# An even-parity precession protocol for detecting nonclassicality and
entanglement

Jinyan Chen  Centre for Quantum Technologies, National University of
Singapore, 3 Science Drive 2, Singapore 117543  Jackson Tiong  Department of
Physics, National University of Singapore, 2 Science Drive 3, Singapore 117542
Lin Htoo Zaw  Centre for Quantum Technologies, National University of
Singapore, 3 Science Drive 2, Singapore 117543  Valerio Scarani  Centre for
Quantum Technologies, National University of Singapore, 3 Science Drive 2,
Singapore 117543  Department of Physics, National University of Singapore, 2
Science Drive 3, Singapore 117542

###### Abstract

We introduce an even-parity precession protocol that can detect
nonclassicality of some quantum states using only measurements of a uniformly-
precessing variable at different points in time. Depending on the system under
study, the protocol may detect the Wigner negativity of a single quantum
harmonic oscillator or of a single spin $j\geq 2$; the non-Gaussian
entanglement of two harmonic oscillators; or genuine multipartite entanglement
of a spin ensemble, whose total spin is integer. Unlike other nonclassicality
tests, simultaneous or sequential measurements are not required. Our protocol
can also detect states that commute with the parity operator, which were
missed by similar protocols built from Tsirelson’s original precession
protocol. This work also closes a long-standing gap by showing the possibility
of detecting the Greenberger–Horne–Zeilinger entanglement of an even number of
qubits using only collective spin measurements.

##  I Introduction

Uniform precession is found in the dynamics of many physical systems, from the
orbits of large celestial bodies, to the harmonic motion of mesoscopic objects
in optomechanical traps, to the rotation of microscopic spins in a uniform
magnetic field. As a uniformly-precessing observable has the same dynamics in
both classical and quantum theory, one would not expect the simple observation
of its values at different times to reveal any nonclassical signatures. Yet,
in an unpublished preprint [1] Tsirelson showed this expectation to be wrong
for the harmonic oscillator, whose dynamics is a precession in phase space. He
proved that the probability of finding the oscillator with position $x>0$ at
one out of three times exceeds the maximal value allowed by classical
mechanics for some suitable quantum states. This is an example of a mechanical
task with a demonstrable quantum advantage, in the same vein as the
probability backflow of a freely-evolving particle [2] and the enhanced
distance travelled by quantum projectiles [3]. In these tasks, simultaneous or
sequential measurements are not required, in contrast to other single-system
nonclassicality tests, like contextuality or Leggett–Garg inequalities [4, 5].

Recently, Tsirelson’s original protocol has been extended to what we will call
the _odd-parity precession protocol_ , which certifies nonclassicality (in the
form of Wigner negativity [6]) by probing a uniformly-precessing observable at
one of $K$ suitably chosen times, where $K$ is odd [7]. In the same work, it
was also noticed that the precession does not need to be in phase space: it
could be in real space, if it involves non-commuting observables. Thus, a
precession of position (circular orbit) cannot show any non-classicality; but
a precession of angular momentum can, and actually does. This led to the
extension of the Tsirelson idea to discrete variables, specifically to the
precession of spins.

The odd-parity precession protocol was further extended to witness
entanglement, by applying it to collective degrees of freedom. Notably, some
non-Gaussian entangled states of two harmonic oscillators can be detected
using only measurements of their center-of-mass position, and without the risk
of false positives typical of witnesses based on uncertainty relations [8]. In
spin ensembles, genuine multipartite entanglement can be detected measuring
only total angular momentum. While similar witnesses of this type were known,
this is the first family that detects also Greenberger–Horne–Zeilinger (GHZ)
states [9].

Like any linear non-classicality witness, the odd-parity precession protocol
cannot detect all states of interest (be they Wigner-negative or entangled).
In particular, it misses all states that commute with the parity operator, as
well as even-number GHZ states. In this paper, we introduce an _even_ -parity
precession protocol, which requires only measurements of a uniformly-
precessing observable at one of $K$ suitably chosen times, with $K$ now even.
This new protocol can be implemented in the same way as the odd-parity one,
but requires a different expression for the score, as checking for positivity
of the observable leads to trivial scores by symmetry. We show that this
protocol detects Wigner negativity and entanglement in close analogy to the
odd-parity protocol, for a different set of states. In particular, when it
comes to spin ensembles, we close the problem of detecting GHZ states using
collective spin measurements by exhibiting a protocol that does it for an even
number of spins (the case of odd number of spins having been solved in [9]).

