, for example, $p^\mu \srb \sr$.)  And after contracting one index of the tensor with a momentum vector, it will also behave as a vector.

The momenta terms available are $p$ and $p'$.  However, the terms constructed from them are not independent, because each must separately obey current conservation.  So whatever terms go into $\Gamma^\mu$, $q_\mu \srb \Gamma^\mu \sr = 0$.  The unique scalar term will be
\beq
	\frac{p^\mu + p'^\mu}{2m} \srb \sr,
\eeq     
which vanishes because $(p'+p) \cdot (p'-p) = p'^2 - p^2 = m^2- m^2=0$.

The unique tensor term allowed will be
\beq
	\frac{q_\nu}{2m} \srb \sigma^{\mu\nu} \sr,
\eeq
which conserves current because $\sigma^{\mu\nu}$ is antisymmetric.  So from these considerations, the general form of the vertex will have three terms, each with a momentum dependent coefficient:
\beq \label{eq:Sh:Gamma}
	\srb \Gamma^\mu \sr 
		= 	c_1(p, p') \frac{p^\mu + p'^\mu}{2m} \srb \sr
			+ c_2(p, p') \srb \gamma^\mu \sr
			+ c_3(p, p') \frac{q_\mu}{2m} \srb \sigma^{\mu\nu} \sr.
\eeq

However, there is one more consideration.  From the Dirac equation can be derived the Gordon identity, which relates these three terms:
\beq \label{eq:Sh:gordon}
	\srb \gamma^\mu \sr = \srb \left( \frac{p^\mu + p'^\mu }{2m} + \frac{i \sigma^{\mu\nu}q_\nu}{2m} \right ) \sr.
\eeq

With this, any one of the terms in \eqref{eq:Sh:Gamma} can be rewritten as some combination of the other two, and its coefficient effectively absorbed into the other two.  To write down the most general form of the vertex, one need only choose two of the three terms.  Which two to choose might depend on the nature of the calculation; at any rate, there are always three paths to go down.

In the calculation of scattering off an external field an agnostic approach will be taken at first.  The scattering amplitude is some combination of the three bilinears.  The goal is to rewrite the scattering amplitude in nonrelativistic language.  So, each bilinear in turn will be rewritten in this manner.

The vertex can be written in each of three ways.  The form factors will be defined with respect to the particular combination of $\gamma^\mu$ and $\sigma^{\mu\nu}$ terms.

\beq
	\Gamma^\mu = \gamma^\mu F_1(q^2) + i \frac{\sigma^{\mu\nu}q_\nu}{2m} F_2 (q^2).
\eeq

Using the Gordon identity to write this in terms of the scalar and vector bilinears, the result is
\beq
	\Gamma^\mu = \gamma^\mu [F_1(q^2) + F_2(q^2) ]  -  \frac{p^\mu  + p'^\mu }{2m}F_2 (q^2).
\eeq
And writing in terms of the scalar and tensor:
\beq
	\Gamma^\mu = \frac{p^\mu  + p'^\mu }{2m} F_1(q^2) + i \frac{\sigma^{\mu\nu}q_\nu}{2m} [F_1(q^2) + F_2(q^2) ] .
\eeq

%TODO move this section up a bit?
The momentum dependence can be uniquely written in terms of $q^2$.  The factors must be Lorentz invariant, so only such combinations of momenta can be considered.  From the momenta $p$ and $p'$, only three such quantities can be constructed.  But since the external fermions are on mass-shell, $p^2 = p'^2 = m^2$.  That leaves only $p \cdot p' = m^2 + p \cdot q$, and this can be related to $q^2$ by noting that since $p^2 = p'^2$,
\beq
	p^2 = p^2 + 2p\cdot q + q^2  \to q^2 = -\frac{1}{2} p \cdot q.
\eeq


Given that there exist these three ways to write the vertex, there are three ways to perform the calculation.  The scattering amplitude will have the form
\beq	
	i M = 	A_\mu \srb(p') \Gamma^\mu(p', p) \sr(p),
\eeq
and no matter which way $\Gamma^\mu$ is written, a nonrelativistic expansion in terms of $\phis$ will be needed.  For now an agnostic approach will be taken, and the expressions for each of the three possible bilinears found.


\subsection{Nonrelativistic expressions for the bilinears}
To compare with the NRQED scattering amplitude, everything needs to be written with consistent language.  We start with the relativistic bilinears, each of which behaves like a four vector, and appears in the amplitude dotted with the external field $A$.  Any dot products $a \cdot b$ should instead be written as $a_0 b_0 - \v{a} \cdot \v{b}$.  To this end, it might be necessary to treat the spatial and time-like components of the four-vector bilinears separately.

The bispinors $u$ will be first written in terms of upper and lower components $\eta$, $\chi$, and then in terms of the nonrelativistic wave spinors $w$.  The relationship between the two sets are
\begin{eqnarray}
	\label{eq:Sh:eta-w} \eta(p) &=& \left( 1 - \frac{\v{p}^2}{8m^2} \right ) \wx(p),	\\
	\label{eq:Sh:chi-w}  \chi(p)	&=& \frac{ \sigdot{p} }{2m} \left(1  - \frac{3\v{p}^2}{8m^2} \right ) \wx(p).
\end{eqnarray}
In writing in terms of the upper and lower components,  the explicit expressions of the $\gamma$ matrices, as well as $\sigma^{\mu\nu}$ will be needed.  

