# An Area Law for Entanglement Entropy in Particle Scattering

Ian Low${}^{\,a,b}$, Zhewei Yin${}^{\,b}$ a High Energy Physics Division,
Argonne National Laboratory, Lemont, IL 60439, USA
b Department of Physics and Astronomy, Northwestern University, Evanston, IL
60208, USA

###### Abstract

The scattering cross section is the effective area of collision when two
particles collide. Quantum mechanically, it is a measure of the probability
for a specific process to take place. Employing wave packets to describe the
scattering process, we compute the entanglement entropy in 2-to-2 scattering
of particles in a general setting using the $S$-matrix formalism. Applying the
optical theorem, we show that the linear entropy $\mathcal{E}_{2}$ is given by
the elastic cross section $\sigma_{{\text{el}}}$ in unit of the transverse
size $L^{2}$ of the wave packet,
$\mathcal{E}_{2}\sim\sigma_{{\text{el}}}/L^{2}$, when the initial states are
not entangled. The result allows for dual interpretations of the entanglement
entropy as an area and as a probability. Since $\sigma_{{\text{el}}}$ is
generally believed, and observed experimentally, to grow with the collision
energy $\sqrt{s}$ in the high energy regime, the result suggests a “second
law” of entanglement entropy for high energy collisions. Furthermore, the
Froissart bound places an upper limit on the entropy growth.

##  I Introduction

Entropy is among the most fundamental concepts in physics and quantifies the
number of possible microstates giving rise to a particular macrostate. It is
directly related to probability – a macrostate containing more microstates has
a higher probability to occur and hence a higher entropy. A physical system
naturally evolves into more probable states, thereby increasing the entropy,
which defines an arrow of time.

Entropy can also be understood as a measure of randomness, or disorder, since
a more probable state is considered to be more disordered. As such the entropy
is also used to quantify the amount of information in a physical system, since
a highly disordered system is capable of carrying a large amount of
information. Indeed, a sentence consists of a single alphabet is very ordered
but conveys little information. In classical information theory the Shannon
entropy, which measures the average amount of information in an event drawn
from a particular probability distribution Shannon (1948); Nielsen and Chuang
(2012), features prominently.

One usually refers to the entropy defined in classical systems as coarse-
grained entropy, which focuses on a subset of simple observables without
knowing the full microscopic configuration space. Even though the underlying
dynamics is deterministic, the ignorance of the microstates necessitates the
use of probability. Quantum mechanics, on the other hand, is inherently
probabilistic and introduces intrinsic randomness in a quantum system. Such
randomness is described by the quantum, or fine-grained, entropy Witten
(2020); Almheiri et al. (2021). The most prominent quantum entropy is the von
Neumann entropy Nielsen and Chuang (2012), which can be generalized to two
classes of quantum entropies: the Rényi entropy Rényi (1960) and the Tsallis
entropy Tsallis (1988).

Similar to their classical counterpart, quantum entropies also quantify the
amount of information present in a quantum system, as well as quantum
correlations, or entanglement, between systems Nielsen and Chuang (2012). They
are computed from the density matrix and also called entanglement entropies.
While by definition the entropy of a closed system is invariant under unitary
transformation, and hence time evolution, this is not always the case for an
open quantum system interacting with an environment. In fact, one often
divides a closed system into a bipartite system whose Hilbert space is ${\cal
H}={\cal H}_{\text{A}}\otimes{\cal H}_{\text{B}}$. The subsystem entropy for A
is then computed from the reduced density matrix $\rho_{\text{A}}$ by tracing
over B, and can be nonzero even when the total system has vanishing entropy.

In many-body systems the entropy is usually an extensive quantity and scales
with the size of the system. Therefore it is quite mysterious that in
gravitational systems, the black hole in particular, the Bekenstein-Hawking
entropy scales with the area of horizon Bekenstein (1973); Hawking (1975).
Since then there have been more examples of area laws for entropy both in
gravitational and non-gravitational many-body systems Srednicki (1993); Ryu
and Takayanagi (2006); Eisert et al. (2010); Dvali (2018), some of which
involves holography Maldacena (1998); Almheiri et al. (2021). In these
examples the areas involved are often computed from a “boundary” between two
“regions.”

In this work we will consider the entanglement entropy in a very different
setting: The 2-to-2 scattering of particles in quantum field theories, using
the $S$-matrix formalism. This is hardly a many-body system, and it’s not
obvious at all if there is a “boundary” for which to define an area.
Nevertheless, using the wave packet formalism to describe the particle
scattering, we show that the linear entropy is given by the elastic cross
section in unit of the transverse size of the wave packet when the initial
states are not entangled. Recently the entanglement entropy in particle
scattering has attracted considerable attention, although the calculation has
mostly been done in specific quantum field theories and differing contexts
Dvali et al. (2015); Cervera-Lierta et al. (2017); Fan and Li (2018); Beane et
al. (2019); Tomaras and Toumbas (2020); Dvali (2021); Peschanski and Seki
(2019); Aoude et al. (2020); Low and Mehen (2021); Dvali and Venugopalan
(2022); Dvali and Sakhelashvili (2022); Müller and Schäfer (2022); Fedida and
Serafini (2023); Liu et al. (2023); Hentschinski et al. (2023); Cheung et al.
(2023); Carena et al. (2024); Sakurai and Spannowsky (2024); Hu et al. (2024);
Aoude et al. (2024); Kowalska and Sessolo (2024). Here we will not commit to a
specific theory and will instead use the $S$-matrix formalism to make general
statements which are non-perturbative in the couplings. Importantly, the wave
packet formalism gives a clear physical interpretation of the outcome.

Our result allows for novel, dual interpretations of the entanglement entropy
as an area, in the cross-sectional sense, and a probability, in the
information-theoretic sense. Moreover, it is generally accepted that the
elastic cross section grows with the collision centre-of-mass (CM) energy
$\sqrt{s}$ Cheng and Wu (1969, 1970), which has been verified experimentally
Block and Cahn (1985); Pancheri and Srivastava (2017). This suggests a version
of the second law of thermodynamics where the entanglement entropy grows with
energy, and the Froissart bound Froissart (1961); Martin (1963) now gives an
upper limit on the growth.

##  II The setup

We consider the scattering of two distinguishable particles: A(lice) and
B(ob). The kinematic data are labelled by the momentum
$p_{{\text{A}}/{\text{B}}}$, which can be either massive or massless. All
other non-kinematic quantum numbers are denoted by $\\{i\\}$ for A and
$\\{\bar{i}\\}$ for B, respectively. These could include spacetime quantum
numbers, such as the spin, or the internal quantum numbers like the isospin.