##  II The Even-parity Precession Protocol and Its Classical Bound

Figure 1: The classical score $s_{K,\Delta}^{c}$ given given that the initial
state is $(A_{x}(0),A_{y}(0))$, for (a) $K=4$, (b) $K=6$, and (c) $K=8$. We
find that $s_{K,\Delta}^{c}$ only takes the values $0$ or $\pm 1/K$, which
implies the classical bound $|s_{K,\Delta}^{c}|\leq 1/K=:\mathbf{s}_{K}^{c}$.

We shall call a physical observable $A_{x}(\theta)$ uniformly precessing with
respect to a parameter $\theta$ if it satisfies

$\displaystyle A_{x}(\theta)$
$\displaystyle=\cos(\theta)A_{x}(0)+\sin(\theta)A_{y}(0),$ (1) $\displaystyle
A_{y}(\theta)$ $\displaystyle=\cos(\theta)A_{y}(0)-\sin(\theta)A_{x}(0).$

where $A_{\hat{n}}(\theta)$ is the value of $A_{\hat{n}}$ along the angle
$\theta$ in the classical case, and an operator in the Heisenberg picture in
the quantum case. In earlier works, uniform precession was usually taken to
have arisen from the dynamics of the system, as with the position and momentum
of a harmonic oscillator with angular frequency $\omega$ at time
$t=\theta/\omega$. Alternately, it could similarly arise from spatial
rotations, as with the $x$ component of any vector
$\vec{A}=(A_{x},A_{y},A_{z})^{T}$ rotated about the $z$ axis. For the rest of
the paper, we shall mostly use the terminology of angles rather than time,
although it should not be forgotten that the precession protocol can be
applied in the same way to both.

The even-parity precession protocol is defined for a uniformly-precessing
observable $A_{x}(\theta)$, a real number $\Delta\geq 0$, and an even integer
$K\geq 4$, and is carried out by performing many independent rounds. In each
round:

  1. 1.

A value $\theta_{k}=\pi k/K$ for $k\in\\{0,1,\dots,K-1\\}$ is chosen;

  2. 2.

$A_{x}(\theta_{k})$ is measured at the chosen angle $\theta_{k}$;

  3. 3.

The score $(-1)^{k}$ is assigned if $|A_{x}(\theta_{k})|\leq\Delta/2$, and the
score $0$ is assigned otherwise .

After many rounds, the average score is calculated as

$s_{K,\Delta}:=\frac{1}{K}\sum_{k=0}^{K-1}(-1)^{k}\Pr[\absolutevalue{A_{x}(\theta_{k})}\leq\frac{\Delta}{2}].$
(2)

In classical mechanics, the score $s^{c}_{K,\Delta}$ of a pure state is fully
determined by its initial configuration $(A_{x}(0),A_{y}(0))$. The classical
scores for $K=2,4,6$ for different initial states are plotted in Fig. 1, where
it is observed that $s_{K,\Delta}^{c}\in\\{0,\pm 1/K\\}$. Indeed, a formal
geometric proof in Appendix A shows this to be true for every $K$. As general
classical states are probability distributions on classical pure states, this
in turn implies by convexity that the classical bound is

$\absolutevalue{s_{K,\Delta}^{c}}\leq\frac{1}{K}=:\mathbf{s}_{K}^{c},$ (3)

which is independent of $\Delta$.

##  III Quantum Violations of the Classical Bound

We shall now study the violation of the classical bound with two particular
examples of uniformly-precessing quantum systems: the continuous variable case
of the quantum harmonic oscillator, and the discrete variable case of spin
angular momentum.

###  III.1 Quantum Harmonic Oscillator

The quantum harmonic oscillator is specified by the Hamiltonian
$H=P^{2}/(2m)+m\omega^{2}X^{2}/2$, where the position $X$ and momentum $P$ are
operators that satisfy the uncertainty relation
$\commutator{X}{P}=i\hbar\mathbbm{1}$. The evolution of these observables in
time is given in the Heisenberg picture as

$\displaystyle X(t)$ $\displaystyle=\cos(\omega t)X+\sin(\omega t)P/(m\omega)$
(4) $\displaystyle P(t)$ $\displaystyle=\cos(\omega t)P-\sin(\omega t)m\omega
X.$

Alternatively, the observable $\sqrt{m\omega/\hbar}X(\theta/\omega)$ is the
quadrature along the angle $\theta$ associated with homodyne detection in
quantum optics [10]. This observable complies with Eq. (1), so the position of
a harmonic oscillator, equivalently the quadrature in optical systems, is
uniformly precessing with respect to $\theta$.