Of course the above expressions for the spinors in terms of $w$ is approximate.  In the NRQED Lagrangian, the terms we wish to fix involve $A_0$ with up to two powers of momentum (such as $\grad \cdot \v{E}$), and $A_i$ with up to three (as in $\sigdot{B} \v{p}^2$).  Because only the case of a constant magnetic field is needed, in any term which will explicitly contain $B$ higher derivatives may be ignored.  Throwing away unneeded terms, whatever is left can be used to calculate the scattering amplitude and compare with NRQED.

This same general procedure will be followed for each of the bilinears.

\subsubsection{Terms involving the scalar bilinear}
Start with the first bilinear, a scalar coupled with a momentum four-vector.  Rewrite it in terms of $\eta$ and $\chi$.
\beq
	(p + p')^\mu \srb \sr  = (p + p')^\mu \left( \eta^\dagger \eta - \chi^\dagger \chi \right ).
\eeq
Now express in terms of $\wx$, replacing $\eta$ and $\chi$ according to \eqref{eq:Sh:eta-w} and \eqref{eq:Sh:chi-w}.
\beq
	(p + p')^\mu \srb \sr = (p + p')^\mu \left \{
		\wxd \left( 1 - \frac{ \v{p'}^2}{8m^2} \right )  \left( 1 - \frac{ \v{p}^2}{8m^2} \right ) \wx
		- \wxd \left( \frac{ \gv{\sigma} \cdot \v{p'}}{2m} \frac{ \gv{\sigma} \cdot \v{p}}{2m} \right ) \wx \right \}.
\eeq 
Combining the terms, and dropping terms beyond the order needed:
\beq
	(p + p')^\mu \srb \sr = (p + p')^\mu \wxd \left( 
		1 -  \frac{ \v{p}^2 + \v{p'}^2}{8m^2} - \frac{ \gv{\sigma} \cdot \v{p'} \gv{\sigma} \cdot \v{p} }{4m^2} 
		\right ) \wx.
\eeq
The term quadratic in $\sigma$ can be simplified, using $\sigdot{p'} \sigdot{p} = \v{p} \cdot \v{p'} + i\gv{\sigma} \cdot \v{q} \times \v{p}$.
\beq \label{eq:Sh:Si}
	(p + p')^\mu \srb \sr  = (p + p')^\mu \wxd \left( 
		1 -  \frac{ \v{p}^2  +2 \v{p} \cdot \v{p'} +  \v{p'}}{8m^2} - \frac{i\gv{\sigma} \cdot \v{q} \times \v{p} }{4m^2} 
		\right ) \wx.
\eeq
The spatial part is straightforward, but the time-like part is 
\beq
	(p + p')^0 \srb \sr  = (p + p')_0 \wxd \left( 
		1 -  \frac{ \v{p}^2  +2 \v{p} \cdot \v{p'} +  \v{p'}^2 }{8m^2} - \frac{i\gv{\sigma} \cdot \v{q} \times \v{p} }{4m^2} 
		\right ) \wx.
\eeq
Approximating the relativistic energies gives $p_0 = m + \v{p}^2 / (2m) $.  So
\beq
	p_0 + p'_0 \approx 2m + \frac{ \v{p}^2 + \v{p'}^2 }{2m} = 2m\left( 1 +  \frac{ \v{p}^2 + \v{p'}^2 }{4m^2} \right ).
\eeq
Then using this correction to the leading order term gives the time like component:
\beq
	(p + p')^0 \srb \sr  = 2m  \wxd \left( 
		1 +  \frac{ \v{p}^2   - 2 \v{p} \cdot \v{p'} +  \v{p'} }{8m^2} - \frac{i\gv{\sigma} \cdot \v{q} \times \v{p} }{4m^2} 
		\right ) \wx,
\eeq
\beq \label{eq:Sh:S0}
	(p + p')^0 \srb \sr  \approx 2m  \wxd \left( 
		1 +  \frac{ \v{q}^2 }{8m^2} - \frac{i\gv{\sigma} \cdot \v{q} \times \v{p} }{4m^2} 
		\right ) \wx.
\eeq




\subsubsection{Terms involving the vector bilinear}
For the term $\srb \gamma^\mu \sr$ it'll be necessary to treat the spatial/time-like indices separately, since they have different spinor structure.

The time-like part is:
\beq
	\srb \gamma^0 u = u^\dagger u,
\eeq
which in terms of $\eta$ and $\chi$ is just
\beq
	= \eta^\dagger \eta + \chi^\dagger \chi.
\eeq
Then rewritten with $\wx$
\beq
	= 	\wxd \left( 1 - \frac{ \v{p'}^2}{8m^2} \right )  \left( 1 - \frac{ \v{p}^2}{8m^2} \right ) \wx
		+ \wxd \left( \frac{ \gv{\sigma} \cdot \v{p'}}{2m} \frac{ \gv{\sigma} \cdot \v{p}}{2m} \right ) \wx 
\eeq
Which is, at the order needed
\beq
	=	 \wxd \left( 
		1 -  \frac{ \v{p}^2 + \v{p'}^2}{8m^2} + \frac{ \gv{\sigma} \cdot \v{p'} \gv{\sigma} \cdot \v{p} }{4m^2} 
		\right ) \wx.
\eeq
Simplifying the last term, using $\sigma_i \sigma_j = \delta_{ij} + i\epsilon_{ijk}\sigma_k$, gives
\beq \label{eq:Sh:V0}
	\srb \gamma^0 \sr = 
			\wxd \left( 
		1 -  \frac{ \v{q}^2 }{8m^2} + \frac{i \gv{\sigma} \cdot \v{q} \times \v{p} }{4m^2} 
		\right ) \wx.
\eeq





