text:
\section{\label{sec:main:main}Tools, Calculations and 4-Point Amplitudes of QED}

\subsection{\label{sec:main:MA}The Massless Limit of a Massive Theory}
Although we were able to find the amplitudes for $e\gamma\bar{e}h$ and $e\bar{e}\gamma\gamma$ using the $x$-factor vertices of \cite{Arkani-Hamed:2017jhn}, as we will see, we were unable to find the amplitude for $e\bar{e}\mu\bar{\mu}$ using this method (see App.~\ref{sec:Constructive Massless eemm}.)  However, we were able to find all the amplitudes if we replaced the massless photon with a massive photon and took the massless limit at the end.  In fact, we have never been unsuccessful calculating an amplitude if all the particles, both internal and external, are massive.  Moreover, when all the particles are massive, we have not needed to analytically continue the momenta or shift the spinors to get the correct results in any of our calculations so far (including \cite{Christensen:2019mch} as well as the amplitudes in this section).  On the other hand, when using a massive photon, we must calculate and add all the possible diagrams and there aren't any shortcuts that bypass diagrams as there are using a massless photon from the beginning.  Since this is the only method that always gives correct results, we detail it first and discuss the $x$-factor method using a massless photon in the next section.

In this section, we will describe these massive calculations and demonstrate how to calculate the amplitudes of this paper using the massless limit.
Most of the apparatus to do this is already present as we will see, but first we must modify the photon vertices to their massive form.  The electron- (muon-) photon vertex is just like the $Z$-boson vertex, but non chiral, and given by
\begin{equation}
    \mathcal{M}_{\textrm{eeA}} =
    \frac{e}{M_A}\left(\langle\mathbf{31}\rangle\lbrack\mathbf{23}\rbrack+\lbrack\mathbf{31}\rbrack\langle\mathbf{23}\rangle\right)\ .
    \label{eq:M_eeA massive vertex}
\end{equation}
There are two unphysical aspects introduced by this form of the vertex.  The first is that the photon is now taken as a spin-1 object and therefore has one extra (unphysical) degree of freedom, somewhat akin to the 2 extra (unphysical) degrees of freedom in field theory.  Furthermore, this unphysical degree of freedom must vanish by the end of the calculation for physical amplitudes, just like it does for field theory (and is proven by the Ward identities in field theory).  Though we don't have a proof that this always occurs in this spinor formulation, we have found that they do cancel in the calculations we have thus far performed.  That is, the final physical amplitudes agree with the results from Feynman diagrams.

Unfortunately, by adding an unphysical degree of freedom, it seems that we have ruined some of the motivation for this constructive formulation of particle physics.  Indeed, although this is true, we believe there might still be some reasons to pursue this approach anyway.  First of all, we see that we have only introduced one unphysical degree of freedom rather than two, so this formulation is still an improvement in principle, although it is still not fully satisfactory.  But, moreover, we might still find that the final resulting amplitudes are superior in their economy and insight compared to the results of Feynman diagrams.  This last point will only be known after many more amplitudes, including loop amplitudes, are worked out and compared to their field-theory equivalents.  At this point, we can only say that it looks promising, but is not conclusive.

The second point is that this vertex introduces division by the mass of the photon, which will be taken to zero, potentially introducing a singularity.  Once again, as in the case of the extra unphysical degree of freedom, we believe that this singularity is removable and that it will always exactly cancel by the end of the calculation.  As with the unphysical degree of freedom, we do not have a proof of this for all calculations, but we have found it to be true for the physical calculations of this paper, as we will see.

In order to calculate the amplitude, we begin by calculating the massive amplitude using the usual rules.  We then Taylor expand the photon spinors using the formulas given in \cite{Arkani-Hamed:2017jhn} and \cite{Christensen:2019mch}.  For convenience, we copy the required expansions here to quadratic order in the photon mass over its energy,
\begin{equation}
 \langle\mathbf{ij}\rangle
 \langle\mathbf{ik}\rangle =
    \left[\begin{array}{c}
    \Bigl( 1 - m_i^2/(4E_i^2) \Bigr)
    \langle i\mathbf{j}\rangle\langle i\mathbf{k}\rangle  \\
    m_i
    \Bigl(\langle i\mathbf{j}\rangle\langle \zeta_i^-\mathbf{k}\rangle
         +\langle \zeta_i^-\mathbf{j}\rangle\langle i\mathbf{k}\rangle
    \Bigr) /(2\sqrt{2E_i})  \\
     m_i^2
    \Bigl( \langle \zeta_i^-\mathbf{j}\rangle\langle \zeta_i^-\mathbf{k}\rangle \Bigr)/(2E_i^2)
    \end{array}\right]

    \ ,
    \label{eq:<ij><ik> expansion}
\end{equation}
\begin{equation}
    \langle \mathbf{ij} \rangle
    \lbrack \mathbf{ik} \rbrack =
    \left[\begin{array}{c}
    -m_i
       \langle i\mathbf{j}\rangle
       \lbrack \tilde{\zeta}_i^+\mathbf{k}\rbrack / ( \sqrt{2E_i} ) \\ \\
     \left( 1 - m_i^2/(4E_i^2) \right)
        \langle i\mathbf{j} \rangle
        \lbrack i\mathbf{k} \rbrack \\
    -m_i^2
        \langle \zeta_i^- \mathbf{j} \rangle
        \lbrack \tilde{\zeta}_i^+ \mathbf{k} \rbrack / ( 2 E_i ) \\ \\
    -m_i
        \langle \zeta_i^-\mathbf{j}\rangle
        \lbrack i\mathbf{k}\rbrack / ( \sqrt{2E_i} )
    \end{array}\right]

    \ ,
    \label{eq:<ij>[ik] expansion}
\end{equation}
and
\begin{equation}
    \lbrack \mathbf{ij} \rbrack
    \lbrack \mathbf{ik} \rbrack =
    \left[\begin{array}{c}
    m_i^2 \Bigl(
    \lbrack\tilde{\zeta}_i^+\mathbf{j}\rbrack
    \lbrack\tilde{\zeta}_i^+\mathbf{k}\rbrack \Bigr)/(2 E_i^2) \\
    m_i
      \left( \lbrack i\mathbf{j}\rbrack
             \lbrack\tilde{\zeta}_i^+\mathbf{k}\rbrack +
             \lbrack\tilde{\zeta}_i^+\mathbf{j}\rbrack
             \lbrack i\mathbf{k}\rbrack \right)/(2\sqrt{2E_i}) \\
    \left( 1 - m_i^2/(4E_i^2) \right)
       \lbrack i\mathbf{j}\rbrack
       \lbrack i\mathbf{k}\rbrack
    \end{array}\right]

    \ ,
    \label{eq:[ij][ik] expansion}
\end{equation}
where particle $i$ is the photon, the top rows are for the negative helicity case, the middle are for 0 helicity and the bottom are for positive helicity.  As we can see from this expression, if we can simply unbold a spinor and have the right helicity, there is no need for a $\zeta$ or $\tilde{\zeta}$.  If unbolding produces the wrong helicity, on the other hand, we need to replace one or both brackets with a $\zeta$ or $\tilde{\zeta}$, since they act like the opposite helicity.  Furthermore, each $\zeta$ or $\tilde{\zeta}$ also comes with a power of $M/\sqrt{2E}$ and increases the order of the leading term in the expansion.

Once we have Taylor expanded the amplitude, we algebraically simplify the expressions at each order in the mass of the photon, $M_A$.  At this point, all division by $M_A$ drops out of the expression.  We then take the $M_A\to0$ limit giving our final amplitude.  In the case of unphysical photon helicities, we expect that the amplitude is identically zero in this limit.  We will now describe the outline of these steps for the major amplitudes of this paper and give greater detail in App.~\ref{sec:Massive Photon}.

The process $e\bar{e}\mu\bar{\mu}$ is begun by multiplying the two photon vertices, symmetrizing the photon spin index, contracting the photon spin index and dividing by the propagator denominator.  We have\begin{align}
    \mathcal{M
    &=
    \frac{e^2}{2 M_A^2\left(s\!-\!M_A^2\right)}
    \left(  \langle\mathbf{1P_{34}}^{I}\rangle
            \lbrack\mathbf{2P_{34}}^{J}\rbrack \!+\!
            \lbrack\mathbf{1P_{34}}^{I}\rbrack
            \langle\mathbf{2P_{34}}^{J}\rangle
    \right) \nonumber \\
    &\times
     \Big(
       \langle\mathbf{3P_{12}}_{I}\rangle
       \lbrack\mathbf{4P_{12}}_{J}\rbrack +
       \lbrack\mathbf{3P_{12}}_{I}\rbrack
       \langle\mathbf{4P_{12}}_{J}\rangle \nonumber \\
    & \qquad
      +\langle\mathbf{3P_{12}}_{J}\rangle
       \lbrack\mathbf{4P_{12}}_{I}\rbrack +
       \lbrack\mathbf{3P_{12}}_{J}\rbrack
       \langle\mathbf{4P_{12}}_{I}\rangle
     \Big) ,
\end{align}
where $P_{ij}=p_i+p_j$.
We next expand this expression, reverse the momentum $P_{34}=-P_{12}$ and use the contraction rules from \cite{Christensen:2019mch} to replace the spinors containing $P_{12}$ with the momentum.  We follow this with a series of Schouten identities, mass identities and momentum conservation.  At the end, the amplitude reduces to
\begin{equation}
  \mathcal{M
    =
    e^2\frac{
    \langle\mathbf{13}\rangle\lbrack\mathbf{24}\rbrack
    +\lbrack\mathbf{13}\rbrack\langle\mathbf{24}\rangle
    +\lbrack\mathbf{14}\rbrack\langle\mathbf{23}\rangle
    +\langle\mathbf{14}\rangle\lbrack\mathbf{23}\rbrack
    }{\left(s-M_A^2\right)}
    \ .
\end{equation}
Finally, we take the massless limit $M_A\to0$ to obtain the final amplitude for this process
\begin{equation}
    \mathcal{M
    =
    \frac{e^2}{s}\left(
    \langle\mathbf{13}\rangle\lbrack\mathbf{24}\rbrack
    +\lbrack\mathbf{13}\rbrack\langle\mathbf{24}\rangle
    +\lbrack\mathbf{14}\rbrack\langle\mathbf{23}\rangle
    +\langle\mathbf{14}\rangle\lbrack\mathbf{23}\rbrack
    \right)\ .
    \label{eq:main:eemm massive}
\end{equation}
Further details can be found in App.~\ref{sec:eemm spinor amp calculation}.  We will discuss this amplitude further in Sec.~\ref{sec:eemm}.

We would next like to discuss a process with one external photon.  In order to do this, we must introduce another neutral particle.  Although this isn't strictly a QED amplitude, we will add the Higgs boson and calculate $e\gamma\bar{e}h$.  In order to do this, we will introduce the Higgs vertex from \cite{Christensen:2018zcq} which is $\mathcal{M}_{eeh} = -\frac{m_e}{v}\left(\langle\mathbf{12}\rangle+\lbrack\mathbf{12}\rbrack\right)$,
where $v=2M_Ws_W/e$, $M_W$ is the mass of the W boson and $s_W$ is the sin of the Weinberg angle.  This amplitude has two diagrams, with the $s$- and $t$-channel propagators.  Combining these with the propagator denominators, contracting the spin indices and using the spin-contraction identities gives us
\begin{align}
  \mathcal{M}^{(s)} &= \frac{e \, m_e}{v(s-m_e^2)}
    \Biggl(
    \frac{m_e}{M_A}
          \left(  \lbrack\mathbf{12}\rbrack\langle\mathbf{23}\rangle +
                  \langle\mathbf{12}\rangle\lbrack\mathbf{23}\rbrack
          \right) \nonumber \\
    &+\frac{
         \left(   \lbrack\mathbf{12}\rbrack
                  \langle\mathbf{2}\lvert p_1\rvert\mathbf{3}\rbrack +
                  \langle\mathbf{12}\rangle
                  \lbrack\mathbf{2}\lvert p_1\rvert\mathbf{3}\rangle
         \right)
         }{M_A} \nonumber \\
    &+
         \left(   \lbrack\mathbf{12}\rbrack
                  \lbrack\mathbf{23}\rbrack +
                  \langle\mathbf{12}\rangle
                  \langle\mathbf{23}\rangle \right)
    \Biggr
\end{align}
\begin{align}
    \mathcal{M}^{(u)} &= \frac{e \, m_e}{v M_A (u-m_e^2)} \Biggl(
    2 m_e\left(
      \langle\mathbf{23}\rangle\lbrack\mathbf{12}\rbrack
     +\langle\mathbf{12}\rangle \lbrack\mathbf{23}\rbrack \right) \nonumber\\
    &+ \lbrack\mathbf{1}\lvert p_4\rvert\mathbf{2}\rangle
       \lbrack\mathbf{23}\rbrack +
       \langle\mathbf{1}\lvert p_4\rvert\mathbf{2}\rbrack
       \langle\mathbf{23}\rangle
    \Biggr) \ .
\end{align}
At this point, we must choose a helicity for the photon.  We will describe the positive helicity, but the negative helicity is analogous.  We Taylor expand the spinors using Eqs.~(\ref{eq:<ij><ik> expansion}) through (\ref{eq:[ij][ik] expansion}) and combine the two diagrams.  We follow this by a series of identities and simplification to obtain
\begin{align}
    \mathcal{M}^+_{eAeh}
    &=
    \frac{e m_e}
         {v\left(s-m_e^2\right)\left(u-m_e^2\right)}
    \Big(
      m_h ^{2} \lbrack\mathbf{1}2\rbrack
               \lbrack2\mathbf{3}\rbrack \nonumber \\
    &- m_e \lbrack \mathbf{1}2 \rbrack
           \lbrack 2\lvert p_{4} \rvert \mathbf{3} \rangle
     + m_e \lbrack 2\mathbf{3}\rbrack
           \lbrack 2\lvert p_4 \rvert \mathbf{1} \rangle \nonumber \\
    &- \langle \mathbf{1} \mathbf{3} \rangle
       \lbrack 2\lvert p_{3} p_{4} \rvert 2 \rbrack \Big) \ .
    \label{eq:main:M_eAeh}
\end{align}
We might also wonder what happens to the amplitude for the unphysical helicity-0 photon.  In principle, it should vanish.  In fact, after Taylor expansion and application of the identities, all the non-vanishing terms exactly cancel and we are left with 0 in the massless limit.
Further details for this amplitude can be found in App.~\ref{sec:eAeh spinor amp calculation}.  We will discuss this amplitude further in Sec.~\ref{sec:M_eAeh}.

