\sektion{11}{October 28}
\descriptioneleven
\section*{\uline{{\sc Feynman Diagrams}} in Model 3}
{\small \[\langle0|(S-1)|0\rangle=\langle\vec{p}|(\underset{
\substack{\text{one-nucleon}}}{S-1)|\vec{p}'\rangle}=\langle\vec{k}|
(\underset{\substack{\text{one-meson}}}{S-1)|\vec{k}'\rangle} \] }
Let's look at nucleon-nucleon scattering at $O(g^2)$. That is the
first order at which there is a contribution to
\[\langle \underset{\text{two-nucleon states}}{\underbrace{p_1'p_2'}
|(S-1)|\underbrace{p_1p_2}}\rangle\]
The $-1$ in $S-1$ is there because we aren't really interested in the
no scattering process, $p_1=p_1'$ and $p_2=p_2'$ or $p_1=p_2'$ and
$p_2=p_1'$, which comes from 1, the $O(g^0)$ term in
\[ S=Te^{-ig\int d^4x\left(\psi^*\psi\phi-\frac{b}{2g}\phi^2-\frac{c}
{g}\psi^*\psi\right)} \]
(Since the power series for $b$ and $c$ begin at order $g^1$ at the
earliest (they are zero if $g=0$), it is not misleading to pull the
$g$ out in front of the whole interaction and talk about the $O(g^0)$
contribution to $S$.)
There are no arrows over $p_1$, $p_2$, $p_1'$ and $p_2'$ because we
are going to use the states that transform nicely under Lorentz
transformations
\begin{align*}
U(\Lambda)|p_1,p_2\rangle&=|\Lambda p_1,\Lambda p_2\rangle\\
|p_1,p_2\rangle &=b^\dagger(p_1)b^\dagger(p_2)|0\rangle\\
b^\dagger(p)&=(2\pi)^{3/2}\sqrt{2\omega_{\vec{p}}}b_{\vec{p}}
\end{align*}
So we don't have to worry about Bose statistics demand $p_1\neq p_2$
and $p_1'\neq p_2'$. We can recover what we've lost by building wave
packets concentrated \uline{around} $p_1=p_2$.
The term in $S$ with two factors of the interaction is
\[\frac{(-ig)^2}{2!}\int d^4x_1d^4x_2
T[\psi^*\psi\phi(x_1)\psi^*\psi\phi(x_2)] \]
After all the hoopla about the turning on and off function, we are
abandoning it, being sloppy: The only term in the Wick expansion of
this term in $S$ that can contribute to two nucleons goes to two
nucleons is
\[\frac{(-ig)^2}{2!}\int d^4x_1d^4x_2:\psi^*\psi\wick{1}{<1\phi(x_1)
\psi^*\psi>1\phi(x_2)}: \]
$\wick{1}{<1\phi(x_1)>1\phi(x_2)}$ is some number. Let's look at
\begin{equation}
\label{eq:11-page2}
\langle p_1'p_2'|:\psi^*(x_1)\psi(x_1)\psi^*(x_2)\psi(x_2):|p_1p_2
\rangle
\end{equation}
The nucleon annihilation terms in $\psi(x_1)$ and $\psi(x_2)$ have to
be used to annihilate two incoming nucleons. The nucleon creation
terms in $\psi^*(x_1)$ and $\psi^*(x_2)$ have to be used to create two
nucleons, so as not to get zero inner product (alternatively I could
say they have to be used to ``annihilate two nucleons on the left'').