The spatial part is
\beq
	\srb \gamma^i \sr = \sr^\dagger \gamma^0 \gamma^i \sr.
\eeq
Writing the matrices explicitly,
\beq
	 \srb^\dagger \begin{pmatrix}
		0 & \sigma_i \\ \sigma_i & 0 		
	\end{pmatrix} \sr,
\eeq
which in terms of spinors is
\beq
	 \eta^\dagger \sigma_i \chi + \chi^\dagger \sigma_i \eta.
\eeq
Replacing the spinors with $w$ gives
\beq
	 \wxd \left\{
		\left( 1 - \frac{ \v{p'}^2}{8m^2}   \right ) \frac{ \sigma_i \sigdot{p} }{2m} \left( 1 - \frac{3\v{p}^2}{8m^2} \right )
			+ \left( 1 - \frac{ 3\v{p'}^2}{8m^2} \right )  \frac{  \sigdot{p'} \sigma_i }{2m} \left( 1 - \frac{\v{p}^2}{8m^2} \right )
	\right\} \wx.
\eeq
Using $\v{p'}^2 = \v{p}^2$ gives
\beq
	= \wxd \left\{
		\frac{ \sigma_i \sigdot{p}  + \sigdot{p'} \sigma_i }{2m} \left( 1 -\frac{ \v{p}^2}{2m^2} \right )
	\right\} \wx.
\eeq
Then $\sigma_i \sigma_j = \delta_{ij} + i\epsilon_{ijk} \sigma_k$
\beq
	= \wxd \left\{
		\frac{ p_i + \epsilon_{ijk} p_j \sigma_k + p'_i  + \epsilon_{jik} p'_j \sigma_k }{2m} \left( 1 -\frac{ \v{p}^2}{2m^2} \right )
	\right\} \wx.
\eeq
So finally
\beq \label{eq:Sh:Vi}
	\srb \gamma^i \sr  = \wxd \left\{
		\frac{ p_i + p'_i - \epsilon_{ijk} q_j \sigma_k }{2m} \left( 1 -\frac{ \v{p}^2}{2m^2} \right )
	\right\} \wx
\eeq



\subsubsection{Terms involving the tensor bilinear}
The tensor term is subject to an additional simplification.  Because the process under considering is elastic scattering off an external static field, terms involving $q_0$ can be dropped.  Under this approximation $q_\nu \sigma^{\mu\nu} \approx  q_j \sigma^{\mu j}$.

Dealing first with the case where $\mu=i$, write the tensor structure explicitly as a matrix:  
\beq
	\srb  \frac{i}{2m} q_j \sigma^{ij} \sr 
		=  \frac{i \epsilon_{ijk} q_j}{2m} \srb \Mblock{\sigma_k}{0}{0}{\sigma_k} \sr.
\eeq
Write in terms of spinors this is
\beq
	 \frac{i \epsilon_{ijk} q_j}{2m} \left( \eta^\dagger \sigma_k \eta - \chi^\dagger \sigma_k \chi \right ),
\eeq
and then in terms of $\wx$
\beq
	= \frac{i \epsilon_{ijk} q_j}{2m} \wxd \left \{
		\left( 1 - \frac{\v{p'}^2}{8m^2} \right ) \sigma_k \left( 1 - \frac{\v{p'}^2}{8m^2} \right )- \frac{ \gv{\sigma} \cdot \v{p'} \sigma_k \gv{\sigma} \cdot \v{p} }{4m^2} \wx
	\right \}.
\eeq
There now appears a term with a triple product of $\sigma$ matrices.  That can be simplified with the following expression:
\beq
	\sigma_a \sigma_b \sigma_c = \sigma_a (\delta_{bc} + i\epsilon_{bcd}\sigma_d)
		=	\sigma_a \delta_{bc} - \sigma_b \delta_{ca} + \sigma_c \delta_{ab} + i \epsilon_{abc}.
\eeq
Using that identity, 
\beq
	\srb  \frac{i}{2m} q_j \sigma^{ij} \sr 
		= \frac{i \epsilon_{ijk} q_j}{2m} \wxd \left \{
			\sigma_k \left( 1 - \frac{\v{p'}^2  + \v{p}^2}{8m^2} \right ) - \frac{ \gv{\sigma} \cdot (\v{p} + \v{p'})p_k - \sigma_k \v{p} \cdot \v{p'} + i \epsilon_{akc} q_a p_c }{4m^2} 
		\right \} \wx .
\eeq
This can be further simplified by combining the like terms $\sigma_k (\v{p'}^2 + \v{p}^2 ) - 2 \sigma_k \v{p} \cdot \v{p'} = \sigma_k \v{q}^2$.
\beq
	\srb  \frac{i}{2m} q_j \sigma^{ij} \sr 
		= \frac{i \epsilon_{ijk} q_j}{2m} \wxd \left \{
			\sigma_k \left( 1 - \frac{\v{q}^2}{8m^2} \right ) - \frac{ \gv{\sigma} \cdot (\v{p} + \v{p'})p_k + i \epsilon_{akc} q_a p_c }{4m^2} 
		\right \} \wx .
\eeq
Now it is necessary to consider exactly what derivatives of the field $A_i$ are to be kept.  The assumption is that $B$ is constant and so $\partial_i B_j = 0$.  Contracted with $A_i$ above, $\epsilon_{ijk} A_i q_j \sim B_k$.  So besides the leading factor, no terms with $q$ are needed.  Applying this simplification,
\beq \label{eq:Sh:Ti}
	\srb  \frac{i}{2m} q_j \sigma^{ij} \sr 
		= \frac{i \epsilon_{ijk} q_j}{2m} \wxd \left \{
			\sigma_k - \frac{ \gv{\sigma} \cdot \v{p} p_k  }{2m^2} 
		\right \} \wx.
\eeq