The scattering process we focus on is
${\text{A}}{\text{B}}\to{\text{A}}{\text{B}}$, from which we construct a
bipartite system ${\cal H}_{{\text{A}}{\text{B}}}={\cal
H}_{{\text{A}}}\otimes{\cal H}_{{\text{B}}}$, where ${\cal
H}_{{\text{A}}/{\text{B}}}$ is the Hilbert space associated with
particle-${\text{A}}/{\text{B}}$. For a general incoming state
$|\rm{in}\rangle$, the density matrix is
$\rho^{\text{i}}=|\rm{in}\rangle\langle\rm{in}|$ and the linear entropy ${\cal
E}_{2}$ for A is computed from the reduced density matrix
$\rho_{{\text{A}}}^{\text{i}}={\rm Tr}_{{\text{B}}}\ \rho^{\text{i}}$,

${\cal E}_{2}=1-{\rm Tr}(\rho_{\text{A}}^{\text{i}})^{2}\ ,$ (1)

which is the $n=2$ case of the Tsallis entropy ${\cal E}_{n}=(1-{\rm
Tr}\,\rho^{n})/(n-1)$ Tsallis (1988). From the von Neumann entropy ${\cal
E}_{\text{vN}}=-{\rm Tr}(\rho\log\rho)$, ${\cal E}_{2}$ can be obtained by
expanding with respect to $\rho=1$ and keeping the leading term.

We will start with a pure initial state and comment on the mixed initial state
later. We will also make the realistic assumption that there is no
entanglement between $p_{\text{A}}$, $p_{\text{B}}$, and between the momenta
and the other quantum numbers. The initial state is then written as

$\displaystyle|\text{in}\rangle=\sum_{i,\bar{i}}\Omega_{i\bar{i}}|\psi_{{\text{A}}}\rangle\otimes|i\rangle\otimes|\psi_{{\text{B}}}\rangle\otimes|\bar{i}\rangle,$
(2)

where $|i\rangle$ and $|\bar{i}\rangle$ are normalized such that $\langle
i|j\rangle=\delta^{ij}$,
$\langle\bar{i}|\bar{j}\rangle=\delta^{\bar{i}\bar{j}}$, and $\Omega$
describes the non-kinematic configuration of the initial state, which is
normalized such that $\text{Tr}(\Omega\Omega^{\dagger})=1$. When the quantum
numbers of two particles are not entangled,
$\Omega_{i\bar{i}}=\omega_{i}\omega^{\prime}_{\bar{i}}$ and
$|\omega|^{2}=|\omega^{\prime}|^{2}=1$. On the other hand,
$|\psi_{\text{A}}\rangle$ and $|\psi_{\text{B}}\rangle$ are described by the
momentum wave packets of A and B, respectively,

$\displaystyle|\psi_{{\text{A}}/{\text{B}}}\rangle=\int_{p}\psi_{{\text{A}}/{\text{B}}}(p)\,|p\rangle,\qquad\int_{p}\equiv\int\frac{d^{3}\vec{p}}{(2\pi)^{3}\sqrt{2E_{p}}}\,,$
(3)

where the one-particle state is normalized as follows,

$\displaystyle\langle
p|q\rangle=(2\pi)^{3}\,2E_{p}\,\delta^{3}(\vec{p}-\vec{q}).$ (4)

The momentum wave packets are normalized such that

$\displaystyle\langle\psi|\psi\rangle=\int\frac{d^{3}\vec{p}}{(2\pi)^{3}}\
|\psi(p)|^{2}=1.$ (5)

With Eq. (5), the initial density matrix $\rho^{\text{i}}$ is properly
normalized, ${\rm Tr}\,\rho^{\text{i}}=1$. If we had used the one-particle
states $|{k}_{\text{A}}\rangle$ and $|{k}_{\text{B}}\rangle$ in place of
$|\psi_{\text{A}}\rangle$ and $|\psi_{\text{B}}\rangle$, the $\delta$-function
normalization in Eq. (4) would have introduced divergences in ${\rm
Tr}\,\rho$, which is often regularized by introducing a finite box in
spacetime. We will not be using such a box normalization. In the CM frame,
$\vec{k}_{\text{A}}=-\vec{k}_{\text{B}}\equiv\vec{k}$.

In an experimental setup the incoming particles are prepared with fairly well-
defined momenta, subject to experimental resolution. Therefore we expect
$\psi(p)_{{\text{A}}/{\text{B}}}$ to be sharply peaked at $\pm\vec{k}$. That
is, if ${\delta_{\text{p}}}$ characterizes the linear width of the momentum
wave packet, we expect

${{\delta_{\text{p}}}}/{|\vec{k}|}\ll 1\ ,\qquad\psi({k})\sim
1/{\delta_{\text{p}}}^{3/2}\ .$ (6)

We compute the entanglement entropy in the limit ${\delta_{\text{p}}}\to 0$
and to the first non-vanishing order in ${\delta_{\text{p}}}/|\vec{k}|$.

Now let us consider the final state produced by the scattering process, which
is given by $|\text{out}\rangle\equiv S|\text{in}\rangle$, where $S=1+iT$ is
the scattering matrix and $T$ is the transition matrix, which is related to
the scattering amplitude $M$:

$\displaystyle\langle\\{k_{\text{f}}\\},f_{\text{f}}|T|\\{k_{\text{i}}\\},f_{\text{i}}\rangle$
$\displaystyle=(2\pi)^{4}\delta^{4}\left(\sum k_{\text{f}}-\sum
k_{\text{i}}\right)M_{f_{\text{i}},f_{\text{f}}}(\\{k_{\text{i}}\\};\\{k_{\text{f}}\\}),$
(7)

where $f_{\text{i}/\text{f}}$ labels the discrete quantum numbers of the
initial and final states. Unitarity of the $S$-matrix implies
$2\,\text{Im}\,T=T^{\dagger}T$, which gives rise to the optical theorem.

In general, $|{\rm out}\rangle$ is a linear superposition of all possible
outcome from the collision of A and B. Since we will focus on the elastic
scattering ${\text{A}}{\text{B}}\to{\text{A}}{\text{B}}$, the outgoing state
is also consisted of A and B, which can be selected by manually applying a
projection operator $P_{{\text{A}}{\text{B}}}$ on $|\text{out}\rangle$, $|{\rm
out}\rangle_{\rm el}=P_{{\text{A}}{\text{B}}}|\text{out}\rangle$. The density
matrix $\rho^{\text{f}}$ constructed out of $|{\rm out}\rangle_{\rm el}$,
however, is not properly normalized,

$\displaystyle\text{Tr}\left(P_{{\text{A}}{\text{B}}}|\text{out}\rangle\langle\text{out}|P_{{\text{A}}{\text{B}}}\right)=\langle\text{out}|P_{{\text{A}}{\text{B}}}|\text{out}\rangle=1-\mathcal{P}_{{\text{inel}}}\,,\
$ (8)

where

$\displaystyle{\mathcal{P}}_{{\text{inel}}}=\langle\text{out}|1-P_{{\text{A}}{\text{B}}}|\text{out}\rangle=\langle\text{in}|T^{\dagger}(1-P_{{\text{A}}{\text{B}}})T|\text{in}\rangle$
(9)

is the probability for AB to scatter inelastically into anything but AB. The
properly normalized $\rho^{\text{f}}$ is now

$\displaystyle\rho^{\text{f}}=\frac{1}{1-{\mathcal{P}}_{{\text{inel}}}}|\text{out}\rangle_{\rm
el}\ {}_{\rm el}\langle\text{out}|\ ,$ (10)

which we compute to leading order in ${\delta_{\text{p}}}/|\vec{k}|$.