In equations
\[\langle p_1'p_2'|:\psi^*\psi(x_1)\psi^*\psi(x_2):|p_1p_2
\rangle=\langle p_1'p_2'|\psi^*(x_1)\psi^*(x_2)|0\rangle\langle0|\psi
(x_1)\psi(x_2):|p_1p_2 \rangle \]
You can easily show that the two contributions to the second matrix
element are (the c.c.~equation is also used)
\begin{equation}\label{eq:11-botpage2}
\langle0|\psi(x_1)\psi(x_2)|p_1p_2\rangle=\underset{\uparrow\text{
$p_1$ absorbed at $x_1$\quad\quad\quad}}{e^{-ip_1\cdot
x_1-ip_2\cdot x_2}+e^{-ip_1\cdot x_2-ip_2\cdot x_1}}
\end{equation}
So there are four contributions to our matrix element
\begin{align}
\langle p_1'p_2'|:\psi^*\psi(x_1)\psi^*\psi(x_2):|p_1p_2\rangle &=
\bigl(e^{ip_1'\cdot x_1+ip_2'\cdot x_2}+e^{ip_1'\cdot x_2+ip_2'\cdot
x_1}\bigr)\bigl(e^{ip_1\cdot x_1+ip_2\cdot x_2}+e^{ip_1\cdot x_2+ip_2
\cdot x_1}\bigr)\nonumber\\
&=\underset{F}{e^{ip_1'\cdot x_1+ip_2'\cdot x_2-ip_1\cdot x_1-ip_2
\cdot x_2}}+\underset{L}{e^{ip_1'\cdot x_2+ip_2'\cdot x_1-ip_1\cdot
x_2-ip_2 \cdot x_1}}\nonumber\\
&+\underset{I}{e^{ip_1'\cdot x_2+ip_2'\cdot x_1 -ip_1\cdot x_1-ip_2
\cdot x_2}}+\underset{O}{e^{ip_1'\cdot x_1+ip_2' \cdot x_2-ip_1\cdot
x_2-ip_2\cdot x_1}}
\end{align}
Notice that the pair of terms on the first line differ only by
$x_1\leftrightarrow x_2$ and that the pair of terms on the second
line only differ by $x_1\leftrightarrow x_2$. Since $x_1$ and $x_2$
are to be integrated over and since $\wick{1}{<1\phi(x_1)>1
\phi(x_2)}$ is symmetric under $x_1\leftrightarrow x_2$ these pairs
give identical contributions to the matrix element. We'll just write
one of the pairs, which cancels the $\frac{1}{2!}$. We have,
\begin{multline}
(-ig)^2\int d^4x_1 d^4x_2 \wick{1}{<1\phi(x_1)>1\phi(x_2)}\bigl(
e^{ip_1'\cdot x_1+ip_2'\cdot x_2 -ip_1\cdot x_1-ip_2 \cdot x_2}+e^{i
p_1'\cdot x_2+ip_2'\cdot x_1 -ip_1 \cdot x_1-ip_2 \cdot x_2}\bigr)\\
=(-ig)^2\int d^4x_1d^4x_2\int\frac{d^4k}{(2\pi)^4}\bigl[e^{ix_1\cdot
(p_1'-p_1+k)}e^{ix_2\cdot(p_2'-p_2-k)}\\
+e^{ix_1\cdot (p_2'-p_1+k)}e^{ix_2\cdot(p_1'-p_2-k)}\bigr]\frac{i}
{k^2-\mu^2+i\epsilon}
\end{multline}
I have used the expression for $\wick{1}{<1\phi(x_1)>1\phi(x_2)}$ and
grouped all the exponential factors by spacetime point.
With these two integrals (there are two terms in the integrand) go two
pictures
\begin{align}\label{eq:11-page4}
&\underset{(a)}{\Diagram{
\momentum{fV}{p_1'} & \momentum{fV}{p_1} \\
& \momentum[ulft]{fv}{k\downarrow}\\
\momentum{fV}{p_2'} & \momentum{fV}{p_2}}}
& &\substack{N+N\rightarrow N+N\\ \text{Feynman diagrams @ $O(g^2)$}}
&
&\underset{(b)}{\Diagram{
\momentum{fV}{p_2'} & \momentum{fV}{p_1} \\
& \momentum[ulft]{fv}{k\downarrow}\\
\momentum{fV}{p_1'} & \momentum{fV}{p_2}}}
\quad\substack{\text{Notice external lines each}\\
\text{have an associated momentum.}\\
\text{The vertices are not numbered.}}
\end{align}
\noindent and two stories\footnote{$\begin{array}{cc}
\\ \text{Conventions:}& \text{in on right}\\ &\text{out on left}
\end{array}$}. The story that goes with picture (a) is this. A
nucleon with momentum $p_1$ comes in and interacts. Out of the
interaction point comes a nucleon with momentum $p_1'$ and a ``virtual
meson.'' This ``virtual meson'' then interacts with a nucleon with
momentum $p_2$ and out of the interaction point comes a nucleon with
momentum $p_2'$. The interaction points $x_1$ and $x_2$ can occur
anywhere, and so they are integrated over. Furthermore, this
``virtual meson'' can have any momentum $k$ and this is integrated
over, although you can see from the factor $\frac{i}{k^2-\mu^2
+i\epsilon}$, ``Feynman's propagator'' that it likes to be on the
meson mass shell, although with $k^0=\pm\sqrt{\vec{k}^2+\mu^2}$.