The case with $\mu=0$ goes as follows.
\beq
	\srb  \frac{i}{2m} q_j \sigma^{0j} \sr
		=	 - \frac{q_j}{2m} \srb \gamma^0 \gamma^j \sr 		
 		=  - \frac{q_j}{2m} \sr^\dagger \gamma^j \sr.
\eeq
Explicitly in matrix form 
\beq
	\srb  \frac{i}{2m} q_j \sigma^{0j} \sr  = - \frac{q_j}{2m} \sr^\dagger \Mblock{0}{\sigma_j}{\sigma_j}{0} \sr. 
\eeq
 Written in terms of spinors, and Galilean three-vector $q^j= -q_j$.
 \beq
 	\srb  \frac{i}{2m} q_j \sigma^{0j} \sr = \frac{q^j}{2m}  \left( \eta^\dagger \sigma_j \chi - \chi^\dagger \sigma_j \eta \right ).
 \eeq
 And then in terms of $w$:
\beq
	=  \frac{q^j}{2m}  \wxd \left \{
		\left(1 - \frac{\v{p'}^2}{8m^2} \right )  \frac{ \sigma_j \gv{\sigma} \cdot \v{p} }{2m} \left(1 - \frac{3\v{p}^2}{8m^2} \right )
		- \left(1 - \frac{3 \v{p'}^2}{8m^2} \right ) \frac{\gv{\sigma} \cdot \v{p'} \sigma_j  }{2m} \left(1 - \frac{\v{p}^2}{8m^2} \right )
	\right \} \wx.
\eeq

This bilinear is contracted with $A_0$, and so actually, only terms involving two additional powers of momenta need be kept.
\beq
	\srb  \frac{i}{2m} q_j \sigma^{0j} \sr =  \frac{q^j}{2m}  \wxd \left \{
		  \frac{ \sigma_j \gv{\sigma} \cdot \v{p} }{2m} 
		-  \frac{\gv{\sigma} \cdot \v{p'} \sigma_j  }{2m} 
	\right \} \wx,
\eeq
or simplifying using the commutator of $\sigma$ matrices
\beq 
	\srb  \frac{i}{2m} q_j \sigma^{0j} \sr =   \frac{q^j}{2m}  \wxd \left \{
		\frac{ 2 i\epsilon_{jik} p_i \sigma_k  - q_i - i\epsilon_{ijk} q_j \sigma_k }{2m} 
	\right \} \wx.
\eeq
One term above vanishes because of symmetry
\beq \label{eq:Sh:T0}
\srb  \frac{i}{2m} q_j \sigma^{0j} \sr
	=  -  \wxd \left \{
			\frac{  \v{q}^2  }{4m^2} 
			- \frac{  i\gv{\sigma} \cdot \v{q} \times \v{p}   }{2m^2} 
	\right \} \wx.
\eeq


\subsubsection{Contraction of bilinears with $A_0$}
The scattering amplitude is composed of bilinears coupled to the field $A$.  The transferred momentum $\v{q}$ becomes a derivative of this field, $\v{q} \to -i\grad$ or equivalently $\grad \to i\v{q}$.  Before finally writing down the full vertex, the coupling of $A$ and the bilinears should be worked out, including appropriate transformations involving $\v{q}$.

First consider terms with $A_0(q)$.  The relation between $\v{q}$ and $\v{E}$ is that $\v{E} = - \grad A_0 = i\v{q} A_0$.  Terms quadratic in $q$ give $\v{q}^2 A_0 = -\grad^2 A_0 =  \grad \cdot \v{E}$.


The term with $A_0$ coupled with the scalar bilinear \eqref{eq:Sh:S0} gives
\beqaL  \label{eq:Sh:SA0}
	eA_0  (p + p')^0 \srb \sr &\approx&
		eA_0 \wxd \left( 
		1 +  \frac{ \v{q}^2 }{8m^2} - \frac{i\gv{\sigma} \cdot \v{q} \times \v{p} }{4m^2} 
		\right ) \wx	\\
	& \approx &
			 \wxd \left( 
				eA_0  +  \frac{ e \grad \cdot \v{E} }{8m^2} - \frac{e \gv{\sigma} \cdot \v{E} \times \v{p} }{4m^2} 
				\right ) \wx.
\eeqaL
The term involving the vector bilinear term \eqref{eq:Sh:V0} is
\beqaL    \label{eq:Sh:VA0}
	eA_0 \srb \gamma^0 u &=& 
			eA_0 \wxd \left( 
		1 -  \frac{ \v{q}^2 }{8m^2} + \frac{i \gv{\sigma} \cdot \v{q} \times \v{p} }{4m^2} 
		\right ) \wx	\\
	&=& 
		 \wxd \left( 
		eA_0  - \frac{ e \grad \cdot \v{E} }{8m^2} + \frac{ e\gv{\sigma} \cdot \v{E} \times \v{p} }{4m^2} 
		\right ) \wx.
\eeqaL

