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PITT-PACC-2401

# Optimizing Entanglement and Bell Inequality Violation
in Top Anti-Top Events

Kun Cheng<EMAIL_ADDRESS>School of Physics, Peking University, Beijing,
100871, China  PITT PACC, Department of Physics and Astronomy,
University of Pittsburgh, 3941 O’Hara St., Pittsburgh, PA 15260, USA  Tao Han
<EMAIL_ADDRESS>PITT PACC, Department of Physics and Astronomy,
University of Pittsburgh, 3941 O’Hara St., Pittsburgh, PA 15260, USA  Matthew
Low<EMAIL_ADDRESS>PITT PACC, Department of Physics and Astronomy,
University of Pittsburgh, 3941 O’Hara St., Pittsburgh, PA 15260, USA

###### Abstract

A top quark and an anti-top quark produced together at colliders have
correlated spins. These spins constitute a quantum state that can exhibit
entanglement and violate Bell’s inequality. In realistic collider experiments,
most analyses allow the axes, as well the Lorentz frame to vary event-by-
event, thus introducing a dependence on the choice of event-dependent basis
leading us to adopt “fictitious states,” rather than genuine quantum states.
The basis dependence of fictitious states allows for an optimization
procedure, which makes the usage of fictitious states advantageous in
measuring entanglement and Bell inequality violation. In this work, we show
analytically that the basis which diagonalizes the spin-spin correlations is
optimal for maximizing spin correlations, entanglement, and Bell inequality
violation. We show that the optimal basis is approximately the same as the
fixed beam basis (or the rotated beam basis) near the $t\bar{t}$ production
threshold, while it approaches the helicity basis far above threshold. Using
this basis, we present the sensitivity for entanglement and Bell inequality
violation in $t\bar{t}$ events at the LHC and a future $e^{+}e^{-}$ collider.
Since observing Bell inequality violation appears to be quite challenging
experimentally, and requires a large dataset in collider experiments, choosing
the optimal basis is crucially important to observe Bell inequality violation.
Our method and general approach are equally applicable to other systems beyond
$t\bar{t}$, including interactions beyond the Standard Model.

###### Contents

  1. I Introduction
  2. II Quantum States in Different Bases and Frames
    1. II.1 Entanglement in Different Spin Bases
    2. II.2 Bell Inequality Violation in Different Spin Bases
    3. II.3 Entanglement and Bell Inequality Violation in Different Lorentz Frames
  3. III Fictitious States
    1. III.1 Fictitious States From Angular-Averaged Sums
    2. III.2 Fictitious States From Event-Dependent Lorentz Frame Choices
    3. III.3 Basis Dependence of Spin Correlation and Entanglement
  4. IV The Optimal Basis for Entanglement and Bell Inequality Violation
    1. IV.1 Concurrence
    2. IV.2 Bell Inequality
    3. IV.3 Alternative Bell Variable
    4. IV.4 Asymmetric Correlation Matrix
  5. V Phenomenological Analyses for $t\bar{t}$ Events
    1. V.1 Characteristics of QCD Sub-Processes
      1. V.1.1 $q\bar{q}\to t\bar{t}$ Process
      2. V.1.2 $gg\to t\bar{t}$ Process
    2. V.2 Optimization Procedure at the LHC
    3. V.3 Optimization Procedure at Lepton Colliders
  6. VI Conclusion and Discussion
  7. A Quantum States at Colliders and Basis Transformations
    1. A.1 Physical Spin State
    2. A.2 Transformations between Spin Bases
  8. B Entanglement of Fictitious States
  9. C Generalized Proof of the Optimal Basis
  10. D Angular Momentum Conservation
  11. E Top Anti-top Produced from Gluon and Photon Scattering

##  I Introduction

One of the most direct manifestations of quantum mechanics at high-energy
colliders is the correlation between the spins of outgoing particles. The
$t\bar{t}$ system, where the spin of the top quark ($t$) and the spin of the
anti-top quark ($\bar{t}$) are correlated [1], is one of the most studied and
most experimentally promising systems. The spin of particles is not measured
directly at high-energy colliders, but rather can be inferred by the
statistical distributions of decay products [2, 3]. Moving beyond correlations
between spins, a two-particle final state at colliders can be treated as a
quantum state [4] (see also [5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
18, 13, 12, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38, 39, 40]). For such a quantum state, it would be very
interesting to determine the entanglement between the top and anti-top and to
measure whether Bell’s inequality is violated. This would be a clear
establishment of quantum information in high-energy collisions.

The magnitude of the spin correlation between outgoing particles depends on
the choice of the quantization axis for the particle spin measurement. The
simplest choice would be to use the $z$-axis of the beam direction as the
quantization axis to measure the spin of the top, $S_{z}^{t}$, and the spin of
the anti-top $S_{z}^{\bar{t}}$, the spin correlation is then

$\mathscr{A}=\langle{S_{z}^{t}}\otimes S_{z}^{\bar{t}}\rangle.$ (1)

If we perform the measurement using a different basis the correlation would
generally be different. It has been customary to take the momentum direction
of the top quark in the $t\bar{t}$ center-of-mass frame as the spin
quantization axis. Using this basis would lead to a different correlation and
therefore, the spin correlation as in Eq. (1) is basis-dependent.

The basis-dependence of spin correlations in the $t\bar{t}$ system is well
understood for fixed initial states in hadronic collisions. For the $q\bar{q}$
initial state, Mahlon and Parke calculated the maximal spin correlation to be
100% and analytically determined the basis that achieves this [2]. For the
$gg$ initial state, Uwer, Mahlon and Parke found that an optimal basis is also
obtainable. In this case, however, the maximal spin correlation does not reach
100% and varies over phase space [41, 42].

The entanglement of a quantum state, unlike the correlations between a pair of
spins, is an intrinsic property of the state and does not depend on the basis
choice of an observable. To treat the $t\bar{t}$ system as a quantum state, we
first assume that a sufficient number of events are produced such that the
$t\bar{t}$ phase space can be divided into infinitesimal bins. The ideal way
to measure the entanglement would be to only consider events from a single
phase space bin. Combining these events would result in a well-defined quantum
state and its entanglement and Bell inequality violation would be basis-
independent.

In order for repeated measurements to be usable to reconstruct a quantum
state, we need an ensemble of identically-prepared quantum states. One example
of identical preparation is only considering $t\bar{t}$ events from a fixed
phase space point. For the generalization to a region of phase space, as would
be required in practice at colliders, identical preparation is still possible
provided that the same quantization axis is used for all events in the phase
space region. This identically prepared spin state is obtained by projecting
the underlying quantum state to a fixed phase space point, and we call this
state a quantum sub-state.

If measurements are performed with variable preparation, then the
reconstruction does not lead to a quantum state. Measuring the spin of the top
along its momentum axis $-$ usually called the helicity basis $-$ where the
top momentum axis points in a different direction in each event and
corresponds to a different quantization axis for each event. Rather than
measuring a quantum state, one is instead measuring an angular-averaged sum of
quantum sub-states. These angular-averaged sums, when measured in an event-
dependent basis, are not genuine quantum states. They are a convex sum of
quantum sub-states and are not added in a way that corresponds to a quantum
state. Ref. [7] labeled these as “fictitious states” and Ref. [30] further
elaborated their properties.

To see the basis-dependence of an angular-averaged sum, consider the density
matrix $\bar{\rho}$ measured in an event-dependent basis

$\bar{\rho}=\sum_{a\in{\rm events}}\rho_{a}\ .$ (2)

Each $\rho_{a}$ is a quantum sub-state corresponding to a single event $a$. To
change basis, one rotates each sub-state by an event-dependent rotation
$U_{a}$

$\bar{\rho}\to\sum_{a\in{\rm events}}U^{\dagger}_{a}\rho_{a}U_{a}\ .$ (3)

Except in special cases, the event-dependent rotation does not correspond to
an overall rotation $U$ of $\bar{\rho}$. In other words $\sum_{a\in{\rm
events}}U^{\dagger}_{a}\rho_{a}U_{a}\neq U^{\dagger}\bar{\rho}U$. Thus, when
observables are computed from $\bar{\rho}$, the results depend on the basis
that is used.

The special case where a common rotation is applied to all events corresponds
to using a fixed basis. When reconstructing the spin density matrix using the
same spin quantization axis in the center-of-mass frame of $t\bar{t}$ for all
events, the angular-averaged sum is the projection of the underlying quantum
state of $t\bar{t}$ into its spin space, and this results in a genuine quantum
state. For a quantum state, the signals of entanglement and Bell inequality
violation are basis-independent.

In practice, the majority of work on entanglement and Bell inequality
violation at colliders utilizes the helicity basis, then the measurements are
performed on fictitious states. For a fictitious state, finding non-zero
entanglement only shows that there exists a quantum sub-state that is
entangled. Likewise, finding a fictitious state that violates a Bell
inequality only shows that there exists a quantum sub-state that violates a
Bell inequality. The usage of fictitious state, however, is still valuable to
determine the existence of entanglement or Bell inequality violation in the
$t\bar{t}$ system.

Since the size of entanglement is basis-dependent for fictitious states, there
must exist an optimal basis that yields the largest entanglement and likewise
for Bell inequality violation. In a previous publication [30], we showed that
the observation of the Bell inequality violation with the fictitious states
will imply the same for the genuine bipartite qubit states. We analytically
demonstrated that the basis that diagonalizes the spin-spin correlation is
optimal for the construction of the fictitious states. In this work, We extend
the previous proofs in great detail in terms of the Lorentz properties and the
collider observables. We show that the basis in which the $t\bar{t}$ spin
correlation matrix is diagonal maximizes both the entanglement and the Bell
inequality violation. We further present comparative results of the $t\bar{t}$
system for different basis choices at the CERN Large Hadron Collider (LHC) as
well at an $e^{+}e^{-}$ collider.

The $t$ and $\bar{t}$ (which we will also call a top pair) produced at the LHC
are mainly entangled near threshold or in the high-$p_{T}$ region. At
sufficiently high energies, the optimal basis is close to the commonly used
helicity basis. For example, at a large invariant mass of the system
$m_{tt}\geq 1.2\ \operatorname{TeV}$, the difference between the predicted
violation of Bell’s inequality between the helicity basis and the optimal
basis is within 30%. Near threshold $m_{tt}\sim 2m_{t}$, however, the optimal
basis is closer to the beam basis [41], while Bell inequality violation is
nearly unobservable if we use the helicity basis. Recently, the ATLAS [43] and
CMS [44, 45] collaborations have observed quantum entanglement in $t\bar{t}$
system with a significance of $5\sigma$, but whether Bell inequality violation
can be measured with statistical significance, even with the full high
luminosity LHC dataset, is unclear [5, 6, 16, 29]. Maximizing the sensitivity
is therefore highly desirable.

Before we conclude this section, a special remark is in order regarding the
generality of the test of the Bell inequality violation in collider
experiments. References [46, 47] pointed out that Bell inequality violation is
not testable at a collider because quantities measured at colliders are
commuting while non-commuting observables are required to violate a Bell
inequality. We would like to reiterate that in this work (and in other recent
works on collider studies), we do not claim to perform a fully general “test
of quantum mechanics” against hidden variable theories. Instead, within the
framework of quantum field theory, we identify a quantum state to begin with
in the collider environment, construct observables that measure the
entanglement or Bell inequality violation, and then optimize these
observables. The non-commutation nature arises from the correlation
information from the quantum density matrix and thus gains access to spins. In
a sense, we look to establish quantum tomography and quantum information in
collider systems rather than question quantum mechanics itself.

The rest of this paper is arranged as follows. In Sec. II, we reiterate the
fact that properties of quantum states are independent of the quantization
axis through explicit calculations. In Sec. III, we introduce the difference
between genuine quantum states and fictitious states and elucidate the basis
dependence of fictitious states. In Sec. IV, we prove that the basis that
diagonalizes the correlation matrix, which is known to maximize spin
correlations, also maximizes entanglement and Bell inequality violation. In
Sec. V, we give a detailed summary of the optimal basis in each Standard Model
(SM) process that produces a top pair, and show how the optimal basis
increases the signals of entanglement and Bell inequality violation in the
$t\bar{t}$ system at the LHC and at a future $e^{+}e^{-}$ collider. Finally,
we conclude in Sec. VI. Some technical details for the basis transformation,
entanglement of fictitious states, a general proof of the optimization basis,
and the equivalence of the $gg\to t\bar{t}$ and $\gamma\gamma\to t\bar{t}$
scattering are provided in a few appendices.

##  II Quantum States in Different Bases and Frames

In this section, we elucidate properties of quantum states, namely that
entanglement and Bell inequality violation do not depend on the basis choice
of spin quantization nor on the Lorentz frame in which the spin is considered.

###  II.1 Entanglement in Different Spin Bases

Consider a bipartite qubit system. The state vector
$\ket{\alpha\bar{\alpha}}\equiv\ket{\alpha}\otimes\ket{\bar{\alpha}}$ is a
direct product of the spin state of each qubit. Its density operator
$\hat{\rho}$ can be written as

$\hat{\rho}=\rho_{\alpha\beta,\bar{\alpha}\bar{\beta}}\ket{\alpha\bar{\alpha}}\bra{\beta\bar{\beta}},\quad\quad\quad\alpha,\bar{\alpha},\beta,\bar{\beta}=1,2~{}(\text{or}~{}\uparrow,\downarrow).$
(4)

Here, the repeated spin indices are summed over 1 and 2 or over $\uparrow$ and
$\downarrow$. The states $\ket{\uparrow}$ and $\ket{\downarrow}$ are the
positive and negative spin eigenstates along the third quantization axis
$\oldvec{\,e}_{3}$, and the relative phase between $\ket{\uparrow}$ and
$\ket{\downarrow}$ is determined by the first quantization axis
$\oldvec{\,e}_{1}$. Then $\rho_{\alpha\bar{\alpha},\beta\bar{\beta}}$ is the
$4\times 4$ matrix representation of $\hat{\rho}$ in the basis
$\ket{\alpha\bar{\alpha}}$.

Next, consider choosing another spin basis
$\ket{\alpha^{\prime}}=\ket{\alpha}U_{\alpha\alpha^{\prime}}$, which is
equivalent to measuring the spin of each qubit along another quantization axis
$\oldvec{\,e}_{3}^{\prime}$. The new spin basis for the bipartite qubit system
is
$\ket{\alpha^{\prime}\bar{\alpha}^{\prime}}=\ket{\alpha\bar{\alpha}}U_{\alpha\alpha^{\prime},\bar{\alpha}\bar{\alpha}^{\prime}}$
with
$U_{\alpha\alpha^{\prime},\bar{\alpha}\bar{\alpha}^{\prime}}=U_{\alpha\alpha^{\prime}}\otimes
U_{\bar{\alpha}\bar{\alpha}^{\prime}}$. The operator $\hat{\rho}$ expressed in
this new spin basis is
$\hat{\rho}=\rho^{\prime}_{\alpha^{\prime}\beta^{\prime},\bar{\alpha}^{\prime}\bar{\beta}^{\prime}}\ket{\alpha^{\prime}\bar{\alpha}^{\prime}}\bra{\beta^{\prime}\bar{\beta}^{\prime}}$.
The transformation of the density matrix from one basis to the other is given
by

$\rho^{\prime}_{\beta\delta,\bar{\beta}\bar{\delta}}=U^{\dagger}_{\beta\alpha,\bar{\beta}\bar{\alpha}}\rho_{\alpha\gamma,\bar{\alpha}\bar{\gamma}}\,U_{\gamma\delta,\bar{\gamma}\bar{\delta}},$
(5)

or $\rho^{\prime}=U^{\dagger}\rho\,U$ with the indices omitted.

Quantum entanglement, which is an intrinsic property of a quantum state,
should not depend on the choice of basis for the spin measurement. The
concurrence is a criterion that provides a quantitative measurement of quantum
entanglement. For a bipartite qubit system, the concurrence can be written as
[48]

$\mathscr{C}(\rho)=\max(0,\ \lambda_{1}-\lambda_{2}-\lambda_{3}-\lambda_{4}),$
(6)

where $\lambda_{i}$ ($i=1,2,3,4$) are the eigenvalues of
$\sqrt{\sqrt{\rho}\tilde{\rho}\sqrt{\rho}}$ ordered in decreasing magnitude,
$\tilde{\rho}=(\sigma_{2}\otimes\sigma_{2})\rho^{*}(\sigma_{2}\otimes\sigma_{2})$
with $\sigma_{2}$ the second Pauli matrix, and $\rho^{*}$ is the complex
conjugate of $\rho$. A quantum state is entangled if and only if
$~{}\mathscr{C}(\rho)>0$. When we express the density matrix $\rho$ in a
different basis as $\rho^{\prime}$, we have $\rho^{\prime}=U^{\dagger}\rho\,U$
and it follows that

$\displaystyle\sqrt{\rho^{\prime}}$
$\displaystyle=\sqrt{U^{\dagger}\rho\,U}=\sqrt{(U^{\dagger}\sqrt{\rho}\;U)(U^{\dagger}\sqrt{\rho}\;U)}=U^{\dagger}\sqrt{\rho}\;U,$
(7) $\displaystyle\tilde{\rho}^{\prime}$
$\displaystyle=(\sigma_{2}\otimes\sigma_{2})U^{T}\rho^{*}U^{*}(\sigma_{2}\otimes\sigma_{2})=U^{\dagger}(\sigma_{2}\otimes\sigma_{2})\rho^{*}(\sigma_{2}\otimes\sigma_{2})U=U^{\dagger}\tilde{\rho}\;U.$
(8)

Therefore, the eigenvalues of $\sqrt{\sqrt{\rho}\tilde{\rho}\sqrt{\rho}}$ are
invariant under the basis transformation, and the concurrence of a quantum
state is basis-independent, i.e.,

$\mathscr{C}(\rho)=\mathscr{C}(U^{\dagger}\rho\;U).$ (9)

###  II.2 Bell Inequality Violation in Different Spin Bases

To discuss spin correlation and Bell inequality violation, it is convenient to
decompose the density operator as

$\displaystyle\hat{\rho}=\frac{1}{4}\left(\hat{I}_{2}\otimes\hat{I}_{2}+B_{i}^{+}\hat{\sigma}_{i}\otimes\hat{I}_{2}+B_{i}^{-}\hat{I}_{2}\otimes\hat{\sigma}_{i}+C_{ij}\hat{\sigma}_{i}\otimes\hat{\sigma}_{j}\right).$
(10)

Here and after, repeated indices are summed from 1 to 3,
$\hat{I}_{2}=\ket{\uparrow}\bra{\uparrow}+\ket{\downarrow}\bra{\downarrow}$ is
the identity operator on a qubit, and the spin operators
$\hat{\sigma}_{i}=\ket{\alpha}\sigma_{i}^{\alpha\beta}\bra{\beta}$ are

$\displaystyle\hat{\sigma}_{1}=\ket{\uparrow}\bra{\downarrow}+\ket{\downarrow}\bra{\uparrow},\quad\quad\hat{\sigma}_{2}=-i\ket{\uparrow}\bra{\downarrow}+i\ket{\downarrow}\bra{\uparrow},\quad\quad\hat{\sigma}_{3}=\ket{\uparrow}\bra{\uparrow}-\ket{\downarrow}\bra{\downarrow},$
(11)

where $\sigma_{i}$ are the Pauli matrices. The density matrix $\hat{\rho}$ is
parameterized by 15 real parameters, $B^{\pm}_{i}$ and $C_{ij}$, after
choosing the spin operators.