Fairy tales like this helped Feynman discover and think about quantum
electrodynamics. In our formalism, they are little more than fairy
tales, but in a formulation of quantum particle mechanics called the
path integral formulation they gain some justification. The words not
only match the pictures, they parallel the mathematics.
The $x_1$ and $x_2$ integrations are easy to do. We get
\begin{multline}\label{eq:11-page5}
(ig)^2\int\frac{d^4k}{(2\pi)^4}\frac{i}{k^2-\mu^2+i\epsilon}\bigl[
(2\pi)^4\delta^{(4)}(p_1'-p_1+k)(2\pi)^4\delta^{(4)}(p_2'-p_2+k)\\
+(2\pi)^4\delta^{(4)}(p_2'-p_1+k)(2\pi)^4\delta^{(4)}(p_1'-p_2+k)
\bigr]
\end{multline}
Because the interaction is spacetime translationally invariant, after
integrating the interaction point over all space-time we get delta
functions which enforce energy-momentum conservation at every vertex.
All the features of this computation generalize to more complicated
$S$ matrix element contributions. I'll give a set of rules for
writing down these integral expressions for contributions to $S$
matrix elements. First, I'll explain in general why there are no
combinatoric factors in model 3 to worry about, no symmetry numbers in
the integral expressions.
Take a given operator in the Wick expansion, which has an associated
Wick diagram
\[ D, \text{ a diagram }\longleftrightarrow \frac{:O(D):}{n(D)!} \]
Designate which of the lines leading out of the diagram annihilates
each incoming particle, and which of the lines creates each outgoing
particle.
This is one contribution to the $S$ matrix element.
Now consider summing over the permutation of the numbered points in
the Wick diagram. While only $\frac{n(D)!}{S(D)}$ of these
permutations actually correspond to different terms in the Wick
expansion, in model 3, all $n(D)!$ of these permutations correspond to
contributions to the $S$ matrix element. This cancels the
$\frac{1}{n(D)!}$ exactly.
There are other possible designations (in general) for the way the
external lines connect to the vertices of the Wick diagram. If they
differ just by a permutation of the vertices then we have already
counted them (by cancelling the $\frac{1}{n(D)!}$). If they don't
differ by just a permutation of the vertices then they correspond to a
different Feynman diagram (the difference between (a) and (b) in %*page
%4*).
Eq.~(\ref{eq:11-page4})
Only in certain theories, like Model 3, do the $n(D)!$ permutations of
the vertices al make different contributions to the $S$ matrix, and in
fact this is true in model 3 only when a diagram\footnote{in fact,
each connected part has to have at least one external line} has at
least one external line. In that case there is an unambiguous way of
identifying each vertex in the diagram. Contributions to
$\langle0|S|0\rangle$, which have no external lines, can have symmetry
factors. The unambiguous labelling statement for diagram (a) is
\begin{align*}
&\Diagram{
\momentum{fV}{p_1'} & \momentum{fV}{p_1} \\
& \momentum[ulft]{fv}{k\downarrow}\\
\momentum{fV}{p_2'} & \momentum{fV}{p_2}} &
\substack{\text{The upper vertex is uniquely labelled as the one where
$p_1$ is absorbed.}\\ \\ \\ \text{The lower vertex is the one
connected to the upper vertex by a muon line.}}
\end{align*}
In this theory, as soon as one vertex is labelled (by an external
line) they are all uniquely labelled.