The process $e\bar{e}\gamma\gamma$ has two diagrams, a $t$- and a $u$-channel diagram.  Connecting the photon vertex to an electron line in the two ways, and contracting the spin indices gives the contributions
\begin{align}
  \mathcal{M}^{(t)}
   &= -\frac{e^2 M_A
       \left( \langle\mathbf{1}\mathbf{3}\rangle
              \lbrack\mathbf{2}\mathbf{4}\rbrack
              \langle\mathbf{3}\mathbf{4}\rangle
            + \lbrack\mathbf{1}\mathbf{3}\rbrack
              \langle\mathbf{2}\mathbf{4}\rangle
              \lbrack\mathbf{3}\mathbf{4}\rbrack
       \right)}{M_A^2 \left(t-m_e^2 \right)} \nonumber \\
    & -\frac{e^2 m_e
       \left( \langle\mathbf{1}\mathbf{3}\rangle
              \langle\mathbf{2}\mathbf{4}\rangle
              \lbrack\mathbf{3}\mathbf{4}\rbrack
            + \lbrack\mathbf{1}\mathbf{3}\rbrack
              \lbrack\mathbf{2}\mathbf{4}\rbrack
              \langle\mathbf{3}\mathbf{4}\rangle
       \right)}{M_A^2 \left(t-m_e^2 \right)} \nonumber \\
    & -\frac{e^2
       \left( \langle\mathbf{1}\mathbf{3}\rangle
              \lbrack\mathbf{2}\mathbf{4}\rbrack
              \lbrack\mathbf{3}\lvert p_{1} \rvert \mathbf{4}\rangle
            + \lbrack\mathbf{1}\mathbf{3}\rbrack
              \langle\mathbf{2}\mathbf{4}\rangle
              \lbrack\mathbf{4}\lvert p_{1} \rvert \mathbf{3}\rangle
       \right)}{M_A^2 \left(t-m_e^2 \right)} \nonumber \\
  \mathcal{M}^{(u)}
    &= \frac{e^2 M_A
       \left( \langle\mathbf{1}\mathbf{4}\rangle
              \lbrack\mathbf{2}\mathbf{3}\rbrack
              \langle\mathbf{3}\mathbf{4}\rangle
            + \lbrack\mathbf{1}\mathbf{4}\rbrack
              \langle\mathbf{2}\mathbf{3}\rangle
              \lbrack\mathbf{3}\mathbf{4}\rbrack
       \right)}{M_A^2 \left(u-m_e^2 \right)} \nonumber \\
    & +\frac{e^2 m_e
       \left( \langle\mathbf{1}\mathbf{4}\rangle
              \langle\mathbf{2}\mathbf{3}\rangle
              \lbrack\mathbf{3}\mathbf{4}\rbrack
            + \lbrack\mathbf{1}\mathbf{4}\rbrack
              \lbrack\mathbf{2}\mathbf{3}\rbrack
              \langle\mathbf{3}\mathbf{4}\rangle
       \right)}{M_A^2 \left(u-m_e^2 \right)} \nonumber \\
    & -\frac{e^2
       \left( \langle\mathbf{1}\mathbf{4}\rangle
              \lbrack\mathbf{2}\mathbf{3}\rbrack
              \lbrack\mathbf{4}\lvert p_{1} \rvert \mathbf{3}\rangle
            + \lbrack\mathbf{1}\mathbf{4}\rbrack
              \langle\mathbf{2}\mathbf{3}\rangle
              \lbrack\mathbf{3}\lvert p_{1} \rvert \mathbf{4}\rangle
       \right)}{M_A^2 \left(u-m_e^2 \right)}.
\end{align}
Taylor expanding, combining, applying identities, simplifying and taking the massless limit brings the amplitude to the final form.  For the $++$ helicity amplitude, the amplitude reduces to only one term.  It is
\begin{equation}
    \mathcal{M}^{++} =
    \frac{e^2  m_e \lbrack34\rbrack ^{2} \langle\mathbf{1}\mathbf{2}\rangle }
    {\left(t-m_e^2\right)\left(u-m_e^2\right)} \ .
    \label{eq:main:M_eeA+A+}
\end{equation}
In the case of the $+-$ helicity, the final amplitude is nearly as simple.  It is given by
\begin{equation}
    \mathcal{M}_{eeAA}^{+-} =
    \frac{e^2 \left(\langle\mathbf{2}4\rangle\lbrack\mathbf{1}3\rbrack +\langle\mathbf{1}4\rangle\lbrack\mathbf{2}3\rbrack \right)\lbrack3\lvert p_{1} \rvert 4\rangle}{\left(t-m_e^2\right)\left(u-m_e^2\right)}\ .
    \label{eq:main:M_eeA+A- massive}
\end{equation}
Although we are not interested in the helicity-0 amplitudes since they are not physical, it would be nice for these unphysical amplitudes to vanish.  In the case of the helicity-$+0$ amplitude, it does indeed vanish in the massless limit.  Although we also tried to show the same for the helicity-$00$ amplitude, we were unable to achieve complete cancellation of all the non-vanishing terms.  We give much greater details of all these calculations in App.~\ref{sec:Massive eeAA}.  We further discuss these amplitudes in Secs.~\ref{sec:M_eeA+A+} and \ref{sec:M_eeA+A-}.

Although we were able to successfully calculate all the physical amplitudes using the massless limit of a massive-photon theory, there are several reasons we would like achieve a fully successful massless theory that bypasses the massive intermediate.  As we already mentioned at the opening of this section, using a massive photon requires adding an unphysical degree of freedom, which must cancel or vanish in the massless limit.  It would be much preferable to only involve physical features throughout the calculation and that is the hope of the constructive amplitude approach.  Additionally, we see that when we calculate the amplitude with a massive photon, we are required to include every diagram that we would include in a Feynman-diagram calculation and there is no improvement in the economy of the calculation.  When using a massless photon from the beginning, on the other hand, as we will see in the coming sections, it only requires one of the diagrams.  In fact, each diagram gives the same final result and the multiple diagrams are redundant.  This is similar to the same feature in purely massless theories, such as gluodynamics, where only one diagram is required to obtain the final amplitude.  In fact, examples like these suggest that, in some cases, the requirement of multiple diagrams with the same internal state is a shortcoming of Feynman diagrams, even for partly massive theories, and of constructive theories with intermediate massive particles, such as the massive photon discussed in this section.  It appears that adding unphysical degrees of freedom has a consequence of adding unnecessary complications and unphysical aspects that must be cancelled.  Therefore, an approach that never introduces these unphysical aspects would be preferable.

\subsection{\label{sec:main:x}The $x$ Factor}

The $x$ factor was a required introduction for 3-point vertices that contained one massless particle and two massive particles of the same mass as described in \cite{Arkani-Hamed:2017jhn}.  We review its definition and properties in App.~\ref{app:x}.  In order to compare with experiments or Feynman diagrams, we must replace $x$ with expressions involving spinors, masses and other momentum invariants.

In the context of QED, the photon electron vertex is given by
\begin{align}
\mathcal{M}_{e\bar{e}\gamma^+} = x_{12}\langle\mathbf{12}\rangle
&&
\mathcal{M}_{e\bar{e}\gamma^-} = \tilde{x}_{12}\lbrack\mathbf{12}\rbrack
\ .
\end{align}
The $x$ and the $\tilde{x}$ transform as helicity-$+1$ and -$-1$ objects respectively.  They are defined in terms of the symmetrized identities
\begin{align}
    x_{12}|3\rangle &= \frac{1}{2m}\left(p_{2}-p_{1}\right)|3\rbrack
    \label{eq:main:x_12|3>=...|3]}
    \\
    \tilde{x}_{12}|3\rbrack &= \frac{1}{2m}\left(p_2-p_1\right)|3\rangle
    \label{eq:main:xt_12|3]=...|3>}
    \ .
\end{align}
By multiplying on the left by $(p_2-p_1)/2m$, we can show that $\tilde{x}=1/x$.
Further, a reference spinor $\langle\xi\rvert$ or $\lbrack\xi\rvert$ can be multiplied on the left to obtain the symmetrized
\begin{align}
    x_{12} &= \frac{\langle\xi\lvert\left(p_2-p_1\right)\rvert3\rbrack}{2m\langle\xi3\rangle}
    \label{eq:main:x_{12} = }
    \\
    \tilde{x}_{12} &=  \frac{\lbrack\xi\lvert\left(p_2-p_1\right)\rvert3\rangle}{2m\lbrack\xi3\rbrack} \ .
\end{align}
However, if the reference spinor is not physical, it must cancel before the end of the calculation.  Sometimes this is straight forward, but not always.  In other cases, a physicsal helicity spinor can be used as the references spinor.  This turns out to be useful when the photon is an external particle.  In any case, we would like to find identities that remove $x$ and $\tilde{x}$ that bypass the reference spinor and replace them directly with physical spinors, masses and momentum invariants.

When the photon is on an internal line, we obtain expressions of the form $x_{ij}\tilde{x}_{kl}\langle\mathbf{ij}\rangle\lbrack\mathbf{kl}\rbrack+\tilde{x}_{ij}x_{kl}\lbrack\mathbf{ij}\rbrack\langle\mathbf{kl}\rangle$, where $i$ and $j$ are the particles on one side of the propagator and $k$ and $l$ are on the other side.  The first term is for the negative helicity photon and includes the vertices from both sides of the propagator while the second term is similarly for the positive helicity photon.  This can be simplified by using the identities
\begin{eqnarray}
    x_{ij}\tilde{x}_{kl}\langle\mathbf{ij}\rangle &=& x_{ij}\tilde{x}_{kl}\lbrack\mathbf{ij}\rbrack
    \label{eq:main:xxt<ij>=xxt[ij]}\\
    x_{ij}\tilde{x}_{kl}\langle\mathbf{kl}\rangle &=& x_{ij}\tilde{x}_{kl}\lbrack\mathbf{kl}\rbrack
    \label{eq:main:xxt<kl>=xxt[kl]}
\end{eqnarray}
to turn angle brackets into square brackets or vice versa.  We give greater detail in Apps.~\ref{app:x_ij<ij>=x_ij[ij]+...} and \ref{app:xxtilde<>[]}, but we can already see that this introduces an ambiguity of the final form into any expression using them.  Nevertheless, using these identities, the spinor products in each term can be transformed into a common form and factored out of the expression.  Once this is done, we need to deal with $\left(x_{ij}\tilde{x}_{kl}+\tilde{x}_{ij}x_{kl}\right)$.  For this, we use the fully symmetrized identity
\begin{equation}
    x_{12}\tilde{x}_{34}+\tilde{x}_{12}x_{34} =
    -\frac{(p_2-p_1)\cdot (p_4-p_3)}{2m_2m_4}\ .
    \label{eq:main:xxt+xtx}
\end{equation}
We review the derivation of this identity in App.~\ref{app:xxt+xtx} that uses the reference spinor.  We show alternate derivations of the same identity which do not use a reference spinor in App.~\ref{app:xxt+xtx revisited}.  In this appendix, we further show the importance of using the fully symmetrized version of these identities as inconsistent results can be obtained if the fully symmetrized versions are not used.  Unfortunately, although we have these identities for internal photons, we do not have an amplitude using them that agrees with Feynman diagrams, as we show in App.~\ref{sec:Constructive Massless eemm}.  It appears that further ingredients are required for internal massless particles.

When external particles are massless, the amplitude expression contains an $x$ for every positive-helicity photon and $\tilde{x}$ for every negative-helicity photon.  Although we could use a reference spinor, we have found that $x$ and $\tilde{x}$ on external lines can be immediately replaced with physical expressions in the following way.
We multiply $x$ or $\tilde{x}$ by the propagator denominator of the other diagram divided by itself.  In the numerator, we write the propagator denominator in terms of a spinor product and use Eqs.~(\ref{eq:main:x_12|3>=...|3]}) and (\ref{eq:main:xt_12|3]=...|3>}) on it.
For example, let's consider the case of one of the diagrams in this article, the s-channel diagram for $e\gamma^+\bar{e}h$.  The expression for this amplitude contains $x_{1,34}$.  The other diagram is in the u channel, therefore, we multiply by $\lbrack2\lvert p_3\rvert2\rangle/(u-m^2)$.  This is the identity because $\lbrack2\lvert p_3\rvert2\rangle=2p_2\cdot p_3=u-m^2$.  However, we can now use $x_{1,34}\lvert2\rangle=\frac{1}{2m}\left(p_3+p_4-p_1\right)\lvert2\rbrack$.  Putting this together and simplifying, we obtain
\begin{equation}
    x_{1,34}
    = x_{1,34}\frac{\lbrack2\lvert p_3\rvert2\rangle}{(u-m^2)}
    = \frac{\lbrack2\lvert p_3p_4\rvert2\rbrack}{m(u-m^2)}
    \ .
\end{equation}
However, we can also obtain the same result by using the reference spinor $\lbrack\xi\rvert = \lbrack2\rvert p_3$ in Eq.~(\ref{eq:main:x_{12} = }).  Greater detail and more examples can be found in App.~\ref{sec:x = alternate}.  We successfully use these identities to obtain the amplitudes for $e\gamma\bar{e}h$ in App.~\ref{sec:Constructive Massless heAe} and for $e\bar{e}\gamma\gamma$ in Apps.~\ref{app:eeA+A+} and \ref{app:eeA+A-}.  Our expressions agree with Feynman diagrams.