When reparametrizing the density operator in another basis
$\ket{\alpha^{\prime}}=\ket{\alpha}U_{\alpha\alpha^{\prime}}$, the spin
operators are

$\displaystyle\hat{\sigma}_{1}^{\prime}=\ket{\uparrow^{\prime}}\bra{\downarrow^{\prime}}+\ket{\downarrow^{\prime}}\bra{\uparrow^{\prime}},\quad\quad\hat{\sigma}_{2}^{\prime}=-i\ket{\uparrow^{\prime}}\bra{\downarrow^{\prime}}+i\ket{\downarrow^{\prime}}\bra{\uparrow^{\prime}},\quad\quad\hat{\sigma}_{3}^{\prime}=\ket{\uparrow^{\prime}}\bra{\uparrow^{\prime}}-\ket{\downarrow^{\prime}}\bra{\downarrow^{\prime}}.$
(12)

The spin operators in two different bases satisfy the following transformation
rule (see App. A.2 for details)

$\displaystyle\hat{\sigma}_{i}^{\prime}=R_{ji}\hat{\sigma}_{j},\quad\quad\quad
R_{ij}=\frac{1}{2}\operatorname{Tr}(U^{\dagger}\sigma_{i}U\sigma_{j}).$ (13)

Therefore, in terms of the parameters $B_{i}^{\pm}$ and $C_{ij}$, the
transformation rule of the density matrix in Eq. (5) can be rewritten as

$B^{\prime\pm}_{i}=R_{ij}^{T}B^{\pm}_{j},\quad\quad\quad
C^{\prime}_{ij}=R_{ik}^{T}C_{k\ell}R_{\ell j}.$ (14)

From the orthogonal condition of spin operators, we have that

$\langle\hat{\sigma}_{i}\otimes\hat{\sigma}_{j}\rangle=C_{ij},\quad\quad\quad\langle\hat{\sigma}_{i}\otimes\hat{I}_{2}\rangle=B_{i}^{+},\quad\quad\quad\langle\hat{I}_{2}\otimes\hat{\sigma}_{i}\rangle=B_{i}^{-},$
(15)

where the expectation value of an operator $\hat{\mathcal{O}}$ is
$\braket{\hat{\mathcal{O}}}\equiv\operatorname{Tr}(\hat{\rho}\hat{\mathcal{O}})$.
Therefore, $B^{\pm}_{i}$ parameterizes the polarization of each particle and
$C_{ij}$ parameterizes the spin correlation matrix. The spin correlation,
corresponding to an entry of the spin correlation matrix, is clearly basis-
dependent from the above transformation rule. For a general unpolarized state,
the maximum of the spin correlation, $\mathscr{A}=\langle S_{z}^{t}\otimes
S_{z}^{\bar{t}}\rangle$, is achieved when the quantization axis
$\oldvec{\,e}_{3}$ is an eigenvector of $C_{ij}$ with the largest eigenvalue
[41].111The spin correlation is taken as
$\langle\sigma_{i}\otimes\sigma_{j}\rangle$ because this corresponds to a
measurable differential cross section $d^{2}\sigma/d\ell_{i}d\bar{\ell}_{j}$,
where $d\ell_{i}$ is an angle of one of the decay products of the top relative
to the $i$th axis and $d\bar{\ell}_{j}$ is an angle of one of the decay
products of the anti-top relative to the $j$th axis. Using the component along
the $z$-direction for both $t$ and $\bar{t}$ is conventional. In QCD, the
$t\bar{t}$ production is $CP$-conserving and the correlation matrix $C_{ij}$
is symmetric so that the diagonal basis always exists. Following Ref. [7], we
will refer to the eigenbasis of $C_{ij}$ as the diagonal basis. Note that
traditionally the diagonal basis for the $q\bar{q}\to t\bar{t}$ process is
called the off-diagonal basis because the amplitudes are only non-zero when
the $t$ and $\bar{t}$ have unlike spins (i.e., off-diagonal in the space of
spins) [49, 50]. For general processes, a basis that makes the amplitudes off-
diagonal is not guaranteed to exist.

Moving onto Bell inequalities, for local theories, the Clauser-Horne-Shimony-
Holt (CHSH) inequality states that [51]

$\displaystyle\left|\vec{a}_{1}\cdot
C\cdot(\oldvec{\,b}_{1}-\vec{b}_{2})+\vec{a}_{2}\cdot
C\cdot(\oldvec{\,b}_{1}+\vec{b}_{2})\right|\leq 2,$ (16)

where $\oldvec{\,a}_{1}$, $\oldvec{\,a}_{2}$, $\oldvec{\,b}_{1}$, and
$\oldvec{\,b}_{2}$ are four normalized spatial directions to measure the spins
of $t$ and $\bar{t}$, and $C$ is the correlation matrix from Eq. (10).
Classically this inequality is always satisfied while quantum theories may
violate the inequality. In this bipartite system, as formulated, the CHSH
inequality can be violated. The choice of the set of spatial directions
determines the value of the left-hand side of Eq. (16). Finding the largest
violation of Bell’s inequality corresponds to choosing the set of spatial
directions that maximizes the left-hand side of Eq. (16). We define the left-
hand side of Eq. (16) as the “Bell variable” $\mathscr{B}(\rho)$ and the
maximization corresponds to [52]

$\mathscr{B}(\rho)=\max_{\oldvec{\,a}_{1},\oldvec{\,a}_{2},\oldvec{\,b}_{1},\oldvec{\,b}_{2}}\left|\vec{a}_{1}\cdot
C\cdot(\oldvec{\,b}_{1}-\vec{b}_{2})+\vec{a}_{2}\cdot
C\cdot(\oldvec{\,b}_{1}+\vec{b}_{2})\right|=2\sqrt{\mu_{1}^{2}+\mu_{2}^{2}},$
(17)

where $\mu_{1}^{2}$ and $\mu_{2}^{2}$ are the largest two eigenvalues of
$C^{T}C$ (if $C$ is symmetric, $\mu_{1,2}$ are the two largest magnitude
eigenvalues of $C$.) The eigenvalues of $C^{T}C$ are invariant under the basis
transformation $C\to R^{T}CR$, therefore the violation of Bell’s inequality is
also basis independent

$\mathscr{B}(\rho)=\mathscr{B}(U^{\dagger}\rho\,U).$ (18)

###  II.3 Entanglement and Bell Inequality Violation in Different Lorentz
Frames

In the case that top and anti-top are not simultaneously at rest, the entries
of the spin correlation matrix measured in different Lorentz frames can be
different. This is because the spin of a spin-1/2 particle measured in two
different Lorentz frames is related by a Wigner rotation which acts on a
particle in its rest frame and the top and anti-top will have different Wigner
rotations.

Consider a top quark that is in a momentum eigenstate. The state vector is
defined in both momentum space and spin space, so it can be written as

$\ket{t}=\ket{p_{t}}\otimes\ket{\phi}=\ket{p_{t}}\otimes\ket{\alpha}\phi_{\alpha},$
(19)

where $\ket{p_{t}}$ denotes the momentum state with four-momentum $p_{t}$,
$\ket{\phi}=\ket{\alpha}\phi_{\alpha}$ is the spin state in top quark rest
frame, and $\alpha=\pm 1$ is the spin index. The spin density matrix of
concern is
$\rho_{\alpha\alpha^{\prime}}=\phi_{\alpha}\phi_{\alpha^{\prime}}^{*}$.

The state vector observed in another Lorentz frame is

$\ket{t}=\ket{p_{t}^{\prime}}\otimes\ket{\phi^{\prime}}=\ket{p_{t}^{\prime}}\otimes\ket{\alpha}\phi^{\prime}_{\alpha}.$
(20)

The state vector observed in two different Lorentz frames satisfies the
following relation [53]

$p_{t}^{\prime\mu}=\Lambda^{\mu}_{\nu}p_{t}^{\nu},\quad\quad\quad\phi^{\prime}_{\alpha^{\prime}}=U(\Lambda)_{\alpha^{\prime}\alpha}^{\dagger}\phi_{\alpha},$
(21)

where $\Lambda$ is the Lorentz transformation that connects these two frame
choices, and the unitary matrix $U(\Lambda)$ is the spinor representation of a
Wigner rotation that depends on $\Lambda$ and the momentum of the top quark;
see Eq. (35) for the explicit form. Therefore, the relationship between the
spin density matrix of a moving top quark observed in two different Lorentz
frames is

$\rho^{\prime}_{\alpha^{\prime}\beta^{\prime}}=U(\Lambda)^{\dagger}_{\alpha^{\prime}\alpha}\rho_{\alpha\beta}U(\Lambda)_{\beta\beta^{\prime}}.$
(22)

For the top pair system, both the top and the anti-top are moving but with
different momenta. Similarly, the density matrix of a quantum sub-state
observed in one Lorentz frame is related to the density matrix in a different
Lorentz frame by [53]

$\rho^{\prime}=(U_{t}^{\dagger}\otimes U_{\bar{t}}^{\dagger})\rho(U_{t}\otimes
U_{\bar{t}}),\quad\quad\quad U_{t}\neq U_{\bar{t}}.$ (23)

Here, the Wigner rotations for $t$ and $\bar{t}$ are different because their
momenta are different.

From Eqs. (7) and (8), we find that the concurrence of this state observed in
two different Lorentz frames is still the same, i.e.,
$\mathscr{C}(\rho)=\mathscr{C}(\rho^{\prime})$. Combining Eq. (23) and Eq.
(10), we find that the spin correlation matrix measured in different Lorentz
frames satisfies

$C^{\prime}=R_{t}^{T}CR_{\bar{t}},$ (24)

where
$(R_{t})_{ij}=\operatorname{Tr}(U_{t}^{\dagger}\sigma_{i}U_{t}\sigma_{j})/2$
and
$(R_{\bar{t}})_{ij}=\operatorname{Tr}(U_{\bar{t}}^{\dagger}\sigma_{i}U_{\bar{t}}\sigma_{j})/2$.
Therefore, the eigenvalues of $C^{T}C$ are invariant under Lorentz
transformations. The size of Bell inequality violation, $\mathscr{B}(\rho)$,
is similarly invariant under Lorentz transformations.

A special remark is that here we only consider quantum sub-states that are
momentum eigenstates. In the case where a state is entangled in spin and
momentum space, the entanglement defined in the spin space is basis-dependent;
see Ref. [53] for details.

Although the independence of the choice of frames and axes is illustrated in a
bipartite qubit system, the basic properties relevant for entanglement and
Bell inequality violation also hold in more general quantum systems.

Figure 1: Illustration of the spins in a $t\bar{t}$ event produced at the LHC
measured (a) in the lab frame and (b) the center-of-mass frame. The two frames
are related by a Lorentz boost along the beamline direction. The light blue
arrows denote the moving directions of $t$ ($\bar{t}$) in the corresponding
frame, and the hollow arrows represent the purposely chosen spin directions of
$t$ ($\bar{t}$) in its rest frame.

##  III Fictitious States

The density matrix $\rho$ of a quantum state as discussed above is
reconstructed from the expectation value of a series of measurements on a
large event ensemble. Denoting the quantum sub-state of each event as
$\rho_{a}$, the reconstructed density matrix is an average of all $N$ quantum
sub-states

$\bar{\rho}=\frac{1}{N}\sum_{a=1}^{N}\rho_{a}.$ (25)

In collider experiments, each event, specified by the differential cross
section at a multi-dimensional phase space point $d\sigma/d\Omega$, is a
measurement of an underlying quantum state, and Eq. (25), in the continuum
limit, becomes

$\bar{\rho}=\frac{1}{\sigma}\int\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}\rho(\Omega).$ (26)

To obtain a genuine quantum state – or a “physical state” – from the average
Eq. (25), we need to write the density matrix $\rho_{a}$ in an event-
independent basis, i.e., use the same axis $\hat{e}_{i}$ to measure the spin
of all events $a$:

$\displaystyle\bar{\rho}^{\oldvec{\,e}_{i}}$
$\displaystyle=\frac{1}{N}\sum_{a=1}^{N}\rho_{a}^{\oldvec{\,e}_{i}}.$ (27)

Here, $\rho_{a}^{\oldvec{\,e}_{i}}$ denotes the density matrix of the quantum
sub-state of $a^{\rm th}$ event measured in the reference frame
$(\oldvec{\,e}_{1},\oldvec{\,e}_{2},\oldvec{\,e}_{3})$. Note that all the
physical states are equivalent, i.e., the physical state averaged in two
different bases are equal with respect to a unitary rotation,
$\bar{\rho}^{\oldvec{\,e}_{i}}=U^{\dagger}\bar{\rho}^{\oldvec{\,e}^{\prime}_{i}}U$.

When measuring the spin of each qubit in an event $a$ in a different basis
$\oldvec{\,e}_{i,a}$, however, the average in Eq. (25) gives

$\bar{\rho}[\oldvec{\,e}_{i,a}]=\frac{1}{N}\sum_{a=1}^{N}\rho_{a}^{\oldvec{\,e}_{i,a}}=\frac{1}{N}\sum_{a=1}^{N}U^{\dagger}_{a}\rho_{a}^{\oldvec{\,e}_{i}}U_{a}.$
(28)

A state averaged in an event-dependent basis is generally not equivalent to
the physical state, i.e., there is no guarantee of a unitary transformation
$U$, such that
$\bar{\rho}[\oldvec{\,e}_{i,a}]=U^{\dagger}\bar{\rho}^{\oldvec{\,e}_{i}}U$.
Therefore, the density matrix averaged in an event-dependent basis is a
“fictitious state,” rather than a genuine quantum state. Moreover, the density
matrix averaged in two different event-dependent bases $\oldvec{\,e}_{i,a}$
and $\oldvec{\,e}_{i,a}^{\prime}$ are not equivalent, i.e., there is no
guarantee of a unitary transformation $U$, such that
$\bar{\rho}[\hat{e}_{i,a}]=U^{\dagger}\bar{\rho}[\hat{e}^{\prime}_{i,a}]U$. In
other words, the fictitious states are basis-dependent.

We can define a fictitious state as the result of constructing a state
according to Eq. (25) but using an event-dependent basis. As shown above, this
is not a quantum state because it does not retain all of the properties that a
quantum state has. We still call it a state (rather than simply a sum),
however, because it does retain some properties of quantum states. One example
is the existence of Bell inequality violation where a fictitious state that
has Bell inequality violation implies the existence of a quantum sub-state
that has Bell inequality violation.

The usage of event-dependent bases and fictitious states may seem unnatural
for low-energy experiments. At high-energy experiments, however, both event-
dependent spin axis and Lorentz frame choices are conventionally used. We next
show that either event-dependent spin axis or event-dependent Lorentz frame
choices lead to fictitious states.

###  III.1 Fictitious States From Angular-Averaged Sums

Experimentally, the polarization of top quarks cannot be measured directly, so
the spin density matrix is reconstructed from the angular distributions of the
final state particles. For example, in the di-lepton decay mode, the
correlation matrix is

$C_{ij}=-9\braket{\cos\theta^{+}_{i}\cos\theta^{-}_{j}},\quad\quad\quad\cos\theta^{\pm}_{i}=\oldvec{\,p}_{\ell^{\pm}}\cdot\oldvec{\,e}_{i},$
(29)

where $\oldvec{\,p}_{\ell^{\pm}}$ is the normalized three-momentum of the
lepton $\ell^{+}$ ($\ell^{-}$) in the rest frame of the $t$ ($\bar{t}$), and
$\oldvec{\,e}_{i}$ is the quantization axis used to determine the lepton
momentum direction. Ideally, with an infinite number of events, one can study
the $t\bar{t}$ system produced at a fixed phase space point $\Omega$, so that
all the events are measured with the same quantum sub-states, determined by
$C_{ij}(\Omega)$. In reality, each measured event corresponds to a top pair
produced at a different phase space point. Therefore, with Eq. (29), one
obtains the correlation matrix of an angular-averaged sum $\bar{\rho}$ in Eq.
(26):222As a reminder, we use angular-averaged sum as the general description
which may correspond to either a quantum state or a fictitious state.

$\bar{C}=\frac{1}{\sigma}\int\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}C(\Omega).$ (30)

As discussed in Ref. [30], to obtain a physical quantum state we must choose a
set of $\oldvec{\,e}_{i}$ that are independent of the top quark momentum with
respect to the average in Eq. (30).

In principle, the lab frame in any collision is a genuine event-independent
frame convenient for measurements and analyses. The center-of-mass frame of
$t\bar{t}$ system, however, is the most commonly used frame that captures the
underlying dynamics when studying $t\bar{t}$ production at hadron colliders.
As illustrated in Fig. 1, the lab frame in hadronic collisions and the center-
of-mass frame of the $t\bar{t}$ system are, in general, related by a Lorentz
boost along the beam direction. The center-of-mass frame is an event-dependent
frame, governed by the interaction and kinematics of the initial state
partons.