\section*{Feynman Rules for Model 3}
You should convince yourself by taking some other low order terms in
the Wick expansion of $S$ and looking at some simple matrix elements
they contribute to that the following set of rules applied to the
diagram always gives you the correct contribution to the $S$ matrix
element.
For external lines $\begin{Bmatrix} \text{incoming}\\
\text{outgoing}\end{Bmatrix}$ momenta are directed $\begin{Bmatrix}
\text{in}\\ \text{out}\end{Bmatrix}$.
Assign a directed momentum to every internal line.
\begin{align*}
&\text{For every} & &\text{Write}\\
& \begin{array}{c}
\text{internal meson line}\\
\Diagram{ \momentum[bot]{f}{\leftarrow k}}
\end{array}
& & \int\frac{d^4k}{(2\pi)^4}\frac{i}{k^2-\mu^2+i\epsilon}\\
& \begin{array}{c}
\text{internal nucleon line}\\
\Diagram{ \momentum[bot]{fV}{\leftarrow p}}
\end{array}
& & \int\frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2+i\epsilon}\\
& \begin{array}{c}
\text{vertex}\\
\Diagram{ \momentum[bot]{fV}{\leftarrow p'} &
\momentum[bot]{fV}{\leftarrow p} \\
& \momentum[bot]{fd}{k\nwarrow}}\end{array}
& & (-ig)(2\pi)^4\delta^{(4)}(p'-p-k)\\
& \begin{array}{cc}
\text{meson vacuum counterterm}\\
\bullet& \substack{(2\pi)^4\delta^{(4)}(0)\text{ would turn into
the}\\
\text{volume of all spacetime if}\\
\text{the system were in a box. This}\\
\text{c.t.~diagram is designed to cancel}\\
\text{diagrams without external lines}\\
\text{which you will see also have a}\\
\text{factor of } \delta^{(4)}(0)\\
}\end{array}
& & ia(2\pi)^4\delta^{(4)}(0)\\
& \begin{array}{cc}
\text{meson mass counterterm}\\
\Diagram{ \momentum[bot]{f}{\leftarrow k'} x
\momentum[bot]{f}{\leftarrow k}}
& \substack{\text{although the meson mass}\\
\text{counterterm had a }\frac{1}{2}\text{ in}\\
\text{the Lagrangian, there is no}\\
\frac{1}{2}\text{ here because there are}\\
\text{two ways to do the interaction}\\
}\end{array}
& & ib(2\pi)^4\delta^{(4)}(k-k')\\
& \begin{array}{c}
\text{nucleon mass counterterm}\\
\Diagram{ \momentum[bot]{fV}{\leftarrow p'} &x
&\momentum[bot]{fV}{\leftarrow p}}
\end{array}
& & ic(2\pi)^4\delta^{(4)}(p-p')
\end{align*}
A catalog of all Feynman diagrams in model 3 up to $O(g^2)$ (except
those related by $C$ or $T$ will not be written down twice)
Order $g$
% define "elevenLcount" as a counter
\newcounter{elevenLcount}
% set the "default" label to print counter as a Roman numeral
\begin{list}{(\arabic{elevenLcount})}
% inform the list command to use this counter
{\usecounter{elevenLcount}
% set rightmargin equal to leftmargin
\setlength{\rightmargin}{\leftmargin}}
%\setlength{\listparindent}{\labelwidth}}
% we can now begin the "items"
\item $\Diagram{fdV \\
& f \\
fuA} = 0 $ if $\mu<2m$ by energy momentum conservation.