##  III Entanglement entropy

The initial state reduced density matrix for A is

$\displaystyle\rho^{\text{i}}_{\text{A}}$ $\displaystyle=$
$\displaystyle\sum_{i,j,\bar{i},\bar{j}}\Omega_{i\bar{i}}(\Omega^{\dagger})_{\bar{j}j}\
\langle\psi_{B},\bar{j}|\psi_{B},\bar{i}\rangle\
|\psi_{A},i\rangle\langle\psi_{A},j|$ (11) $\displaystyle=$
$\displaystyle\sum_{i,j}(\Omega\Omega^{\dagger})_{ij}|\psi_{A},i\rangle\langle\psi_{A},j|,$

and the entanglement entropy is

$\displaystyle\mathcal{E}^{\text{i}}_{2}=1-\text{Tr}(\Omega^{\dagger}\Omega)^{2}.$
(12)

Clearly, if there is no entanglement between the initial states,
$(\Omega^{\dagger}\Omega)^{2}=\Omega^{\dagger}\Omega$ and the entropy
$\mathcal{E}^{\text{i}}_{2}=0$.

(a) In momentum space

(b) In position space

Figure 1: The wave packet in momentum space and in position space,
respectively. In the limit ${\delta_{\text{p}}}L\gg 1$ and
${\delta_{\text{p}}}\to 0$, it approaches the plane wave in $z$-direction in
the position space.

To compute the final state reduced density matrix from Eq. (10), we choose the
momentum-space wave packet to be a cube with a side length of
$2{\delta_{\text{p}}}$, as shown in Fig. 1a. The wave packet is chosen to be
uniform in the $p_{z}$-direction but not in the $p_{x,y}$-directions as we
would like the wave packet to approach the plane wave in position space.
Specifically, we define the regularized delta functions

$\displaystyle\tilde{\delta}^{3}(\vec{p})$ $\displaystyle=$
$\displaystyle\tilde{\delta}_{0}(p_{x})\tilde{\delta}_{0}(p_{y})\tilde{\delta}_{z}(p_{z})\,,$
(13) $\displaystyle\tilde{\delta}_{0}(k)$ $\displaystyle=$
$\displaystyle\frac{\pi}{2}\frac{\Theta(k+\delta_{\text{p}})-\Theta(k-\delta_{\text{p}})}{\text{Si}({\delta_{\text{p}}}L/2)}\frac{L}{2\pi}{\text{sinc\,}}\frac{kL}{2},$
(14) $\displaystyle\tilde{\delta}_{z}(k)$ $\displaystyle=$
$\displaystyle\frac{\Theta(k+\delta_{\text{p}})-\Theta(k-\delta_{\text{p}})}{2\delta_{\text{p}}}\,,$
(15)

where ${\text{sinc\,}}(k)=(\sin k)/k$ is the Fourier transform of a uniform
rectangular function and ${\rm Si}(x)=\int_{0}^{x}dy\ {\text{sinc\,}}(y)$. The
parameter $L$ characterizes the linear size of the wave packet in the position
space in the $xy$-plane. We assume ${\delta_{\text{p}}}L\gg 1$ such that
$\lim_{{\delta_{\text{p}}}\to 0}\tilde{\delta}^{(3)}(k)=\delta^{(3)}(k)$. The
correctly normalized wave function is

$\displaystyle\psi_{{\text{A}}/{\text{B}}}(\vec{p})=\frac{8\sqrt{\pi^{5}{\delta_{\text{p}}}}}{L}\tilde{\delta}^{3}(\vec{p}-\vec{k}_{{\text{A}}/{\text{B}}}).$
(16)

The wave packet in the position space is shown in Fig. 1b, which is uniform
inside a cross-sectional square of size $L^{2}$, confined within the shaded
sides, up to $\mathcal{O}(1/({\delta_{\text{p}}}L))$ corrections; in the
$z$-direction it is unbounded, with the width of the peak characterized by
$1/{\delta_{\text{p}}}$. It is useful to set
$1/({\delta_{\text{p}}}L)\lesssim{\delta_{\text{p}}}/|\vec{k}|$, so that
taking the plane wave limit amounts to expanding around the single small
parameter ${\delta_{\text{p}}}/|\vec{k}|$. This requirement is equivalent to
$L\lambda\gg 1/{\delta_{\text{p}}}^{2}$, where $\lambda$ is the Compton
wavelength for momentum $k$. For plane waves this is certainly satisfied.

With details provided in the supplementary material, one can show

$\displaystyle{\mathcal{P}}_{{\text{inel}}}=I_{0}(|\vec{k}|)\left[\sigma_{{\text{inel}}}+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\right],$
(17)

where $\sigma_{{\text{inel}}}$ is the inelastic cross section for
${\text{A}}{\text{B}}\to$ anything but AB, and

$\displaystyle I_{0}(|\vec{k}|)$ $\displaystyle=$ $\displaystyle
4|\vec{k}|\sqrt{s}\int_{p_{1},p_{2},q_{1},q_{2}}\psi_{\text{A}}(p_{1})\psi_{\text{B}}(p_{2})\psi_{\text{A}}^{*}(q_{1})$
(18)
$\displaystyle\times\psi_{\text{B}}^{*}(q_{2})(2\pi)^{4}\delta^{4}(q_{1}+q_{2}-p_{1}-p_{2})$

For our choice of wave packets in Eq. (16), we show in the Supplementary
Materials that

$\displaystyle
I_{0}(|\vec{k}|)=\frac{1}{L^{2}}\left(1+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\right).$
(19)

In other words, $I_{0}(|\vec{k}|)$ is the inverse of the cross-sectional area
of the wave packet at the leading order in ${\delta_{\text{p}}}\to 0$, in
which limit $L\gg 1/{\delta_{\text{p}}}$. The physical picture here is that
the wave packet is traveling at a high speed in the $z$-direction, which is
Lorentz-contracted, so in the CM frame the two wave packets look like two thin
pancakes whose linear size in the transverse direction is much bigger than
that in the longitudinal direction. In the strict plane wave limit, when
$L\to\infty$, the two incoming wave packets are so spread out in the
transverse direction in position space that the probability of them
interacting with each other is very small. This consideration suggests we can
expand ${\mathcal{P}}_{{\text{inel}}}$ in Eq. (10) as a small parameter.