The third type of term, with the tensor bilinear appearing:
\beqaL  \label{eq:Sh:TA0}
e A_0 \srb  \frac{i}{2m} q_j \sigma^{0j} \sr
	&=&  - e A_0 \wxd\left \{
			\frac{  \v{q}^2  }{4m^2} 
			- \frac{  i\gv{\sigma} \cdot \v{q} \times \v{p}   }{2m^2} 
	\right \} \wx	\\
	&=&  -   \wxd \left \{
			\frac{e  \grad \cdot \v{E} }{4m^2} - \frac{e \gv{\sigma} \cdot \v{E} \times \v{p} }{2m^2} 
	\right \} \wx.
\eeqaL

\subsubsection{Contraction of bilinears with $A_i$}

Terms with $\v{A}$ and $\v{q}$ give rise to terms with $\v{B} = \grad \times \v{A}$.  There are some that are simple to transform, containing $\epsilon_{ijk} iq_i A_j \to B_k$.  In others, the way $B$ arises is less clear.  

Consider the expression $(\v{p} + \v{p'})\cdot \v{A}  (\gv{\sigma} \cdot \v{p} \times i\v{q})$.  Kinematic constraints will help in expressing this term.  First, it is convenient to develop an identity for $iq_i (\v{p} + \v{p'}) \cdot \v{A}$.  Using $\epsilon_{ijk} B_k = i(q_i A_j - q_j A_i)$:
\beq
	(p + p')_i \epsilon_{ijk} B_k = (p + p')_i  \left( i q_i A_j - i q_j A_i \right ).
\eeq
There are additional conditions coming from the fact that this is elastic scattering.  Since in this case $ ( \v{p'} + \v{p}) \cdot \v{q} = \v{p'}^2 - \v{p}^2 = 0$, the $(p+p')_i q_i$ term vanishes, leaving
\beq
	(p + p')_i \epsilon_{ijk} B_k = -i (p + p')_i q_j A_i = -i q_j(\v{p} + \v{p'}) \cdot \v{A} .
\eeq

This gives the identity
\beq \label{eq:Sh:pqA}
i q_j (\v{p} + \v{p'}) \cdot \v{A}  =  - \epsilon_{ijk} (p + p')_i  B_k.
\eeq
This identity helps reduce the more complicated term:
\beqa
(\v{p} + \v{p'}) \cdot \v{A} ( \gv{\sigma} \cdot \v{p} \times i\v{q} )
	&=&	 \sigma_i p_j  \epsilon_{ijk} [i q_k (\v{p} + \v{p'}) \cdot \v{A} ]	\\
	&=&	- \sigma_i p_j \epsilon_{ijk} [ \epsilon_{\ell k m} (p + p')_\ell B_ m]	\\
	&=&	 \sigma_i p_j \epsilon_{ijk} [ \epsilon_{\ell m k} (p + p')_\ell B_ m]	\\
	&=&	\sigma_i p_j (p+p')_\ell B_m [\delta_{i \ell} \delta_{jm} - \delta_{i m} \delta_{j \ell}]	\\
	&=&	\gv{\sigma} \cdot  (\v{p} + \v{p'}) \v{B} \cdot \v{p} - \gv{\sigma} \cdot \v{B} (\v{p} + \v{p'}) \cdot \v{p}.
\eeqa
For the case of a constant $\v{B}$, any terms of the type $q_i B_j$ vanish.  So the above reduces to
\beq \label{eq:Sh:A-p-id}
(\v{p} + \v{p'}) \cdot \v{A}  (\gv{\sigma} \cdot \v{p} \times i\v{q} )
	= 2 \{  (\gv{\sigma} \cdot  \v{p})( \v{B} \cdot \v{p}) - \v{p}^2 \gv{\sigma} \cdot \v{B} \}.
\eeq