\subsection{\label{sec:main:constructive no Shift}Constructive Calculations}
Now that we have the ingredients for the constructive amplitude calculations, we will calculate the QED amplitudes containing photons.  We begin with the process $e\bar{e}\mu\bar{\mu}$ which has an internal photon in the s channel.  The vertices are $e x_{12}\langle\mathbf{12}\rangle$ on the left and $e \tilde{x}_{34}\lbrack\mathbf{34}\rbrack$ on the right for a negative helicity photon and
$e \tilde{x}_{12}\lbrack\mathbf{12}\rbrack$ and $e x_{34}\langle\mathbf{34}\rangle$ for a positive helicity photon.  Multiplying and dividing by the propagator denominator gives us
\begin{equation}
    \mathcal{M} =
    \frac{e^2}{s}  \left(
    x_{34} \tilde{x}_{12} \lbrack\mathbf{1}\mathbf{2}\rbrack
    \langle\mathbf{3}\mathbf{4}\rangle
    +x_{12} \tilde{x}_{34}
    \langle\mathbf{1}\mathbf{2}\rangle \lbrack\mathbf{3}\mathbf{4}\rbrack  \right)\ .
\end{equation}
We next use the identities in Eqs.~(\ref{eq:main:xxt<ij>=xxt[ij]}) and (\ref{eq:main:xxt<kl>=xxt[kl]}) to convert the spinors in each term into a common spinor term that can be factored out.  These identities give a lot of ambiguity whether angle or square brackets should be in the final result.  Therefore, we might think a linear combination of the forms might be appropriate and write the amplitude as
\begin{align}
\mathcal{M} =
  -\frac{e^2}{s}
   \frac{t-u}{2m_em_\mu}
   \big( & A \langle\mathbf{12}\rangle
             \langle\mathbf{34}\rangle
          +B \lbrack\mathbf{12}\rbrack
             \langle\mathbf{34}\rangle \nonumber \\
         +&C \langle\mathbf{12}\rangle
             \lbrack\mathbf{34}\rbrack
          +D \lbrack\mathbf{12}\rbrack
             \lbrack\mathbf{34}\rbrack \big),
    \label{eq:main:M_eemm massless no shift}
\end{align}
where we have also used the identity in Eq.~(\ref{eq:main:xxt+xtx}) to replace $x_{12} \tilde{x}_{34} +  \tilde{x}_{12} x_{34}$ and $A+B+C+D=1$ are the coefficients for each choice of the form of the spinor products.  As far as we know at this point, any values of $A,B,C$ and $D$ are allowed so long as they add to one.  This amplitude does not agree with Feynman diagrams for any value of these ambiguous coefficients and so we think further constructive ingredients are necessary.  We will consider momentum and spinor shifts in Secs.~\ref{sec:main:spinor shifts}, \ref{sec:main:large z} and \ref{sec:main:constructive}.  Unfortunately, we will see that spinor shifts do not help.  Further details for this calculation can be found in App.~\ref{sec:Constructive Massless eemm no Spinor Shift}.

We next turn to the process $e\gamma^+\bar{e}h$, with one external photon.  For this process, there are potentially two diagrams that could be used.  It turns out that, when using a massless photon and the $x$ factor along with a momentum shift, we only need one of the diagrams and not both.  They both give the identical result.
For the u-channel diagram, we have the Higgs-electron vertex $\frac{m_e}{v}\left(\langle\mathbf{1P}_{23\mathrm{I}}\rangle+\lbrack\mathbf{1P}_{23\mathrm{I}}\rbrack\right)$ and the photon-electron vertex $e x_{14,3}\langle\mathbf{P}^{\mathrm{I}}_{14}\mathbf{3}\rangle$.  Multiplying and dividing by the propagator denominator, we have
\begin{equation}
    \mathcal{M}_u^+ =
    \frac{em_e}{v}\frac{x_{14,3}
    (\langle\mathbf{1P}_{23\mathrm{I}}\rangle+\lbrack\mathbf{1P}_{23\mathrm{I}}\rbrack)
    \langle\mathbf{P}_{14}^{\mathrm{I}}\mathbf{3}\rangle}
    {u-m_e^2}
    \ .
\end{equation}
After using the spin-contraction, mass and momentum conservation identities, we find
\begin{equation}
    \mathcal{M}^+_u
    = \frac{em_e}{v}\frac{x_{14,3}
    (2m_e\langle\mathbf{13}\rangle+\lbrack\mathbf{1}\lvert p_4\rvert\mathbf{3}\rangle)
    }
    {u-m_e^2}\ .
\end{equation}
Our next task is to remove the $x_{14,3}$.  Following the rules discussed in App.~\ref{sec:x = alternate}, we find
\begin{equation}
    x_{14,3} = \frac{\lbrack2\lvert p_1p_3\rvert2\rbrack}{m_e(s-m_e^2)}\ ,
\end{equation}
where we have used the form that contains the propagator denominator from the other diagram that we are not analyzing.  This makes obtaining the correct result easier.  Plugging this in, we find
\begin{equation}
    \mathcal{M}^+_u
    = \frac{e}{v}\frac{\lbrack2\lvert p_1p_3\rvert2\rbrack
    (2m_e\langle\mathbf{13}\rangle+\lbrack\mathbf{1}\lvert p_4\rvert\mathbf{3}\rangle)
    }
    {\left(s-m_e^2\right)\left(u-m_e^2\right)}\ .
\end{equation}
At this point, we perform a series of Schouten, mass and momentum conservation identities to simplify this expression to
\begin{align}
    \mathcal{M}_u^+ &=
    -\frac{e m_e}{v\left(s-m_e^2\right)\left(u-m_e^2\right)}
      \Big( m_h ^{2} \lbrack\mathbf{1}2\rbrack\lbrack2\mathbf{3}\rbrack
    - m_e \lbrack\mathbf{1}2\rbrack\lbrack2\lvert p_{4} \rvert \mathbf{3}\rangle
    \nonumber\\
    &+ m_e \lbrack2\mathbf{3}\rbrack \langle\mathbf{1}\lvert p_{4} \rvert 2\rbrack
    - \langle\mathbf{1}\mathbf{3}\rangle \lbrack2\lvert p_{3} p_{4} \rvert 2\rbrack
    \Big),
\end{align}
which agrees with Feynman diagrams.
Further details for this process can be found in App.~\ref{sec:Constructive Massless heAe no Shift}.

Turning to processes with two external photons, we begin with the process $e\bar{e}\gamma^+\gamma^+$.  Once again, there are two diagrams, this time in the t and u channels.  Once again, we can use either diagram and do not need both since they both give identical final results.  We will demonstrate the u channel, but have also done the t channel with full agreement.  The vertices are $e x_{1,23}\langle\mathbf{1P}_{23\mathrm{I}}\rangle$ and $e x_{14,2}\langle\mathbf{P}_{14}^{\mathrm{I}}\mathbf{2}\rangle$, giving us the amplitude
\begin{equation}
    \mathcal{M}_u^{++} =
    -\frac{e^2 m_e x_{1,23}x_{14,2}\langle\mathbf{12}\rangle}{u-m_e^2}
    \ ,
\end{equation}
where we have already simplified the contracted spinor index.  As in the previous process, we must now replace the $x$ factors.  There are multiple ways we could do this, including the use of Eqs.~(\ref{eq:x_1,23}) and (\ref{eq:x_14,2}), similar to our calculation of the previous process.  However, since we have two photons in this process, we can also use forms for $x_{1,23}$ and $x_{14,2}$ that involve the helicity spinor of the other photon as the reference spinor.  In particular, we can take
\begin{align}
    x_{1,23} = -\frac{\langle3\lvert p_1\rvert4\rbrack}{m_e\langle34\rangle}
    \quad \mbox{and}\quad
    x_{14,2} = \frac{\langle4\lvert p_2\rvert3\rbrack}{m_e\langle43\rangle}
    \ ,
\end{align}
where, we can show that these forms are equivalent to Eqs.~(\ref{eq:x_1,23}) and (\ref{eq:x_14,2}) on shell by use of the usual identities.  However, in some cases, a more efficient route to the final amplitude is possible by use of these simpler expressions.  In order to get the amplitude into the standard form where there are no spinors in the denominator, we multiply the numerator and denominator of these identies with $\lbrack34\rbrack$ to obtain
\begin{align}
    x_{1,23} = \frac{\langle3\lvert p_1\rvert4\rbrack\lbrack34\rbrack}{m_e s}
    \quad \mbox{and}\quad
    x_{14,2} = \frac{\langle4\lvert p_2\rvert3\rbrack\lbrack34\rbrack}{m_e s}
    \ .
\end{align}
We don't simplify the numerator of these because we know we want the $\lbrack34\rbrack^2$ in the final form of the amplitude.  Plugging these identities into the amplitude, we have
\begin{equation}
\mathcal{M}_u^{++} =
    \frac{e^2 \langle4\lvert p_2\rvert3\rbrack \langle3\lvert p_1\rvert4\rbrack
    \lbrack34\rbrack^2\langle\mathbf{12}\rangle}{m_e s^2\left(u-m_e^2\right)}
    \ .
\end{equation}
In the denominator, we can use momentum conservation and the on-shell condition to modify $s = -t-u+2m_e^2 = -(t-m_e^2)$.  However, we only do this with one of them because we use the standard identities to simplify $\langle4\lvert p_2\rvert3\rbrack \langle3\lvert p_1\rvert4\rbrack = m_e^2 s$ and one $s$ cancels.  Our final result is
\begin{equation}
   \mathcal{M}_u^{++} =
    -\frac{e^2 m_e \langle\mathbf{12}\rangle \lbrack34\rbrack^2 }
    {\left(t-m_e^2\right)\left(u-m_e^2\right)} \ ,
\end{equation}
which agrees with Feynman diagrams.  Further details for this process can be found in App.~\ref{app:eeA+A+ no shift}.

For our final process, we consider $e\bar{e}\gamma^+\gamma^-$.
Of the two diagrams, in the t and u channels, this time we will demonstrate the t-channel diagram.  As before, we have done the calculation with both diagrams and find exactly the same final amplitude with either one.  The vertices are $e x_{1,24}\langle\mathbf{1P}_{24\mathrm{I}}\rangle$ and $e \tilde{x}_{13,2}\lbrack\mathbf{P}_{13}^{\mathrm{I}}\mathbf{2}\rbrack$, giving the initial amplitude
\begin{equation}
    \mathcal{M}_t^{+-} =
    \frac{e^2 x_{1,24}\tilde{x}_{13,2}
    \left(m_e\lbrack\mathbf{12}\rbrack
    +\langle\mathbf{1}\lvert p_3\rvert\mathbf{2}\rbrack
    \right)}
    {t-m_e^2}
    \ ,
\end{equation}
where we have used the spin-index contraction and mass identities.  Similar to the $++$-helicity case, we must replace the $x$.  This time, since we have two photons \textit{and} we have one $x$ and one $\tilde{x}$, we get a particularly simple form if use $x_{1,24}=-\langle4\lvert p_1\rvert3\rbrack/m_e\langle43\rangle$ and $\tilde{x}_{13,2}=\lbrack3\lvert p_2\rvert4\rangle/m_e\lbrack34\rbrack$ with the product
\begin{equation}
    x_{1,24}\tilde{x}_{13,2}
    =
    \frac{\lbrack3\lvert p_1\rvert4\rangle^2}{m_e^2 s}
    \ ,
\end{equation}
after using momentum conservation.  Since, on shell, we have $s=-(u-m_e^2)$, we can plug this into the amplitude to obtain
\begin{equation}
    \mathcal{M}_t^{+-} =
    -\frac{e^2 \lbrack3\lvert p_1\rvert4\rangle^2
    \left(m_e\lbrack\mathbf{12}\rbrack
    +\langle\mathbf{1}\lvert p_3\rvert\mathbf{2}\rbrack
    \right)}
    {m_e^2\left(t-m_e^2\right)\left(u-m_e^2\right)}
    \ .
\end{equation}
We then perform a series of simplifications that use both the usual identities and the on-shell condition to end with
\begin{equation}
    \mathcal{M}_t^{+-} =
    \frac{e^2 \lbrack3\lvert p_1\rvert4\rangle
    \left(
    \langle\mathbf{2}4\rangle\lbrack\mathbf{1}3\rbrack
    + \langle\mathbf{1}4\rangle\lbrack\mathbf{2}3\rbrack
    \right)
    }
    {\left(t-m_e^2\right)\left(u-m_e^2\right)}
    \ ,
\end{equation}
in agreement with Feynman diagrams.
Further details can be found in App.~\ref{app:eeA+A- no shift}.

\subsection{\label{sec:main:spinor shifts}Spinor Shifts}
In order to extend the constructive methods from purely massless theories to theories with mass, we should generalize the spinor shifts.  They might be required to obtain the correct amplitudes in some cases.  We include much greater detail in App.~\ref{app:analytic continuation}.

In purely massless theories, we choose two of the external momenta that lie on opposite sides of the propagator and analytically continue them as in
\begin{align}
    \hat{p}_i &= p_i + z q\\
    \label{eq:main:phat = p+zq}
    \hat{p}_j &= p_j - z q,
\end{align}
where $z$ is a complex number and $q$ is a complex momentum.  It can be shown \cite{Britto:2005fq} that if the amplitude vanishes in the large $z$ limit, it can be built up recursively by stitching together smaller on-shell amplitudes and dividing by propagator denominators.  The on-shell condition for the internal line is accomplished by choosing a value for $z$ where the internal line goes on shell and relaxing this property at the end of the calculation.