Examples of event-dependent bases are shown in Fig. 2, where we have the fixed
beam basis in Fig. 2(a) with an arbitrary azimuthal angle orientation, the
rotated beam basis in Fig. 2(b) in line with the vectors $\oldvec{\,p}_{t}$
and $\oldvec{\,z}$ forming the scattering plane, the helicity basis in Fig.
2(c) with the spin along the $\oldvec{\,p}_{t}$ direction, and the diagonal
basis in Fig. 2(d) to be worked out later.333To construct the rotated beam
basis and the helicity basis, only knowledge of the event-by-event kinematics
is required, whereas the diagonal basis also requires knowledge of the
production density matrix. At leading order, in the helicity basis the
correlation matrix $C_{ij}$ only has one off-diagonal entry $C_{kr}$ such that
only one rotational angle $\xi$ in the $\oldvec{\,k}$-$\oldvec{\,r}$ plane is
needed to determine the diagonal basis. This angle is shown in Fig. 2.
Depending on the specific physical process, a certain basis may be
advantageous. For instance, when the invariant mass of the $t\bar{t}$ system
is large, the helicity basis is most commonly used.

In practice, event-dependent bases are commonly used when averaging the lepton
momentum. The basis $\oldvec{\,e}_{i}$ in Eq. (29) is replaced with an event-
dependent basis $\oldvec{\,e}_{i}(\Omega)$, and the average in Eq. (29) yields

$\bar{C}^{\rm fic}=\frac{1}{\sigma}\int\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}R^{T}(\Omega)C(\Omega)R(\Omega),$ (31)

where $R(\Omega)$ is the event-dependent rotation that rotates the fixed basis
to $\oldvec{\,e}_{i}(\Omega)$, and $\bar{C}^{\rm fic}$ is the correlation
matrix of the fictitious state constructed in Eq. (28).

Figure 2: (a) Fixed beam basis $(\oldvec{\,x},\oldvec{\,y},\oldvec{\,z})$. (b)
Rotated beam basis
$(\oldvec{\,x}^{\prime},\oldvec{\,y}^{\prime},\oldvec{\,z})$. (c) Helicity
basis $(\oldvec{\,r},\oldvec{\,n},\oldvec{\,k})$. (d) Diagonal basis
$(\oldvec{\,e}_{1}^{\,\rm diag},\oldvec{\,e}_{2}^{\,\rm
diag},\oldvec{\,e}_{3}^{\,\rm diag})$. The rotation angle $\xi$ of the
diagonal basis depends on the production process, e.g.,
$\tan\xi=(1/\gamma)\tan\theta$ for $q\bar{q}\to t\bar{t}$ [50].

###  III.2 Fictitious States From Event-Dependent Lorentz Frame Choices

In addition to event-dependent spin quantization axis choices, an event-
dependent Lorentz frame choice also results in a fictitious state, such as a
bipartite $t\bar{t}$ state constructed in the center-of-mass frame, in
contrast to that in the lab frame. We start from one event in which the four-
momentum of top, anti-top, and their decayed leptons in the lab frame are
$p_{t}^{\rm lab}$, $p_{\bar{t}}^{\rm lab}$, $p_{\ell^{+}}^{\rm lab}$, and
$p_{\ell^{-}}^{\rm lab}$, respectively. When reconstructing the top pair spin
correlation matrix in the rest frames of the top and the anti-top, we boost
the lepton momenta in the lab frame to the rest frames of top and anti-top
quark, respectively,

$\displaystyle p_{\ell^{+}}$ $\displaystyle=L(p_{t}^{\rm
lab})p_{\ell^{+}}^{\rm lab},\quad\quad\quad p_{\ell^{-}}=L(p_{\bar{t}}^{\rm
lab})p_{\ell^{-}}^{\rm lab},$ (32)

where $L(p)$ is a pure Lorentz boost from the lab frame to the rest frame of
the particle with momentum $p$. Next, consider reconstructing the correlation
matrix in another (event-dependent) frame, where the four momentum of each
particle are $p_{t}^{\Lambda}$, $p_{\bar{t}}^{\Lambda}$
$p_{\ell^{+}}^{\Lambda}$, and $p_{\ell^{-}}^{\Lambda}$. This frame is related
to the lab frame by a Lorentz transformation $\Lambda$, so that
$p^{\Lambda}=\Lambda p^{\rm lab}$. Boosting the lepton momenta measured in
this frame to the rest frames of the $t$ and the $\bar{t}$ yields

$\displaystyle p_{\ell^{+}}^{\prime}$
$\displaystyle=L(p_{t}^{\Lambda})p_{\ell^{+}}^{\Lambda},\quad\quad\quad
p_{\ell^{-}}^{\prime}=L(p_{\bar{t}}^{\Lambda})p_{\ell^{-}}^{\Lambda}.$ (33)

Combining Eqs. (32) and (33), we find that $p_{\ell}$ and $p_{\ell}^{\prime}$
are related by a Lorentz transformation

$\displaystyle p_{\ell^{+}}^{\prime}$
$\displaystyle=L(p_{t}^{\Lambda})p_{\ell^{+}}^{\Lambda}=L(\Lambda p_{t}^{\rm
lab})\Lambda p_{\ell^{+}}^{\rm lab}=L(\Lambda p_{t}^{\rm lab})\Lambda
L^{-1}(p_{t}^{\rm lab})\;p_{\ell^{+}},$ $\displaystyle p_{\ell^{-}}^{\prime}$
$\displaystyle=L(p_{\bar{t}}^{\Lambda})p_{\ell^{-}}^{\Lambda}=L(\Lambda
p_{\bar{t}}^{\rm lab})\Lambda p_{\ell^{-}}^{\rm lab}=L(\Lambda
p_{\bar{t}}^{\rm lab})\Lambda L^{-1}(p_{\bar{t}}^{\rm lab})\;p_{\ell^{-}}.$
(34)

The Lorentz transformation $L(\Lambda p)\Lambda L^{-1}(p)$ is a pure rotation,
called a Wigner rotation

$L(\Lambda p)\Lambda L^{-1}(p)=\begin{pmatrix}1&0\\\
0&R(\Lambda,p)\end{pmatrix}.$ (35)

Consequently, the reconstructed lepton momenta with these two different
methods are related by the Wigner rotation,

$\oldvec{\,p}_{\ell^{+}}^{\prime}=R(\Lambda,p_{t}^{\rm
lab})\oldvec{\,p}_{\ell^{+}},\quad\quad\quad\oldvec{\,p}_{\ell^{-}}^{\prime}=R(\Lambda,p_{\bar{t}}^{\rm
lab})\oldvec{\,p}_{\ell^{-}}.$ (36)

Here, these two $3\times 3$ rotations depend on the four-momentum of top or
anti-top quark and the Lorentz transformation $\Lambda$ between two different
frames. We abbreviate the Wigner rotation $R(\Lambda,p_{t}^{\rm lab})$ and
$R(\Lambda,p_{\bar{t}}^{\rm lab})$ as $R_{t}(\Lambda)$ and
$R_{\bar{t}}(\Lambda)$ in the following.

Therefore, as introduced in Sec. II.3, the correlation matrix of a quantum
sub-state measured in two different Lorentz frames satisfies the relation

$C^{\prime}(\Omega)=R_{t}^{T}(\Lambda_{\Omega})C(\Omega)R_{\bar{t}}(\Lambda_{\Omega}).$
(37)

We find that measuring the correlation matrix in a different Lorentz frame is
equivalent to measuring the correlation matrix in the original Lorentz frame
but with a different spin axis of $\oldvec{\,\,e}_{i,\Omega}^{\,t}$ and
$\oldvec{\,\,e}_{i,\Omega}^{\,\bar{t}}$ for $t$ and $\bar{t}$, respectively,
where
$\oldvec{\,\,e}_{i,\Omega}^{\,t}=R_{t,ji}(\Lambda_{\Omega})\oldvec{\,e}_{j}$
and
$\oldvec{\,\,e}_{i,\Omega}^{\,\bar{t}}=R_{\bar{t},ji}(\Lambda_{\Omega})\oldvec{\,e}_{j}$.
Here $\oldvec{\,e}_{j}$ is the spin axis used in the original Lorentz frame.

Thus our formalism of fictitious states in Eq. (28) applies equally for event-
dependent Lorentz frame choices. When reconstructing Eq. (29) with event-
dependent Lorentz frames, the resulting correlation matrix is

$\displaystyle\bar{C}^{\rm
fic}=\frac{1}{\sigma}\int\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}R_{t}(\Lambda_{\Omega})C(\Omega)R_{\bar{t}}(\Lambda_{\Omega})^{T},$
(38)

where $\Lambda_{\Omega}$ denotes the Lorentz transformation from the lab frame
to an event-dependent Lorentz frame. Similar to Eq. (31), the averaged
correlation matrix $\bar{C}^{\rm fic}$ also depends on the event-by-event
rotations $R_{t}(\Lambda_{\Omega})$ and $R_{\bar{t}}(\Lambda_{\Omega})$.

###  III.3 Basis Dependence of Spin Correlation and Entanglement

Due to the choice of event-dependent bases, fictitious states are
conventionally utilized when discussing quantum observables of a “state”
produced at colliders. While the entanglement and Bell inequality violation of
quantum states are basis-independent as we demonstrated earlier, observables
computed using fictitious states and fictitious states themselves are basis-
dependent in general. For example, the entanglement and Bell inequality
violation of a fictitious state becomes basis-dependent. Consequently, this
gives us the freedom to find a basis that optimizes the measurement of
entanglement and Bell inequality violation.

Before digging into details of optimizing the entanglement and Bell inequality
violation, it is worth pointing out that the basis dependence of the
entanglement of fictitious states is different from the basis dependence of
spin correlations, which has been thoroughly studied for the $t\bar{t}$ system
[49, 50, 41].

The spin correlation only depends on the quantization axis chosen to measure
the spin, e.g., defining $\ket{\uparrow}$ and $\ket{\downarrow}$ with respect
to the third quantization axis $\oldvec{\,e}_{3}$. The size of spin
correlation does not depend on the phases of $\ket{\uparrow}$ and
$\ket{\downarrow}$. On the other hand, angular-averaged sums do depend on
these phases. Equivalently, they depend on all three quantization axes
$\oldvec{\,e}_{i}(\mathbf{k})$. For example, in the high-$p_{T}$ region, a
maximally-entangled Bell triplet state is produced from both the $q\bar{q}$
and the $gg$ initial states

$\ket{\Psi(\phi)}=\frac{e^{-i\phi}\ket{\uparrow\uparrow}+e^{i\phi}\ket{\downarrow\downarrow}}{\sqrt{2}}.$
(39)

Here, the relative phase of the state is calculated in fixed beam basis. The
azimuthal angular average of the density matrix in the fixed beam basis is

$\bar{\rho}^{\rm fixed}_{\rm
triplet}=\frac{1}{2\pi}\int\operatorname{d\phi}\ket{\Psi(\phi)}\bra{\Psi(\phi)}=\frac{\ket{\uparrow\uparrow}\bra{\uparrow\uparrow}+\ket{\downarrow\downarrow}\bra{\downarrow\downarrow}}{2},$
(40)

As the azimuthal angle in the rotated beam basis is always chosen to be zero,
the azimuthal angular average of the density matrix in rotated beam basis is

$\displaystyle\bar{\rho}^{\rm rotated}_{\rm triplet}$
$\displaystyle=\frac{1}{2\pi}\int\operatorname{d\phi}\ket{\Psi(0)}\bra{\Psi(0)}=\ket{\Psi(0)}\bra{\Psi(0)},$
(41)

where
$\ket{\Psi(0)}=\ket{\Psi(\phi=0)}=(\ket{\uparrow\uparrow}+\ket{\downarrow\downarrow})/\sqrt{2}$.
The angular-averaged sum in the fixed basis $\bar{\rho}^{\rm fixed}_{\rm
triplet}$ is a separable state, while $\bar{\rho}^{\rm rotated}_{\rm triplet}$
is still a maximally entangled state, although the spin correlation $\langle
S_{3}^{t}\otimes S_{3}^{\bar{t}}\rangle=1$ is the same for both angular-
averaged sums.

##  IV The Optimal Basis for Entanglement and Bell Inequality Violation

In this section, we prove that the diagonal basis maximizes the entanglement
and the Bell inequality violation for fictitious states. For simplicity, we
first present our proof assuming the spin correlation matrix is symmetric,
which is true in the center-of-mass frame of the $t\bar{t}$ system when the
$t\bar{t}$ events are produced from $CP$-conserving processes (as in the SM).

We start with an arbitrary event-dependent basis choice
$\oldvec{\,e}_{i,\Omega}$. The correlation matrix in this basis is

$\bar{C}_{ij}=\frac{1}{\sigma}\int\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}C_{ij,\Omega},\quad\quad\quad{\rm with}\
(\bar{c}_{1},\bar{c}_{2},\bar{c}_{3})\ {\rm as}\ {\rm eigenvalues\ of}\
\left[\bar{C}\right].$ (42)

Here, we omit the basis choice label for simplicity when referring to a
general event-dependent basis.

When the correlation matrix for each sub-state is symmetric, it can be
diagonalized with an event-dependent rotation:444 Without loss of generality,
the eigenvalues are ordered as
$\mu_{1,\Omega}\geq\mu_{2,\Omega}\geq\mu_{3,\Omega}$.

$C^{\rm
diag}_{\Omega}=R_{\Omega}C_{\Omega}R^{T}_{\Omega}=\begin{pmatrix}\mu_{1,\Omega}&0&0\\\
0&\mu_{2,\Omega}&0\\\ 0&0&\mu_{3,\Omega}\\\ \end{pmatrix}.$ (43)

This diagonal spin correlation matrix is obtained if one measures the spin of
$t$ and $\bar{t}$ with the following quantization axis

$\oldvec{\,\,e}_{i,\Omega}^{\,\rm diag}=R_{ij}\oldvec{\,e}_{j},$ (44)

where $\oldvec{\,e}_{i,\Omega}^{\,\rm diag}$ are three eigenvectors of
$C_{\Omega}$ and they are referred to as the diagonal basis.

In the diagonal basis, the eigenvalues of the angular-averaged spin
correlation matrix are

$\bar{c}^{\rm diag}_{i}=\frac{1}{\sigma}\int\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}\mu_{i,\Omega}\equiv\bar{\mu}_{i}.$ (45)

In any other basis, the eigenvalues of angular-averaged spin correlation
matrix $\bar{C}$, namely $\bar{c}_{i}$, are generally not equal to the average
of $\mu_{i}$, but they satisfy the following relations [30],

$\displaystyle\bar{c}_{1}+\bar{c}_{2}+\bar{c}_{3}=\bar{\mu}_{1}+\bar{\mu}_{2}+\bar{\mu}_{3}=\operatorname{Tr}(\bar{C}),$
(46) $\displaystyle\bar{\mu}_{1}\geq\bar{c}_{i}\geq\bar{\mu}_{3}.$ (47)

Eq. (46) is a direct consequence of taking the trace of Eqs. (42) and (43). To
prove Eq. (47), which states that $\bar{c}_{i}$ are bounded by
$\bar{\mu}_{i}$, we first denote the three eigenvectors of $\overline{C}$ as
$\hat{v}_{i}$. The corresponding eigenvalue is then

$\displaystyle\bar{c}_{i}=\hat{v}_{i}\cdot\overline{C}\cdot\hat{v}_{i}=\frac{1}{\sigma}\int\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}\left(\hat{v}_{i}\cdot
C_{\Omega}\cdot\hat{v}_{i}\right).$ (48)

Applying Eq. (43) and $C_{\Omega}=R_{\Omega}C^{\rm
diag}_{\Omega}R_{\Omega}^{T}$, we find that

$\hat{v}_{i}\cdot
C_{\Omega}\cdot\hat{v}_{i}=\sum_{\ell=1}^{3}\left|(R_{\Omega}\cdot\hat{v}_{i})_{\ell}\right|^{2}\mu_{\ell,\Omega},$
(49)

is a convex sum of $\mu_{i,\Omega}$. This leads to

$\mu_{1,\Omega}\geq\hat{v}_{i}\cdot
C_{\Omega}\cdot\hat{v}_{i}\geq\mu_{3,\Omega}.$ (50)

Therefore, Eq. (47) holds as a convex sum of Eq. (50).

With Eqs. (46) and (47), we are ready to prove that the diagonal basis
maximizes both $\mathscr{B}(\bar{\rho})$ and $\mathscr{C}(\bar{\rho})$.

###  IV.1 Concurrence

We focus on the unpolarized $t\bar{t}$ system at leading order where the
polarization vectors $B_{i}^{\pm}$ are zero. In this case $\tilde{\rho}=\rho$,
as can be seen from the construction below Eq. (6). After some explicit
calculation [4, 7], the concurrence is expressed as follows, with the largest
eigenvalue labelled $\lambda(\rho)$:

$\mathscr{C}(\rho)=\max(2\lambda(\rho)-1,0).$ (51)

To compare the concurrence of the angular-averaged density matrix in different
bases, we only need to compare their largest eigenvalues. Substituting the
correlation matrix into Eq. (10), we find the four eigenvalues of an angular-
averaged density matrix, $\bar{\rho}$, are

$\frac{1-\operatorname{Tr}(\bar{C})}{4},\quad\frac{1+\operatorname{Tr}(\bar{C})-\bar{c}_{1}}{4},\quad\frac{1+\operatorname{Tr}(\bar{C})-\bar{c}_{2}}{4},\quad\frac{1+\operatorname{Tr}(\bar{C})-\bar{c}_{3}}{4}.$
(52)

Thus
$\lambda(\bar{\rho})=\frac{1}{4}\max[1-\operatorname{Tr}(\bar{C}),1+\operatorname{Tr}(\bar{C})-\bar{c}_{i}]$.

In the diagonal basis, the largest eigenvalue of $\bar{\rho}^{\rm diag}$ is
$\lambda(\bar{\rho}^{\rm
diag})=\frac{1}{4}\max[1-\operatorname{Tr}(\bar{C}),1+\operatorname{Tr}(\bar{C})-\bar{\mu}_{i}]$.
From Eq. (47), we have $\bar{c}_{i}\geq\bar{\mu}_{3}$ for any $i$. Then,
$\lambda(\bar{\rho}^{\rm diag})\geq\lambda(\bar{\rho})$ and
$\mathscr{C}(\bar{\rho}^{\rm diag})\geq\mathscr{C}(\bar{\rho})$. Therefore,
the diagonal basis maximizes the concurrence of angular-averaged states.