Figure 2: The scaled maximum score $s^{\infty}_{K,\Delta}$ against $\Delta$
for the quantum harmonic oscillator, for (a) selected values of $K$ and (b) a
detailed look at $K=4$ where the approximate values of $\Delta$ at the two
peaks are also shown. The scores were obtained by constructing the matrix
$S^{\infty}_{K,\Delta}$ in the energy eigenbasis of the harmonic oscillator up
to some truncation, then numerically calculating its maximum eigenvalue.

Hence, we can perform the precession protocol on $X(\theta/\omega)$, where the
expected score is given by

$\displaystyle s_{K,\Delta}^{\infty}$
$\displaystyle=\frac{1}{K}\sum_{k=0}^{K-1}(-1)^{k}\Pr[\sqrt{\frac{m\omega}{\hbar}}\absolutevalue{X\pqty{\frac{\theta_{k}}{\omega}}}\leq\frac{\Delta}{2}]$
(5) $\displaystyle=\tr(\rho S_{K,\Delta}^{\infty}).$

Here, the density operator $\rho$ specifies the initial state of the quantum
system, and

$S_{K,\Delta}^{\infty}:=\frac{1}{K}\sum_{k=0}^{K-1}(-1)^{k}e^{i\theta_{k}\frac{H}{\hbar\omega}}\int_{-\sqrt{\frac{\hbar\Delta^{2}}{2m\omega}}}^{\sqrt{\frac{\hbar\Delta^{2}}{2m\omega}}}\differential{x}\outerproduct{x}{x}e^{-i\theta_{k}\frac{H}{\hbar\omega}},$
(6)

where $\ket{x}$ are the improper eigenvectors of $X$ with eigenvalue $x$. The
infinity symbol in the superscript pegs these scores to the quantum harmonic
oscillator, to differentiate from the finite-dimensional discrete variable
system we will study later.

Obtaining the maximum quantum score
$\mathbf{s}^{\infty}_{K,\Delta}:=\max_{\rho}{\tr}(\rho S_{K,\Delta}^{\infty})$
is therefore the same as solving for the largest eigenvalue of
$S^{\infty}_{K,\Delta}$. In order to obtain numerical estimates of the quantum
score, we construct $S_{K,\Delta}^{\infty}$ as a matrix in a chosen
orthonormal basis up to some truncation, and solve for its eigenvalues with
standard numerical tools (see Appendix B for details).

The maximum quantum scores $\mathbf{s}_{K,\Delta}^{\infty}$ obtained using
this method are plotted in Fig. 2 against $\Delta$ for some choices of $K$.
The dependence of $\mathbf{s}_{K,\Delta}^{\infty}$ on $\Delta$ is highly
nontrivial. Nonetheless, it can be eked out from Fig. 2 that (i) the classical
bound is saturated or violated in most cases; (ii) the first and largest peak
occurs at $0<\Delta<1$; and (iii) the maximum quantum score seems to approach
the classical bound for large $\Delta$.

We show in Appendix C that
$\forall\Delta\in(0,\infty):\mathbf{s}_{K,\Delta}^{\infty}\geq\mathbf{s}_{K}^{c}$,
which proves observation (i). We do not have analytical proofs for the other
two observations; but we note that performing the precession protocol with
$\Delta<1$ probes features of the quantum state within the
$|x|<\sqrt{\hbar/(4m\omega)}$ and $|p|<\sqrt{\hbar m\omega/4}$ region. Thus,
observation (ii) points to sub-Planck structure of Wigner function, i.e.
features found within an area $\Delta x\Delta p<\hbar$ in phase space [11].

Figure 3: The Wigner function of the quantum harmonic oscillator state that
achieves the score $\max_{\Delta}\mathbf{s}_{K,\Delta}^{\infty}$ for (a)
$K=4$, (b) $K=6$, and (c) $K=8$. A comparison with Fig. 1 shows that the
negativities are concentrated in the region where
$\mathbf{s}_{K,\Delta}^{c}=-1/K$. Notice also that the states are unchanged
under the parity transformation $x\to-x$ and $p\to-p$, which shows that the
even-parity precession protocol can detect states that commute with the parity
operator.