\item $\Diagram{ f0 flSA flSu fs0 f} =0 $\\
\end{list}
Order $g^2$
\begin{list}{(\arabic{elevenLcount})}
% inform the list command to use this counter
{\usecounter{elevenLcount}
\setcounter{elevenLcount}{2}
% set rightmargin equal to leftmargin
\setlength{\rightmargin}{\leftmargin}}
%\setlength{\listparindent}{\labelwidth}}
% we can now begin the "items"
\item $\begin{array}{c}
\Diagram{fdV \\
& f \\
fuA}\\\Diagram{fdV \\
& f \\
fuA} \end{array} =0^2$ if $\mu<2m$
\item $\left.\begin{array}{lc}
\text{(a)} & \Diagram{f flV fluA f}\\
\\
\\
\text{(b)} & \Diagram{flV flu f0 & f & f0 flV flu}\\
\\
\\
\text{(c)} & \begin{array}{c}\bullet\\
\substack{\uparrow\\ \text{vacuum energy c.t.~to
$O(g^2)$}}\end{array}\end{array}\right\}\substack{\text{\normalsize
Because we demand there be no corrections to}\\\text{\normalsize
$\langle0|S|0\rangle$ these sum to zero. This fixes the
vacuum}\\ \text{\normalsize energy c.t.~to $O(g^2)$}}$
\end{list}
In 4(c) think of the c.t.~as $O(g^2)$. Its value is
\underline{determined} by the fact that it has to cancel some $O(g^2)$
contributions to the vacuum-to-vacuum $S$ matrix element.
\begin{list}{(\arabic{elevenLcount})}
% inform the list command to use this counter
{\usecounter{elevenLcount}
\setcounter{elevenLcount}{4}
% set rightmargin equal to leftmargin
\setlength{\rightmargin}{\leftmargin}}
%\setlength{\listparindent}{\labelwidth}}
% we can now begin the "items"
\item $\left.\begin{array}{lc}
\text{(a)} & \Diagram{fV f a fl flu f fV }\\
\\
\\
\text{(b)} & \Diagram{fV x fV}\\
& \substack{\uparrow\\ \text{nucleon % corrected from "meson"
mass c.t.~to
$O(g^2)$}}\end{array} \right\}\substack{\text{\normalsize
Sum to zero because we demand that}\\\text{\normalsize
there are no corrections to
$\langle\!\!\!\!\!\!\!\!\underbrace{\vec{p}|S|\vec{p}'}_{\substack{\text{one
nucleon each}}}\!\!\!\!\!\!\!\!\rangle$}
}$ \\
\\
\\
$\begin{array}{lc}
\text{(c)} &\begin{matrix}\!\!\!\!\!\!\!\!\!\!\!\!\Diagram{f0 flV flu}\\ \\ \Diagram{fV fv
fV}\end{matrix}\end{array}$ this diagram
comes out zero for the same reason as (2) does.
\\
\\
\item $\left.\begin{array}{lc}
\text{(a)} & \Diagram{f f0 flA fluV f0 f }\\
\\
\\
\text{(b)} & \Diagram{f x f}\end{array}\right\}\substack{\text{\normalsize
Sum to zero because we demand that}\\\text{\normalsize there are
no corrections to
$\langle\!\!\!\!\!\!\underbrace{\vec{k}|S|\vec{k}'}_{\substack{\text{one
meson each}}}\!\!\!\!\!\!\rangle$}
}$ \end{list}
The remaining order $g^2$ diagrams are more interesting. They
contribute to the following processes
\begin{list}{(\arabic{elevenLcount})}
% inform the list command to use this counter
{\usecounter{elevenLcount}
\setcounter{elevenLcount}{6}
% set rightmargin equal to leftmargin
\setlength{\rightmargin}{\leftmargin}}
%\setlength{\listparindent}{\labelwidth}}
% we can now begin the "items"
\item $N+N\longrightarrow N+N$ (connected by $C$ to
$\bar{N}+\bar{N}\longrightarrow\bar{N}+\bar{N}$)
\item $N+\bar{N}\longrightarrow N+\bar{N}$
\item $N+\phi\longrightarrow N+\phi$ (connected by $C$ to
$\bar{N}+\phi\longrightarrow\bar{N}+\phi$)
\item $N+\bar{N}\longrightarrow \phi+\phi$ (connected by $T$ to
$\phi+\phi\longrightarrow N+\bar{N}$)
\end{list}
Although $\phi+\phi\longrightarrow\phi+\phi$ appears in this theory,
indeed it must appear, it does not do so until $O(g^4)$. The diagram
is
\[ \Diagram{ fd & & & fu \\
& fA \\
&fvA & fvV\\
& fV \\
fu & & & fd } \]
We have already written down the contributions to process (7). The
diagrams are
\begin{align*}
&\underset{(a)}{\Diagram{
\momentum{fV}{p_1'} & \momentum{fV}{p_1} \\
& \momentum[ulft]{fv}{k\downarrow}\\
\momentum{fV}{p_2'} & \momentum{fV}{p_2}}}
\quad \text{\normalsize and }\quad
\underset{(b)}{\Diagram{
\momentum{fV}{p_2'} & \momentum{fV}{p_1} \\
& \momentum[ulft]{fv}{k\downarrow}\\
\momentum{fV}{p_1'} & \momentum{fV}{p_2}}}
\quad\text{\normalsize a.k.a.} % how to do the line crossing??