Putting everything together, the final state entanglement entropy in the
${\delta_{\text{p}}}\to 0$ limit is

$\displaystyle\mathcal{E}^{\text{f}}_{2}$ $\displaystyle=$ $\displaystyle
1-\text{Tr}(\Omega^{\dagger}\Omega)^{2}(1+2{\mathcal{P}}_{{\text{inel}}})$
(20)
$\displaystyle+\frac{I_{0}(|\vec{k}|)}{|\vec{k}|\sqrt{s}}\left\\{\text{Im}\
\text{Tr}\left[\Omega^{\dagger}\Omega\Omega^{\dagger}M^{\text{F}}(\Omega)\right]+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\right\\}$
$\displaystyle=$ $\displaystyle
1-\text{Tr}(\Omega^{\dagger}\Omega)^{2}+\frac{I_{0}(|\vec{k}|)}{|\vec{k}|\sqrt{s}}\bigl{\\{}\text{Im}\
\text{Tr}\left[\Omega^{\dagger}\Omega\Omega^{\dagger}M^{\text{F}}(\Omega)\right]$
$\displaystyle-2|\vec{k}|\sqrt{s}\,\sigma_{{\text{inel}}}\
\text{Tr}(\Omega^{\dagger}\Omega)^{2}+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\bigr{\\}},$

where
$[M^{\text{F}}(\Omega)]_{i\bar{i}}=\sum_{j,\bar{j}}\Omega_{j\bar{j}}M_{j\bar{j},i\bar{i}}^{\text{F}}$
and

$\displaystyle M_{i\bar{i},j\bar{j}}^{\text{F}}$ $\displaystyle=$
$\displaystyle
M_{i\bar{i},j\bar{j}}(k_{\text{A}},k_{\text{B}};k_{\text{A}},k_{\text{B}})$
(21)

is the elastic forward scattering amplitude for
${\text{A}}(i){\text{B}}(\bar{i})\to{\text{A}}(j){\text{B}}(\bar{j})$. We
demonstrate in the Supplementary Materials that terms dropped in Eq. (20) are
higher orders in ${\delta_{\text{p}}}/|\vec{k}|$. Now consider the case where
the initial states are not entangled; specifically,
$\Omega_{i\bar{i}}=\omega_{i}\omega^{\prime}_{\bar{i}}$, where

$\displaystyle\omega=(1,0,0,\cdots),\quad\omega^{\prime}=(1,0,0,\cdots).$ (22)

Then

$\displaystyle\mathcal{E}^{\text{f}}_{2}$ $\displaystyle=$ $\displaystyle
I_{0}(|\vec{k}|)\frac{\text{Im}(M^{\text{F}}_{1\bar{1},1\bar{1}})-2|\vec{k}|\sqrt{s}\,\sigma_{{\text{inel}}}+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)}{|\vec{k}|\sqrt{s}}$
(23) $\displaystyle=$ $\displaystyle
2I_{0}(|\vec{k}|)\left[\sigma_{\text{tot}}-\sigma_{{\text{inel}}}+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\right]$
$\displaystyle=$ $\displaystyle
2\frac{\sigma_{{\text{el}}}}{L^{2}}+\mathcal{O}({\delta_{\text{p}}}^{5}/|\vec{k}|^{5})\,,$

where we have used the optical theorem to rewrite
$\text{Im}(M^{\text{F}}_{1\bar{1},1\bar{1}})$ in terms of the total cross
section $\sigma_{\text{tot}}$, and $\sigma_{{\text{el}}}$ is now the elastic
cross section for ${\text{A}}{\text{B}}\to{\text{A}}{\text{B}}$. It is
interesting to observe that ${\sigma_{{\text{el}}}}/{L^{2}}$ (at the leading
order) is nothing but the probability for the two-to-two elastic scattering to
occur,

$\displaystyle\frac{\sigma_{{\text{el}}}}{L^{2}}+\mathcal{O}({\delta_{\text{p}}}^{5}/|\vec{k}|^{5})={\mathcal{P}}_{{\text{el}}}=\langle\text{in}|T^{\dagger}P_{{\text{A}}{\text{B}}}T|\text{in}\rangle,\
\ $ (24)

which can be seen from the definition of cross section,
$\sigma_{{\text{el}}}=N_{{\text{el}}}/(d_{\text{A}}l_{\text{A}}d_{\text{B}}l_{\text{B}}A)$,
where $d_{{\text{A}}/{\text{B}}}$ are the number density of uniform beams,
$l_{{\text{A}}/{\text{B}}}$ are the length of these beams, $A$ is the cross
sectional area of both beams and $N_{{\text{el}}}$ is the expectation value
for the number of elastic scattering events. In our case we are scattering two
single particles head-to-head, thus
$d_{\text{A}}l_{\text{A}}A=d_{\text{B}}l_{\text{B}}A=1$, $A=L^{2}$ and
$N_{{\text{el}}}={\mathcal{P}}_{\text{el}}$. The relation
$\mathcal{E}_{2}\sim{\mathcal{P}}_{\text{el}}$ is actually insensitive to the
details of the wave packets, as long as we stay in the plane wave limit Low
and Yin (2024).

The factor of $2$ in Eq. (23) clearly comes from the fact that the linear
entropy is computed from the second power of the density matrix. Then the
computation can be directly generalized to e.g. the $n$-Tsallis entropy
$\mathcal{E}_{n}(\rho)=(1-\text{Tr}\,\rho^{n})/(n-1)$ where $n\geq 2$ is an
integer:

$\displaystyle\mathcal{E}_{n}(\rho^{\text{f}}_{\text{A}})$ $\displaystyle=$
$\displaystyle\frac{n}{n-1}\frac{\sigma_{{\text{el}}}}{L^{2}}+\mathcal{O}({\delta_{\text{p}}}^{5}/|\vec{k}|^{5}).$
(25)

We have been focusing on pure initial states; now we discuss briefly the case
of mixed initial states. Let us consider unentangled states characterized by
the following density matrix:

$\displaystyle\rho^{\text{i}}=\rho^{\text{i}}_{f_{\text{A}}}\otimes\rho^{\text{i}}_{p_{\text{A}}}\otimes\rho^{\text{i}}_{f_{\text{B}}}\otimes\rho^{\text{i}}_{p_{\text{B}}},$
(26)

where $\rho^{\text{i}}_{f_{{\text{A}}/{\text{B}}}}$ and
$\rho^{\text{i}}_{p_{{\text{A}}/{\text{B}}}}$ are density matrices for the
discrete and momentum subsystems of the two initial particles, respectively.
Let us consider mixed states in discrete quantum numbers, such as the spin:

$\displaystyle\rho^{\text{i}}_{f_{\text{A}}}=\sum_{i}{\mathbf{f}}_{i}\
|i\rangle\langle
i|,\qquad\rho^{\text{i}}_{f_{\text{B}}}=\sum_{\bar{i}}\bar{{\mathbf{f}}}_{\bar{i}}\
|\bar{i}\rangle\langle\bar{i}|,$ (27)

where ${\mathbf{f}}_{i}\geq 0$, $\bar{{\mathbf{f}}}_{\bar{i}}\geq 0$, and
$\sum_{i}{\mathbf{f}}_{i}=\sum_{\bar{i}}\bar{{\mathbf{f}}}_{\bar{i}}=1$.
Suppose the momentum states are still pure:
$\rho^{\text{i}}_{p_{\text{A}}}=|\psi_{\text{A}}\rangle\langle\psi_{\text{A}}|$,
$\rho^{\text{i}}_{p_{\text{B}}}=|\psi_{\text{B}}\rangle\langle\psi_{\text{B}}|$,
the subsystem entropy of A is