The Wigner function of the state that maximally violates the precession
protocol is plotted in Fig. 3 for $K=4,6,8$ . Comparing this to Fig. 1, we
find that the Wigner negativity is concentrated in the regions where the
classical score is $s_{K,\Delta}^{c}=-1/K$, and $W(x,p)$ take on large
positive values when $s_{K,\Delta}^{c}=1/K$. The relationship between Wigner
negativity and the violation of the classical bound is further quantified in
Appendix D. In particular, we prove the lower bound

$\displaystyle K(|s_{K,\Delta}^{\infty}|-\mathbf{s}_{K}^{c})$
$\displaystyle\leq 2\mathcal{N}_{V}$ (7)

on the Wigner negativity volume

$\displaystyle\mathcal{N}_{V}:=\int\differential{x}\int\differential{p}[|W(x,p)|-W(x,p)]/2\,,$
(8)

which quantifies Wigner negativity as a resource [12, 13, 14].

Finally, from its definition in Eq. (6), it is straightforward to see that
$S^{\infty}_{K,\Delta}$ commutes with the parity operator $\Pi$, which has the
action $\Pi X\Pi=-X$ and $\Pi P\Pi=-P$. As such, $S^{\infty}_{K,\Delta}$ and
$\Pi$ can be simultaneously diagonalized, which means that there are
eigenvectors $\ket{\lambda}$ of $S^{\infty}_{K,\Delta}$ satisfying
$\commutator{\outerproduct{\lambda}{\lambda}}{\Pi}=0$ that violate the
classical bound. A particular example are the states in Fig. 3. This solves
the limitation of the odd-parity precession protocol, which cannot detect
states that commute with the parity operator [7].

###  III.2 Spin Angular Momentum

In quantum mechanics, the components of the angular momentum vector
$\vec{J}=(J_{x},J_{y},J_{z})$ satisfy the commutation relation
$\commutator{J_{x}}{J_{y}}=i\hbar J_{z}$, where $\hbar$ is the reduced Planck
constant. Let $\ket{j,m_{\hat{n}}}$ denote the simultaneous eigenvector of
$\lvert\vec{J}\rvert^{2}$ and $J_{\hat{n}}$, where $\hat{n}\in\\{x,y,z\\}$,
with associated eigenvalues $\hbar^{2}j(j+1)$ and $\hbar m$ respectively.
Here, $j$ is a nonnegative integer or half-integer, and
$m\in\\{-j,-j+1,\dots,j-1,j\\}$.

Generally, $\vec{J}=\oplus_{j}\vec{J}^{(j)}$ is a direct sum of irreducible
blocks $\vec{J}^{(j)}$ spanned by the eigenvectors of
$\lvert\vec{J}\rvert^{2}$ with eigenvalue $\hbar^{2}j(j+1)$ [15]. For
simplicity, we shall consider each block $\vec{J}^{(j)}$, each with a fixed
$j$, separately.

Now, $J_{x}^{(j)}$ evolving under a rotation generated by the Hamiltonian
$H\propto-J_{z}^{(j)}$ results in the uniformly-precessing observable

$J_{x}^{(j)}(\theta)=e^{-i\theta_{k}\frac{J_{z}^{(j)}}{\hbar}}J_{x}^{(j)}e^{i\theta_{k}\frac{J_{z}^{(j)}}{\hbar}}=\cos(\theta)J_{x}^{(j)}+\sin(\theta)J_{y}^{(j)}.$
(9)

If the even-parity precession protocol is performed on $J_{x}^{(j)}/\hbar$,
the expected score will be $s_{K,\Delta}^{(j)}={\tr}(\rho S_{K,\Delta}^{(j)})$
given that the initial state of the system is $\rho$, where

$S_{K,\Delta}^{(j)}:=\frac{1}{K}\sum_{k=0}^{K-1}(-1)^{k}e^{-i\theta_{k}\frac{J_{z}^{(j)}}{\hbar}}\Pr(\frac{\absolutevalue\big{J_{x}^{(j)}}}{\hbar}\leq\frac{\Delta}{2})e^{i\theta_{k}\frac{J_{z}^{(j)}}{\hbar}},$
(10)

with ${\Pr}(|J_{x}^{(j)}|/\hbar\leq\Delta/2)$ defined in terms of the
eigenvectors of $J_{x}^{(j)}$ as