For leading-order $t\bar{t}$ production, the direction $\oldvec{\,n}$ in Fig.
2 is always an eigenvector of $C_{ij}$ (see App. D for details). Whenever
$\operatorname{Tr}(C)<0$ or $C_{nn}$ is the smallest eigenvalue of $C_{ij}$,
the angular-averaged density matrix has exactly the same concurrence in the
rotated beam basis, the helicity basis, and the diagonal basis. For $t\bar{t}$
production in QCD, this is the case for most channels (see Table 1) and is the
case in the near threshold region measured by ATLAS [43] and by CMS [44].
Therefore, accidentally, the concurrence is usually not sensitive to basis
choice as long as one axis is chosen as the normal direction of the scattering
plane.

###  IV.2 Bell Inequality

We show next that the diagonal basis is also the one that maximizes Bell
inequality violation for fictitious states. Since the spin correlation matrix
$\bar{C}$ of the angular-averaged sum $\bar{\rho}$ is symmetric, the
eigenvalues of $(\bar{C})^{T}\bar{C}$ are $\bar{c}_{i}^{2}$. In accordance
with Eq. (17), we have

$\mathscr{B}(\bar{\rho})=2\max_{i\neq
j}\sqrt{\bar{c}_{i}^{2}+\bar{c}_{j}^{2}}.$ (53)

To show that the diagonal basis is the optimal basis for measuring the Bell
inequality, i.e., $\mathscr{B}(\bar{\rho}^{\rm
diag})\geq\mathscr{B}(\bar{\rho})$, we need to prove that for any $i\neq j$,
there exist $k\neq\ell$ satisfying

$\bar{c}_{i}^{2}+\bar{c}_{j}^{2}~{}\leq~{}\bar{\mu}_{k}^{2}+\bar{\mu}_{\ell}^{2},$
(54)

where $\bar{\mu}_{i}$ are the eigenvalues of $\bar{C}^{\rm diag}$, see Eqs.
(43) and (45).

The relative signs of $\bar{\mu}_{1}$, $\bar{\mu}_{2}$, and $\bar{\mu}_{3}$
can be divided into the following three cases, and Eq. (54) can be proved case
by case (see Ref. [30] for details):

  1. (a)

$\bar{\mu}_{1}\geq\bar{\mu}_{2}\geq\bar{\mu}_{3}\geq
0\qquad\Longrightarrow\qquad\bar{c}_{i}^{2}+\bar{c}_{j}^{2}\leq\bar{\mu}_{1}^{2}+\bar{\mu}_{2}^{2}$,

  2. (b)

$0\geq\bar{\mu}_{1}\geq\bar{\mu}_{2}\geq\bar{\mu}_{3}\qquad\Longrightarrow\qquad\bar{c}_{i}^{2}+\bar{c}_{j}^{2}\leq\bar{\mu}_{2}^{2}+\bar{\mu}_{3}^{2}$,

  3. (c)

$\bar{\mu}_{1}\geq
0\geq\bar{\mu}_{3}\qquad\Longrightarrow\qquad\bar{c}_{i}^{2}+\bar{c}_{j}^{2}\leq\bar{\mu}_{1}^{2}+\bar{\mu}_{3}^{2}$.

Therefore, the fictitious state obtained using the diagonal basis gives the
largest violation of Bell’s inequality, $\mathscr{B}(\overline{\rho}^{\rm
diag})\geq\mathscr{B}(\overline{\rho})$.

###  IV.3 Alternative Bell Variable

The variable $\mathscr{B}(\rho)$ is the theoretical upper limit of the Bell
inequality violation as it is obtained by considering all possible values for
the directions $\oldvec{\,a}_{i}$ and $\oldvec{\,b}_{i}$ to maximize Eq. (16).
In practice, for a simpler estimation of statistical uncertainties,555It is
argued that the variable $\mathscr{B}(\rho)$ has asymmetric uncertainties
which, if not accounted for properly, lead to a biased estimate [6]. it is
preferred to a priori fix the directions $\oldvec{\,a}_{i}$ and
$\oldvec{\,b}_{i}$ with respect to the quantization axes as

$\oldvec{\,a}_{1}=\oldvec{\,e}_{i},\quad\oldvec{\,a}_{2}=\oldvec{\,e}_{j},\quad\oldvec{\,b}_{1}=\frac{\oldvec{\,e}_{i}\pm\oldvec{\,e}_{j}}{\sqrt{2}},\quad
b_{1}=\frac{\oldvec{\,e}_{i}\mp\oldvec{\,e}_{j}}{\sqrt{2}},\quad(i\neq j).$
(55)

As a result, the observable for Bell inequality violation, Eq. (16), is

$\mathscr{B}^{\prime}(\rho)=\sqrt{2}\max_{i\neq j}|C_{ii}\pm C_{jj}|.$ (56)

In the diagonal basis, the diagonal terms of the correlation matrix
$\bar{C}_{ij}^{\rm diag}$ of the angular-averaged sum are simply the
eigenvalues of $\bar{C}_{ij}^{\rm diag}$, i.e., $\bar{C}_{ii}^{\rm
diag}=\bar{\mu}_{i}$. In any other basis, as with Eqs. (46) and (47), we have

$\displaystyle\bar{C}_{11}+\bar{C}_{22}+\bar{C}_{33}=\bar{\mu}_{1}+\bar{\mu}_{2}+\bar{\mu}_{3}=\operatorname{Tr}(\bar{C}),$
(57) $\displaystyle\bar{\mu}_{1}\geq\bar{C}_{ii}\geq\bar{\mu}_{3}.$ (58)

Here, Eq. (58) is satisfied because $\bar{C}_{ii}$ are bounded by
$\bar{c}_{i}$, the eigenvalues of $\bar{C}$, while $\bar{c}_{i}$ are bounded
by $\bar{\mu}_{i}$. Therefore the diagonal basis is also the one that
maximizes the variable $\mathscr{B}^{\prime}(\bar{\rho})$.

###  IV.4 Asymmetric Correlation Matrix

If a $t\bar{t}$ event is constructed in an arbitrary Lorentz frame other than
the $t\bar{t}$ center-of-mass frame, or if the production involves
$CP$-violating interactions, the spin correlation matrix, $C_{\Omega}$, for
each quantum sub-state is generally not symmetric. Still, the spin correlation
matrix $C_{\Omega}$ can be diagonalized with an event-dependent singular value
decomposition:

$C^{\rm
diag}_{\Omega}=R_{\Omega}^{t}C_{\Omega}\left(R^{\bar{t}}_{\Omega}\right)^{T}=\begin{pmatrix}\mu_{1,\Omega}&0&0\\\
0&\mu_{2,\Omega}&0\\\ 0&0&\mu_{3,\Omega}\\\ \end{pmatrix}.$ (59)

From $C_{\Omega}$, the diagonal matrix $C^{\rm diag}_{\Omega}$ can be obtained
by using the singular vectors of $C_{\Omega}$ to determine the two rotations,
$R_{\Omega}^{t}$ and $R^{\bar{t}}_{\Omega}$, needed to rotate $C_{\Omega}$
into $C^{\rm diag}_{\Omega}$. Let the three left (right) singular vectors of
$C_{\Omega}$ be $\oldvec{\,e}_{i,\Omega}^{\,t}$
($\oldvec{\,e}_{i,\Omega}^{\,\bar{t}}$), then

$\oldvec{\,\,e}_{i,\Omega}^{\,t}=R_{ij,\Omega}^{t}\oldvec{\,e}_{j},\quad\quad\quad\oldvec{\,\,e}_{i,\Omega}^{\,\bar{t}}=R_{ij,\Omega}^{\bar{t}}\oldvec{\,e}_{j}.$
(60)

Alternatively, within this Lorentz frame, one could measure the spins of $t$
and $\bar{t}$ with two different sets of quantization axes,
$\oldvec{\,\,e}_{i,\Omega}^{\,t}$ and $\oldvec{\,\,e}_{i,\Omega}^{\,\bar{t}}$,
to directly measure $C^{\rm diag}_{\Omega}$.

Due to the transformation between $C_{\Omega}$ and $C^{\rm diag}_{\Omega}$ in
Eq. (59), both spin correlation matrices have the same singular values.
Consequently, in any Lorentz frame the spin correlation matrix will have the
same singular values and lead to the same matrix $C^{\rm diag}_{\Omega}$ in
the diagonal basis meaning that this generalized definition of the diagonal
basis is independent of choice of Lorentz frame. Our proof can be adjusted to
include asymmetric correlation matrices (see App. C for details) and thus we
have shown that the diagonal basis is the optimal basis for general processes.

##  V Phenomenological Analyses for $t\bar{t}$ Events

In theory and in practice, it is perhaps most convenient to start the analysis
for $t\bar{t}$ events in the helicity basis in the center-of-mass frame. The
diagonal basis can be obtained from the helicity basis by a single rotation in
the scattering plane. This is because the vector $\oldvec{\,n}$ is always an
eigenvector of the spin correlation matrix. This angle is shown visually in
Fig. 2(d) and relates the bases via

$\oldvec{\,\,e}_{1}^{\,\rm
diag}=\oldvec{\,r}\cos\xi+\oldvec{\,k}\sin\xi,\quad\quad\quad\oldvec{\,\,e}_{2}^{\,\rm
diag}=\oldvec{\,n},\quad\quad\quad\oldvec{\,\,e}_{3}^{\,\rm
diag}=\oldvec{\,k}\cos\xi-\oldvec{\,r}\sin\xi.$ (61)

The rotation angle $\xi$ depends on the production process and on the phase
space point of $t\bar{t}$. For simplicity, when we summarize the optimal
(diagonal) basis for different SM processes in the following, all the
correlation matrices are expressed in the helicity basis.

###  V.1 Characteristics of QCD Sub-Processes

We first consider the parton-level processes in QCD of $q\bar{q}\to t\bar{t}$
and $gg\to t\bar{t}$. We express the $2\to 2$ unpolarized production cross
section at the partonic level as

$\operatorname{d}\sigma(t\bar{t})=\frac{\beta}{64\pi^{2}\hat{s}}\
\overline{\sum}|\mathcal{M}|^{2}\
\operatorname{d}\Omega,\quad\quad\quad\operatorname{d}\Omega=\operatorname{d}\phi\
\operatorname{d}\cos\theta.$ (62)

With respect to the $t\bar{t}$ spin states labeled by $\alpha,\beta$, the
scattering matrix element and the spin density matrix in Eq. (10) are related
in the $4\times 4$ matrix form as

${\sum}|\mathcal{M}|^{2}=\operatorname{Tr}(\mathcal{M}_{\alpha\bar{\alpha}}\mathcal{M}^{*}_{\beta\bar{\beta}})\equiv\operatorname{Tr}(\mathcal{R}_{\alpha\beta,\bar{\alpha}\bar{\beta}}),\quad\quad\quad\rho_{\alpha\beta,\bar{\alpha}\bar{\beta}}=\frac{\cal{R}_{\alpha\beta,\bar{\alpha}\bar{\beta}}}{\operatorname{Tr}(\cal{R})}.$
(63)

Further details can be found in App. A. As seen in Eq. (10), $B^{\pm}_{i}$
terms do not contribute for an unpolarized $t\bar{t}$ system, and the density
matrix is thus obtained by summing over the correlation matrix elements
$C_{ij}$. The squared matrix elements, spin correlation matrices $C_{ij}$, and
the rotation angle $\xi$ of the optimization for each process are listed in
Table 1, where the velocity factor in the center-of-mass frame of $t\bar{t}$
is $\beta=(1-4m_{t}^{2}/\hat{s})^{1/2}$ and the boost factor is
$\gamma=\sqrt{\hat{s}}/(2m_{t})$, where $\sqrt{\hat{s}}$ is the center-of-mass
energy of $t\bar{t}$. The QCD dynamical pre-factors from the tree-level matrix
element calculations are

$\kappa_{q}=g_{s}^{2}\frac{(N^{2}-1)}{N^{2}},\quad\quad\quad\kappa_{g}=\frac{2g_{s}^{2}}{(\beta^{2}c_{\theta}^{2}-1)^{2}}\frac{N^{2}(\beta^{2}c_{\theta}^{2}+1)-2}{N(N^{2}-1)},$
(64)

where $N=3$ is the number of colors, and $c_{\theta}=\cos\theta$ is the
scattering angle of the top quark in the $t\bar{t}$ center-of-mass frame; see
Fig. 2. The spin configurations of $t\bar{t}$ listed in Table 1 are
illustrated in Fig. 3 and Fig. 4, where the $g_{L}g_{R}\to t\bar{t}$ process
is $\beta^{3}$-suppressed near threshold and therefore omitted in Fig. 3(a).
The $t\bar{t}$ spins in Fig. 4(a) have no preferred orientation.

QCD production | $\overline{\sum}|\mathcal{M}|^{2}$ | spin correlation matrix $C_{ij}$ | $\xi$
---|---|---|---
$q\bar{q}\to t\bar{t}$ | $\kappa_{q}\left(2-\beta^{2}s_{\theta}^{2}\right)$ | $\begin{pmatrix}\frac{(2-\beta^{2})s_{\theta}^{2}}{2-\beta^{2}s_{\theta}^{2}}&0&-\frac{2c_{\theta}s_{\theta}\sqrt{1-\beta^{2}}}{2-\beta^{2}s_{\theta}^{2}}\\\ 0&\frac{-\beta^{2}s_{\theta}^{2}}{2-\beta^{2}s_{\theta}^{2}}&0\\\ -\frac{2c_{\theta}s_{\theta}\sqrt{1-\beta^{2}}}{2-\beta^{2}s_{\theta}^{2}}&0&\frac{2c_{\theta}^{2}+\beta^{2}s_{\theta}^{2}}{2-\beta^{2}s_{\theta}^{2}}\\\ \end{pmatrix}$ |  $\tan\xi=\frac{1}{\gamma}\tan\theta$
$g_{L}g_{R}\to t\bar{t}$ | $\kappa_{g}\beta^{2}s_{\theta}^{2}(2-\beta^{2}s_{\theta}^{2})$ |  $\tan\xi=\frac{1}{\gamma}\tan\theta$
$g_{L}g_{L}/g_{R}g_{R}\to t\bar{t}$ | $\kappa_{g}(1-\beta^{4})$ | $\begin{pmatrix}\frac{\beta^{2}-1}{\beta^{2}+1}&0&0\\\ 0&\frac{\beta^{2}-1}{\beta^{2}+1}&0\\\ 0&0&-1\\\ \end{pmatrix}$ | $\xi=0$
Table 1: The squared matrix element, spin correlation matrix $C_{\Omega}$, and
rotation angle to the diagonal basis for $t\bar{t}$ production for the initial
states of $q\bar{q}$, unlike-helicity gluons ($g_{L}g_{R}$), and like-helicity
gluons ($g_{L}g_{L}/g_{R}g_{R}$). The spin correlation matrix is expressed in
the helicity basis ($\oldvec{\,r},\oldvec{\,n},\oldvec{\,k}$). Figure 3: Spin
configurations of $t\bar{t}$ produced from unlike-helicity initial states: (a)
for $q\bar{q}\to t\bar{t}$ near threshold, with the cross section proportional
to $\beta$; and (b) for $q\bar{q},\ g_{L}g_{R}\to t\bar{t}$ in the boosted
region. Figure adapted from Ref. [42]. Figure 4:  Spin configurations of
$t\bar{t}$ produced from like-helicity gluons near and above threshold. The
cross section is proportional to $\beta$. Figure adapted from Ref. [42].

####  V.1.1 $q\bar{q}\to t\bar{t}$ Process

For $q\bar{q}\to t\bar{t}$, the differential value of the spin correlation
matrix in the diagonal basis
$(\oldvec{\,e}_{1}^{\rm\,diag},\oldvec{\,e}_{2}^{\rm\,diag},\oldvec{\,e}_{3}^{\rm\,diag})$
is

$C^{\rm
diag}=\begin{pmatrix}\frac{\beta^{2}s_{\theta}^{2}}{2-\beta^{2}s_{\theta}^{2}}&0&0\\\
0&-\frac{\beta^{2}s_{\theta}^{2}}{2-\beta^{2}s_{\theta}^{2}}&0\\\ 0&0&1\\\
\end{pmatrix}.$ (65)

The spins of the top and anti-top along the third direction
$\oldvec{\,e}_{3}^{\rm\,diag}$ are 100% correlated. This reproduces the well-
known result in Ref. [2]. Such constructed $t\bar{t}$ quantum sub-states
produced from $q\bar{q}\to t\bar{t}$ at any fixed phase space point are
entangled. In particular, in the high-$p_{T}$ region where $\beta\approx 1$
and $s_{\theta}\approx 1$, the spins of the top and the anti-top are maximally
correlated along all three directions, resulting in a maximally entangled
state. The maximal violation of the Bell inequality is given by the largest
two eigenvalues of the spin correlation matrix as seen in Eq. (53).

For $q\bar{q}\to t\bar{t}$, the angle between the diagonal basis and the
helicity basis (see Fig. 2) is $\xi=(\tan\theta)/\gamma$. Near threshold
$\gamma\to 1$ and the rotation angle $\xi$ is approximately the same as the
scattering angle $\theta$, which means that the diagonal basis approaches the
rotated beam basis. In the highly-boosted region, the rotation angle $\xi$
becomes zero and the diagonal basis coincides with the helicity basis. To show
how the optimal (diagonal) basis maximizes the Bell inequality violation, we
average the quantum sub-states over all of phase space, and compare the Bell
inequality violation of the states averaged in different bases.

We present our numerical results in Fig. 5 in the rotated beam basis, the
helicity basis, and the diagonal basis. We recall that Bell inequality
violation is inferred when $\mathscr{B}>2$. We see that the optimal (diagonal)
basis usually leads to a considerable increase of Bell inequality violation,
reaching ${\mathscr{B}}\approx 2.24$ far above threshold. In the boosted
region or near threshold, the commonly-used helicity basis or rotated beam
basis, respectively, are already nearly optimal. We also evaluated the density
matrix in the fixed beam basis with the azimuthal angle averaged, but the
effects for Bell inequality violation are hardly visible in this basis, as
explained in detail in App. D.