$\displaystyle\mathcal{E}^{\text{f}}_{2,{\text{A}}}=2I_{0}(|\vec{k}|)\sum_{i,\bar{i}}\,{\mathbf{f}}_{i}^{2}\,\bar{{\mathbf{f}}}_{\bar{i}}\,\left[(\sigma_{{\text{el}}})_{i\bar{i}}+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\right].$
(28)

where $(\sigma_{{\text{el}}})_{i\bar{i}}$ is the total elastic cross section
where the initial discrete quantum numbers are $i$ and $\bar{i}$. On the the
hand, the elastic scattering probability for colliding mixed states is

$\displaystyle{\mathcal{P}}_{{\text{el}}}$ $\displaystyle=$
$\displaystyle\frac{1}{L^{2}}\sum_{i,\bar{i}}{\mathbf{f}}_{i}\bar{{\mathbf{f}}}_{\bar{i}}(\sigma_{{\text{el}}})_{i\bar{i}}+\mathcal{O}({\delta_{\text{p}}}^{5}/|\vec{k}|^{5})$
(29) $\displaystyle\equiv$
$\displaystyle\frac{\overline{\sigma_{{\text{el}}}}}{L^{2}}+\mathcal{O}({\delta_{\text{p}}}^{5}/|\vec{k}|^{5})\,,$

from which we see that, when the density matrix satisfies
$(\rho^{\text{i}}_{f_{\text{A}}})^{2}\propto\rho^{\text{i}}_{f_{\text{A}}}$,
the subsystem entropy can again be interpreted as a cross section,

$\displaystyle\mathcal{E}^{\text{f}}_{2,{\text{A}}}=\frac{2}{n_{\text{A}}}\frac{\overline{\sigma_{{\text{el}}}}}{L^{2}}+\mathcal{O}({\delta_{\text{p}}}^{5}/|\vec{k}|^{5}).$
(30)

The condition
$(\rho^{\text{i}}_{f_{\text{A}}})^{2}\propto\rho^{\text{i}}_{f_{\text{A}}}$ is
satisfied when ${\mathbf{f}}_{i}$ is zero or the non-zero entries are all
given by the same constant value, ${\mathbf{f}}_{i}=1/n_{\text{A}}$, where
$n_{\text{A}}$ counts the number of nonzero entries in ${\mathbf{f}}_{i}$. The
same criteria was proposed in Ref. Cheung et al. (2023), which ensures the
subsystem entropy to not decrease upon time evolution. A realistic scenario
fulfilling this criteria is the unpolarized scattering, in which case
$\overline{\sigma_{{\text{el}}}}$ is simply the unpolarized elastic cross
section.

##  IV Discussions and Conclusion

Having shown the entanglement entropy in 2-to-2 particle scattering is given
by the elastic cross section in units of the transverse size of the wave
packets, several interesting observations follow.

Since the cross section is the effective area of collision when particles
collide, we have essentially discovered an area law for the entanglement
entropy in a two-body system. In other examples of area laws for the entropy,
the area is often given by the boundary of a spatial region Bekenstein (1973);
Hawking (1975); Srednicki (1993); Ryu and Takayanagi (2006); Eisert et al.
(2010); Dvali (2018). In the present case it is not clear if a similar
interpretation could be given. We speculate that, perhaps, the cross section
could be viewed as the area of the space-like surface connecting the past and
the future light cones of the two particle scattering. On the other hand,
cross section measures the probability that a quantum mechanical process would
happen, in accordance with the thermodynamic and information-theoretic
interpretations of the entropy as probability.

Scaling properties of cross sections, in particular the total and elastic
cross sections, with respect to the collision energy $\sqrt{s}$, have been
studied intensively for high energy collisions many decades ago. Froissart and
Martin showed that the total cross section is bounded from the above by
$\log^{2}s$ Froissart (1961); Martin (1963) in the $S$-matrix formalism. Cheng
and Wu argued very generally in quantum field theory that both the total and
the elastic cross sections rise with $\sqrt{s}$ Cheng and Wu (1969, 1970),
which is a somewhat surprising result because cross sections for elementary
processes at a fixed order in perturbation always decrease with $\sqrt{s}$.
However, in a high energy scattering process there is abundant energy to
create radiations and splittings of soft and collinear particles, which
necessitates the resummation of higher order diagrams, akin to the soft photon
resummation in quantum electrodynamics. These effects enhance the cross
section to increase with the energy in the form of $\log s$. In fact, Ref.
Cheng and Wu (1970) proposed a “Black Disk” picture of a total absorption area
of radius $\log s$ for high energy collisions, which is similar in spirit to
the wave packet adopted in this work. The growth of the cross section with
respect to $\sqrt{s}$ has been measured over several orders of magnitude in
$\sqrt{s}$ in $pp$ and $p\bar{p}$ collisions at the Tevatron and the Large
Hadron Collider Block and Cahn (1985); Pancheri and Srivastava (2017). (See,
for example, Fig. 5.40 in Ref. Pancheri and Srivastava (2017).) Even though
many of the studies are based on hadron-hadron collisions, it has become clear
recently in the study of a multi-TeV muon collider that radiations and
splittings are also important in high energy collisions of leptons Chen et al.
(2017); Han et al. (2021); Ruiz et al. (2022). Therefore, similar
considerations for cross sections for lepton-lepton collisions should apply.

These observations lead to a “second law” of entanglement entropy with respect
to the CM energy $\sqrt{s}$, in which the area (cross section) grows with the
energy and so does the entanglement entropy. The Froissart bound now
constrains the growth of the entropy with respect to energy. It would be
interesting to explore whether there is a holographic understanding of the
relation between the entanglement entropy and the cross section.

Last but not least, certain other cross sections, for instance the Deep
Inelastic Scattering-like cross section, can also be related to entanglement
entropies corresponding to different ways to construct a bipartite system in
2-to-2 scatterings Low and Yin (2024). It will be interesting to see if all
cross sections can be interpreted this way.

##  V Acknowledgement

Discussions with Tao Han on the scaling properties of cross sections are
gratefully acknowledged. We also thank Gia Dvali and Minglei Xiao for comments
on the manuscript. This work is supported in part by the U.S. Department of
Energy, Office of High Energy Physics, under contract DE-AC02-06CH11357 at
Argonne, as well as by the U.S. Department of Energy, Office of Nuclear
Physics, under grant DE-SC0023522 at Northwestern.