When the momenta are analytically continued in this way, the spinors are also impacted as they are functions of the momenta.  Therefore, we must also find the accompanying shifts of the spinors in order to work out the expressions for the amplitudes.  In purely massless theories, this has already been done.
Given two momenta $p_i$ and $p_j$, there are two ways we can shift the spinors.  We can either do an $\lbrack i,j\rangle$ shift, where
\begin{align}
    \lbrack\hat{i}\rvert &= \lbrack i\rvert + z \lbrack j\rvert\\
    |\hat{j}\rangle &= |j\rangle - z |i\rangle\ ,
    \label{eq:main:|jhat> = |j>-z|i>}
\end{align}
and $\lvert j\rbrack$ and $\lvert i\rangle$ are not shifted,
or the reverse.  It can be shown that this shift preserves momentum conservation as well as the on-shell property for both particles $i$ and $j$.  Additionally, if the special value of $z= -\lbrack ik\rbrack/\lbrack jk\rbrack$,
where $k$ represents all the other momenta on the same side of the propagator as $i$ is chosen, then the internal line also goes on shell as well.  For convenience of our calculations, we can insert this value of $z$ directly into our spinor shifts, to obtain
\begin{eqnarray}
    \lbrack\hat{i}\rvert &=& \lbrack i\rvert
    -\frac{s_{ik}}{\langle i\lvert p_k\rvert j\rbrack} \lbrack j\rvert
    \\
    \lvert\hat{j}\rangle &=& \lvert j\rangle
    +\lvert i\rangle\frac{s_{ik}}{\langle i\lvert p_k\rvert j\rbrack} \ ,
\end{eqnarray}
where $s_{ik}=(p_i+p_k)^2$, and we have used $z=-\lbrack ik\rbrack\langle ki\rangle/\lbrack jk\rbrack\langle ki\rangle = s_{ik}/\langle i\lvert p_k\rvert j\rbrack$ for convenient comparison with the massive expressions.

The simplest way to generalize this spinor shift is when the internal line becomes massive but the shifted spinors remain massless helicity spinors.  In this case, only the internal-line on-shell condition changes.  This results in the special value of $z=\left(M^2-s_{ik}\right)/\langle i\lvert p_k\rvert j\rbrack$, where $M$ is the mass of the internal particle, and the final shifts become
\begin{align}
    \lbrack\hat{i}\rvert &= \lbrack i\rvert
    -\frac{\left(s_{ik}-M^2\right)}{\langle i\lvert p_k\rvert j\rbrack} \lbrack j\rvert
    \label{eq:main:[ihat|: 0,0,M}
    \\
    \lvert\hat{j}\rangle &= \lvert j\rangle
    +\lvert i\rangle\frac{\left(s_{ik}-M^2\right)}{\langle i\lvert p_k\rvert j\rbrack} \ .
    \label{eq:main:|jhat>: 0,0,M}
\end{align}
We can see that this trivially reduces to the all-massless case in the massless limit.  Further details can be found in App.~\ref{app:[i,j> massless massless}

We must next consider spinor shifts where one of the shifted spinors is a massive spin spinor.  Let's begin with a $\lbrack\mathbf{i},j\rangle$ shift where particle $i$ is massive but particle $j$ remains massless.  In this case, the shifted spinor has an additional spin index $\lbrack \mathbf{i}_{\mathrm{I}}\rvert$ and, in order to shift it using the helicity spinor $\lbrack j\rvert$, this requires the complex number to generalize to a complex number with its own spin index $z^{\mathrm{I}}$.  In other words, the shift becomes
\begin{eqnarray}
    \lbrack\mathbf{\hat{i}}_\mathrm{I}| &=& \lbrack\mathbf{i}_\mathrm{I}| + z_\mathrm{I} \lbrack j|
    \\
    |\hat{j}\rangle &=& |j\rangle - z_\mathrm{I} |\mathbf{i}^\mathrm{I}\rangle
    \ .
\end{eqnarray}
It is straight forward to see that momentum is still conserved and that particle $j$ remains massless, but the on-shell condition for particle $j$ requires the new constraint
\begin{equation}
    z_{\mathrm{I}}\lbrack\mathbf{i}^{\mathrm{I}}j\rbrack = 0
    \ .
\end{equation}
Additionally, requiring the internal line to be on shell, gives the constraint
\begin{equation}
    z_{\mathrm{I}}\langle \mathbf{i}^{\mathrm{I}} \lvert p_k\rvert j\rbrack =
    M^2 - s_{ik} \ ,
\end{equation}
where, once again, $M$ is the mass of the internal particle.  Solving these constraints simultaneously gives
\begin{equation}
    z_I = \frac{\lbrack j\mathbf{i}_I\rbrack}{
    \lbrack j\lvert p_i  p_k\rvert j\rbrack}
    \left(M^2 - s_{ik}\right)
    \ ,
\end{equation}
and this reduces to the all-massless result in the massless limit.  This can be plugged back into the spinor shifts to give our final shifts
\begin{eqnarray}
    \lbrack\mathbf{\hat{i}}| &=& \lbrack\mathbf{i}| + \frac{\lbrack \mathbf{i} j\rbrack\lbrack j\rvert}{
    \lbrack j\lvert p_i  p_k\rvert j\rbrack}
    \left(s_{ik}-M^2\right)
    \label{eq:main:[ihat|: mi,0,M}
    \\
    |\hat{j}\rangle &=& |j\rangle - \frac{p_i\lvert j\rbrack}{
    \lbrack j\lvert p_i  p_k\rvert j\rbrack}
    \left(s_{ik}-M^2\right) \ .
    \label{eq:main:|jhat>: mi,0,M}
\end{eqnarray}
Further details can be found in App.~\ref{app:[i,j> massive massless}.

The opposite case, the shift $\lbrack i,\mathbf{j}\rangle$, where particle $i$ is massless and particle $j$ is massive, is analogous.  The final shifts are given by
\begin{eqnarray}
    \lbrack\hat{i}| &=& \lbrack\mathrm{i}| - \frac{\langle i \rvert p_j}{
    \langle i \lvert p_jp_k\rvert i\rangle}
    \left(s_{ik}-M^2\right)
    \label{eq:main:[ihat|: 0,mj,M}
    \\
    |\mathbf{\hat{j}}\rangle &=& |\mathbf{j}\rangle + \frac{\lvert i\rangle\langle i \mathbf{j}\rangle}{
    \langle i \lvert p_jp_k\rvert i\rangle}
    \left(s_{ik}-M^2\right) \ .
    \label{eq:main:|jhat>: 0,mj,M}
\end{eqnarray}
Further details can be found in App.~\ref{app:[i,j> massless massive}.

Our next case is the shift $\lbrack\mathbf{i,j}\rangle$, where both particles $i$ and $j$ are massive.  The most general case where all external particles and the internal line are massive turns out to be quite complicated.  On the other hand, if every particle in an amplitude, including the internal particles, were massive, we appear not to need the shift of the spinors, nor of the momenta.  So, the all-massive case does not appear to be of interest and a lack of simple formulas for spinors in this case does not seem important.  On the other hand, there are still cases where both shifted spinors are massive, even if other particles in the amplitude are not.  If the internal line is massive, but one or more of the unshifted external particles are massless, we might in principle need the formulas for the shifted spinors.  This could be important, for example, for the processes $e\gamma\bar{e}h$ and $e\bar{e}\gamma\gamma$.  However, it turns out that these shifts are also extremely complicated.  Moreover, because we can still calculate these processes using one of the previous shifts where at least one of the shifted spinors is massless, the lack of simple shifts in this case also does not appear important.  Therefore, we also skip it, but further details can be found in App.~\ref{app:[i,j> both massive massive internal line}.

We have one further case that we must consider, the shift $\lbrack\mathbf{i,j}\rangle$, where both spinors are massive, but the internal line is massless.  This case is important, for example, for the process $e\bar{e}\mu\bar{\mu}$, where we cannot use one of the previous spinor shifts, since all the external particles are massive.  In this case, both spinors have a spin index and this requires the complex number to have two spin indices.  We begin with the spinor shift
\begin{eqnarray}
    \lbrack\mathbf{\hat{i}}_\mathrm{I}| &=& \lbrack\mathbf{i}_\mathrm{I}| + z^{\ \mathrm{J}}_\mathrm{I} \lbrack\mathbf{j}_\mathrm{J}|
    \\
    |\mathbf{\hat{j}}^\mathrm{J}\rangle &=& |\mathbf{j}^\mathrm{J}\rangle - z^{\ \mathrm{J}}_\mathrm{I} |\mathbf{i}^\mathrm{I}\rangle\ .
\end{eqnarray}
Momentum conservation is still straight forward, but now the on-shell condition for both external particles are non-trivial and are given by
\begin{align}
    z_\mathrm{I}^{\ \mathrm{J}}\left(
    2\lbrack\mathbf{i}^{\mathrm{I}}\mathbf{j}_{\mathrm{J}}\rbrack
    - z_{\ \mathrm{J}}^{\mathrm{I}}m_j
    \right) &= 0
    \\
    z^{\ \mathrm{J}}_{\mathrm{I}}\left(
    2\langle\mathbf{i}^{\mathbf{I}}\mathbf{j}_\mathrm{J}\rangle
    - z^{\mathrm{I}}_{\ \mathrm{J}}m_i
    \right) &= 0\ .
\end{align}
Finally, we must combine this with the on-shell condition for the internal line, which is
\begin{equation}
    z_{\mathrm{I}}^{\ \mathrm{J}}\langle \mathbf{i}^{\mathrm{I}} \lvert p_k\rvert\mathbf{j}_{\mathrm{J}}\rbrack =
     -s_{ik}\ .
\end{equation}
The simplest solution to these three constraints is
\begin{equation}
    z_I^{\ J} = -\frac{\left(m_i\lbrack \mathbf{i}_{\mathrm{I}}\mathbf{j}^{\mathrm{J}} \rbrack + \langle \mathbf{i}_{\mathrm{I}}\lvert p_k\rvert \mathbf{j}^{\mathrm{J}}\rbrack\right)s_{ik}}{\left(m_i\lbrack \mathbf{i}_{\mathrm{K}}\mathbf{j}^{\mathrm{L}} \rbrack + \langle \mathbf{i}_{\mathrm{K}}\lvert p_k\rvert \mathbf{j}^{\mathrm{L}}\rbrack\right)\langle \mathbf{i}^{\mathrm{K}} \lvert p_k\rvert\mathbf{j}_{\mathrm{L}}\rbrack}\ ,
\end{equation}
which, it turns out, can be simplified to
\begin{equation}
    z_I^{\ J} = \frac{m_i\lbrack \mathbf{i}_{\mathrm{I}}\mathbf{j}^{\mathrm{J}} \rbrack + \langle \mathbf{i}_{\mathrm{I}}\lvert p_k\rvert \mathbf{j}^{\mathrm{J}}\rbrack}{m_im_j}\ .
\end{equation}
If we plug this back into our spinor shifts, we get a particularly simple form for the spinor shifts, namely
\begin{align}
    \lbrack\mathbf{\hat{i}}| &=
    - \frac{1}{m_i}\langle \mathbf{i}\lvert p_k
    \label{eq:main:[ihat|: mi,mj,0}
    \\
    |\mathbf{\hat{j}}\rangle
    &=
    \frac{1}{m_j}p_l\lvert\mathbf{j} \rbrack\ .
    \label{eq:main:|jhat>: mi,mj,0}
\end{align}
The momentum shifts turn out to be even simpler.  They are
\begin{align}
    \hat{p}_i &= -p_k
    \\
    \hat{p}_j &= -p_l
    \ ,
\end{align}
where $p_k$ represents all the other momenta on the same side of the propagator as $p_i$ and $p_l$ represents all the other momenta on the same side as $p_j$.  In a 4-point amplitude, they are just the other two momenta.  In a higher-point amplitude, they are the sum of the other momenta.
Further details can be found in App.~\ref{app:[i,j> both massive}.
Although we were successful finding spinor shifts for this case, the only process in this paper to which they apply, in principle, is to the process $e\bar{e}\mu\bar{\mu}$.  Unfortunately, we do not obtain the correct amplitude in this case.  However, we do not believe this failure is due to the spinor shifts being incorrect.  Rather, we think it is due to the amplitude not satisfying the large-$z$ condition for this method to work.  We will discuss the large-$z$ behavior next.

\subsection{\label{sec:main:large z}Large $\mathbf{z}$ Limit}
The theorem proving that the amplitude can be written as a product of smaller amplitudes divided by the propagator denominator depends on the amplitude vanishing in the large $z$ limit after the momentum shift $\hat{p}_i = p_i + z q$ and $\hat{p}_j = p_j - z q$ \cite{Britto:2005fq}.  This was proven for a purely massless theory, but it should be true for a partly massive theory as well.  On the other hand, the theorem only states that it is possible in this limit.  It does not state that the amplitude does not split up this way in the absence of an asymptotically vanishing amplitude.  Therefore, we might wonder whether we can still write a massive amplitude in some cases as a product of smaller amplitudes divided by a propagator denominator even when the amplitude does not vanish in the large $z$ limit.   We find that the answer is yes, sometimes, but not always, we can.  We present examples of each case in this article.