Figure 5: Bell inequality violation of angular-averaged states as a function
of center-of-mass energy. Bell inequality violation occurs when
$\mathscr{B}[\rho]>2$.

####  V.1.2 $gg\to t\bar{t}$ Process

The process $gg\to t\bar{t}$ can be divided into two sub-processes: unlike-
helicity gluon scattering ($g_{L}g_{R}$, $P$-wave, with angular momentum
$|S_{z}|=2$) and like-helicity gluon scattering ($g_{L}g_{L},g_{R}g_{R}$,
$S$-wave, with angular momentum $|S_{z}|=0$), as shown in Table 1, Fig. 3, and
Fig. 4.

The top pair produced from unlike-helicity gluons has exactly the same spin
correlation as in the $q\bar{q}\to t\bar{t}$ process. Consequently, the $t$
and $\bar{t}$ usually have like-spins, e.g., with $\oldvec{\,k}$ as spin
quantization axis, the $t\bar{t}$ spin state is
$(\ket{\uparrow\uparrow}+\ket{\downarrow\downarrow})/\sqrt{2}$, in the
high-$p_{T}$ region. The $t\bar{t}$ produced from like-helicity gluons, on the
other hand, usually have unlike-spins. These like-helicity gluons produce the
$t$ and $\bar{t}$ in a spin singlet state
$(\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow})/\sqrt{2}$. This is a
maximally-entangled configuration, especially near threshold.

While the top pair produced from either like-helicity or unlike-helicity
gluons are entangled as spin singlet or triplet configuration, the complete
process of $gg\to t\bar{t}$ with unpolarized gluons does not always result in
an entangled $t\bar{t}$ system. In general, the top pair is produced in a
mixed state with a spin correlation of

$C_{gg\to
t\bar{t}}=\frac{|\mathcal{M}_{\text{unlike}}|^{2}C_{\text{unlike}}+|\mathcal{M}_{\text{like}}|^{2}C_{\text{like}}}{|\mathcal{M}_{\text{unlike}}|^{2}+|\mathcal{M}_{\text{like}}|^{2}}\
.$ (66)

The squared amplitudes $|\mathcal{M}_{\text{(un)like}}|$ and spin correlation
matrices $C_{\text{(un)like}}$ for like-helicity and unlike-helicity gluon
scattering are listed in Table 1. Comparing the squared amplitudes, we find
that like-helicity gluon scattering dominates when $\beta\ll 1$ near the
production threshold, while unlike-helicity gluon scattering dominates when
$\gamma s_{\theta}\gg 1$, far above threshold. Consequently, top pairs
produced from gluon scattering are entangled in these two separated kinematic
regions: near threshold and high-$p_{T}$ region.

We present numerical results for Bell inequalities at the partonic level for
$gg\to t\bar{t}$ in Fig. 6 for various basis choices. Near threshold, the top
pair produced from like-helicity gluon scattering approximately forms a spin
singlet. The correlation matrix is proportional to the identity matrix at
leading order, $C_{ij}={\rm diag}(-1,-1,-1)+\mathcal{O}(\beta^{2})$, which
leads to maximal Bell inequality violation reaching ${\cal B}=2\sqrt{2}$ when
$\beta\to 0$, as seen in Fig. 6(a). The entanglement of $t\bar{t}$ produced
from $gg$ is not sensitive to the basis choice in this case. We note that
$gg\to t\bar{t}$ is significantly more sensitive than $q\bar{q}\to t\bar{t}$
in measuring entanglement and Bell inequality violation near threshold.

In the high-$p_{T}$ region where $\gamma s_{\theta}\gg 1$ ($\gamma\gg 1$ and
$c_{\theta}\approx 0$), the top pair produced from unlike-helicity gluon
scattering forms a spin triplet, and the spin correlation matrix is
$C_{ij}={\rm
diag}(1,-1,1)+\mathcal{O}(c_{\theta}^{2})+\mathcal{O}(\gamma^{-2})$. The spin
correlation matrix is approximately invariant under any rotation in the
scattering plane. After fixing one quantization axis
$\oldvec{\,e}_{2}=\oldvec{\,n}$, the entanglement is not sensitive to the
choice of the other two axes. As a result, the fictitious states obtained in
both the helicity basis and the rotated beam basis both yield nearly the
optimal signal of Bell inequality violation, reaching ${\cal B}\approx 2.6$ at
large $\gamma$, as seen in Fig. 6(b). Similar to $q\bar{q}\to t\bar{t}$, in
the high-$p_{T}$ region, no Bell inequality violation can be seen if the
density matrix is averaged over the azimuthal angle in the fixed beam basis.

It is important to note that the like-helicity and unlike-helicity gluons
contribute oppositely to the spin correlation of the top pair, i.e., top pairs
produced from like-helicity gluons ($g_{L}g_{L},g_{R}g_{R}$) tend to have
like-spins while top pairs produced from unlike-helicity gluons ($g_{L}g_{R}$)
tend to have unlike-spins; see Figs. 3 and 4 for illustration. As a result, in
the phase space region where the contribution like-helicity and unlike-
helicity gluons are comparable, the spin correlation of $t\bar{t}$ is largely
canceled and there is no entanglement in $t\bar{t}$ system. Similarly, the
spin correlation of $t\bar{t}$ produced from $q\bar{q}$ and like-helicity
gluons also cancel with each other.

Figure 6: Bell inequality violation of $gg\to t\bar{t}$. (a) Near threshold;
the density matrix is angular-averaged. (b) High-$p_{T}$ region with a
selection cut $|\cos\theta|<0.5$ imposed.

###  V.2 Optimization Procedure at the LHC

At hadron colliders, in reality, the $t\bar{t}$ quantum state is a mixed state
due to contributions from both $q\bar{q}\to t\bar{t}$ and $gg\to t\bar{t}$

$\rho^{t\bar{t}}=\omega^{q\bar{q}}\rho^{q\bar{q}\to
t\bar{t}}+\omega^{gg}\rho^{gg\to t\bar{t}}.$ (67)

At a given center-of-mass energy $\sqrt{\hat{s}}$, the weights
$\omega^{q\bar{q}}$ and $\omega^{gg}$ are determined by the parton
luminosities and production rates

$\omega^{I}=\frac{L_{I}|\mathcal{M}_{I\to
t\bar{t}}|^{2}}{L_{q\bar{q}}|\mathcal{M}_{q\bar{q}\to
t\bar{t}}|^{2}+L_{gg}|\mathcal{M}_{gg\to t\bar{t}}|^{2}},\quad\quad\quad
I=q\bar{q},\ gg.$ (68)

Taking the spin correlation matrices from Table 1 into account, we obtain the
diagonal basis for realistic hadronic processes

$\tan(2\xi)=\frac{(L_{qq}\kappa_{q}+L_{gg}\kappa_{g}\beta^{2}s_{\theta}^{2})s_{2\theta}\sqrt{1-\beta^{2}}}{(L_{qq}\kappa_{q}+L_{gg}\kappa_{g}\beta^{2}s_{\theta}^{2})(c_{2\theta}+\beta^{2}s_{\theta}^{2})+L_{gg}\kappa_{g}\beta^{2}(1-\beta^{2})},$
(69)

where $\kappa_{q}$ and $\kappa_{g}$ are the QCD factors defined in Eq. (64).
This result generalizes the basis of the $gg\to t\bar{t}$ process obtained in
Refs. [41, 42]. In Fig. 7, we show the relation between the angles $\xi$ and
$\theta$ at the $13~{}\operatorname{TeV}$ LHC for various values of the
$t\bar{t}$ invariant mass. Near threshold at low $m_{tt}$ and $\beta\to 0$,
$\tan(2\xi)\approx\tan(2\theta)$; thus the rotated beam basis aligns with the
optimal (diagonal) basis. Far above threshold when $\beta\to 1$,
$\tan(2\xi)\to 0$ and the helicity basis aligns with the optimal (diagonal)
basis.

Figure 7: The rotation angle $\xi$ that obtains the optimal (diagonal) basis
as a function of scattering angle $\theta$ for different values of
$m_{t\bar{t}}$.

At the LHC energies, $t\bar{t}$ production is dominated by gluon fusion. Both
the near-threshold and the high-$p_{T}$ regions contribute uniquely to the
quantum information properties. Near threshold, top pairs produced from gluon
fusion are entangled in a spin singlet configuration. However, the
contribution of $q\bar{q}$ can have a sizable effect on the Bell inequality
violation of $t\bar{t}$. As illustrated in Figs. 3 and 4, the spin correlation
of a $t\bar{t}$ pair produced from unlike-helicity gluons and $q\bar{q}$ have
opposite signs and cancel with each other. This cancellation greatly reduces
the signal of entanglement and makes measuring Bell inequality violation near
threshold challenging at the LHC. We show the Bell variable both near the
threshold versus a upper cut on the speed $\beta$, in Fig. 8(a), and above
threshold versus a lower cut on $\gamma$, in Fig. 8(b), which are equivalently
to an upper and a lower bound on the invariant mass $m_{t\bar{t}}$ (labeled by
the upper horizontal axes), respectively. We calculate the parton luminosity
with CT18 [54] and demonstrate the results in Fig. 8(a), where the Bell
inequality violation decreases to about 2.11 at the energy of the LHC even
though the gluon fusion still dominates $t\bar{t}$ production
($L_{q\bar{q}}/L_{gg}\approx 0.11$). For illustration, we also calculate the
expectation at a future 100 TeV hadron collider (FCC-hh), where we achieve a
result of about 2.55 near threshold because the $q\bar{q}$ contribution is
further suppressed relative to the LHC ($L_{q\bar{q}}/L_{gg}\approx 0.04$). It
is also important to note that the optimal (diagonal) basis approaches the
rotated beam basis near threshold. In particular, when using the Bell variable
$\mathcal{B}^{\prime}$ defined in Eq. (56), the violation of Bell inequalities
near threshold is only measurable in the beam basis but not in the helicity
basis. This explains the simulation results in Ref. [9].

Figure 8: Bell inequality violation $\mathcal{B}^{\prime}$ of $pp\to
t\bar{t}$. (a) Results near threshold versus upper cut of $m_{t\bar{t}}$; the
density matrix is fully angular-averaged. Note that
$\mathcal{B}^{\prime}[\bar{\rho}^{\rm hel}]<2$ for the state averaged in
helicity basis at LHC. (b) Result in high $p_{T}$ region versus lower cut of
$m_{t\bar{t}}$; an angular selection cut $|\cos\theta|<0.5$ is chosen.

In the high-$p_{T}$ region, top pairs produced at hadron colliders are
entangled as the $t\bar{t}$ are in a spin triplet configuration. Both quark
initial states and unlike-helicity gluon initial state contribute to the spin
triplet, while the like-helicity gluon scattering contaminates the spin
correlation as summarized at the end of Sec. V.1. Therefore, more $q\bar{q}\to
t\bar{t}$ contribution makes the Bell inequality violation larger, but
generally the result is not highly sensitive to $q\bar{q}$ contribution as the
contribution from like-helicity gluon scattering is small at high-$p_{T}$
region; see Fig. 8(b).

###  V.3 Optimization Procedure at Lepton Colliders

At a lepton collider such as an $e^{+}e^{-}$ collider, top pair production not
far above the threshold proceeds through the $s$-channel process
$e^{+}e^{-}\to t\bar{t}$ via the $\gamma/Z$ exchange. This process generally
produces the same spin correlations as the process $q\bar{q}\to t\bar{t}$. The
diagonal basis was first explored in $e^{+}e^{-}\to t\bar{t}$ in Ref. [49],
where it was called the “off-diagonal basis.”

The difference between electroweak processes and QCD processes is that the $Z$
boson couplings to fermions are chiral so the electroweak process
$e^{+}e^{-}\to t\bar{t}$ always results in a certain degree of polarization
even if the incoming beams are unpolarized.

Including both the vector and axial current interactions, a general
$s$-channel amplitude can be parametrized as

$\mathcal{M}_{t\bar{t}}=\frac{e^{2}g_{\mu\nu}}{s}J_{\rm in,\pm
1}^{\mu}\left(f_{V}J_{{\rm out},V}^{\nu}+f_{A}J_{{\rm out},A}^{\nu}\right).$
(70)

Here, $J^{\mu}_{\rm in,\pm 1}=\frac{\sqrt{\hat{s}}}{2}(0,1,\pm i,0)$ is the
current of initial states, $J^{\mu}_{{\rm out},V}=\bar{t}\gamma^{\mu}t$ and
$J^{\mu}_{{\rm out},A}=\bar{t}\gamma^{\mu}\gamma^{5}t$ are vector and axial
currents of the final states, respectively. $f_{V/A}$ parametrizes the
interaction strength of the vector/axial-vector current. More specifically,
the amplitudes of $e^{+}e^{-}\to t\bar{t}$ are written as

$\displaystyle\mathcal{M}_{e^{-}_{L}e^{+}_{R}\to t\bar{t}}$
$\displaystyle=\frac{e^{2}g_{\mu\nu}}{s}J_{\rm
in,-1}^{\mu}\left(f^{L}_{V}J_{{\rm out},V}^{\nu}+f^{L}_{A}J_{{\rm
out},A}^{\nu}\right),$ (71) $\displaystyle\mathcal{M}_{e^{-}_{R}e^{+}_{L}\to
t\bar{t}}$ $\displaystyle=\frac{e^{2}g_{\mu\nu}}{s}J_{\rm
in,1}^{\mu}\left(f^{R}_{V}J_{{\rm out},V}^{\nu}+f^{R}_{A}J_{{\rm
out},A}^{\nu}\right),$ (72)

and the coupling strengths are

$\displaystyle f_{V}^{L}$
$\displaystyle=Q_{e}Q_{t}+\frac{(I^{3}_{e}-Q_{e}s_{W}^{2})(I_{t}^{3}-2Q_{t}s_{W}^{2})}{2c_{W}^{2}s_{W}^{2}}\frac{\hat{s}}{\hat{s}-m_{Z}^{2}},$
(73) $\displaystyle f_{A}^{L}$
$\displaystyle=-\frac{(I^{3}_{e}-Q_{e}s_{W}^{2})I_{t}^{3}}{2c_{W}^{2}s_{W}^{2}}\frac{\hat{s}}{\hat{s}-m_{Z}^{2}},$
(74) $\displaystyle f_{V}^{R}$
$\displaystyle=Q_{e}Q_{t}-\frac{Q_{e}(I_{t}^{3}-2Q_{t}s_{W}^{2})}{2c_{W}^{2}}\frac{\hat{s}}{\hat{s}-m_{Z}^{2}},$
(75) $\displaystyle f_{A}^{R}$
$\displaystyle=\frac{Q_{e}I_{t}^{3}}{2c_{W}^{2}}\frac{\hat{s}}{\hat{s}-m_{Z}^{2}},$
(76)

where $Q_{e}=-1$, $Q_{t}=2/3$, $I_{e}^{3}=-1/2$, and $I_{t}^{3}=1/2$. The
results above can be directly generalized to the pair production of any
fermion. For instance, for the $\tau^{-}\tau^{+}$ system, we simply replace
$Q_{t}$ with $Q_{\tau}=-1$ and $I_{t}^{3}$ with $I^{3}_{\tau}=-1/2$.

Since the tops are produced with some polarization, the $B_{i}^{\pm}$’s in Eq.
(10) are non-zero, and the concurrence cannot be simply expressed by the
eigenvalue of $C_{ij}$ as in section IV.1. It is thus important to measure all
the parameters $B_{i}^{\pm}$ and $C_{ij}$, to achieve quantum tomography for
the $t\bar{t}$ bipartite state. On the other hand, Bell’s inequality only
depends on the correlation matrix $C_{ij}$ and our discussion in Sec. IV.2 is
still valid.

For the $e^{-}_{R}e^{+}_{L}\to t\bar{t}$ and $e^{-}_{L}e^{+}_{R}\to t\bar{t}$
processes, respectively, the spin correlation matrix of $t\bar{t}$ is given in
terms of the chiral couplings and the scattering angle $\theta$ in the center-
of-mass frame in the helicity basis as

$\displaystyle
C_{ij}=\frac{1}{f_{V}^{2}(2-\beta^{2}s_{\theta}^{2})+f_{A}^{2}\beta^{2}(1+c_{\theta}^{2})\pm
4f_{V}f_{A}\beta c_{\theta}}\times$ (77)
$\displaystyle\begin{pmatrix}s_{\theta}^{2}\left(f_{V}^{2}(2-\beta^{2})-f_{A}^{2}\beta^{2}\right)&0&-2(f_{V}^{2}c_{\theta}\pm
f_{V}f_{A}\beta)s_{\theta}\sqrt{1-\beta^{2}}\\\
0&(f_{A}^{2}-f_{V}^{2})\beta^{2}s_{\theta}^{2}&0\\\ -2(f_{V}^{2}c_{\theta}\pm
f_{V}f_{A}\beta)s_{\theta}\sqrt{1-\beta^{2}}&0&f_{V}^{2}(2c_{\theta}^{2}+\beta^{2}s_{\theta}^{2})+f_{A}^{2}\beta^{2}(1+c_{\theta}^{2})\pm
4f_{V}f_{A}\beta\end{pmatrix}.$

The eigenvalues $\mu_{i}$ of the spin correlation matrix are

$\mu_{1}=-\mu_{2}=\frac{\beta^{2}s_{\theta}^{2}(f_{V}^{2}-f_{A}^{2})}{f_{V}^{2}(2-\beta^{2}s_{\theta}^{2})\pm
4\beta
f_{V}f_{A}c_{\theta}+\beta^{2}f_{A}^{2}(1+c_{\theta}^{2})},\quad\quad\quad\mu_{3}=1,$
(78)

and the maximal value of $|\mu_{1}|$ is reached at an angle
$\cos\theta=\mp\beta f_{A}/f_{V}$. The spin correlation matrices of $t\bar{t}$
produced from $e_{R}^{-}e_{L}^{+}$ and $e_{L}^{-}e_{R}^{+}$ are related by
$f^{R}_{V}\leftrightarrow f^{L}_{V}$ and $f^{R}_{A}\leftrightarrow-f^{L}_{A}$.
These two processes with different electron polarizations lead to similar
$t\bar{t}$ angular distributions and spin correlations [49] because
numerically $f^{R}_{V}/f^{R}_{A}\sim-f^{L}_{V}/f^{L}_{A}$. Additionally, these
ratios have a mild dependence on the collision energy, as shown in Fig. 9(a).
As a result, the quantum entanglement of a top pair produced at $e^{+}e^{-}$
colliders is not very sensitive to electron beam polarization. We show this by
plotting the Bell variable $\mathcal{B}[\rho]$ for quantum sub-states produced
at different scattering angles in Fig. 9(b). On the other hand, the ratio
between vector and axial current coupling strength can be changed by new
physics (such as models with new vector bosons), and the entanglement of top
pairs at $e^{+}e^{-}$ colliders is sensitive to such new physics.