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##  Appendix A Supplementary Materials: Wave Packet Treatment

Here we detail the computations involving wave packets. First, let us see that
in general, how the universal factor $I_{0}(|\vec{k}|)$ appears at the leading
order in the relevant computations. For example, consider
${\mathcal{P}}_{{\text{inel}}}$ in Eq. (9), which is given by

$\displaystyle{{\mathcal{P}}}_{{\text{inel}}}$ $\displaystyle=$
$\displaystyle\int_{p_{1},p_{2},q_{1},q_{2}}\psi_{\text{A}}(p_{1})\psi_{\text{B}}(p_{2})\psi_{\text{A}}^{*}(q_{1})\psi_{\text{B}}^{*}(q_{2})$
(31)
$\displaystyle\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\times(2\pi)^{4}\delta^{4}(q_{1}+q_{2}-p_{1}-p_{2})\,{\cal
F}_{{\text{inel}}}(p_{1},p_{2},q_{1},q_{2})\,,\,$

with

$\displaystyle(2\pi)^{4}\delta^{4}(q_{1}+q_{2}-p_{1}-p_{2})\,{\cal
F}_{{\text{inel}}}(p_{1},p_{2},q_{1},q_{2})$ (32) $\displaystyle=$
$\displaystyle\sum_{i,\bar{i},j,\bar{j}}\Omega_{i\bar{i}}\Omega^{*}_{j\bar{j}}\langle
q_{1},j;q_{2},\bar{j}|T^{\dagger}(1-P_{{\text{A}}{\text{B}}})T|p_{1},i;p_{2},\bar{i}\rangle,\quad\,\,\,$

which can be expressed in terms of inelastic amplitudes using Eq. (7). Now,
without specifying the detail of the wave packets, we require that in momentum
space the wave functions are bounded within an
$\mathcal{O}({\delta_{\text{p}}})$ radius, thus

$\displaystyle{\cal F}_{{\text{inel}}}(p_{1},p_{2},q_{1},q_{2})$
$\displaystyle=$ $\displaystyle{\cal
F}_{{\text{inel}}}(k_{\text{A}},k_{\text{B}},k_{\text{A}},k_{\text{B}})$ (33)
$\displaystyle\qquad\times[1+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)]\,$

can be pulled out of the integral in Eq. (31):

$\displaystyle{{\mathcal{P}}}_{{\text{inel}}}={\cal
F}_{{\text{inel}}}(k_{\text{A}},k_{\text{B}},k_{\text{A}},k_{\text{B}})\left[1+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\right]$
$\displaystyle\times\int_{p_{1},p_{2},q_{1},q_{2}}\psi_{\text{A}}(p_{1})\psi_{\text{B}}(p_{2})\psi_{\text{A}}^{*}(q_{1})\psi_{\text{B}}^{*}(q_{2})$
$\displaystyle\times(2\pi)^{4}\delta^{4}(q_{1}+q_{2}-p_{1}-p_{2})$
$\displaystyle=I_{0}(|\vec{k}|)\left[\sigma_{{\text{inel}}}+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\right],$
(34)

and we arrive at Eq. (17). In general, we see in computations other wave
packet integrals similar to $I_{0}(|\vec{k}|)$. However, we will argue that
all other integrals encountered are higher in ${\delta_{\text{p}}}$ and can be
neglected.

Now let us consider the case of uniform wave functions in the transverse
directions of the position space. Specifically, we define
$\tilde{\delta}^{3}(\vec{p})$ as in Eq. (13) and the wave packet as in Eq.
(16). The wave function is chosen in the following spirit: We first bound the
wave function strictly inside a cube of size $(2{\delta_{\text{p}}})^{3}$ in
the momentum space, as shown in Fig. 1a. Next, we treat the transverse
directions separately compared to the $z$ direction, as in Fig. 1b: We make
the wave function in the $z$ direction uniform within the
$2{\delta_{\text{p}}}$ sized window in the momentum space. On the other hand,
we require that the wave function is bounded and uniform within a square of
size $L^{2}$ in the $x$\- and $y$-direction of the position space, up to
$\mathcal{O}(1/({\delta_{\text{p}}}L))$ corrections; we choose $L\gg
1/{\delta_{\text{p}}}$, with $1/{\delta_{\text{p}}}$ characterizing the width
of the peak along the $z$ direction in the position space. We also set
$1/({\delta_{\text{p}}}L)\lesssim{\delta_{\text{p}}}/|\vec{k}|$ so there is a
single small parameter ${\delta_{\text{p}}}/|\vec{k}|$ to expand.

Let us first consider the general integral below:

$\displaystyle\\!\\!\\!\\!\\!\\!\tilde{I}_{0}=\int
d^{3}\vec{p}_{1}d^{3}\vec{p}_{2}d^{3}\vec{q}_{1}d^{3}\vec{q}_{2}\
\delta^{4}(q_{1}+q_{2}-p_{1}-p_{2})$
$\displaystyle\\!\\!\\!\\!\\!\\!\times{\cal F}_{0}(p_{1},p_{2},q_{1},q_{2})$
$\displaystyle\\!\\!\\!\\!\\!\\!\times\tilde{\delta}^{3}(\vec{p}_{1}-\vec{k})\tilde{\delta}^{3}(\vec{p}_{2}+\vec{k})\tilde{\delta}^{3}(\vec{q}_{1}-\vec{k})\tilde{\delta}^{3}(\vec{q}_{2}+\vec{k}).\quad$
(35)

As previously argued, we can pull ${\cal F}_{0}$ out of the integral at the
leading order (assuming ${\cal
F}_{0}(k_{\text{A}},k_{\text{B}},k_{\text{A}},k_{\text{B}})$ to be non-
vanishing):

$\displaystyle\\!\\!\\!\\!\\!\\!\tilde{I}_{0}={\cal
F}_{0}(k_{\text{A}},k_{\text{B}},k_{\text{A}},k_{\text{B}})\left[1+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\right]$
$\displaystyle\\!\\!\\!\\!\\!\\!\times\int
d^{3}\vec{p}_{1}\,d^{3}\vec{p}_{2}\,d^{3}\vec{q}_{1}\,d^{3}\vec{q}_{2}\
\tilde{\delta}^{3}(\vec{p}_{1}-\vec{k})\tilde{\delta}^{3}(\vec{p}_{2}+\vec{k})$
$\displaystyle\\!\\!\\!\\!\\!\\!\times\tilde{\delta}^{3}(\vec{q}_{1}-\vec{k})\tilde{\delta}^{3}(\vec{q}_{2}+\vec{k})\delta^{4}(q_{1}+q_{2}-p_{1}-p_{2}).$
(36)

We can first integrate out the spatial $\delta$-functions along the
$\vec{p}_{1}$ direction, and then integrate out the energy $\delta$-function
in the $(p_{2})_{z}$ direction:

$\displaystyle\delta(E_{p_{1}}+E_{p_{2}}-E_{p_{3}}-E_{p_{4}})$
$\displaystyle\to\left(\frac{E_{k_{\text{A}}}E_{k_{\text{B}}}}{|\vec{k}|\sqrt{s}}+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\right)\delta((p_{2})_{z}-r_{0}),$
(37)

where the factor in front of the $\delta$-function on the right hand side of
the above can be pulled out of the integral like ${\cal F}_{0}$, and $r_{0}$
is the root of $(p_{2})_{z}$ in

$\displaystyle\sqrt{(\vec{q_{1}}+\vec{q_{2}}-\vec{p_{2}})^{2}+m_{\text{A}}^{2}}+\sqrt{\vec{p}_{2}^{2}+m_{\text{B}}^{2}}$
$\displaystyle\qquad\qquad-\sqrt{\vec{q}_{1}^{2}+m_{\text{A}}^{2}}-\sqrt{\vec{q}_{2}^{2}+m_{\text{B}}^{2}}=0.$
(38)