We begin by analyzing the large $z$ behavior of the amplitude for $e\bar{e}\mu\bar{\mu}$.  We obtained its amplitude in Eq.~(\ref{eq:main:eemm massive})
\begin{equation}
    \mathcal{M} = \frac{e^2
    \left(\langle\mathbf{13}\rangle\lbrack\mathbf{24}\rbrack
    +\langle\mathbf{14}\rangle\lbrack\mathbf{23}\rbrack
    +\lbrack\mathbf{13}\rbrack\langle\mathbf{24}\rangle
    +\lbrack\mathbf{14}\rbrack\langle\mathbf{23}\rangle
    \right)}{(p_1+p_2)^2}\ .
\end{equation}
and it has been explicitly checked with Feynman diagrams (this large $z$ analysis could also be performed on squared Feynman diagrams with the same result.)  As we described in the last subsection, the two momenta that are shifted must come from opposite sides of the propagator.  Therefore, there are eight different spinor shifts that are allowed.  Each of them shifts either $p_1$ or $p_2$, but not both.  Therefore, the denominator grows as $z^2$ for every choice.  On the other hand, no matter what choice of particle $i$ and $j$ in the shift $\lbrack\mathbf{i,j}\rangle$ we make, it will always shift two spinors in at least one term in the numerator.  Therefore, the numerator also grows as $z^2$.  Altogether, this amplitude asymptotically approaches a constant in the large $z$ limit.
\begin{equation}
    \lim_{z\to\infty}\mathcal{M} = \mathcal{O}(z^0)
    \ .
\end{equation}
Since this amplitude does not vanish in the limit for any choice of momentum shift, the theorem does not apply.  In fact, as we show in App.~\ref{sec:Constructive Massless eemm}, we are unable to obtain this amplitude using a massless photon with the present set of tools.  However, on the other hand, we are able to find this amplitude beginning with a massive photon and taking the massless limit.  In fact, if we look at the amplitude before taking the massless limit,
\begin{equation}
  \mathcal{M
    =
    e^2\frac{
    \langle\mathbf{13}\rangle\lbrack\mathbf{24}\rbrack
    +\lbrack\mathbf{13}\rbrack\langle\mathbf{24}\rangle
    +\lbrack\mathbf{14}\rbrack\langle\mathbf{23}\rangle
    +\langle\mathbf{14}\rangle\lbrack\mathbf{23}\rbrack
    }{\left(s-M_A^2\right)}
    \ ,
\end{equation}
we can see that it still does not vanish in the large $z$ limit of a momentum shift.  Nevertheless, we are able to find this amplitude as a product of smaller amplitudes divided by a propagator denominator.  Further details for this amplitude can be found in App.~\ref{app:eemm large z}.

For the process $e\gamma^+\bar{e}h$, the amplitude is given by Eq.~(\ref{eq:main:M_eAeh}).  In this case, we have two propagator denominators.  Because of this, some shifts will straddle both propagators and lead to $z^4$ growth in the denominator and others will only straddle one of the propagators and lead to $z^2$ growth in the denominator.  Additionally, there is great variety in the structure of the spinors in the numerator including a complete lack of spinors for the Higgs boson and only square spinors for the photon.  This suggests that shifts including the photon or the Higgs might have lower $z$ growth in the numerator.  All of this results in a variety of large-$z$ behavior for this process depending on the choice of spinor shift.  Of all the shifts, we find that $\lbrack\mathbf{4},2\rangle, \lbrack\mathbf{1,3}\rangle, \lbrack\mathbf{3,1}\rangle, \lbrack2,\mathbf{4}\rangle, \lbrack\mathbf{3},2\rangle$ and $\lbrack\mathbf{1},2\rangle$ lead to a vanishing amplitude in the large $z$ limit.  Our calculation of this amplitude was successful using either a massive photon or using a massless photon as we show in App.~\ref{sec:Constructive Massless heAe}.  We include further details for the large-$z$ behavior for this process in App.~\ref{app:large z:eAeh}.

Turning to $e\bar{e}\gamma^+\gamma^+$, the amplitude is very simple
\begin{equation}
    \mathcal{M}^{++} =
    \frac{e^2  m_e \lbrack34\rbrack ^{2} \langle\mathbf{1}\mathbf{2}\rangle }
    {\left(t-m_e^2\right)\left(u-m_e^2\right)} \ .
\end{equation}
Once again, we have a choice of shifts that straddle both propagators leading to $z^4$ denominator growth and others that only straddle one of the propagators leading to $z^2$ growth in the denominator.  The numerator is equally easy to analyze.  Any spinor shift that only has angle brackets for the photons and ony square brackets for the electron and positron does not lead to any numerator $z$ growth at all.  Therefore, we can choose many spinor shifts leading to a vanishing asymptotic limit.  They are the shifts $\lbrack\mathbf{1,2}\rangle, \lbrack\mathbf{2,1}\rangle, \lbrack3,4\rangle, \lbrack4,3\rangle, \lbrack\mathbf{1},3\rangle, \lbrack\mathbf{1},4\rangle, \lbrack\mathbf{2},3\rangle$ and $\lbrack\mathbf{2},4\rangle$.  We were able to find an amplitude that agreed with Feynman diagram for this process using both a massless photon and a massive photon.
The amplitude for the process $e\bar{e}\gamma^+\gamma^-$ is nearly as simple and is
\begin{equation}
\mathcal{M}^{+-} = \frac{e^2\left(\lbrack\mathbf{1}3\rbrack\langle\mathbf{2}4\rangle+\langle\mathbf{1}4\rangle\lbrack\mathbf{2}3\rbrack\right)\lbrack3\lvert p_1\rvert4\rangle}{(t-M_e^2)(u-M_e^2)} \ .
\end{equation}
In the same way, we can see that this amplitude vanishes in the asymptotic limit for the shifts $\lbrack4,3\rangle, \lbrack\mathbf{2},3\rangle, \lbrack4,\mathbf{2}\rangle, \lbrack\mathbf{1,2}\rangle$ and $\lbrack\mathbf{2,1}\rangle$.
Greater detail for these processes can be found in App.~\ref{app:large z:eeAA}.

\subsection{\label{sec:main:constructive}Constructive Calculations with a Momentum and Spinor Shift}
We have already done the initial steps of the calculation for $e\bar{e}\mu\bar{\mu}$ and begin with the amplitude in the form
\begin{align}
    \mathcal{M} =
    -\frac{e^2}{s}
    \frac{t-u}{2m_em_\mu}
    \big(&
    A\langle\mathbf{12}\rangle\langle\mathbf{34}\rangle
    +B\lbrack\mathbf{12}\rbrack\langle\mathbf{34}\rangle
    \nonumber\\
    &+C\langle\mathbf{12}\rangle\lbrack\mathbf{34}\rbrack
    +D\lbrack\mathbf{12}\rbrack\lbrack\mathbf{34}\rbrack
    \big),
\end{align}
from Eq.~(\ref{eq:main:M_eemm massless no shift}), where $A+B+C+D=1$ are the coefficients for each choice of the form of the spinor products and must add to 1.  This ambiguity in the amplitude is alarming, but we will see that the momentum and spinor shifts remove the ambiguity.

None of the momentum shifts cause the correct amplitude to vanish in the limit of large $z$ (see App.~\ref{app:eemm large z}).  Therefore, we should not expect that a momentum shift must work.  Nevertheless, we attempt it anyway.   There are eight possible momentum shifts that straddle the s-channel propagator.  For example, a $\lbrack\mathbf{1,3}\rangle$ shift has the consequence that $\hat{t}-\hat{u}=2t-2m_e^2-2m_\mu^2$ and the following spinor relations
\begin{align}
    \langle\mathbf{12}\rangle\langle\hat{\mathbf{3}}\mathbf{4}\rangle
    &= \langle\mathbf{12}\rangle\lbrack\mathbf{34}\rbrack
    \\
    \lbrack\hat{\mathbf{1}}\mathbf{2}\rbrack\langle\hat{\mathbf{3}}\mathbf{4}\rangle
    &= \langle\mathbf{12}\rangle\lbrack\mathbf{34}\rbrack
    \\
    \langle\mathbf{12}\rangle\lbrack\mathbf{3}\mathbf{4}\rbrack
    &=  \langle\mathbf{12}\rangle\lbrack\mathbf{34}\rbrack
    \\
    \lbrack\hat{\mathbf{1}}\mathbf{2}\rbrack\lbrack\mathbf{3}\mathbf{4}\rbrack
    &= \langle\mathbf{12}\rangle\lbrack\mathbf{34}\rbrack\ .
\end{align}
Interestingly, all the choices of spinor brackets are transformed into the same $\langle\mathbf{12}\rangle\lbrack\mathbf{34}\rbrack$ by this momentum shift.  This is fortunate, since it removes the ambiguity in the spinor products resulting from the identities in Eqs.~(\ref{eq:main:xxt<ij>=xxt[ij]}) and (\ref{eq:main:xxt<kl>=xxt[kl]}).  That is, the unknown coefficients $A,B,C$ and $D$ all give the same final result after the momentum shift.  The final result is
\begin{equation}
    \mathcal{M}_{\lbrack\mathbf{1},\mathbf{3}\rangle} =
    -\frac{e^2}{s}
    \frac{(-s-u+m_e^2+m_\mu^2)}{m_em_\mu}
    \langle\mathbf{12}\rangle\lbrack\mathbf{34}\rbrack\ .
\end{equation}
However, since the numerator is taken to be on shell, we drop the $s$.   If we try the other momentum shifts, we either get the same result or its conjugate.  All together, we find
\begin{align}
    \mathcal{M}_{\lbrack\mathbf{1},\mathbf{3}\rangle,\lbrack\mathbf{2,4}\rangle,\lbrack\mathbf{1},\mathbf{4}\rangle,\lbrack\mathbf{2,3}\rangle}
    &=
    \frac{e^2}{s}
    \frac{(u-m_e^2-m_\mu^2)}{m_em_\mu}
    \langle\mathbf{12}\rangle\lbrack\mathbf{34}\rbrack
    \label{eq:M_eemm shifted A}
    \\
    \mathcal{M}_{\lbrack\mathbf{3},\mathbf{1}\rangle,\lbrack\mathbf{4,2}\rangle,\lbrack\mathbf{4},\mathbf{1}\rangle,\lbrack\mathbf{3},\mathbf{2}\rangle} &=
    \frac{e^2}{s}
    \frac{(u-m_e^2-m_\mu^2)}{m_em_\mu}
    \lbrack\mathbf{12}\rbrack\langle\mathbf{34}\rangle
    \ .
    \label{eq:M_eemm shifted B}
\end{align}
After squaring, there is only one unique form, since $\lvert\langle\mathbf{12}\rangle\lbrack\mathbf{34}\rbrack \rvert^2=\lvert\lbrack\mathbf{12}\rbrack\langle\mathbf{34}\rangle \rvert^2$.  All of them have removed the ambiguity in the type of spinor product.  Unfortunately, this result is wrong.  In fact, this is easily seen in the high-energy behavior where these amplitudes grow for all energies, due to the $u$ in the numerator, and would violate perturbative unitarity.  Further details for this process can be found in App.~\ref{sec:Constructive Massless eemm}.

We next turn to the process $e\gamma^+\bar{e}h$, with one external photon.  For this process, there are potentially two diagrams that could be used, we only need one of them.  Since we demonstrated the u-channel diagram in Sec.~\ref{sec:main:constructive no Shift}, we will perform the s-channel diagram here.  We have the Higgs-electron vertex  $\frac{m_e}{v}(\langle\mathbf{ P}_{12}^{\mathrm{I}}\mathbf{3}\rangle+\lbrack\mathbf{P}_{12}^{\mathrm{I}}\mathbf{3}\rbrack)$ and the photon-electron vertex $e(x_{1,34}\langle\mathbf{1P}_{34\mathrm{I}}\rangle)$.  Multiplying and dividing by the propagator denominator, we have
\begin{equation}
    \mathcal{M}_s^{+} =
    \frac{em_e}{v}\frac{x_{1,34}
    (2m_e\langle\mathbf{13}\rangle
    -\langle\mathbf{1}\lvert p_4\rvert\mathbf{3}\rbrack)
    }
    {s-m_e^2}\ ,
\end{equation}
after using the spin-contraction, mass and momentum conservation identities.

Our next task is to remove the $x_{1,34}$.  Following the rules discussed in App.~\ref{sec:x = alternate}, we find
\begin{equation}
    x_{1,34} =         \frac{\lbrack2|p_1p_3|2\rbrack}{m_e\left(u-m_e^2\right)}\ ,
\end{equation}
where we have used the form that contains the propagator denominator from the other diagram that we are not analyzing.
Plugging this in, we find
\begin{equation}
    \mathcal{M}_s^{+} =
    \frac{e}{v}\frac{\lbrack2\lvert p_1p_3\rvert2\rbrack
    (2m_e\langle\mathbf{13}\rangle
    -\langle\mathbf{1}\lvert p_4\rvert\mathbf{3}\rbrack)
    }
    {\left(s-m_e^2\right)\left(u-m_e^2\right)}\ .
\end{equation}
We can see that this has directly exposed the positive helicity spinors for the photon, as expected.  We show in App.~\ref{app:large z:eAeh} that the amplitude for this process vanishes in the large-$z$ limit for the shifts $\lbrack\mathbf{4},2\rangle, \lbrack\mathbf{1},\mathbf{3}\rangle, \lbrack\mathbf{3},\mathbf{1}\rangle, \lbrack2,\mathbf{4}\rangle, \lbrack\mathbf{3},2\rangle$ or $ \lbrack\mathbf{1},2\rangle$.  Some of these momentum shifts straddle both propagator denominators, not just the denominator of the u-channel diagram that we are working on.  For this reason, we might wonder whether we should hold $(s-m_e^2)$ fixed, and not shift it, even though it is not the original propagator denominator.  The answer, we find, is that we must hold it fixed, no matter which momentum shift we choose, since it is already correct.  Of the shifts, we only have simple formulas for the shifts where at least one of the shifted momenta are for massless particles.  We have tried the remaining four momentum shifts and have found the same unique result that agrees with Feynman diagrams with all of them except $\lbrack2,\mathbf{4}\rangle$, which we were unable to simplify to the correct form.  We have also found the same success with the u-channel diagram.  For example, if we do the $\lbrack\mathbf{4},2\rangle$ shift, which has momenta on both sides of both the s- and u-channel diagrams, we obtain
\begin{equation}
    \hat{\mathcal{M}}_s^{+} =
    \frac{e}{v}\frac{\lbrack2\lvert p_1p_3\rvert2\rbrack
    (2m_e\langle\mathbf{13}\rangle
    -\langle\mathbf{1}\lvert \hat{p}_4\rvert\mathbf{3}\rbrack)
    }
    {\left(s-m_e^2\right)\left(u-m_e^2\right)}\ ,
\end{equation}
where only the momentum $p_4$ is shifted in the numerator.  We then plug in the shift
\begin{equation}
    \langle\mathbf{1}\lvert \hat{p}_4\rvert\mathbf{3}\rbrack = \langle\mathbf{1}\lvert p_4\rvert\mathbf{3}\rbrack
    + \langle\mathbf{1}\lvert p_4\rvert 2\rbrack\lbrack2\mathbf{3}\rbrack
    \frac{(s-m_e^2)}{\lbrack 2\lvert p_4  p_3\rvert 2\rbrack}
    \ .
\end{equation}
After simplifying by applying the usual identities multiple times, we finally obtain
\begin{eqnarray}
    \mathcal{M}^{+} &=&
    \frac{e m_e}{v\left(s-m_e^2\right)\left(u-m_e^2\right)}
      \Big( m_h ^{2} \lbrack\mathbf{1}2\rbrack\lbrack2\mathbf{3}\rbrack
      \nonumber\\
    &&- m_e \lbrack\mathbf{1}2\rbrack\lbrack2\lvert p_{4} \rvert \mathbf{3}\rangle
    + m_e \lbrack2\mathbf{3}\rbrack \langle\mathbf{1}\lvert p_{4} \rvert 2\rbrack
    \nonumber\\
    &&- \langle\mathbf{1}\mathbf{3}\rangle \lbrack2\lvert p_{3} p_{4} \rvert 2\rbrack
    \Big)\ .
\end{eqnarray}
Further details for this process can be found in App.~\ref{sec:Constructive Massless heAe}.