Now consider $A_i$ coupled with each of the particular bilinear expressions.  (Remember that here $A_i = -\v{A}_i$.)  With the term involving a scalar bilinear \eqref{eq:Sh:Si}
\beqaL \label{eq:Sh:SAi}
	\frac{A_i (p + p')^i}{2m} \srb \sr  &=& - \frac{\v{A} \cdot (\v{p} + \v{p'} ) }{2m} \wxd \left( 
		1 -  \frac{ \v{p}^2  +2 \v{p} \cdot \v{p'} +  \v{p'}}{8m^2} - \frac{i\gv{\sigma} \cdot \v{q} \times \v{p} }{4m^2} 
		\right ) \wx	\\
	 &=&	- \wxd  \left [ \frac{e \v{A} \cdot (\v{p} + \v{p'} ) }{2m} \left(1 - \frac{\v{p}^2}{4m^2} \right )  \right ] \wx
	- \wxd  \left [  \frac{ 	 e (\gv{\sigma} \cdot  \v{p})( \v{B} \cdot \v{p}) - e \v{p}^2 \gv{\sigma} \cdot \v{B} }{4m^2} \right ] \wx . \nonumber
\eeqaL
The term involving a vector bilinear coupled to $A_i$ comes from \eqref{eq:Sh:Vi}
\beqaL \label{eq:Sh:VAi}
A_i \srb \gamma^i \sr  &=& \wxd \left\{
		A_i \frac{ p_i + p'_i - \epsilon_{ijk} q_j \sigma_k }{2m} \left( 1 -\frac{ \v{p}^2}{2m^2} \right )
	\right\} \wx \\
	&=&- \wxd \left \{
		 \frac{ \v{A} \cdot (\v{p} + \v{p'}) + \gv{\sigma} \cdot \v{B} }{2m} \left(1 - \frac{\v{p}}{2m^2} \right ) 
	\right\} \wx.
\eeqaL
Finally the term with a tensor gives
\beqaL  \label{eq:Sh:TAi}
	A_i \srb  \frac{i}{2m} q_j \sigma^{ij} \sr 
		&=& \frac{i \epsilon_{ijk} A_i q_j}{2m} \wxd \left \{
			\sigma_k - \frac{ \gv{\sigma} \cdot \v{p} p_k  }{2m^2} 
		\right \} \wx \\ 
		&=&  - \wxd \left \{ 
				\frac{ \gv{\sigma} \cdot \v{B} }{2m} - \frac{ (\gv{\sigma} \cdot \v{p}) (\v{B} \cdot \v{p}) }{4m^2}
			\right \} \wx.
\eeqaL

% Start of comparison
\subsubsection{Full vertex}

Now that the three bilinears have been calculated, the complete scattering amplitude can be written down.  Each of the three forms should prove equivalent.  To simplify comparison, the coupling to $A_0$ and $A_i$ can be treated separately.

The first has $	\Gamma^\mu = \gamma^\mu F_1(q^2) + i \frac{\sigma^{\mu\nu}q_\nu}{2m} F_2 (q^2)$.  Using \eqref{eq:Sh:VA0} and \eqref{eq:Sh:TA0}
\beq
	eA_0 \srb  \Gamma^0 \sr = 
		 F_1   \wxd \left( 
		eA_0  - \frac{ e \grad \cdot \v{E} }{8m^2} + \frac{ e\gv{\sigma} \cdot \v{E} \times \v{p} }{4m^2} 
		\right ) \wx
		- F_2    \wxd \left \{
			\frac{e  \grad \cdot \v{E} }{4m^2} - \frac{e \gv{\sigma} \cdot \v{E} \times \v{p} }{2m^2} 
	\right \} \wx.
\eeq
The second is	$\Gamma^\mu = \gamma^\mu [F_1(q^2) + F_2(q^2) ]  -  \frac{p^\mu  + p'^\mu }{2m}F_2 (q^2)$.  Using \eqref{eq:Sh:VA0} and \eqref{eq:Sh:SA0}.
\beq
	eA_0 \srb  \Gamma^0 \sr = 
		 (F_1 + F_2)  \wxd \left( 
		eA_0  - \frac{ e \grad \cdot \v{E} }{8m^2} + \frac{ e\gv{\sigma} \cdot \v{E} \times \v{p} }{4m^2} 
		\right ) \wx
		- F_2   \wxd \left( 
		eA_0  +  \frac{ e \grad \cdot \v{E} }{8m^2} - \frac{e \gv{\sigma} \cdot \v{E} \times \v{p} }{4m^2} 
		\right ) \wx.
\eeq
The third combination is $	\Gamma^\mu = \frac{p^\mu  + p'^\mu }{2m} F_1(q^2) + i \frac{\sigma^{\mu\nu}q_\nu}{2m} [F_1(q^2) + F_2(q^2) ] $.  Using \eqref{eq:Sh:SA0} and \eqref{eq:Sh:TA0}
\beq
	 A_0 \srb  \Gamma^0 \sr = 
		F_1 \wxd \left( 
		eA_0  +  \frac{ e \grad \cdot \v{E} }{8m^2} - \frac{e \gv{\sigma} \cdot \v{E} \times \v{p} }{4m^2} 
		\right ) \wx
		- [F_1 + F_2] \wxd \left \{
			\frac{e  \grad \cdot \v{E} }{4m^2} - \frac{e \gv{\sigma} \cdot \v{E} \times \v{p} }{2m^2} 
	\right \} \wx.
\eeq
Taking any of these three results and collecting like terms gives the result
\beq
		 A_0 \srb  \Gamma^0 \sr = \wxd \left(  
			F_1  eA_0
			+ [F_1 + 2 F_2] \left [ 
				\frac{e \gv{\sigma} \cdot \v{E} \times \v{p} }{4m^2}  - \frac{e  \grad \cdot \v{E} }{8m^2}
			\right ]
		\right ) \wx,
\eeq
so the calculations are consistent with the Gordon identity.