Figure 9:  (a) The ratio between vector and axial coupling strength as a
function of energy. (b) The Bell variable $\mathcal{B}[\rho]$ as a function of
$\cos\theta$ at 1 TeV $e^{+}e^{-}\to t\bar{t}$ collision.

Figure 10: (a) The rotation angle $\xi$ that obtains the optimal (diagonal)
basis as a function of scattering angle $\theta$ for different values of
$\sqrt{\hat{s}}$. (b) Bell inequality violation of fictitious states averaged
in different bases. A selection cut of $\cos\theta<0$ is used to choose the
mostly entangled phase space.

Figure 11: Significance of measuring $\mathscr{B}(\rho)>2$ at 1 TeV
$e^{+}e^{-}$ colliders. (a) Different basis choices for unpolarized
scattering. (b) Optimal (diagonal) basis for different beam polarizations with
the dashed curve for $(e^{+},e^{-}=+80\%,-80\%)$ and dotted curve for
$(e^{+},e^{-}=-80\%,+80\%)$, where the sign $+(-)$ refers to right-handed
(left-handed) beam polarization.

The rotation angle to the diagonal basis is

$\tan\xi=\frac{1}{\gamma}\frac{f_{V}s_{\theta}}{f_{V}c_{\theta}\pm
f_{A}\beta},$ (79)

for $e^{-}_{R}e^{+}_{L}$ and $e^{-}_{L}e^{+}_{R}$ collisions, respectively,
which reproduces the maximally-correlated basis studied in Ref. [49]. We plot
the angle relations in Fig. 10(a) in the $\xi$-$\theta$ plane. Similar to
$q\bar{q}\to t\bar{t}$, the helicity basis or the rotated beam basis aligns
with the optimal (diagonal) basis in the boosted region or near threshold,
respectively. The difference for the electroweak processes is that the spin
correlation of $t\bar{t}$ produced from $e^{+}e^{-}$ annihilation is not
symmetric under $\theta\to\pi-\theta$, and the $t\bar{t}$ are mainly entangled
in the backward region with $\cos\theta<0$; see Fig. 9(b). Figure 10(b) shows
the Bell inequality violation of the state in angular averaged in different
bases versus the boost factor. We see that the optimal (diagonal) basis also
leads to a considerable increase in Bell inequality violation in the boosted
region. To provide a quantitative measurement of the Bell inequality
violation, we define a “significance” as the ratio of the deviation and its
error

$\mathcal{S}=\frac{\mathcal{B}-2}{\delta\mathcal{B}},$ (80)

where the uncertainty $\delta\mathcal{B}$ is estimated from the statistical
uncertainty of correlation matrix $-$ see e.g. Ref. [29] for detail. We found
that the optimal basis choice greatly increases the significance of measuring
Bell inequality violation. We illustrate this for a 1 TeV lepton collider as
shown in Fig. 11(a). Although the $t\bar{t}$ spin correlation is not sensitive
to initial state beam polarization, it is more advantageous to measure Bell
inequality violation with a left-hand polarized electron beam and right-hand
polarized positron, due to a larger production cross section, as shown in Fig.
11(b).

The $\gamma\gamma\to t\bar{t}$ process becomes comparable to $q\bar{q}\to
t\bar{t}$ around 4 TeV [55], and can be dominant at higher energies. As shown
in Appendix E, this process produces exactly the same spin state and
correlations as the $gg\to t\bar{t}$ process after summing over the color
indices, even though their interactions and Feynman diagrams differ.
Therefore, the optimal basis choice for $\gamma\gamma\to t\bar{t}$ process is
exactly the same as the basis of $gg\to t\bar{t}$ process described in Sec.
V.1, and the Bell inequality violation of $\gamma\gamma\to t\bar{t}$ produces
nearly the same results as in Fig. 6. Unlike at the LHC where one cannot
separate the $t\bar{t}$ events produced from $q\bar{q}$ and $gg$, another
advantage of lepton colliders is that the annihilation process $e^{+}e^{-}\to
t\bar{t}$ and the fusion process $\gamma\gamma\to t\bar{t}$ can be effectively
separated by the invariant mass $m_{t\bar{t}}$, making it possible to measure
the spin singlet Bell state at high-energy lepton colliders.

##  VI Conclusion and Discussion

Studying the entanglement and Bell inequality violation of $t\bar{t}$ requires
interpreting $t\bar{t}$ as a bipartite quantum state. For each physical
quantum state, quantum entanglement is an intrinsic property and intuitively
does not depend on the quantization axis of spin measurements. In realistic
experiments at colliders, however, $t\bar{t}$ events are analyzed at different
phase space points in different Lorentz frames and coordinates. This situation
means that event-dependent bases are often more useful, however, their use
yields a fictitious state instead of a genuine quantum state. The entanglement
and Bell inequality violation of fictitious state are no longer an intrinsic
property of $t\bar{t}$ produced at colliders, and they depend on the basis
choice of the measurements.

In this work, we point out that the usage of fictitious states is not only
valuable to determine the existence of the entanglement and Bell inequality
violation, but also advantageous because it allows for an optimization
procedure. We showed that the basis-dependence allows us to find an optimal
basis, such that the fictitious state averaged in this basis has the largest
signal of entanglement or Bell inequality violation. By proving that the
diagonal basis maximizes both entanglement and Bell inequality violation of an
angular-averaged state, we gave the optimal basis choice to study the
$t\bar{t}$ system at colliders. We summarize our findings as follows:

  * •

We derived an optimal basis choice on an event-by-event basis for a given
process, and found that it is approximately the same as the fixed beam basis
or the rotated beam basis near the $t\bar{t}$ production threshold, while it
approaches the helicity basis far above threshold.

  * •

For $t\bar{t}$ events produced from purely $s$-channel processes, namely
$e^{+}e^{-}\to t\bar{t}$ and $q\bar{q}\to t\bar{t}$, the entanglement and Bell
inequality violation are very sensitive to the choice of basis. Using the
optimal basis is especially crucial. For instance, at a $1\
\operatorname{TeV}\ e^{+}e^{-}$ collider, the signal of Bell inequality
violation in the optimal basis is 40% larger than the signal of Bell
inequality violation in the helicity basis.

  * •

At high-energy hadron colliders, the top pairs are dominantly produced from
the $gg\to t\bar{t}$ process. The detection of Bell inequality violation at
the LHC is very challenging because of the rather weak signal near the
$t\bar{t}$ threshold and the low event rate far above threshold. The situation
is improved at higher energies such as at the FCC-hh.

  * •

Near the $t\bar{t}$ production threshold at the LHC, using the optimal basis
leads to a considerable increase in the signal of Bell inequality violation.
In the high-$p_{T}$ region at the LHC, e.g. $m_{tt}>1.2$ TeV and
$\cos\theta_{t}<0.5$, the improvement of Bell inequality violation measurement
from the commonly-used helicity basis to the optimal basis is 30%. Realistic
simulations and optimization would be needed to quantify the observability.

We would like to reiterate that, although the presentation in this work is for
the $t\bar{t}$ events, our method and general approach are also applicable to
other systems. Given a quantity calculated from a quantum state, an optimized
fictitious state which maximizes the value of the specified quantity can be
found following our procedure, even including interactions beyond the Standard
Model. This is invaluable as part of the program of quantum tomography for a
vast variety of quantum states produced in collider experiments. We anticipate
that the program will lead us to a deeper understanding of the quantum world.

###### Acknowledgements.

We thank Yoav Afik, Alan Barr, Juan de Nova, and Arthur Wu for discussions.
This work was supported in part by the U.S. Department of Energy under grant
No. DE-SC0007914 and in part by the Pitt PACC. TH would like to thank the
Aspen Center for Physics, where part of this work is complete, which is
supported by the National Science Foundation (NSF) grant PHY-1607611. ML is
also supported by the National Science Foundation under grant No. PHY-2112829.
KC is supported in part by the National Science Foundation of China under
grant No. 12235001.

##  Appendix A Quantum States at Colliders and Basis Transformations

###  A.1 Physical Spin State

In this appendix, we define a physical spin density matrix of quantum states
at colliders and discuss basis transformations of the spin density matrix. We
consider the process $I\to t\bar{t}$ where the initial state $I$ is
$e^{+}e^{-}$, $q\bar{q}$, or $gg$. In the center-of-mass frame of $t\bar{t}$,
the momentum of $t$ and $\bar{t}$ are

$p_{t}^{\mu}=(E,|p|\mathbf{k}),\quad\quad\quad
p_{\bar{t}}^{\mu}=(E,-|p|\mathbf{k}),$ (81)

respectively, where
$\mathbf{k}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ is the
normalized direction of the momentum of $t$.

The total quantum state $\ket{t\bar{t}}$ from the process $I\to t\bar{t}$ can
be expanded in momentum and spin (and color) eigenstates
$\ket{\mathbf{k},\alpha\bar{\alpha}}$. Here, the bipartite particle state
$\ket{\mathbf{k},\alpha\bar{\alpha}}$ is defined as
$\ket{\mathbf{k},\alpha\bar{\alpha}}=\ket{\mathbf{k},\alpha}\otimes\ket{-\mathbf{k},\bar{\alpha}}$,
where $\ket{\mathbf{k},\alpha}$ denotes the top quark state with spin $\alpha$
and momentum $\mathbf{k}$, and $\ket{-\mathbf{k},\bar{\alpha}}$ denotes the
anti-top quark state with spin $\bar{\alpha}$ and momentum $-\mathbf{k}$. The
quantum state $\ket{t\bar{t}}$ expanded in terms of
$\ket{\mathbf{k},\alpha\bar{\alpha}}$ is

$\ket{t\bar{t}}\propto\int\operatorname{d\mathbf{k}}\sum_{\alpha\bar{\alpha}}\ket{\mathbf{k},\alpha\bar{\alpha}}\bra{\mathbf{k},\alpha\bar{\alpha}}T\ket{I,\lambda}=\int\operatorname{d\mathbf{k}}\sum_{\alpha\bar{\alpha}}\mathcal{M}^{\lambda}_{\alpha\bar{\alpha}}(\mathbf{k})\ket{\mathbf{k},\alpha\bar{\alpha}},$
(82)

where $\lambda$ denotes the quantum number of the initial state, $T$ is the
transition matrix and $\mathcal{M}^{\lambda}_{\alpha\bar{\alpha}}(\mathbf{k})$
is the corresponding amplitude. Typically, the initial state is a mixed state,
i.e.,
$\rho^{I}=\sum_{\lambda,\lambda^{\prime}}\rho^{I}_{\lambda\lambda^{\prime}}\ket{I,\lambda}\bra{I,\lambda^{\prime}}$.

The production density matrix is

$\hat{\rho}^{tt}\propto\int\operatorname{d\mathbf{k}}\operatorname{d\mathbf{k}}^{\prime}\sum_{\alpha\bar{\alpha}}\sum_{\alpha^{\prime}\bar{\alpha}^{\prime}}\mathcal{M}^{\lambda}_{\alpha\bar{\alpha}}(\mathbf{k})\rho^{I}_{\lambda\lambda^{\prime}}(\mathcal{M}^{\lambda^{\prime}}_{\alpha^{\prime}\bar{\alpha}^{\prime}}(\mathbf{k}^{\prime}))^{*}\ket{\mathbf{k},\alpha\bar{\alpha}}\bra{\mathbf{k}^{\prime},\alpha^{\prime}\bar{\alpha}^{\prime}},$
(83)

and this complete density matrix is defined in spin space, color space,666The
color index is omitted here for simplicity. and momentum space, but we only
focus on the entanglement in spin space. To obtain a spin density matrix, we
first need to trace over the state in the color space, then we can either
project the state into an eigenstate in the momentum space or trace the state
over the momentum space.

By projecting the quantum state into a momentum eigenstate, i.e., selecting
the $t\bar{t}$ with fixed scattering angle, we obtain a quantum sub-state for
$t\bar{t}$ produced at a fixed phase space point

$\hat{\rho}(\mathbf{k})=\bra{\mathbf{k}}\hat{\rho}\ket{\mathbf{k}}\\\
=\rho(\mathbf{k})_{\alpha\alpha^{\prime},\bar{\alpha}\bar{\alpha}^{\prime}}\ket{\alpha\bar{\alpha}}\bra{\alpha^{\prime}\bar{\alpha}^{\prime}},$
(84)

where
$\rho(\mathbf{k})_{\alpha\bar{\alpha},\alpha^{\prime}\bar{\alpha}^{\prime}}$
is the density matrix of the state $\hat{\rho}(\mathbf{k})$ expressed in the
basis $\ket{\alpha\bar{\alpha}}$.

By tracing the quantum state over the momentum space, we obtain the angular
average of $\rho(\mathbf{k})$ in a phase space region $\Pi$,

$\displaystyle\hat{\rho}_{\Pi}$
$\displaystyle=\operatorname{Tr}_{\mathbf{k}\in\Pi}\Big{(}\hat{\rho}\ket{\mathbf{k}}\bra{\mathbf{k}}\Big{)}$
$\displaystyle=\frac{1}{\sigma_{\Pi}}\int_{\Omega\in\Pi}\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}\rho(\mathbf{k})_{\alpha\alpha^{\prime},\bar{\alpha}\bar{\alpha}^{\prime}}\ket{\alpha\bar{\alpha}}\bra{\alpha^{\prime}\bar{\alpha}^{\prime}}.$
(85)

Here, $\sigma_{\Pi}=\int_{\Omega\in\Pi}\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}$ is the integrated cross section in the
region $\Omega\in\Pi$. To obtain the density matrix for the physical state
$\hat{\rho}_{\Pi}$, we need to average the density matrix
$\rho(\mathbf{k})_{\alpha\bar{\alpha},\alpha^{\prime}\bar{\alpha}^{\prime}}$
of the quantum sub-states in an event-independent basis.

###  A.2 Transformations between Spin Bases

To represent the spin density matrix $\rho$ with a $4\times 4$ matrix
$\rho_{\alpha\alpha^{\prime},\bar{\alpha}\bar{\alpha}^{\prime}}$, we need to
choose a basis for the spin measurements. The bases of spin eigenstates
$\ket{\alpha\bar{\alpha}}=\ket{\uparrow\uparrow},\ket{\uparrow\downarrow},\ket{\downarrow\uparrow},\ket{\downarrow\downarrow}$
can be defined based on the quantization axes $\oldvec{\,e}_{i}$.
Specifically, $\ket{\uparrow}$ or $\ket{\downarrow}$ denote the state with
positive or negative spin along the third direction $\oldvec{\,e}_{3}$ in the
rest frame of (anti-)top quark, and we have

$\begin{split}\oldvec{\,\sigma}\cdot\oldvec{\,e}_{1}&=\ket{\uparrow}\bra{\downarrow}+\ket{\downarrow}\bra{\uparrow},\\\
\oldvec{\,\sigma}\cdot\oldvec{\,e}_{2}&=-i\ket{\uparrow}\bra{\downarrow}+i\ket{\downarrow}\bra{\uparrow},\\\
\oldvec{\,\sigma}\cdot\oldvec{\,e}_{3}&=\ket{\uparrow}\bra{\uparrow}-\ket{\uparrow}\bra{\uparrow},\end{split}$
(86)

with
$\oldvec{\,\sigma}\cdot\oldvec{\,e}_{3}\ket{\uparrow}=\ket{\uparrow},\oldvec{\,\sigma}\cdot\oldvec{\,e}_{3}\ket{\downarrow}=-\ket{\downarrow}$.
A different choice of quantization axis yields the spin eigenstates

$\begin{split}\oldvec{\,\sigma}\cdot\oldvec{\,e}_{1}^{\prime}&=\ket{\uparrow}^{\prime}\bra{\downarrow}^{\prime}+\ket{\downarrow}^{\prime}\bra{\uparrow}^{\prime},\\\
\oldvec{\,\sigma}\cdot\oldvec{\,e}_{2}^{\prime}&=-i\ket{\uparrow}^{\prime}\bra{\downarrow}^{\prime}+i\ket{\downarrow}^{\prime}\bra{\uparrow}^{\prime},\\\
\oldvec{\,\sigma}\cdot\oldvec{\,e}_{3}^{\prime}&=\ket{\uparrow}^{\prime}\bra{\uparrow}^{\prime}-\ket{\uparrow}^{\prime}\bra{\uparrow}^{\prime}.\end{split}$
(87)

Here, two different quantization axis choices $\oldvec{\,e}_{i}$ and
$\oldvec{\,e}_{i}^{\prime}$ can be related by an SO(3) rotation

$\oldvec{\,e}_{i}^{\prime}=R_{ji}\oldvec{\,e}_{j},$ (88)

and two spin bases are related by an SU(2) transformation

$\ket{\beta}^{\prime}=U_{\alpha\beta}\ket{\alpha}.$ (89)

Substituting Eq. (89) into Eq. (87) and we obtain

$R_{ij}=\frac{1}{2}\operatorname{Tr}(U^{\dagger}\sigma_{i}U\sigma_{j}).$ (90)

We can also express the explicit form of $R_{ij}$ and $U_{\alpha\beta}$ in
terms of Euler angles. The spin bases and quantization axes shown in Fig. 2
are related by

$\displaystyle\oldvec{\,e}_{i}^{\rm rotated}$
$\displaystyle=R(0,\phi)_{ji}\oldvec{\,e}_{j}^{\rm fixed},$ (91a)
$\displaystyle\oldvec{\,e}_{i}^{\rm helicity}$
$\displaystyle=R(\theta,\phi)_{ji}\oldvec{\,e}_{j}^{\rm fixed},$ (91b)
$\displaystyle\oldvec{\,e}_{i}^{\rm diag}$
$\displaystyle=R(\theta-\xi,\phi)_{ji}\oldvec{\,e}_{j}^{\rm fixed},$ (91c)

and

$\displaystyle\ket{\alpha\bar{\alpha}}^{\rm rotated}$
$\displaystyle=U(0,\phi)_{\beta\alpha,\bar{\beta}\bar{\alpha}}\ket{\beta\bar{\beta}}^{\rm
fixed},$ (92a) $\displaystyle\ket{\alpha\bar{\alpha}}^{\rm helicity}$
$\displaystyle=U(\theta,\phi)_{\beta\alpha,\bar{\beta},\bar{\alpha}}\ket{\beta\bar{\beta}}^{\rm
fixed},$ (92b) $\displaystyle\ket{\alpha\bar{\alpha}}^{\rm diag}$
$\displaystyle=U(\theta-\xi,\phi)_{\beta\alpha,\bar{\beta}\bar{\alpha}}\ket{\beta\bar{\beta}}^{\rm
fixed},$ (92c)

where $R(\theta,\phi)=e^{-i\theta S_{y}}e^{-i\phi S_{z}}$,
$U_{\beta\alpha,\bar{\beta}\bar{\alpha}}=U_{\beta\alpha}\otimes
U_{\bar{\beta}\bar{\alpha}}$, and
$U(\theta,\phi)\equiv\left(e^{-i\theta\frac{\sigma_{y}}{2}}e^{-i\phi\frac{\sigma_{z}}{2}}\right)\otimes\left(e^{-i\theta\frac{\sigma_{y}}{2}}e^{-i\phi\frac{\sigma_{z}}{2}}\right)$.