Turning to processes with two external photons, we begin with the process $e\bar{e}\gamma^+\gamma^+$.  Once again, there are two diagrams, this time in the t and u channels.  Once again, we can use either diagram and do not need both since they both give identical final results.  We will demonstrate the t channel, but have also done the u channel with full agreement.  The vertices are $e x_{1,24}\langle\mathbf{1P}_{24\mathrm{I}}\rangle$ and $e x_{13,2}\langle\mathbf{P}_{13}^{\mathrm{I}}\mathbf{2}\rangle$, giving us the amplitude
\begin{equation}
    \mathcal{M}_t^{++} =
    \frac{e^2 m_e x_{1,24}x_{13,2}\langle\mathbf{12}\rangle}{t-m_e^2}
    \ ,
\end{equation}
where we have already simplified the contracted spinor index.  As in the previous process, we must now replace the $x$ factors.  Now that we have two $x$ factors, we must decide which propagator denominators to use in each of them.  It is convenient to use the other propagator denominator for at least one of them, but each choice is related by momentum conservation on shell.  We will demonstrate the shift $\lbrack\mathbf{1},4\rangle$, and replace $x$ with
\begin{align}
    x_{1,24}
    &= \frac{\lbrack3\lvert p_1p_2\rvert3\rbrack}{m(u-m^2)}
    \\
    x_{13,2}
    &= \frac{\lbrack4\lvert p_1p_2\rvert4\rbrack}{m(u-m^2)}
    \ ,
\end{align}
giving us
\begin{equation}
    \hat{\mathcal{M}}_t^{++} =
    \frac{e^2 \langle\mathbf{12}\rangle \lbrack3\lvert \hat{p}_1p_2\rvert3\rbrack\lbrack4\lvert \hat{p}_1p_2\rvert4\rbrack }{m_e\left(t-m_e^2\right)\left(u-m_e^2\right)^2} \ .
\end{equation}
As before, since the denominator is already what we want, we only shifted the numerator.  Furthermore, we can see again that the positive helicity spinors for the two positive-helicity photons are immediately resolved.  Our next step is to plug in the momentum shifts, which after use of the standard identities, is
\begin{align}
    \lbrack3\lvert \hat{p}_1p_2\rvert3\rbrack\lbrack4\lvert \hat{p}_1p_2\rvert4\rbrack
    &= -m_e^2\left(u-m_e^2\right)\lbrack34\rbrack^2.
\end{align}
We see that the extra propagator denominator $(u-m_e^2)$ cancels and we are left with
\begin{equation}
   \mathcal{M}_t^{++} =
    -\frac{e^2 m_e \langle\mathbf{12}\rangle \lbrack34\rbrack^2 }
    {\left(t-m_e^2\right)\left(u-m_e^2\right)} \ ,
\end{equation}
which agrees with Feynman diagrams.  (Presumably, if we had been doing a QCD calculation with quarks and gluons, since all 3 diagrams contribute, we would need to include both the other propagator denominators and they would not cancel.)
Further details for this process can be found in App.~\ref{app:eeA+A+}.

For our final process, we consider $e\bar{e}\gamma^+\gamma^-$.
Of the two diagrams, in the t and u channels, this time we will demonstrate the u-channel diagram.  As before, we have done the calculation with both diagrams and find exactly the same final amplitude with either one.  The vertices are $e \tilde{x}_{1,23}\lbrack\mathbf{1P}_{23\mathrm{I}}\rbrack$ and $e x_{14,2}\langle\mathbf{P}_{14}^{\mathrm{I}}\mathbf{2}\rangle$, giving the initial amplitude
\begin{equation}
    \mathcal{M}_u^{+-} =
    \frac{e^2 \tilde{x}_{1,23}x_{14,2}
    \left(
    m_e\langle\mathbf{12}\rangle
    +\lbrack\mathbf{1}\lvert p_4\rvert \mathbf{2}\rangle
    \right)}
    {u-m_e^2}
    \ ,
\end{equation}
where we have used the spin-index contraction and mass identities.  Similar to the $++$-helicity case, we must replace one $x$ with a form containing $(t-m_e^2)$, but momentum conservation and the on-shell condition relates the choices.  We see in App.~\ref{app:large z:eeAA} that there are five momentum shifts that cause the amplitude to vanish at large $z$.  Of these, we have simple formulas for $\lbrack4,3\rangle, \lbrack\mathbf{2},3\rangle$ and $\lbrack4,\mathbf{2}\rangle$.  This time we will do the shift $\lbrack4,3\rangle$, which will make the second shift including the propagator denominator $s$ convenient.  Therefore, we use
\begin{align}
    \tilde{x}_{1,23} &=
    \frac{\langle4\lvert p_1p_3\rvert4\rangle}{m_e\ s}
    \\
    x_{14,2} &=
    \frac{\lbrack3\lvert p_1p_2\rvert3\rbrack}{m_e(t-m_e^2)}
    \ .
\end{align}
Making these replacements and shifting the momenta gives
\begin{equation}
    \hat{\mathcal{M}}_u^{+-} =
    \frac{e^2
    \lbrack3\lvert p_1p_2\rvert3\rbrack
    \langle4\lvert p_1\hat{p}_3\rvert4\rangle
    \left(
    m_e\langle\mathbf{12}\rangle
    +\lbrack\mathbf{1}\lvert \hat{p}_4\rvert \mathbf{2}\rangle
    \right)}
    {m_e^2\ s \left(t-m_e^2\right)\left(u-m_e^2\right)}
    \ .
\end{equation}
Our momentum shift is
\begin{equation}
    \hat{p}_3 =
    p_3 + \lvert3\rbrack\langle4\rvert \frac{u-m_e^2}{\langle4\lvert p_1\rvert 3\rbrack}
    \ ,
\end{equation}
which gives a $\langle44\rangle$, causing $\langle4\lvert p_1\hat{p}_3\rvert4\rangle=\langle4\lvert p_1p_3\rvert4\rangle$, due to the masslessness of particle $4$.  We also need \begin{equation}
    \lbrack\mathbf{1}\lvert \hat{p}_4\rvert \mathbf{2}\rangle = \lbrack\mathbf{1}\lvert p_4\rvert \mathbf{2}\rangle - \lbrack\mathbf{1}3\rbrack\langle4\mathbf{2}\rangle 2p_1\cdot p_4/\langle4\lvert p_1\rvert3\rbrack
    \ .
\end{equation}
After a series of identities and simplifications, $s$ can be factored out and cancel the $s$ in the denominator.  The final form can be brought to
\begin{equation}
    \mathcal{M}_u^{+-} =
    \frac{e^2 \lbrack3\lvert p_1\rvert4\rangle
    \left(
    \langle\mathbf{2}4\rangle\lbrack\mathbf{1}3\rbrack
    + \langle\mathbf{1}4\rangle\lbrack\mathbf{2}3\rbrack
    \right)
    }
    {\left(t-m_e^2\right)\left(u-m_e^2\right)}
    \ .
\end{equation}
This form agrees with Feynman diagrams.  We have also done this calculation with the other momentum shifts and for both diagrams and get the same result.
Further details can be found in App.~\ref{app:eeA+A-}.

\subsection{\label{sec:main:amplitudes}The Amplitudes}
In this section, we add further comments about the spinor amplitudes for each process.

\subsubsection{\label{sec:eemm}$\mathbf{e,\bar{e},\bar{\mu},\mu}$ and $\mathbf{e,\bar{e},\bar{e},e}$}

We begin with the process $e\bar{e}\to\mu\bar{\mu}$.  As we saw in Sec.~\ref{sec:main:MA} and in greater detail in App.~\ref{sec:eemm spinor amp calculation}, we can calculate this amplitude using a intermediate massive photon and take the massless limit.  Unfortunately, as we saw in Sec.~\ref{sec:main:constructive} and in greater detail in App.~\ref{sec:Constructive Massless eemm}, we do not yet have the ingredients necessary to calculate the correct spinor amplitude using the constructive technique with massless photons.

Before giving the result, let us note that the general structure of the amplitude allows us to enumerate the possible spinor terms that could appear.  In particular, since all four external particles are massive spin-$1/2$ particles, each term must be composed of one spin spinor for each particle.  Each of these can be either angle or square spinors.  There are ten possible combinations with no momenta sandwiched between the spinors, namely     $\langle\mathbf{12}\rangle\lbrack\mathbf{34}\rbrack$,
$\langle\mathbf{13}\rangle\lbrack\mathbf{24}\rbrack$,
$\langle\mathbf{14}\rangle\lbrack\mathbf{23}\rbrack$,
$\lbrack\mathbf{12}\rbrack\langle\mathbf{34}\rangle$,
$\lbrack\mathbf{13}\rbrack\langle\mathbf{24}\rangle$,
$\lbrack\mathbf{14}\rbrack\langle\mathbf{23}\rangle$,
$\langle\mathbf{13}\rangle\langle\mathbf{24}\rangle$,
$\langle\mathbf{14}\rangle\langle\mathbf{23}\rangle$,
$\lbrack\mathbf{13}\rbrack\lbrack\mathbf{24}\rbrack$ and
$\lbrack\mathbf{14}\rbrack\lbrack\mathbf{23}\rbrack$.  (We do not include
$\langle\mathbf{12}\rangle\langle\mathbf{34}\rangle$ and
$\lbrack\mathbf{12}\rbrack\lbrack\mathbf{34}\rbrack$ because they can be obtained from the others by a Schouten transformation.)  We further note that, due to the vectorial nature of the photon, we expect a symmetry between angle and square brackets.  Therefore, we expect the amplitude to contain some combination of $\langle\mathbf{12}\rangle\lbrack\mathbf{34}\rbrack+\lbrack\mathbf{12}\rbrack\langle\mathbf{34}\rangle, \langle\mathbf{13}\rangle\lbrack\mathbf{24}\rbrack+\lbrack\mathbf{13}\rbrack\langle\mathbf{24}\rangle, \langle\mathbf{14}\rangle\lbrack\mathbf{23}\rbrack+\lbrack\mathbf{14}\rbrack\langle\mathbf{23}\rangle, \langle\mathbf{13}\rangle\langle\mathbf{24}\rangle+\lbrack\mathbf{13}\rbrack\lbrack\mathbf{24}\rbrack$ and $\langle\mathbf{14}\rangle\langle\mathbf{23}\rangle+\lbrack\mathbf{14}\rbrack\lbrack\mathbf{23}\rbrack$.  We could also consider spinor products with a momentum, however, it turns out not to be necessary in this case and we prefer to present the simplest possible form of the amplitude.  In the absence of the massless-limit approach, we could try different combinations of these terms until we find one that agrees with Feynman diagrams.  In fact, this is how we initially found this amplitude.  As we saw in Sec.~\ref{sec:main:MA}, the correct amplitude only contains two of these, both with the same coefficient.  The amplitude is
\begin{equation}
    \mathcal{M} = \frac{e^2}{s}
    \left(\langle\mathbf{13}\rangle\lbrack\mathbf{24}\rbrack \!+\!
          \langle\mathbf{14}\rangle\lbrack\mathbf{23}\rbrack \!+\!
          \lbrack\mathbf{13}\rbrack\langle\mathbf{24}\rangle \!+\!
          \lbrack\mathbf{14}\rbrack\langle\mathbf{23}\rangle
    \right).
    \label{eq:M_eemm}
\end{equation}
We have checked explicitly and our result agrees analytically and exactly with Feynman diagrams at all energies and all angles.  In particular, following the procedures outlined in \cite{Christensen:2019mch}, we have squared our amplitude and checked this expression against the Feynman diagram result output by CalcHEP\cite{Belyaev:2012qa}.

As we can see, this amplitude does not contain any momenta in the numerator.  Further, every term is composed of an electron or positron paired with a muon or anti-muon in all possible combinations where one spinor product is an angle product and the other is a square product.

We also note that the published version of this amplitude \cite{Arkani-Hamed:2017jhn} is given by
\begin{equation}
    \mathcal{M}^{AHH} = \frac{e^2}{s}
    \frac{\langle\mathbf{12}\rangle\langle\mathbf{34}\rangle}
         {m}
    \left(p_1-p_2\right) \!\cdot\! p_3 \ .
    \label{eq:M_eemm-Arkani}
\end{equation}
One may initially suspect that these amplitudes are equivalent after some sort of transformation, such as some combination of momentum conservation and Schouten identities.   However, this is not the case and, as we show in App.~\ref{sec:Constructive Massless eemm}, momentum shifts do not correct this amplitude.  In fact, we can see that this expression grows for all energies, violating perturbative unitarity, unlike the correct amplitude given above.