\end{align*}
They give (see Eq.~(\ref{eq:11-page5}))
\begin{multline*}
(-ig)^2\frac{i}{(p_1-p_1')^2-\mu^2+i\epsilon}(2\pi)^4\delta^{(4)}
(p_1'+p_2'-p_1-p_2)\\
+(-ig)^2\frac{i}{(p_1-p_2')^2-\mu^2+i\epsilon}(2\pi)^4\delta^{(4)}
(p_1'+p_2'-p_1-p_2)
\end{multline*}
Note that the $\frac{1}{(2\pi)^4}$ associated with $d^4k$ exactly
cancels with the $(2\pi)^4$ associated with the $\delta$ functions
used to do the integral. All our formulas have been arranged so that
$2\pi$'s always go with $\delta$'s and $\frac{1}{2\pi}$'s always go
with $\int dk$'s.
Note that you can shortcut these trivial integrations over $\delta$
functions by just assigning internal momenta so as to conserve
momentum whenever an internal momentum is determined by the other
momenta at a vertex.
Finally, note that performing the trivial integrals over $\delta$
functions always gives you a factor
\begin{align*}
(2\pi)^4\delta^{(4)}(\!\!\!\!\!\!\!\underbrace{p_f}_{\substack{\text{sum of all}\\
\text{final momenta}}}\!\!\!\!\!\!\!\!-\!\!\!\!\!\!\!\!\overbrace{p_i}^{\substack{\text{sum of
all}\\\text{initial momenta}}}\!\!\!\!\!\!\!\!) \quad\quad\substack{\text{\normalsize
at least when the diagram}\\ \text{\normalsize is of one connected piece}}
\end{align*}
We define $a_{fi}$, the invariant Feynman amplitude by
\[\langle f|(S-1)|i\rangle = ia_{fi}\delta^{(4)}(p_f-p_i)
\]
The factor of $i$ is inserted to match the phase convention of NRQM.
For $N+N\longrightarrow N+N$
\[ia = (-ig)^2\left[\frac{i}{(p_1-p_1')^2-\mu^2+i\epsilon}+
\frac{i}{(p_1-p_2')^2-\mu^2+i\epsilon}\right]\]
Let's look at this in the COM frame.
\begin{align*}
\begin{array}{lc}
p_1=(\sqrt{p^2+m^2},p\!\!\!\!\overbrace{\vec{e}}^{\text{unit
vector}}\!\!\!\!)\\
p_2=(\sqrt{p^2+m^2},-p\vec{e})\\
p_3=(\sqrt{p^2+m^2},p\vec{e}')\\
p_4=(\sqrt{p^2+m^2},-p\vec{e}')\\ % ** there's a diagram here
\end{array}%\quad\quad\substack{\vec{e}\cdot\vec{e}'=\cos\theta \\
%\theta\text{ is the scattering angle}\\ \text{ in the com frame}}
\end{align*}
\begin{center}
\includegraphics[width=5 cm]{11-fig1.eps}\\
$\vec{e}\cdot\vec{e}'=\cos\theta$, $\theta$ is the scattering angle in
the COM frame.
\end{center}
$E_T=2\sqrt{p^2+m^2}$ is often used to characterize collisions. In
the nonrelativistic limit, which we will be taking, $p$ is more
useful.