Now turning to the coupling with $\v{A}$.  The first has $	\Gamma^\mu = \gamma^\mu F_1(q^2) + i \frac{\sigma^{\mu\nu}q_\nu}{2m} F_2 (q^2)$.  Using \eqref{eq:Sh:VAi} and \eqref{eq:Sh:TAi}
\beq
\begin{split}
	e A_i \srb \Gamma^i \sr = & 
		 - e F_1 \wxd \left \{
		 \frac{ \v{A} \cdot (\v{p} + \v{p'}) + \gv{\sigma} \cdot \v{B} }{2m} \left(1 - \frac{\v{p}}{2m^2} \right ) 
	\right\} \wx
	\\& - e F_2  \wxd \left \{ \frac{ \gv{\sigma} \cdot \v{B} }{2m} - \frac{ (\gv{\sigma} \cdot \v{p}) (\v{B} \cdot \v{p}) }{4m^2}
			\right \} \wx.
\end{split}
\eeq
The second is	$\Gamma^\mu = \gamma^\mu [F_1(q^2) + F_2(q^2) ]  -  \frac{p^\mu  + p'^\mu }{2m}F_2 (q^2)$.  Using \eqref{eq:Sh:VAi} and \eqref{eq:Sh:SAi}:
\beq \begin{split}
	e A_i \srb \Gamma^i \sr = & 
		- e [F_1 + F_2] \wxd \left \{
		 \frac{  \v{A} \cdot (\v{p} + \v{p'}) +  \gv{\sigma} \cdot \v{B} }{2m} \left(1 - \frac{\v{p}}{2m^2} \right ) 
	\right\} \wx
	\\& + e F_2 \wxd  \left [ \frac{e \v{A} \cdot (\v{p} + \v{p'} ) }{2m} \left(1 - \frac{\v{p}^2}{4m^2} \right )  \right ] \wx
	+ eF_2 \wxd  \left [  \frac{ 	 e (\gv{\sigma} \cdot  \v{p})( \v{B} \cdot \v{p}) - e \v{p}^2 \gv{\sigma} \cdot \v{B} }{4m^2} \right ] \wx.
\end{split} \eeq
The third combination is $\Gamma^\mu = \frac{p^\mu  + p'^\mu }{2m} F_1(q^2) + i \frac{\sigma^{\mu\nu}q_\nu}{2m} [F_1(q^2) + F_2(q^2) ] $.  Using \eqref{eq:Sh:SAi} and \eqref{eq:Sh:TAi}
\beq \begin{split}
	e A_i \srb \Gamma^i \sr = & 
		 e F_1 \wxd  \left [ \frac{e \v{A} \cdot (\v{p} + \v{p'} ) }{2m} \left(1 - \frac{\v{p}^2}{4m^2} \right )  \right ] \wx
		 + e F_1 \wxd  \left [  \frac{ (\gv{\sigma} \cdot  \v{p})( \v{B} \cdot \v{p}) - e \v{p}^2 \gv{\sigma} \cdot \v{B} }{4m^2} \right ] \wx
	\\&	-[F_1 + F_2] \wxd \left \{ \frac{ \gv{\sigma} \cdot \v{B} }{2m} - \frac{ (\gv{\sigma} \cdot \v{p}) (\v{B} \cdot \v{p}) }{4m^2}
			\right \} \wx.
\end{split} \eeq

For any of these paths, the total result is that
\beq \begin{split}
	e A_i \srb \Gamma^i \sr =  \wxd \Bigg \{
		- \frac{ e\v{A} \cdot (\v{p} + \v{p'} ) }{2m} \left( 1 - \frac{\v{p}^2}{2m^2} \right )
		- [F_1 + F_2] \frac{ e \gv{\sigma} \cdot \v{B} }{2m} 
	\\	+ F_1 \frac{ e \gv{\sigma} \cdot \v{B} \v{p}^2 }{4m^3} 
		+ F_2 \frac{ (\gv{\sigma} \cdot \v{p}) (\v{B} \cdot \v{p}) }{4m^2}
	\Bigg \} \wx.
\end{split}
\eeq

\section{Electron scattering off an external field in NRQED}
Earlier, the Lagrangian \eqref{eq:nr:L-half} for spin one-half particles was calculated.  To calculate the scattering off an external field in this theory, only those terms that include one power of the electromagnetic field are needed.  Call that part of the Lagrangian containing such terms $\mathcal{L}_A$.  This set of terms need not, by themselves, be gauge invariant.  However, previously care was taken to group terms in a gauge invariant way.  So first the terms of $\mathcal{L}_A$ must be disentangled from the gauge invariant terms (involving one or more powers of $\v{D}$).



First take the kinetic terms, involving simple powers of $\v{D}$.  $\v{D} = \grad - ie\v{A}$, so
\beq
	\v{D}^2 = (\nabla_i - ieA_i)(\nabla_i - ieA_i).
\eeq
It is a mixture of terms with two, one, or no powers of the external field $A$.  If just the terms with one power of $\v{A}$ are included, what remains is
\beq
	-ie( A_i \nabla_i + A_i \nabla_i = -ie \{A_i, \nabla_i\}.
\eeq
So from $\v{D}^2/ 2m$ emerges, in $\mathcal{L}_A$,  $-ie \{\nabla_i, A_i\}/2m$.