Now that we have the amplitude for $e\bar{e}\mu\bar{\mu}$, we can construct the amplitude for $e\bar{e}e\bar{e}$.  There are now two diagrams.  The $s$-channel diagram from the previous muon case and a $t$-channel diagram.  The amplitude is
\begin{align}
    \mathcal{M} &= \frac{e^2}{s}
    \left(\langle\mathbf{13}\rangle\lbrack\mathbf{24}\rbrack \!+\!
          \langle\mathbf{14}\rangle\lbrack\mathbf{23}\rbrack \!+\!
          \lbrack\mathbf{13}\rbrack\langle\mathbf{24}\rangle \!+\!
          \lbrack\mathbf{14}\rbrack\langle\mathbf{23}\rangle
    \right)
    \nonumber\\
    &- \frac{e^2}{t}
    \left(\langle\mathbf{12}\rangle\lbrack\mathbf{34}\rbrack \!-\!
          \langle\mathbf{14}\rangle\lbrack\mathbf{23}\rbrack \!+\!
          \lbrack\mathbf{12}\rbrack\langle\mathbf{34}\rangle \!-\!
          \lbrack\mathbf{14}\rbrack\langle\mathbf{23}\rangle
    \right),
    \label{eq:M_eeee}
\end{align}
where we have interchanged $2\leftrightarrow3$.  We then used the antisymmetry property $\langle\mathbf{32}\rangle=-\langle\mathbf{23}\rangle$ and $\lbrack\mathbf{32}\rbrack=-\lbrack\mathbf{23}\rbrack$.  The relative negative sign between diagrams comes from the exchanged identical fermions.  Indeed, we have computed the squared amplitude and compared with Feynman diagrams and found exact agreement.

We note that, even with two diagrams, this amplitude was still reducible to a form with no momenta in the numerator.   Also, as far as we can tell, we do not gain anything by combining the two terms into an expression with a common denominator.

\subsubsection{\label{sec:M_eAeh}$\mathbf{e,\gamma^{\pm},\bar{e},h}$}
We next turn to a process with only a single external photon.  We add the Higgs boson and construct $e\gamma^{\pm}\bar{e}h$.  The structure of this amplitude tells us we must have two helicity spinors for the photon, both square if the photon has positive helicity or both angle if the photon has negative helicity, and one spin spinor each for the electron and positron that can be either angle or square.  For example, let's first consider the process $e\gamma^+\bar{e}h$.  In this case, there is only one possible term with no momenta.  It is $\lbrack\mathbf{1}2\rbrack\lbrack2\mathbf{3}\rbrack$.  This turns out not to be enough and we need to consider terms with a momenta.  With one momentum, we could construct $\langle\mathbf{1}\lvert p_4\rvert2\rbrack\lbrack2\mathbf{3}\rbrack$ or $\lbrack\mathbf{1}2\rbrack\lbrack2\lvert p_4\rvert\mathbf{3}\rangle$.  The only other possibility would be to use $p_3$ or $p_1$ instead of $p_4$, but that would be equivalent to these by momentum conservation and the mass identities.  Finally, it turns out that we also need terms with two momenta.  There are only two unique terms possible.  They are $\langle\mathbf{13}\rangle\lbrack2\lvert p_3p_4\rvert2\rbrack$ and $\lbrack\mathbf{13}\rbrack\lbrack2\lvert p_3p_4\rvert2\rbrack$.  Other forms, such as $\langle\mathbf{1}\lvert p_4\rvert2\rbrack\lbrack2\lvert p_4\rvert\mathbf{3}\rangle$, can be converted to these by use of Schouten and other identities.  Once again, we could try different combinations of these terms until we find agreement with Feynman diagrams and this is how we initially found this amplitude.  Of these five possible terms, the correct amplitude uses four of them and each has a mass coefficient to bring them to the same mass dimension.  It is
\begin{align} \nonumber
\mathcal{M}^+ &=
    \frac{e m_e}
         {v\left(s-m_e^2\right)\left(u-m_e^2\right)} \\ \nonumber
     & \times
      \Big( m_h ^{2}
            \lbrack\mathbf{1}2\rbrack
            \lbrack2\mathbf{3}\rbrack
      - m_e \lbrack\mathbf{1}2\rbrack
            \lbrack 2\lvert p_{4} \rvert \mathbf{3}\rangle \nonumber \\
     &+ m_e \lbrack2\mathbf{3}\rbrack
            \langle\mathbf{1}\lvert p_{4} \rvert 2\rbrack
          - \langle\mathbf{1}\mathbf{3}\rangle
            \lbrack2\lvert p_{3} p_{4} \rvert 2\rbrack \Big).
\label{eq:M_eA+eh}
\end{align}
For the negative-helicity photon case, $e\gamma^-\bar{e}h$, we find
\begin{align} \nonumber
    \mathcal{M}^- = &
    \frac{e m_e}{v\left(s-m_e^2\right)\left(u-m_e^2\right)} \times \qquad \\ \nonumber
    & \Big(
      m_h ^{2} \langle \mathbf{1} 2 \rangle
               \langle 2 \mathbf{3} \rangle
      - m_e \langle \mathbf{1} 2 \rangle
            \langle 2 \lvert p_{4} \rvert \mathbf{3} \rbrack \\
    & + m_e \langle 2 \mathbf{3} \rangle
            \lbrack \mathbf{1} \lvert p_{4} \rvert 2 \rangle
      - \lbrack \mathbf{1} \mathbf{3} \rbrack
        \langle 2 \lvert p_{3} p_{4} \rvert 2 \rangle
   \Big)\ ,
   \label{eq:M_eAeh}
\end{align}
with all angle brackets replaced with square brackets and vice versa.
In addition to trying multiple different forms and testing, we found this form both as the massless limit of a massive photon theory in Sec.~\ref{sec:main:MA} and App.~\ref{sec:eAeh spinor amp calculation} and by a purely massless-photon theory using the $x$ factor in Sec.~\ref{sec:main:constructive no Shift} and ~\ref{sec:Constructive Massless heAe no Shift} and also using momentum shifts in Sec.~\ref{sec:main:constructive} and App.~\ref{sec:Constructive Massless heAe}.

In order to compare this with Feynman diagrams, we calculated the same process using Feynman diagrams and multiplied by the positive-helicity (negative-helicity) polarization vector before squaring for the positive- (negative-)helicity case.  We found exact analytic agreement with our result.  We also summed the squared amplitude over helicities and compared with the output of CalcHEP and found exact analytic agreement.

An interesting feature of the propagator denominators in the cases where there are external photons is that they can be written more compactly and suggestively in spinor form.  For example, $(s-m_e^2)=2p_1\cdot p_2 = \lbrack2\lvert p_1\rvert2\rangle$ and $(u-m_e^2)=2p_2\cdot p_3=\lbrack2\lvert p_3\rvert2\rangle$, where both have the square and angle helicity spinors for the photon and have the electron and positron momenta in the middle.  With this, we could write the positive-helicity amplitude as
\begin{align} \nonumber
\mathcal{M}^+ &=
    \frac{e m_e}
         {v\lbrack2\lvert p_1\rvert2\rangle\lbrack2\lvert p_3\rvert2\rangle} \\ \nonumber
    & \times
    \Big( m_h ^{2}
          \lbrack\mathbf{1}2\rbrack
          \lbrack2\mathbf{3}\rbrack
    - m_e \lbrack\mathbf{1}2\rbrack
          \lbrack2\lvert p_{4} \rvert \mathbf{3}\rangle \nonumber \\
   &+ m_e \lbrack2\mathbf{3}\rbrack
          \langle\mathbf{1}\lvert p_{4} \rvert 2\rbrack
        - \langle\mathbf{1}\mathbf{3}\rangle
          \lbrack2\lvert p_{3} p_{4} \rvert 2\rbrack \Big),
\end{align}
and similarly for the negative-helicity photon.  This denominator likely generalizes in an interesting way to amplitudes with more photons and more electron lines.

\subsubsection{\label{sec:M_eeA+A+}$\mathbf{e,\bar{e},\gamma^{\pm},\gamma^{\pm}}$}
Turning to the process with two same-helicity photons, we consider $e,\bar{e},\gamma^{\pm},\gamma^{\pm}$.  In order to obtain the correct transformation properties, each term of the amplitude must contain the right spinors.  For example, for the $e,\bar{e},\gamma^+,\gamma^+$ case, each term must have two square helicity spinors for particle 3, two square helicity spinors for particle 4, and one spin spinor of either type for particle 1 and also for particle 2.  If we do not need a term with a momentum, there are only a few possibilities.  Particles 3 and 4 can be together in terms like $\lbrack34\rbrack^2\langle\mathbf{12}\rangle$ and $\lbrack34\rbrack^2\lbrack\mathbf{12}\rbrack$.  We could also have particles 1 and 2 together with particles 3 and 4 in terms such as $\lbrack34\rbrack\lbrack\mathbf{1}3\rbrack\lbrack\mathbf{2}4\rbrack$ and $\lbrack34\rbrack\lbrack\mathbf{1}4\rbrack\lbrack\mathbf{2}3\rbrack$, but one of these last two are equivalent to the other three by Schouten identities.  We could, and originally did, try different combinations of these terms until we found the amplitude that agrees with Feynman diagrams.  It is
\begin{align}
\mathcal{M}^{++} &=
   \frac{e^2 m_e \lbrack 34 \rbrack^2 \langle \mathbf{12} \rangle}
        {(t-m_e^2)(u-m_e^2)} .
 \label{eq:M_eeA+A+}
\end{align}
For the double-negative-helicity case, we have
\begin{align}
\mathcal{M}^{--} &=
   \frac{e^2 m_e \langle 34 \rangle^2 \lbrack \mathbf{12} \rbrack}
        {(t-m_e^2)(u-m_e^2)} .
\end{align}
We have also found these amplitudes as the massless limit of a massive-photon theory in Sec.~\ref{sec:main:MA} and App.~\ref{sec:Massive eeAA} and using a purely massless theory in Sec.~\ref{sec:main:constructive no Shift} and App.~\ref{app:eeA+A+ no shift} without the momentum shift and Sec.~\ref{sec:main:constructive} and App.~\ref{app:eeA+A+} with the momentum shift.  We have checked our results with Feynman diagrams and found exact analytic agreement.  We did this by calculating the Feynman diagrams and multiplying by polarization vectors for the photons before squaring.  We also summed over all the helicity combinations after squaring and comparing with the output of CalcHEP.

As for the single-photon amplitude, we could write the denominator in spinor form to obtain
\begin{align}
\mathcal{M}^{++}
 &= \frac{e^2 m_e \lbrack 34 \rbrack^2 \langle \mathbf{12} \rangle}
        {\lbrack 3 \lvert p_1 \rvert3 \rangle \lbrack4 \lvert p_1 \rvert 4 \rangle}
        \label{eq:main:MeeA+A+ alternate den}
        \\
\mathcal{M}^{--}
 &= \frac{e^2 m_e \langle 34 \rangle^2 \lbrack \mathbf{12} \rbrack}
        {\lbrack 3 \lvert p_1 \rvert 3 \rangle \lbrack 4 \lvert p_1 \rvert 4 \rangle} .
        \label{eq:main:MeeA-A- alternate den}
\end{align}
The denominator of both of these could also be written as $\lbrack3\lvert p_1\rvert3\rangle\lbrack3\lvert p_2\rvert3\rangle = \lbrack4\lvert p_2\rvert4\rangle\lbrack4\lvert p_1\rvert4\rangle$, each related by momentum conservation.  There may be some further insight from these forms.  We wonder if perhaps a generalization of these amplitudes for the process $e\bar{e}\gamma^+\cdots\gamma^+$, with $N-2$ positive-helicity photons might be $e^{N-2} m_e\langle\mathbf{12}\rangle\lbrack34\rbrack\lbrack45\rbrack\cdots\lbrack N3\rbrack/\lbrack3\lvert p_1\rvert3\rangle\cdots \lbrack N\lvert p_1\rvert N\rangle$.  We will investigate this in a future publication.

\subsubsection{\label{sec:M_eeA+A-} $\mathbf{e,\bar{e},\gamma^{\pm},\gamma^{\mp}}$}
Our final process has two photons of opposite helicity.  If we think about the process $e\bar{e}\gamma^+\gamma^-$, each term must have two square helicity spinors for particle 3, two angle helicity spinors for particle 4, and one spin spinor of either type for particle 1 and particle 2.  It turns out that there is no way to write a term with no momenta.  We can't have a spinor product between the square bracket of particle 3 and the angle bracket of particle 4 without any momenta in between and there aren't enough spinors from particles 1 and 2 to contract with all the helicity spinors from particles 3 and 4.  So, this amplitude only has terms with one or more momenta.  If we consider the case where each term has one momentum, we must put the momentum in between the helicity spinors for particle 3 and 4.  Therefore, we have $\lbrack\mathbf{1}3\rbrack\langle\mathbf{2}4\rangle\lbrack3\lvert p_1\rvert4\rangle$ and $\lbrack\mathbf{2}3\rbrack\langle\mathbf{1}4\rangle\lbrack3\lvert p_1\rvert4\rangle$.  A simple combination of these two terms gives the correct amplitude as
\begin{align}
\mathcal{M}^{+-} &=
  e^2
   \frac{
     \left( \lbrack\mathbf{1}3\rbrack\langle\mathbf{2}4\rangle
           +\langle\mathbf{1}4\rangle\lbrack\mathbf{2}3\rbrack
     \right)
   \lbrack 3 \lvert p_1 \rvert 4 \rangle}
        {(t-m_e^2)(u-m_e^2)} .
\label{eq:M_eeA+A-}
\end{align}
Similarly,
\begin{align}
\mathcal{M}^{-+}
&= e^2
  \frac{
    \left(
      \langle\mathbf{1}3\rangle\lbrack\mathbf{2}4\rbrack
      +\lbrack\mathbf{1}4\rbrack\langle\mathbf{2}3\rangle
    \right)
    \lbrack4\lvert p_1\rvert3\rangle}
       {(t-m_e^2)(u-m_e^2)} .
\end{align}
We also find these amplitudes as the massless limit of the massive theory in Sec.~\ref{sec:main:MA} and App.~\ref{sec:Massive eeAA} and with the massless photon theory without a momentum shift in Sec.~\ref{sec:main:constructive no Shift} and App.~\ref{app:eeA+A- no shift} and with a momentum shift in Sec.~\ref{sec:main:constructive} and App.~\ref{app:eeA+A-}.  As in past cases, we have compared this with Feynman diagrams and obtain exact analytic agreement for the squared amplitude, both for the individual helicity-amplitudes squared and for their sum over helicities.  As before, the denominator can be written in multiple ways, such as $(t-m_e^2)(u-m_e^2)=\lbrack 3 \lvert p_1 \rvert 3 \rangle \lbrack 4 \lvert p_1 \rvert 4 \rangle = \lbrack3\lvert p_1\rvert3\rangle\lbrack3\lvert p_2\rvert3\rangle = \lbrack4\lvert p_2\rvert4\rangle\lbrack4\lvert p_1\rvert4\rangle$.  It would be interesting to find the amplitude with more than two photons with a variety of helicities.