Then the integration we need in the $(p_{2})_{z}$ direction is

$\displaystyle\int
d(p_{2})_{z}\delta((p_{2})_{z}-r_{0})\tilde{\delta}_{z}((p_{2})_{z}+|\vec{k}|)$
(39) $\displaystyle=$
$\displaystyle\frac{\Theta(r_{0}+|\vec{k}|+{\delta_{\text{p}}})-\Theta(r_{0}+|\vec{k}|-{\delta_{\text{p}}})}{2{\delta_{\text{p}}}},$

which is non-vanishing within a $2\delta_{p}$ sized window if we vary $r_{0}$
in the above. Therefore, to get leading order effects in
${\delta_{\text{p}}}$, we need to solve $r_{0}$ up to linear order in
${\delta_{\text{p}}}$, which turns out to be
$r_{0}=(q_{2})_{z}+\mathcal{O}({\delta_{\text{p}}}^{2}/|\vec{k}|^{2})$.111One
can consider characterizing the size of the wave packet in the $p_{z}$
direction by some $\delta_{z}$, which is independent of ${\delta_{\text{p}}}$.
Then the solution $r_{0}$ needs to be computed up to
$\mathcal{O}(\delta_{z}/|\vec{k}|)$, but it will admit an expansion of
$\max({\delta_{\text{p}}},\delta_{z})$. Therefore, we need to require
$\delta_{z}\gtrsim{\delta_{\text{p}}}$ for our computation to still hold. In
other words, $1/{\delta_{\text{p}}}$ is the largest possible length scale
which characterizes the spread of the wave packet in the direction of the
momentum. As $1/{\delta_{\text{p}}}\ll L$, the wave packet is much more
compressed in the $z$ direction compared to the transverse directions. Then
one sees that the integration of the different spatial dimensions actually
decouples.

In the transverse directions, notice that if we take
$1/({\delta_{\text{p}}}L)\to 0$ in Eq. (14), we have

$\displaystyle\int dk\
\tilde{\delta}_{0}(k)\to\int_{-\infty}^{\infty}d(kL)\frac{1}{2\pi}\,{\text{sinc\,}}\frac{kL}{2}.$
(40)

Therefore, as we are assuming ${\delta_{\text{p}}}\gg 1/L$, at the leading
order the $\Theta$-functions as well as the normalization factor related to
$\text{Si}(x)$ in Eq. (14) can be effectively omitted from our computation:

$\displaystyle\tilde{\delta}_{0}(k)$ $\displaystyle=$
$\displaystyle\frac{\pi}{2}\frac{\Theta(k+\delta_{\text{p}})-\Theta(k-\delta_{\text{p}})}{\text{Si}({\delta_{\text{p}}}L/2)}\frac{L}{2\pi}\,{\text{sinc\,}}\frac{kL}{2}$
(41) $\displaystyle\sim$
$\displaystyle\frac{L}{2\pi}\,{\text{sinc\,}}\frac{kL}{2}\ .$

The above approximation is valid up to
$\mathcal{O}(1/({\delta_{\text{p}}}L))$, and as we would like an expansion of
small ${\delta_{\text{p}}}/|\vec{k}|$, it is then desirable to have
$1/({\delta_{\text{p}}}L)\lesssim{\delta_{\text{p}}}/|\vec{k}|$. The useful
integral in the transverse directions is then

$\displaystyle\int d(p_{1})_{x}\int d(p_{2})_{x}\int d(q_{1})_{x}\int
d(q_{2})_{x}$
$\displaystyle\times\delta\left[(\vec{p}_{1}+\vec{p}_{2}-\vec{q}_{1}-\vec{q}_{2})_{x}\right]{\text{sinc\,}}\frac{L(p_{1})_{x}}{2}{\text{sinc\,}}\frac{L(p_{2})_{x}}{2}$
$\displaystyle\times{\text{sinc\,}}\frac{L(q_{1})_{x}}{2}{\text{sinc\,}}\frac{L(q_{2})_{x}}{2}=\left(\frac{2\pi}{L}\right)^{3}.$
(42)

In the $z$ direction we need

$\displaystyle\int_{k-{\delta_{\text{p}}}}^{k+{\delta_{\text{p}}}}d(p_{1})_{z}\int_{-k-{\delta_{\text{p}}}}^{k+{\delta_{\text{p}}}}d(p_{2})_{z}\int_{k-{\delta_{\text{p}}}}^{k+{\delta_{\text{p}}}}d(q_{1})_{z}$
$\displaystyle\times\int_{-k-{\delta_{\text{p}}}}^{-k+{\delta_{\text{p}}}}d(q_{2})_{z}\delta\left[(\vec{p}_{1}+\vec{p}_{2}-\vec{q}_{1}-\vec{q}_{2})_{z}\right]\delta[(\vec{p}_{2}-\vec{q}_{2})_{z}]$
$\displaystyle\ =4{\delta_{\text{p}}}^{2}.$ (43)

Then

$\displaystyle\tilde{I}_{0}$ $\displaystyle=$
$\displaystyle\frac{1}{16\pi^{2}}\frac{E_{k_{\text{A}}}E_{k_{\text{B}}}}{|\vec{k}|\sqrt{s}}\frac{L^{2}}{{\delta_{\text{p}}}^{2}}$
(44) $\displaystyle\times{\cal
F}_{0}(k_{\text{A}},k_{\text{B}},k_{\text{A}},k_{\text{B}})\left[1+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\right].$

Now we can compute $I_{0}(|\vec{k}|)$ from $\tilde{I}_{0}$. Let’s plug in the
explicit form of the wave packet from Eq. (16) into $I_{0}(|\vec{k}|)$ in Eq.
(18), and compare with $\tilde{I}_{0}$ in Eq. (A), we see $I_{0}(|\vec{k}|)$
is given by

$\displaystyle{\cal
F}_{0}(k_{\text{A}},k_{\text{B}},k_{\text{A}},k_{\text{B}})\to\frac{2^{12}\pi^{10}{\delta_{\text{p}}}^{2}4|\vec{k}|\sqrt{s}}{(2\pi)^{8}2E_{k_{\text{A}}}2E_{k_{\text{B}}}L^{4}}.$
(45)

In the end we arrive at

$\displaystyle
I_{0}(|\vec{k}|)=\frac{1}{L^{2}}\left(1+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\right).$
(46)

We see clearly here that at the leading order, $I_{0}(|\vec{k}|)$ is exactly
the inverse of the cross-sectional area of the wave packet.