##  Appendix B Entanglement of Fictitious States

In this section, we prove that a fictitious state with non-zero entanglement
implies that there exists a quantum sub-state with non-zero entanglement. The
convex sum of quantum sub-states in an event-dependent basis gives a
fictitious state,

$\displaystyle\bar{\rho}_{\alpha\beta,\bar{\alpha}\bar{\beta}}^{\rm fic}$
$\displaystyle=\frac{1}{\sigma_{\Pi}}\int_{\Omega\in\Pi}\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}\rho(\mathbf{k})_{\alpha\beta,\bar{\alpha}\bar{\beta}}.$
(93)

The spin correlation matrix of the fictitious state is also a convex sum of
the spin correlation matrix of quantum sub-states

$\displaystyle\bar{C}^{\rm fic}_{ij}$
$\displaystyle=\frac{1}{\sigma_{\Pi}}\int_{\Omega\in\Pi}\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}C(\mathbf{k})_{ij}.$ (94)

We start from concurrence. From Eq. (6) it is difficult to see the
relationship between the concurrence of quantum sub-states and the concurrence
of their angular average, so we turn to the separability condition. For
bipartite qubit systems, $\mathscr{C}(\rho)=0$ is a necessary and sufficient
condition of separability [48]. A quantum state of $t\bar{t}$ is separable if
it can be written as a convex sum of $\rho^{t}\otimes\rho^{\bar{t}}$ as
follows,

$\rho=\sum_{i}p_{i}\rho^{t}_{i}\otimes\rho^{\bar{t}}_{i}.$ (95)

If all the quantum sub-states $\rho(\mathbf{k})$ are separable, then their
angular average $\bar{\rho}$, a convex sum of $\rho(\mathbf{k})$, is obviously
separable. Therefore, the entanglement in the fictitious states implies the
existence of entanglement in a quantum sub-state.

The violation of Bell inequality is similar. If all the quantum sub-states
satisfy Bell’s inequality Eq. (16) for any four spatial directions
$\oldvec{\,a}_{1},\oldvec{\,a}_{2},\oldvec{\,b}_{1}$, and $\oldvec{\,b}_{2}$,
i.e.,

$\displaystyle\vec{a}_{1}\cdot
C(\mathbf{k})(\oldvec{\,b}_{1}-\vec{b}_{2})+\vec{a}_{2}\cdot
C(\mathbf{k})(\oldvec{\,b}_{1}+\vec{b}_{2})\in[-2,2],$ (96)

then the angular average (convex sum) of Eq. (96) gives

$\displaystyle\frac{1}{\sigma_{\Pi}}\int_{\Omega\in\Pi}\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}\left(\vec{a}_{1}\cdot
C(\mathbf{k})(\oldvec{\,b}_{1}-\vec{b}_{2})+\vec{a}_{2}\cdot
C(\mathbf{k})(\oldvec{\,b}_{1}+\vec{b}_{2})\right)$ $\displaystyle=$
$\displaystyle~{}\vec{a}_{1}\cdot\bar{C}\cdot(\oldvec{\,b}_{1}-\vec{b}_{2})+\vec{a}_{2}\cdot\bar{C}\cdot(\oldvec{\,b}_{1}+\vec{b}_{2})$
(97) $\displaystyle\in$ $\displaystyle~{}[-2,2],$ (98)

and the fictitious state also satisfies Bell’s inequality. Therefore, if the
violation of Bell’s inequality is observed in the fictitious state, there must
be some quantum sub-state violating Bell’s inequality.

##  Appendix C Generalized Proof of the Optimal Basis

We start with an arbitrary event-dependent basis choice. The spin correlation
matrix in this basis is

$\bar{C}_{ij}=\frac{1}{\sigma}\int\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}C_{ij,\Omega}.$ (99)

When $\bar{C}_{ij}$ is not symmetric, the optimal Bell inequality violation is
related to the singular values of $C_{ij}$ rather than the eigenvalues of
$C_{ij}$. Unlike in the main text, in this appendix we use $\bar{c}_{i}$ to
denote the singular value of $\bar{C}_{ij}$,

$(\bar{c}_{1},\bar{c}_{2},\bar{c}_{3})=\operatorname{singular-
value}\left[\bar{C}^{\rm basis}\right].$ (100)

The singular values are all positive. Without loss of generality, we assume
$\bar{c}_{1}\geq\bar{c}_{2}\geq\bar{c}_{3}$.

Similarly, in this appendix we use $\mu_{\Omega}$ to denote the singular
values instead of the eigenvalues of a quantum sub-state. The diagonal basis
is obtained from an event-dependent singular value decomposition:

$C^{\rm
diag}_{\Omega}=R_{\Omega}^{t}C_{\Omega}\left(R^{\bar{t}}_{\Omega}\right)^{T}=\begin{pmatrix}\mu_{1,\Omega}&0&0\\\
0&\mu_{2,\Omega}&0\\\ 0&0&\mu_{3,\Omega}\\\ \end{pmatrix}.$ (101)

In the diagonal basis, the singular values of the angular-averaged spin
correlation matrix are

$\bar{c}^{\rm diag}_{i}=\frac{1}{\sigma}\int\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}\mu_{i,\Omega}\equiv\bar{\mu}_{i}.$ (102)

In Eqs. (46) and (47), we showed that the eigenvalues of a general angular-
averaged spin correlation matrix are bounded by the eigenvalues of
$\bar{C}^{\rm diag}$. Similarly, the singular values of the angular-averaged
spin correlation matrix $\bar{C}$ in a general basis, namely $\bar{c}_{i}$,
satisfy the following relations

$\displaystyle\bar{c}_{i}\leq\bar{\mu}_{1},$ (103)
$\displaystyle\bar{c}_{1}+\bar{c}_{2}\leq\bar{\mu}_{1}+\bar{\mu}_{2}.$ (104)

Eq. (103) holds because an entry of a matrix must be smaller than its largest
singular value. To prove Eq. (104), we start with two lemmas:

Lemma 1: The trace of any real matrix $\mathbf{C}$ is smaller than the sum of
its singular values.
Proof: The singular value decomposition is $C^{\rm
diag}=R_{l}CR_{r}^{T}=\operatorname{diag}(\mu_{1},\cdots\mu_{n})$, so
$\operatorname{Tr}(C)=\operatorname{Tr}(R_{l}^{T}C^{\rm
diag}R_{r})=\operatorname{Tr}(R^{lr}C^{\rm
diag})=\sum_{i=1}^{n}R^{lr}_{ii}\mu_{i}$, where $R^{lr}\equiv
R_{r}^{T}R_{l}^{T}$. Because $R^{lr}$ is an orthogonal rotation, all of its
elements are smaller than one, so
$\operatorname{Tr}(C)=\sum_{i=1}^{n}R^{lr}_{ii}\mu_{i}\leq\sum_{i=1}^{n}\mu_{i}$

Lemma 2: Consider a general matrix $\mathbf{A}$ and the first $2\times 2$
submatrix $\mathbf{A}_{2\times 2}$ of it, the sum of its largest two singular
values is always larger than or equal to the sum of the singular values of
$\mathbf{A}_{2\times 2}$.
Proof: see Ref. [56].

Next, we denote three left-singular (right-singular) vectors of $\bar{C}$ as
$\oldvec{\,u}_{i}$ ($\oldvec{\,v}_{i}$). The corresponding singular value is
then

$\displaystyle\bar{c}_{i}=\oldvec{\,u}_{i}\cdot\overline{C}\cdot\oldvec{\,v}_{i}=\frac{1}{\sigma}\int\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}\left(\oldvec{\,u}_{i}\cdot
C_{\Omega}\cdot\oldvec{\,v}_{i}\right)\equiv\frac{1}{\sigma}\int\operatorname{d\Omega}\frac{\rm
d\sigma}{\operatorname{d\Omega}}c_{i,\Omega},$ (105)

where we have defined $c_{i,\Omega}=\oldvec{\,u}_{i}\cdot
C_{\Omega}\cdot\oldvec{\,v}_{i}$, so that $\bar{c}_{i}$ is a convex sum of
$c_{i,\Omega}$. In addition, we define another matrix $C^{\prime}_{\Omega}$ as

$C^{\prime}_{ij,\Omega}=\oldvec{\,u}_{i}\cdot
C_{\Omega}\cdot\oldvec{\,v}_{j}.$ (106)

The three singular value of $C^{\prime}_{\Omega}$ are $\mu_{i,\Omega}$, and
the trace of first $2\times 2$ submatrix of $C^{\prime}_{\Omega}$ is
$c_{1,\Omega}+c_{2,\Omega}$. From the above two lemmas, we find that
$c_{1,\Omega}+c_{2,\Omega}\leq\mu_{1,\Omega}+\mu_{2,\Omega}$, which shows Eq.
(104).

With Eqs. (103) and (104), we are ready to show that
$\bar{c}_{1}^{2}+\bar{c}_{2}^{2}<\bar{\mu}_{1}^{2}+\bar{\mu}_{2}^{2}$. The
proof is nearly the same with Case (a) in Sec. IV.2.

If $0\leq\bar{c}_{1}\leq\bar{\mu}_{2}$, we have
$\bar{c}_{1}^{2}+\bar{c}_{2}^{2}\leq
2\bar{\mu}_{2}^{2}\leq\bar{\mu}_{1}^{2}+\bar{\mu}_{2}^{2}$ directly.

If $\bar{\mu}_{2}\leq\bar{c}_{1}\leq\bar{\mu}_{1}$, we first have

$\displaystyle-\frac{\bar{\mu}_{1}-\bar{\mu}_{2}}{2}\leq\frac{\bar{\mu}_{1}+\bar{\mu}_{2}}{2}-\bar{c}_{1}\leq$
$\displaystyle\frac{\bar{\mu}_{1}-\bar{\mu}_{2}}{2}.$ (107)

Then, according to $\bar{c}_{1}+\bar{c}_{2}\leq\bar{\mu}_{1}+\bar{\mu}_{2}$,
we have,

$\bar{c}_{1}^{2}+\bar{c}_{2}^{2}\leq\bar{c}_{1}^{2}+\left(\bar{\mu}_{1}+\bar{\mu}_{2}-\bar{c}_{1}\right)^{2}.$
(108)

Next, we define the same function as in Sec. IV.2:

$\displaystyle f(\Delta)$
$\displaystyle=\left(\frac{\bar{\mu}_{1}+\bar{\mu}_{2}}{2}+\Delta\right)^{2}+\left(\frac{\bar{\mu}_{1}+\bar{\mu}_{2}}{2}-\Delta\right)^{2},$
$\displaystyle=\frac{(\bar{\mu}_{1}+\bar{\mu}_{2})^{2}}{2}+2\Delta^{2},$ (109)

which satisfies $f(\Delta_{1})\leq f(\Delta_{2})$ for
$|\Delta_{1}|\leq|\Delta_{2}|$. Then $\bar{\mu}_{1}^{2}+\bar{\mu}_{2}^{2}$ and
the right-hand side of Eq. (108) and can be rewritten using $f(\Delta)$:

$\displaystyle\bar{c}_{i}^{2}+\left(\bar{\mu}_{1}+\bar{\mu}_{2}-\bar{c}_{i}\right)^{2}$
$\displaystyle=f\left(\frac{\bar{\mu}_{1}+\bar{\mu}_{2}}{2}-\bar{c}_{i}\right),$
(110) $\displaystyle\bar{\mu}_{1}^{2}+\bar{\mu}_{2}^{2}$
$\displaystyle=f\left(\frac{\bar{\mu}_{1}-\bar{\mu}_{2}}{2}\right).$ (111)

From Eq. (107), the right-hand side of Eq. (110) is smaller than the right-
hand side of Eq. (111), and it follows that
$\bar{c}_{i}^{2}+\left(\bar{\mu}_{1}+\bar{\mu}_{2}-\bar{c}_{i}\right)^{2}\leq\bar{\mu}_{1}^{2}+\bar{\mu}_{2}^{2}$.
In combination with Eq. (108), we reach our conclusion,
$\bar{c}_{i}^{2}+\bar{c}_{j}^{2}\leq\bar{\mu}_{1}^{2}+\bar{\mu}_{2}^{2}$.

##  Appendix D Angular Momentum Conservation

Unlike with a general density matrix, as parameterized by Eq. (10), the
quantum states produced at a collider are constrained by the angular momentum
conservation of the scattering processes. This leads to some consequences
closely related to the entanglement of $t\bar{t}$ system.

Denote the total spin of the initial states along the beamline as $S_{z}^{I}$,
then the $t\bar{t}$ state produced at the collider must have the following
form

$\ket{t\bar{t}}\propto
e^{iS_{z}^{I}\phi}\left(e^{-i\phi}\mathcal{M}_{\uparrow\uparrow}\ket{\uparrow\uparrow}+\mathcal{M}_{\uparrow\downarrow}\ket{\uparrow\downarrow}+\mathcal{M}_{\downarrow\uparrow}\ket{\downarrow\uparrow}+e^{i\phi}\mathcal{M}_{\downarrow\downarrow}\ket{\downarrow\downarrow}\right),$
(112)

in the fixed beam basis.

At leading order, the polarized amplitudes $\mathcal{M}_{\alpha\bar{\alpha}}$
are real (with respect to a common phase), then the correlation matrix in the
fixed beam basis is

$C^{\rm
fixed}\propto\scalebox{0.8}{\mbox{$\displaystyle\begin{pmatrix}2\left(\mathcal{M}_{\downarrow\downarrow}\mathcal{M}_{\uparrow\uparrow}\cos(2\phi)+\mathcal{M}_{\downarrow\uparrow}\mathcal{M}_{\uparrow\downarrow}\right)&2\mathcal{M}_{\downarrow\downarrow}\mathcal{M}_{\uparrow\uparrow}\sin(2\phi)&2\cos(\phi)\left(\mathcal{M}_{\downarrow\uparrow}\mathcal{M}_{\uparrow\uparrow}-\mathcal{M}_{\downarrow\downarrow}\mathcal{M}_{\uparrow\downarrow}\right)\\\
2\mathcal{M}_{\downarrow\downarrow}\mathcal{M}_{\uparrow\uparrow}\sin(2\phi)&2\mathcal{M}_{\downarrow\uparrow}\mathcal{M}_{\uparrow\downarrow}-2\mathcal{M}_{\downarrow\downarrow}\mathcal{M}_{\uparrow\uparrow}\cos(2\phi)&2\sin(\phi)\left(\mathcal{M}_{\downarrow\uparrow}\mathcal{M}_{\uparrow\uparrow}-\mathcal{M}_{\downarrow\downarrow}\mathcal{M}_{\uparrow\downarrow}\right)\\\
2\cos(\phi)\left(\mathcal{M}_{\uparrow\downarrow}\mathcal{M}_{\uparrow\uparrow}-\mathcal{M}_{\downarrow\downarrow}\mathcal{M}_{\downarrow\uparrow}\right)&2\sin(\phi)\left(\mathcal{M}_{\uparrow\downarrow}\mathcal{M}_{\uparrow\uparrow}-\mathcal{M}_{\downarrow\downarrow}\mathcal{M}_{\downarrow\uparrow}\right)&\mathcal{M}_{\downarrow\downarrow}^{2}-\mathcal{M}_{\downarrow\uparrow}^{2}-\mathcal{M}_{\uparrow\downarrow}^{2}+\mathcal{M}_{\uparrow\uparrow}^{2}\\\
\end{pmatrix}.$}}$ (113)

The spin correlation matrix in the rotated beam basis is

$C^{\rm
rotated}\propto\begin{pmatrix}2\mathcal{M}_{\downarrow\downarrow}\mathcal{M}_{\uparrow\uparrow}+2\mathcal{M}_{\downarrow\uparrow}\mathcal{M}_{\uparrow\downarrow}&0&2\mathcal{M}_{\downarrow\uparrow}\mathcal{M}_{\uparrow\uparrow}-2\mathcal{M}_{\downarrow\downarrow}\mathcal{M}_{\uparrow\downarrow}\\\
0&2\mathcal{M}_{\downarrow\uparrow}\mathcal{M}_{\uparrow\downarrow}-2\mathcal{M}_{\downarrow\downarrow}\mathcal{M}_{\uparrow\uparrow}&0\\\
2\mathcal{M}_{\uparrow\downarrow}\mathcal{M}_{\uparrow\uparrow}-2\mathcal{M}_{\downarrow\downarrow}\mathcal{M}_{\downarrow\uparrow}&0&\mathcal{M}_{\downarrow\downarrow}^{2}-\mathcal{M}_{\downarrow\uparrow}^{2}-\mathcal{M}_{\uparrow\downarrow}^{2}+\mathcal{M}_{\uparrow\uparrow}^{2}\\\
\end{pmatrix}.$ (114)

As a consequence of Eq. (112), the spin correlation matrix of $t\bar{t}$ is
diagonal in the normal direction of the scattering plane.