The authors of \cite{Arkani-Hamed:2017jhn} also calculated this process.  From their Eq.~(5.21), we copy
\begin{equation}
    \mathcal{M}^{-+} = \frac{e^2\langle3|p_1-p_2|4\rbrack}{2(t-m_e^2)(u-m_e^2)}\left(\langle\mathbf{2}3\rangle\lbrack\mathbf{1}4\rbrack+\langle\mathbf{1}3\rangle\lbrack\mathbf{2}4\rbrack\right)\ ,
\end{equation}
where we have interchanged particles $2$ and $4$ to bring them into the same order as our calculation.  We can simplify this by using momentum conservation in the first spinor product.  If we make the replacement $p_2=-p_1-p_3-p_4$, the $p_3$ and $p_4$ terms vanish due to the masslessness of the photons and we are left with
\begin{equation}
    \mathcal{M}^{-+} = \frac{e^2\left(\langle\mathbf{2}3\rangle\lbrack\mathbf{1}4\rbrack+\langle\mathbf{1}3\rangle\lbrack\mathbf{2}4\rbrack\right)\langle3|p_1|4\rbrack}{(t-m_e^2)(u-m_e^2)}\ ,
\end{equation}
which agrees with our results.

\section{\label{sec:conclusions}Summary and Conclusions}
In this paper, we considered the constructive approach to calculating the 4-point scattering amplitudes of QED, namely the processes $e\bar{e}\mu\bar{\mu}$, $e\bar{e}e\bar{e}$, $e\gamma^{\pm}\bar{e}h$, $e\bar{e}\gamma^{\pm}\gamma^{\pm}$ and $e\bar{e}\gamma^{\pm}\gamma^{\mp}$.  We found that, if we begin with a massless photon, the current tools are insufficient for correctly calculating the processes $e\bar{e}\mu\bar{\mu}$ and $e\bar{e}e\bar{e}$, with an internal photon, but that they are sufficient for the processes with an external photon,  $e\gamma^{\pm}\bar{e}h$, $e\bar{e}\gamma^{\pm}\gamma^{\pm}$ and $e\bar{e}\gamma^{\pm}\gamma^{\mp}$.  This, moreover, continues to be true after analytically continuing the momenta and generalizing the spinor shifts to massive spin spinors.  Further refinement of the constructive approach is still necessary.  On the other hand, all of these processes can be obtained if we begin with a massive photon and take the massless limit at the end of the calculation.  These processes are also simple and sufficiently constrained that the amplitudes can be obtained by trial and error.

In greater detail, in Sec.~\ref{sec:main:MA} and App.~\ref{sec:Massive Photon}, we gave the photon a mass $M_A$, updated its vertex in Eqs.~(\ref{eq:M_eeA massive vertex}) and described Taylor expansions of its massive spinors in Eqs.~(\ref{eq:<ij><ik> expansion}) through (\ref{eq:[ij][ik] expansion}).  We then calculated each of the scattering amplitudes of this paper using this massive theory and showed that the correct physical amplitudes were obtained in the massless $M_A\to0$ limit.  For the process $e\bar{e}\mu\bar{\mu}$, the amplitude was given in Eq.~(\ref{eq:main:eemm massive}), for $e\gamma^+\bar{e}h$ in Eq.~(\ref{eq:main:M_eAeh}), for $e\bar{e}\gamma^+\gamma^+$ in Eq.~(\ref{eq:main:M_eeA+A+}) and for $e\bar{e}\gamma^+\gamma^-$ in Eq.~(\ref{eq:main:M_eeA+A- massive}).  Additionally, we looked at the unphysical amplitudes with 0-helicity photons, suggesting that they should vanish in the massless limit.  We successfully showed the vanishing limit for $e\gamma^0\bar{e}h$ and $e\bar{e}\gamma^+\gamma^0$ in Apps.~\ref{app:eA0eh} and \ref{app:eeA+A0}, but were unsuccessful at obtaining a vanishing limit for $e\bar{e}\gamma^0\gamma^0$, as we detail in App.~\ref{app:eeA0A0}.

Although using a massive photon was successful in all the amplitudes calculated here, it would be preferable to have a fully consistent massless photon theory.  Giving the photon a mass requires that its two physical helicities be combined into a a spin-1 object that also contains an unphysical helicity-0 object.  The helicity-0 state must fall out of all physical amplitudes in the massless limit, and does in our calculations, but having an unphysical state in intermediate calculations is a shortcoming that we would like to overcome in constructive calculations.  Furthermore, when using a massive photon, every diagram that would be included in a Feynman-diagram calculation must also be included in the massive spinor calculation.  There is no improvement in the efficiency of the calculation.  Whereas, in a massless-photon theory, where successful, when the photon is external, only one of the diagrams is required to give the full result.  This leads to an improved efficiency of the calculation and suggests that the multiple diagrams are redundant and a shortcoming of Feymman diagrams and of the massive-photon spinor calculation.  For these reasons, although success was achieved with the massive photon, we further explore the massless-photon theory.

In Sec.~\ref{sec:main:x} and App.~\ref{app:x}, we review the x factor.  We derive new useful formulas for $x$ and $\tilde{x}$ when the massless particle is external.  The details can be found App.~\ref{sec:x = alternate}.  In App.~\ref{app:xxt+xtx revisited}, we discuss the importance of using the fully symmetrized identities involving $x$.

In Sec.~\ref{sec:main:constructive no Shift} and App.~\ref{app:Calc AHH Spinor Amps No Shifts}, we calculate the amplitudes using massless photons and the $x$ factor.  We begin with $e\bar{e}\mu\bar{\mu}$ and show that the result [in Eq.~(\ref{eq:main:M_eemm massless no shift})] is both ambiguous and not in agreement with Feynman diagrams.  On the other hand, we also calculate the amplitude for $e\gamma^+\bar{e}h$ and $e\bar{e}\gamma^+\gamma^+$ and $e\bar{e}\gamma^+\gamma^-$ using the standard constructive technique and find agreement with Feynman diagrams as well as with the results already obtained using the massless limit of a massive photon.

In Sec.~\ref{sec:main:spinor shifts} and App.~\ref{app:analytic continuation}, we describe the analytical continuation of two external momenta on opposite sides of the propagator and the associated shift in the spinors.  We begin with a review of the massless case and then generalize this shift to the massive case.  We begin by finding the shift $\lbrack i,j\rangle$ when the two spinors are massless but the internal line is massive in Eqs.~(\ref{eq:main:[ihat|: 0,0,M}) and (\ref{eq:main:|jhat>: 0,0,M}).  We then find the shift $\lbrack\mathbf{i},j\rangle$ when particle $i$ is massive, particle $j$ is massless and the internal line is massive in Eqs.~(\ref{eq:main:[ihat|: mi,0,M}) and (\ref{eq:main:|jhat>: mi,0,M}).  We find the opposite case $\lbrack i,\mathbf{j}\rangle$ when particle $i$ is massless, particle $j$ is massive and the internal line is massive in Eqs.~(\ref{eq:main:[ihat|: 0,mj,M}) and (\ref{eq:main:|jhat>: 0,mj,M}).  Finally, we turn to the case $\lbrack\mathbf{i,j}\rangle$ where both external particles are massive.  We note that the formulas are excessively complicated and not necessary when the internal line is also massive, but work out the expressions when the internal line is massless and give it in Eqs.~(\ref{eq:main:[ihat|: mi,mj,0}) and (\ref{eq:main:|jhat>: mi,mj,0}).

We consider the large $z$ limit of the analytically continued amplitudes in Sec.~\ref{sec:main:large z} and App.~\ref{app:large z behavior}.   This is important because the proof that amplitudes can be split up into smaller amplitudes in this way relies on the amplitude vanishing for asymptotically large $z$ \cite{Britto:2005fq}.  We find that the amplitude for $e\bar{e}\mu\bar{\mu}$ does not vanish for any choice of momentum shift and suggest that this may be part of the reason that the constructive techniques with a massless photon fail to obtain the correct result.  However, we note that not vanishing for large $z$ does not imply that the amplitude cannot be built up in this way.  In fact, we point out that the amplitude for $e\bar{e}\mu\bar{\mu}$ using a massive photon also does not asymptotically vanish.  Nevertheless, as we showed in Sec.~\ref{sec:main:MA}, the constructive method does succeed in finding this amplitude.  On the other hand, we find that for the processes $e\gamma^{\pm}\bar{e}h$, $e\bar{e}\gamma^{\pm}\gamma^{\pm}$ and $e\bar{e}\gamma^{\pm}\gamma^{\mp}$, the amplitude does vanish for multiple choices of momentum shifts.  We take this as confirmation that the techniques should work for these processes and they do.

In Sec.~\ref{sec:main:constructive} and App.~\ref{app:Calc AHH Spinor Amps}, we recalculate the amplitudes where we explicitly shift the momenta and the spinors.  Although the large $z$ limit of the process $e\bar{e}\mu\bar{\mu}$ does not vanish, we considered it instructive to see what occurs when the shift is applied to the constructive amplitude.  On the positive side, the shift removes the ambiguity in the amplitude resulting from the identity relating angle and square spinor products shown in Eqs.~(\ref{eq:xxt<ij>=xxt[ij]}) and (\ref{eq:xxt<kl>=xxt[kl]}).  Unfortuantely, perhaps unsurprisingly, the resulting amplitude in Eqs.~(\ref{eq:M_eemm shifted A}) and (\ref{eq:M_eemm shifted B}) still do not agree with Feynman diagrams.  Using the shifts on the processes $e\gamma^{\pm}\bar{e}h$, $e\bar{e}\gamma^{\pm}\gamma^{\pm}$ and $e\bar{e}\gamma^{\pm}\gamma^{\mp}$ on the other hand, produces the correct final amplitudes, suggesting that the shifts are correct and not the source of the problem with the process $e\bar{e}\mu\bar{\mu}$.

In Sec.~\ref{sec:main:amplitudes}, we further discuss the correct amplitudes for all the processes discussed in this paper.  For each of them, we note that most of the structure can be determined by considering the transformation properties of the amplitude and that the remaining ambiugiuty can be determined by trial and error.  We also note that all the amplitudes have been compared analytically with Feynman diagrams and found to be in agreement for all energies and all angles.  We extend the result for the process $e\bar{e}\mu\bar{\mu}$ to the process $e\bar{e}e\bar{e}$ by exchanging spinor states $2\leftrightarrow3$ and making appropriate sign changes.  Again, we checked this process against Feynman diagrams.  Although this process with an internal photon has two diagrams, we do not see any simplification occurring by combining with a common denominator, unlike the processes with an external photon and two diagrams where the amplitude is much simpler with a common denominator.  For the processes with an external photon, we also note that the propagator denominator can be written in an alternate way, more reminiscent of purely massless amplitudes.  For example, for the process $e\bar{e}\gamma^{\pm}\gamma^{\pm}$,  $(t-m_e^2)= \lbrack3\lvert p_1\rvert3\rangle$ and $(u-m_e^2)=\lbrack4\lvert p_1\rvert4\rangle$, giving us the amplitude in the form of Eqs.~(\ref{eq:main:MeeA+A+ alternate den}) and (\ref{eq:main:MeeA-A- alternate den}).  This amplitude is suggestive for a generalization for the amplitude with an arbitrary number of same-helicity photons.

In the future, the greatest open question is how to calculate scattering amplitudes constructively with an internal massless photon (or gluon).  New ingredients are needed.  Beyond that, we would like to calculate all the 4-point amplitudes of the Standard Model, followed by higher-point and higher-loop amplitudes and compare with Feynman diagrams to determine whether these spinor amplitudes would potentially improve the efficiency of matrix element generators used to compare theory with experiment at the colliders.  We also hope that these structures will lead to improved understanding of the SM.

\section{Acknowledgements}
We would like to thank Yu-tin Huang for suggesting adding a spin index to $z$ in the massive spinor shifts.

redpajama_set_name: RedPajamaArXiv


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The image of introverts in politics is still so off base. Take views of Obama for example. He is a known introvert. His lower key temperament, passion for privacy, preference for focused conversation and considered comments all support this personality preference.
Okay – most of the description sounds like it makes sense based on Heileman's proximity to Obama. But come on. Liking people? There is NO evidence that introverts don't like people. The introvert tendency to keep feelings to themselves can create misunderstandings and a perception gap. What they intend to show is not what is seen. I think this assumption about President Obama is a clear case of the perception gap at work.
This entry was posted in General and tagged Andrew Sullivan, Daily Beast, Introverts, jennifer kahnweiler, Jon Heileman, obama on September 14, 2012 by Jennifer Kahnweiler.