Now let us see concretely other types integrals involving the overlap of wave
packets which appear in the entanglement entropy calculation. We will show
that they are higher order in ${\delta_{\text{p}}}/|\vec{k}|$ and can be
neglected. For example, $\mathcal{E}^{\text{f}}_{2}$ in Eq. (20) contains the
following terms:

$\displaystyle 2\,\text{Re}\ \text{tr}\
(\Omega^{\dagger}\tilde{M}(\Omega))^{2}-2\,\text{\text{Tr}}\left[\Omega^{\dagger}\Omega\tilde{M}^{(2)}_{1}(\Omega)\right],$
(47)

where

$\displaystyle\left[\tilde{M}(\Omega)\right]_{i\bar{i}}=\sum_{j,\bar{j}}\Omega_{j\bar{j}}\int_{p_{1},p_{2},q_{1},q_{2}}\psi_{\text{A}}(p_{1})\psi_{\text{B}}(p_{2})$
$\displaystyle\times\psi_{\text{A}}^{*}(q_{1})\psi_{\text{B}}^{*}(q_{2})(2\pi)^{4}\delta^{4}(q_{1}+q_{2}-p_{1}-p_{2})$
$\displaystyle\times M_{j\bar{j},i\bar{i}}(p_{1},p_{2};q_{1},q_{2})$
$\displaystyle=\frac{1}{4|\vec{k}|\sqrt{s}{L^{2}}}\Big{[}\sum_{j,\bar{j}}\Omega_{j\bar{j}}M_{j\bar{j},i\bar{i}}^{\text{F}}+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\Big{]},$
(48)
$\displaystyle\left[\tilde{M}^{(2)}_{1}(\Omega)\right]_{\bar{i}_{1}\bar{i}_{2}}=\sum_{i,j,\bar{j},k,\bar{k}}\int_{p_{1},p_{2},p_{3},p_{4},q_{1},q_{2},q_{4}}\frac{1}{\sqrt{2E_{p_{3}}}}$
$\displaystyle\Omega^{*}_{j\bar{j}}\Omega_{k\bar{k}}\psi^{*}_{\text{A}}(p_{1})\psi^{*}_{\text{B}}(p_{2})\psi_{\text{B}}(p_{4})\psi_{\text{A}}(q_{1})\psi_{\text{B}}(q_{2})\psi^{*}_{\text{B}}(q_{4})$
$\displaystyle\times(2\pi)^{8}\delta^{4}(p_{3}+p_{4}-p_{1}-p_{2})\delta^{4}(q_{3}+q_{4}-q_{1}-q_{2})$
$\displaystyle\times M^{*}_{j\bar{j},i\bar{i}_{1}}(p_{1},p_{2};p_{3},p_{4})$
$\displaystyle\times\left.M_{k\bar{k},i\bar{i}_{2}}(q_{1},q_{2};q_{3},q_{4})\right|_{\vec{q}_{3}=\vec{q}_{1}+\vec{q}_{2}-\vec{q}_{4}}.$
(49)

Then the first term in Eq. (47) is
$\mathcal{O}(1/{(|\vec{k}|L)^{4}})\lesssim\mathcal{O}({\delta_{\text{p}}}^{8}/|\vec{k}|^{8})$
and can be dropped. (Recall we demand
$1/({\delta_{\text{p}}}L)\lesssim{\delta_{\text{p}}}/|\vec{k}|$.) The second
term in Eq. (47) involves the following integral:

$\displaystyle\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\tilde{I}_{1}=\int
d^{3}\vec{p}_{1}d^{3}\vec{p}_{2}d^{3}\vec{p}_{3}d^{3}\vec{p}_{4}d^{3}\vec{q}_{1}d^{3}\vec{q}_{2}d^{3}\vec{q}_{4}$
$\displaystyle\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\times\delta^{4}(q_{1}+q_{2}-p_{3}-q_{4})\delta^{4}(p_{1}+p_{2}-p_{3}-p_{4})$
$\displaystyle\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\times{\cal
F}_{1}(p_{1},p_{2},p_{3},p_{4},q_{1},q_{2},q_{4})\tilde{\delta}^{3}(\vec{p}_{1}-\vec{k})\tilde{\delta}^{3}(\vec{p}_{2}+\vec{k})$
$\displaystyle\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\times\tilde{\delta}^{3}(\vec{p}_{4}+\vec{k})\tilde{\delta}^{3}(\vec{q}_{1}-\vec{k})\tilde{\delta}^{3}(\vec{q}_{2}+\vec{k})\tilde{\delta}^{3}(\vec{q}_{4}+\vec{k}).\
$ (50)

Here, $p_{3}$ is constrained by the on-shell condition of type-A particles,
and the integrand in the above will constrain it into a range of radius
$\mathcal{O}({\delta_{\text{p}}})$ around $k_{\text{A}}$. Similar to
$\tilde{I}_{0}$, we pull out ${\cal F}_{1}$ first. Then we integrate out
$\delta^{3}(\vec{p_{1}}+\vec{p}_{2}-\vec{p}_{3}-\vec{p_{4}})$ in the direction
of $\vec{p}_{3}$, and the other spatial $\delta$-function in the direction of
$\vec{q}_{1}$. Next, integrating out
$\delta(E_{q_{1}}+E_{q_{2}}-E_{p_{3}}-E_{q_{4}})$ in the $(q_{2})_{z}$
direction, we convert it to $\delta[(\vec{q}_{2}-\vec{q}_{4})_{z}]$;
integrating out $\delta(E_{p_{1}}+E_{p_{2}}-E_{p_{3}}-E_{p_{4}})$ in the
$(p_{4})_{z}$ direction, we convert it to
$\delta[(\vec{p}_{4}-\vec{p}_{2})_{z}]$. Again, the integrals in the spatial
dimensions decouple, and upon integration we have

$\displaystyle\tilde{I}_{1}$ $\displaystyle=$ $\displaystyle{\cal
F}_{1}(k_{\text{A}},k_{\text{B}},k_{\text{A}},k_{\text{B}},k_{\text{A}},k_{\text{B}},k_{\text{B}})\left[1+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\right]$
(51)
$\displaystyle\times\frac{1}{32\pi^{2}}\left(\frac{E_{k_{\text{A}}}E_{k_{\text{B}}}}{|\vec{k}|\sqrt{s}}\right)^{2}\frac{L^{2}}{{\delta_{\text{p}}}^{3}}.$

Therefore,

$\displaystyle\left[\tilde{M}^{(2)}_{1}(\Omega)\right]_{\bar{i}_{1}\bar{i}_{2}}=\left(\frac{1}{4|\vec{k}|\sqrt{s}{L^{2}}}\right)^{2}$
$\displaystyle\times\Big{[}\sum_{i,j,\bar{j},k,\bar{k}}\Omega^{*}_{j\bar{j}}\Omega_{k\bar{k}}M^{\text{F}*}_{j\bar{j},i\bar{i}_{1}}M^{\text{F}}_{k\bar{k},i\bar{i}_{2}}+\mathcal{O}({\delta_{\text{p}}}/|\vec{k}|)\Big{]},\
\ $ (52)

thus the second term in Eq. (47) is also $\mathcal{O}(1/{(|\vec{k}|L)^{4}})$
and can be dropped. Notice that we are dropping these higher order terms
because we are taking the plane wave limit ${\delta_{\text{p}}}\to 0$, instead
of any weak coupling limit as in e.g. Ref. Aoude et al. (2024). Therefore, our
computation is valid to all orders in the coupling strength.