Another consequence of angular momentum conservation is that the azimuthal
angular average in the fixed beam basis usually leads to a complete
cancellation of entanglement. Assume that we have produced a maximally
entangled state $\ket{\Psi_{i}}$ (i.e., a Bell state) at $\phi=0$,

$\displaystyle\ket{\Psi_{0}}$
$\displaystyle=\frac{\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}}{\sqrt{2}},$
(115) $\displaystyle\ket{\Psi_{1}}$
$\displaystyle=\frac{\ket{\uparrow\uparrow}-\ket{\downarrow\downarrow}}{\sqrt{2}},$
(116) $\displaystyle\ket{\Psi_{2}}$
$\displaystyle=i\frac{\ket{\uparrow\uparrow}+\ket{\downarrow\downarrow}}{\sqrt{2}},$
(117) $\displaystyle\ket{\Psi_{3}}$
$\displaystyle=-\frac{\ket{\uparrow\downarrow}+\ket{\downarrow\uparrow}}{\sqrt{2}}.$
(118)

Here, $\ket{\Psi_{0}}$ is a spin singlet state and the other three comprise a
spin triplet. Then we must have

$\displaystyle\ket{\Psi_{0}(\phi)}$
$\displaystyle=\frac{\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}}{\sqrt{2}},$
(119) $\displaystyle\ket{\Psi_{1}(\phi)}$
$\displaystyle=\frac{\ket{\uparrow\uparrow}-e^{2i\phi}\ket{\downarrow\downarrow}}{\sqrt{2}},$
(120) $\displaystyle\ket{\Psi_{2}(\phi)}$
$\displaystyle=i\frac{\ket{\uparrow\uparrow}+e^{2i\phi}\ket{\downarrow\downarrow}}{\sqrt{2}},$
(121) $\displaystyle\ket{\Psi_{3}(\phi)}$
$\displaystyle=-\frac{\ket{\uparrow\downarrow}+\ket{\downarrow\uparrow}}{\sqrt{2}}$
(122)

away from $\phi=0$ (with respect to a global phase). After integrating the
density matrix $\int d\phi\ket{\Psi_{i}(\phi)}\bra{\Psi_{i}(\phi)}$ over the
full range of $\phi$, we cannot observe entanglement for $\ket{\Psi_{1}}$ and
$\ket{\Psi_{2}}$.

In the high-$p_{T}$ region for most SM processes, the entanglement of
$t\bar{t}$ is large from the triplet component $\ket{\Psi_{2}}$. Therefore for
the quantum state, which is necessarily averaged in fixed beam basis, the
azimuthal angular average of the density matrix is

$\bar{\rho}^{\rm fixed}_{\rm
triplet}=\frac{1}{2\pi}\int\operatorname{d\phi}\ket{\Psi_{2}(\phi)}\bra{\Psi_{2}(\phi)}=\frac{\ket{\uparrow\uparrow}\bra{\uparrow\uparrow}+\ket{\downarrow\downarrow}\bra{\downarrow\downarrow}}{2},$
(123)

which is a separable state.

##  Appendix E Top Anti-top Produced from Gluon and Photon Scattering

In this appendix, we show that $\gamma\gamma\to t\bar{t}$ produces exactly the
same spin density matrix as $gg\to t\bar{t}$ after summing over the color
indices, even though their interactions and Feynman diagrams differ. The
amplitudes of $g_{a}(p_{1})g_{b}(p_{2})\to t_{i}(p_{3})\bar{t}_{j}(p_{4})$ can
be parametrized as

$i\mathcal{M}(g_{a}g_{b}\to
t_{i}\bar{t}_{j})=-ig_{s}^{2}\Big{[}(T^{a}T^{b})_{ij}\mathcal{M}_{t}+(T^{b}T^{a})_{ij}\mathcal{M}_{u}+if^{abc}T^{c}_{ij}\mathcal{M}_{s}\Big{]},$
(124)

where $a,b$, and $c$ are the color indexes of gluon, $i$ and $j$ are the color
indexes of quarks, $T^{a}$ are the generators of $SU(N)$, and $f^{abc}$ is the
structure constant. $\mathcal{M}_{s}$, $\mathcal{M}_{t}$, and
$\mathcal{M}_{u}$ are from the amplitude of $s$-channel, $t$-channel, and
$u$-channel diagrams

$\displaystyle\mathcal{M}_{s}$
$\displaystyle=\frac{\bar{u}_{3}\gamma_{\rho}v_{4}}{\hat{s}}\epsilon_{1\mu}\epsilon_{2\mu}\left[(p_{1}-p_{2})^{\rho}g^{\mu\nu}+(p_{2}-k)^{\mu}g^{\nu\rho}+(k-p1)^{\mu}g^{\nu\rho}\right],$
(125) $\displaystyle\mathcal{M}_{t}$
$\displaystyle=\bar{u}_{3}\frac{\\!\not\\!\epsilon_{1}(\\!\not\\!p_{t}-m)\\!\not\\!\epsilon_{2}}{p_{t}^{2}-m^{2}}v_{4},$
(126) $\displaystyle\mathcal{M}_{u}$
$\displaystyle=\bar{u}_{3}\frac{\\!\not\\!\epsilon_{2}(\\!\not\\!p_{u}-m)\\!\not\\!\epsilon_{1}}{p_{u}^{2}-m^{2}}v_{4},$
(127)

where $k=-p_{3}-p_{4}$, $p_{t}=p_{2}-p_{4}$, $p_{u}=p_{1}-p_{4}$, and $m$ is
the mass of the top quark. The denominator of the propagator written in terms
of velocity factor $\beta$ and scattering angle $\theta$ are
$p_{t}^{2}-m^{2}=\frac{\hat{s}}{2}(\beta c_{\theta}-1)$,
$p_{u}^{2}-m^{2}=-\frac{\hat{s}}{2}(\beta c_{\theta}+1)$, and $k^{2}=\hat{s}$.

For the QED processes $\gamma\gamma\to t\bar{t}$, the amplitude is simply

$i\mathcal{M}(\gamma\gamma\to
t\bar{t})=-ie^{2}(\mathcal{M}_{t}+\mathcal{M}_{u}).$ (128)

Before comparing the spin density matrix from Eq. (124) and Eq. (128), we
first list some general identities. Based on
$\\{\gamma^{\mu},\gamma^{\nu}\\}=2g^{\mu\nu}$ and the Dirac equations of the
spinors, we have

$\displaystyle\bar{u}_{3}\\!\not\\!\epsilon_{1}\\!\not\\!p_{3}\\!\not\\!\epsilon_{2}v_{4}$
$\displaystyle=2\bar{u}_{3}\\!\not\\!\epsilon_{2}v_{4}(p_{3}\cdot\epsilon_{1})-m\bar{u}_{3}\\!\not\\!\epsilon_{1}\\!\not\\!\epsilon_{2}v_{4},$
(129a)
$\displaystyle\bar{u}_{3}\\!\not\\!\epsilon_{1}\\!\not\\!p_{4}\\!\not\\!\epsilon_{2}v_{4}$
$\displaystyle=2\bar{u}_{3}\\!\not\\!\epsilon_{1}v_{4}(p_{4}\cdot\epsilon_{2})+m\bar{u}_{3}\\!\not\\!\epsilon_{1}\\!\not\\!\epsilon_{2}v_{4},$
(129b)
$\displaystyle\bar{u}_{3}\\!\not\\!\epsilon_{2}\\!\not\\!p_{3}\\!\not\\!\epsilon_{1}v_{4}$
$\displaystyle=2\bar{u}_{3}\\!\not\\!\epsilon_{1}v_{4}(p_{3}\cdot\epsilon_{2})-m\bar{u}_{3}\\!\not\\!\epsilon_{2}\\!\not\\!\epsilon_{1}v_{4},$
(129c)
$\displaystyle\bar{u}_{3}\\!\not\\!\epsilon_{2}\\!\not\\!p_{4}\\!\not\\!\epsilon_{1}v_{4}$
$\displaystyle=2\bar{u}_{3}\\!\not\\!\epsilon_{2}v_{4}(p_{4}\cdot\epsilon_{1})+m\bar{u}_{3}\\!\not\\!\epsilon_{2}\\!\not\\!\epsilon_{1}v_{4},$
(129d)

which lead to

$\displaystyle\bar{u}_{3}\\!\not\\!\epsilon_{1}(\\!\not\\!p_{3}-\\!\not\\!p_{4})\\!\not\\!\epsilon_{2}v_{4}-$
$\displaystyle(\epsilon_{1}\leftrightarrow\epsilon_{2})$ $\displaystyle=$
$\displaystyle
2\bar{u}_{3}\\!\not\\!\epsilon_{1}v_{4}(k\cdot\epsilon_{2})-2\bar{u}_{3}\\!\not\\!\epsilon_{2}v_{4}(k\cdot\epsilon_{1})-2m\bar{u}_{3}[\\!\not\\!\epsilon_{1},\\!\not\\!\epsilon_{2}]v_{4}.$
(130)

Based on the transverse conditions $p_{1}\cdot\epsilon_{1}=0$ and
$p_{2}\cdot\epsilon_{2}=0$, we have

$\displaystyle\\!\not\\!\epsilon_{1}\\!\not\\!p_{1}\\!\not\\!\epsilon_{2}+\\!\not\\!\epsilon_{2}\\!\not\\!p_{1}\\!\not\\!\epsilon_{1}$
$\displaystyle=2(p_{1}\cdot\epsilon_{2})\\!\not\\!\epsilon_{1}-2\\!\not\\!p_{1}(\epsilon_{1}\cdot\epsilon_{2}),$
(131a)
$\displaystyle\\!\not\\!\epsilon_{1}\\!\not\\!p_{2}\\!\not\\!\epsilon_{2}+\\!\not\\!\epsilon_{2}\\!\not\\!p_{2}\\!\not\\!\epsilon_{1}$
$\displaystyle=2(p_{2}\cdot\epsilon_{1})\\!\not\\!\epsilon_{2}-2\\!\not\\!p_{2}(\epsilon_{1}\cdot\epsilon_{2}),$
(131b)

which lead to

$\displaystyle\\!\not\\!\epsilon_{1}(\\!\not\\!p_{1}-\\!\not\\!p_{2})\\!\not\\!\epsilon_{2}+$
$\displaystyle(\epsilon_{1}\leftrightarrow\epsilon_{2})$ $\displaystyle=$
$\displaystyle
2\\!\not\\!\epsilon_{1}(p_{1}\cdot\epsilon_{2})-2\\!\not\\!\epsilon_{2}(p_{2}\cdot\epsilon_{1})-2(\\!\not\\!p_{1}-\\!\not\\!p_{2})(\epsilon_{1}\cdot\epsilon_{2}).$
(132)

For the QED process, the spin density matrix is directly given by the helicity
amplitudes,

$\rho^{\rm
QED}_{\alpha\bar{\alpha},\alpha^{\prime}\bar{\alpha}^{\prime}}\propto\mathcal{M}_{\alpha\bar{\alpha}}(\gamma\gamma\to
t\bar{t})\mathcal{M}_{\alpha^{\prime}\bar{\alpha}^{\prime}}^{*}(\gamma\gamma\to
t\bar{t}).$ (133)

For the QCD process, we need to sum over color to obtain the spin density
matrix,

$\rho^{\rm
QCD}_{\alpha\bar{\alpha},\alpha^{\prime}\bar{\alpha}^{\prime}}\propto\sum_{a,b,i,j}\mathcal{M}_{\alpha\bar{\alpha}}(g_{a}g_{b}\to
t_{i}\bar{t}_{j})\mathcal{M}_{\alpha^{\prime}\bar{\alpha}^{\prime}}^{*}(g_{a}g_{b}\to
t_{i}\bar{t}_{j}).$ (134)

To compare these two spin density matrices, we decompose the $gg\to t\bar{t}$
process according to the color configuration,

$\displaystyle\mathcal{M}(g_{a}g_{b}\to t_{i}\bar{t}_{j})$
$\displaystyle=g_{s}^{2}\Big{[}(T^{a}T^{b})_{ij}\mathcal{M}_{t}+(T^{b}T^{a})_{ij}\mathcal{M}_{u}+if^{abc}T^{c}_{ij}\mathcal{M}_{s}\Big{]}$
$\displaystyle=g_{s}^{2}\left[\\{T^{a},T^{b}\\}_{ij}\frac{\mathcal{M}_{t}+\mathcal{M}_{u}}{2}+if^{abc}T^{c}_{ij}\left(\frac{\mathcal{M}_{t}-\mathcal{M}_{u}}{2}+\mathcal{M}_{s}\right)\right].$
(135)

Since there is no interference between the gluon color symmetric part and
asymmetric part,

$\sum_{a,b,i,j}f^{abc}T^{c}_{ij}~{}\\{T^{a},T^{b}\\}_{ij}=0,$ (136)

they contribute to the $t\bar{t}$ spin correlation separately. The color
symmetric contribution is exactly the same as the $\gamma\gamma\to t\bar{t}$.

Next, we show that the color asymmetric contribution also produces the same
spin density matrix with $\gamma\gamma\to t\bar{t}$, by proving

$\frac{\mathcal{M}_{t}-\mathcal{M}_{u}}{2}+\mathcal{M}_{s}\propto\mathcal{M}_{t}+\mathcal{M}_{u}.$
(137)

By decomposing $\mathcal{M}_{t}+\mathcal{M}_{u}$ and
$\mathcal{M}_{t}-\mathcal{M}_{u}$ into symmetric and asymmetric part with
respect to $\epsilon_{1}$ and $\epsilon_{2}$, we have

$\displaystyle\mathcal{M}_{t}+\mathcal{M}_{u}=\frac{2}{\hat{s}(\beta^{2}c_{\theta}^{2}-1)}$
$\displaystyle\Bigg{[}\\!\not\\!\epsilon_{1}\left(\frac{\\!\not\\!p_{t}+\\!\not\\!p_{u}+2m}{2}+\frac{\\!\not\\!p_{t}-\\!\not\\!p_{u}}{2}\beta
c_{\theta}\right)\\!\not\\!\epsilon_{2}+(\epsilon_{1}\leftrightarrow\epsilon_{2})$
$\displaystyle+\\!\not\\!\epsilon_{1}\left(\frac{\\!\not\\!p_{t}+\\!\not\\!p_{u}+2m}{2}\beta
c_{\theta}+\frac{\\!\not\\!p_{t}-\\!\not\\!p_{u}}{2}\right)\\!\not\\!\epsilon_{2}-(\epsilon_{1}\leftrightarrow\epsilon_{2})\Bigg{]},$
(138)
$\displaystyle\mathcal{M}_{t}-\mathcal{M}_{u}=\frac{2}{\hat{s}(\beta^{2}c_{\theta}^{2}-1)}$
$\displaystyle\Bigg{[}\\!\not\\!\epsilon_{1}\left(\frac{\\!\not\\!p_{t}+\\!\not\\!p_{u}+2m}{2}\beta
c_{\theta}+\frac{\\!\not\\!p_{t}-\\!\not\\!p_{u}}{2}\right)\\!\not\\!\epsilon_{2}+(\epsilon_{1}\leftrightarrow\epsilon_{2})$
$\displaystyle+\\!\not\\!\epsilon_{1}\left(\frac{\\!\not\\!p_{t}+\\!\not\\!p_{u}+2m}{2}+\frac{\\!\not\\!p_{t}-\\!\not\\!p_{u}}{2}\beta
c_{\theta}\right)\\!\not\\!\epsilon_{2}-(\epsilon_{1}\leftrightarrow\epsilon_{2})\Bigg{]}.$
(139)

Then, we have

$\displaystyle\beta
c_{\theta}(\mathcal{M}_{t}+\mathcal{M}_{u})-(\mathcal{M}_{t}-\mathcal{M}_{u})=$
$\displaystyle\frac{2}{\hat{s}}\bar{u}_{3}\Bigg{[}\\!\not\\!\epsilon_{1}\left(\frac{\\!\not\\!p_{t}-\\!\not\\!p_{u}}{2}\right)\\!\not\\!\epsilon_{2}+\\!\not\\!\epsilon_{2}\left(\frac{\\!\not\\!p_{t}-\\!\not\\!p_{u}}{2}\right)\\!\not\\!\epsilon_{1}$
$\displaystyle+\\!\not\\!\epsilon_{1}\left(\frac{\\!\not\\!p_{t}+\\!\not\\!p_{u}}{2}+m\right)\\!\not\\!\epsilon_{2}-\\!\not\\!\epsilon_{2}\left(\frac{\\!\not\\!p_{t}+\\!\not\\!p_{u}}{2}+m\right)\\!\not\\!\epsilon_{1}\Bigg{]}v_{4}.$
(140)

Here, $p_{t}+p_{u}=p_{3}-p_{4}$ and $p_{t}-p_{u}=p_{1}-p_{2}$. Taking Eqs. (E)
and (E) into account, we have

$\displaystyle\beta
c_{\theta}(\mathcal{M}_{t}+\mathcal{M}_{u})-(\mathcal{M}_{t}-\mathcal{M}_{u})$
$\displaystyle=$
$\displaystyle\frac{2}{\hat{s}}\left[\bar{u}_{3}\\!\not\\!\epsilon_{1}v_{4}(k-p_{1})\cdot\epsilon_{2}+\bar{u}_{3}\\!\not\\!\epsilon_{2}v_{4}(p_{2}-k)\cdot\epsilon_{1}+\bar{u}_{3}(\\!\not\\!p_{1}-\\!\not\\!p_{2})v(\epsilon_{1}\cdot\epsilon_{2})\right],$
$\displaystyle=$ $\displaystyle 2\mathcal{M}_{s}.$ (141)

Therefore,

$\frac{\mathcal{M}_{t}-\mathcal{M}_{u}}{2}+\mathcal{M}_{s}=\frac{\beta
c_{\theta}}{2}(\mathcal{M}_{t}+\mathcal{M}_{u}),$ (142)

holds generally. Substituting Eqs. (142), (E), and (136) into Eq. (134), we
find that the $\gamma\gamma\to t\bar{t}$ and $gg\to t\bar{t}$ processes
produce exactly the same $t\bar{t}$ spin correlation.

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