Define the momentum transfer, $\Delta$, and the crossed momentum
transfer, $\Delta_c$, by
\begin{align*}
(p_1-p_1')^2&=-\Delta^2\\
(p_1-p_2')^2&=-\Delta_c^2
\end{align*}
In our COM variables
\begin{align*}
\Delta^2&=2p^2(1-\cos\theta) & \Delta_c^2&=2p^2(1+\cos\theta)
\end{align*}
The invariant Feynman amplitude is
\[ a=g^2\left[\frac{1}{\Delta^2+\mu^2}+\frac{1}{\Delta_c^2+\mu^2}
\right] \]
We have dropped the $i\epsilon$ because it is unnecessary. For
physically accessible values of $\Delta^2$ and $\Delta_c^2$ the
denominators are never less than $\mu^2$.
The first term is peaked (peaked sharper at higher $p$) in the forward
($\theta\approx0$) direction. The second term produces an identical
peak in the backward direction. Of course when identical particles
collide who is to say what is forward and what is backward.
$\theta=0$ is indistinguishable from a scattering angle of
$\theta=\pi$. The %$\cancel{\text{amplitude}}$
probability had better have come out symmetrical.
People were scattering nucleons off nucleons long before quantum field
theory was around, and at low energies they could describe scattering
processes adequately with NRQM. Let's try to understand our amplitude
in NRQM. First we'll find the NR analog of the first term.
In the COM frame, two body scattering is simplified to the problem of
scattering of a potential (classically and quantum-mechanically).
P.T.~at lowest order gives
\begin{align*}
\langle \vec{k}'|S-1|\vec{k}\rangle &\propto \langle
\vec{k}'|V|\vec{k}\rangle \\
&=\int d^3r V(\vec{r})e^{-i\vec{\Delta}\cdot\vec{r}} & \text{``Born''
approximation}\\
&=\tilde{V}(\vec{\Delta}) & \vec{\Delta}=\vec{k}'-\vec{k}
\end{align*}
To explain the first term in our scattering amplitude using NRQM we
must have
\[
\tilde{V}(\vec{\Delta})\propto\frac{1}{\Delta^2+\mu^2}\Longrightarrow
\tilde{V}(\vec{r})\propto\frac{g^2e^{-\mu r}}{r} \]
Our amplitude, which is characterized by having a simple pole in a
physically unobservable region, at $\Delta^2=-\mu^2$, corresponds to
the Born approx.~to the Yukawa interaction$!$
The second term also has an analog in NRQM. With two identical
particles, the Hamiltonian should contain an exchange potential
\begin{align*}
H&=H_0+\!\!\!\underset{\substack{\\ \text{Yukawa}\\
\text{potential}}}{V}\!\!\!+\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\underbrace{VE}_{\substack{\quad\quad\quad\text{exchange Yukawa
potential}}} &
\underset{\substack{$E$\text{ is the exchange
operator}}}{E|\vec{r}_1,\vec{r}_2\rangle=|\vec{r}_2,\vec{r}_1\rangle}\\
V|\vec{r}_1,\vec{r}_2\rangle&\propto \frac{g^2e^{-\mu
r}}{r}|\vec{r}_1,\vec{r}_2\rangle & r=|\vec{r}_1-\vec{r}_2|
\end{align*}
The exchange Yukawa potential is the source of a simple pole in the
amplitude at $\Delta_c^2=-\mu^2$, also in a physically unobservable
region. The NRQM amplitude is proportional to
$\tilde{V}(\vec{\Delta})+\tilde{V}(\vec{\Delta}_{exch})$
($\vec{\Delta}_{exch}=\vec{k}'+\vec{k}$). In a partial wave expansion
of the amplitude, the exchange potential gives a contribution
$\left\{\begin{array}{c} \text{identical} \\
\text{opposite}\end{array}\right\}$ to the direct potential if $l$ is
$\left\{\begin{array}{c} \text{even} \\
\text{odd}\end{array}\right\}$.
This is because in the COM an eigenstate of angular momentum is an
eigenstate of the exchange operator $E$ with eigenvalue $(-1)^2$.