The second kinetic term is $\v{D}^4$:
\beq
	\v{D}^4 = (\nabla_i - ieA_i)(\nabla_i - ieA_i) (\nabla_j - ieA_j)(\nabla_j - ieA_j),
\eeq
or
\beq
	\v{D}^4 = ( \grad^2 - ie \{A_i, \nabla_i\} - e^2 \v{A}^2)( \grad^2 - ie \{A_j, \nabla_j\} - e^2 \v{A}^2).
\eeq
Keeping again only the terms with a single power of $A$, what remains may be expressed as the double anti-commutator
\beq
	-ie \{ \grad^2, \{\nabla_i, A_i \} \} .
\eeq

There are then several terms involving one or more powers of $D$ combined with either $E$ or $B$.  Since only terms with a single power of the field are of interest, in all such terms $\v{D}$ may be replaced with $\grad$.



So finally, writing all such terms from the original Lagrangian involved in scattering off an external field leaves:
\small
\beq 
\label{eq:Sh:nrL-A}
\begin{split} 
\mathcal{L}_A = \fnrb \Bigg (  -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^{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} \Bigg )\fnr.
\end{split} \eeq
\normalsize



\subsubsection{Calculation of one photon scattering}



The form of the Lagrangian that will account for all interactions with a single photon is now established.  From this, the idea is to calculate a particular process: scattering off an external field, with incoming momentum $\v{p}$, outgoing $\v{p'}$, and $\v{q} = \v{p'} - \v{p}$. 
One diagram could be associated with each term in \eqref{eq:Sh:nrL-A}, but the total amplitude is just going to be the sum of all these one-photon vertices.  These of course can just be read off directly from the Lagrangian.

It is necessary to switch to the language of momentum space.  The recipe is this: replace the fields $\Psi$ with the spinors $\phi$, and any operator $\grad$ acting will become $i\v{p}$ if it acts on the right, $i\v{p'}$ if it is to the left.

The terms originating from $D^2$ are:
\beq
	- ie  \frac{ \{\nabla_i, A_i \} }{2m} ,
\eeq
which in position space become
\beq
	e \frac{ \v{A} \cdot (\v{p} + \v{p'}) }{2m}.
\eeq

While the terms arising from $D^4$
\beq
-ie \frac{ \{\grad^2, \{\nabla_i, A_i \}  \} }{8m^3} ,
\eeq
become
\beq
- e \frac{ \v{A} \cdot (\v{p} + \v{p'}) (\v{p'}^2 + \v{p}^2 ) }{8m^3} .
\eeq


Some of the expressions involving $\grad$ and $\v{E}$ can be simplified.  Because $E_i = -\partial_i \Phi$
\beq
	\v{\grad} \times \v{E} - \v{E} \times \v{\grad} =  - 2 \v{E} \times \v{\grad}.
\eeq
And also
\beq
\v{\grad} \smalldot \v{E} - \v{E} \smalldot \v{\grad} = (\partial_i E_i).
\eeq


After all these considerations, the total scattering amplitude off an external field may be written down, before any assumptions are applied:



 \mbox{
\begin{minipage}{1.0in}
   \includegraphics[scale=0.5]{eps/blob-wave} 
\end{minipage}
=
\begin{minipage}{1.6in}
\scriptsize
\beq
\begin{split}
	 ie\wnrb \Bigg(& - A_0 +    \frac{ \v{A} \cdot (\v{p} + \v{p'}) }{2m} 
		- \frac{  \v{A} \cdot (\v{p} + \v{p'}) \v{p}^2 + \v{p'}^2 \v{A} \cdot (\v{p} + \v{p'}) }{8m^3} 
	\\&	+ c_F  \frac{\v{S} \smalldot \v{B}} {2m}   	
		+ c_D \frac{ ( \partial_i E_i ) }{8m^2}	
		+ c^{1}_S \frac{  \v{E} \times \v{p} }{4m^2}
		- c_{W_1}  \frac{  (\v{S} \smalldot \v{B} ) (\v{p}^2 + \v{p'}^2)  }{8m^3}
	\\&	+ c_{W_2} \frac{  (\v{S} \smalldot \v{B} ) (\v{p} \cdot \v{p'}) }{4m^3}
		-  c_{p'p} \frac{  (\v{S} \smalldot \v{p'}) (\v{B} \smalldot \v{p}) + (\v{B} \smalldot \v{p'}) (\v{S} \smalldot \v{p}) }{8m^3} \Bigg )\wnr .
\end{split}
\eeq 

\end{minipage}
}



This can be simplified somewhat.  Choose the gauge such that $\nabla_i A_i = 0$.  If elastic scattering is specified, then kinematics dictate that $\v{p'}^2 = \v{p}^2$.   Finally, if $\v{B}$ is constant, the $c_W$ terms become indistinguishable, since $[ \nabla_i, B_j] = 0$.    (It is only this last assumption that costs any information, since it becomes impossible to determine $c_{W_1}$ and $c_{W_2}$ separately. )  Then the scattering amplitude, as calculated from $\mathcal{L}_{NRQED}$, is:

\beq 
\begin{split} \label{eq:Sh:nrqedScatter}
	iM =&
		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^{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} \Bigg )\wnr .
\end{split}
\eeq




\section{Compton scattering in QED}
The relevant terms in the NRQED Lagrangian can all be fixed by the previous calculation of scattering off an external field, because even though there are terms involving two powers of the photon field, the requirement of gauge invariance means they share a coefficient with one photon terms.  However, for reasons of self consistency it would be good to check that the coefficients really do work out the same if calculated independently.

The easiest process to calculate involving two photons will be Compton scattering.   

\input{ComptonOneHalf}