$\displaystyle\Pr(\frac{\absolutevalue\big{J_{x}^{(j)}}}{\hbar}\leq\frac{\Delta}{2})$
$\displaystyle=\\!\\!\sum_{m=-j}^{j}\\!\\!\Pr(|m|\leq\frac{\Delta}{2})\outerproduct{j,m_{x}}{j,m_{x}}$
(11)
$\displaystyle=\\!\\!\sum_{m=-(\lfloor\frac{\Delta}{2}+j\rfloor-j)}^{\lfloor\frac{\Delta}{2}+j\rfloor-j}\\!\\!\outerproduct{j,m_{x}}{j,m_{x}}$
$\displaystyle=\Pr(\frac{\absolutevalue\big{J_{x}^{(j)}}}{\hbar}\leq\left\lfloor\frac{\Delta}{2}+j\right\rfloor-j).$

The maximum quantum score $\mathbf{s}_{K,\Delta}^{(j)}:=\max_{\rho}{\tr}(\rho
S_{K,\Delta}^{(j)})$ can be found by using commonly-available eigensolvers on
the finite-dimensional operator $S_{K,\Delta}^{(j)}$.

Figure 4:  The quantum scores $\mathbf{s}_{4,\Delta}^{{j}}$ for spin angular
momentum against $\Delta$. (a) Plots for the five specific values
$j\in\\{2,15,24,50,400\\}$ for the whole score range. (b) Superposition of the
plots for all integer and half-integer values $0\leq j\leq 400$, zoomed in on
the range where the classical bound is violated. The plot of
$\mathbf{s}_{4,\Delta}^{{j}}$ appears to converge towards
$\mathbf{s}_{4,\Delta}^{\infty}$ (Fig. 2) for large $j$.

These scores are plotted for $K=4$ for several values of $j$ in Fig. 4(a). The
stepped behavior is due to the discrete eigenvalues of $J_{x}$, while the
dependence of $\mathbf{s}_{K,\Delta}^{(j)}$ on $\Delta$ is rather complex.
However, if we superimpose such plots for many values of $j$ in Fig. 4(b), the
traces appear to converge to the harmonic oscillator scores plotted in Fig. 2.
This seems to imply that

$\lim_{j\to\infty}\mathbf{s}_{K,\Delta/\sqrt{j}}^{(j)}\stackrel{{\scriptstyle?}}{{=}}\mathbf{s}_{K,\Delta}^{\infty},$
(12)

which parallels a similar relationship between the spin angular momentum and
harmonic oscillator in the odd-parity precession protocol [7], although we
have not been able to formally prove Eq. (12) in the current work.

Figure 5:  $\Delta$ that satisfy
$|\mathbf{s}_{4,\Delta}^{(j)}-\max_{\Delta^{\prime}}\mathbf{s}_{4,\Delta^{\prime}}^{(j)}|\leq\varepsilon$
for $\varepsilon=2^{-8}$ (gray line) and for $\varepsilon=0$ (dark purple
line). We find that they cluster around $\sqrt{3(j-2)}$ (dashed black line)
and $\sqrt{(j-2)/e}$ (dashed yellow line). Comparing to Fig. 4, these two
clusters of $\Delta$ correspond to the largest two peaks of
$\mathbf{s}_{4,\Delta}^{(j)}$. In particular, the maximum violation for fixed
$j$ occurs roughly at
$\operatorname*{argmax}_{\Delta}\mathbf{s}_{4,\Delta}^{(j)}\approx\sqrt{(j-2)/e}$

To determine the choice of $\Delta$ that maximizes the quantum violation, we
plot the values of $\Delta$ that achieve quantum scores close to the maximum
possible score in Fig. 5. That is, they satisfy
$|\mathbf{s}_{4,\Delta}^{(j)}-\max_{\Delta^{\prime}}\mathbf{s}_{4,\Delta^{\prime}}^{(j)}|\leq\varepsilon$
for $\varepsilon=2^{-8}$ (gray line) and for $\varepsilon=0$ (dark purple
line). We find that the optimal values of $\Delta$ tend to cluster around
$\sqrt{3(j-2)}$ and $\sqrt{(j-2)/e}$. Cross-referencing to Fig. 4(b) and Fig.
2, they correspond to the largest two peaks with
$\operatorname*{argmax}_{\Delta}\mathbf{s}_{4,\Delta}^{(j)}\approx\sqrt{(j-2)/e}$,
and has the limiting values $\Delta/\sqrt{j}\to 1/\sqrt{e}$ and
$\Delta/\sqrt{j}\to\sqrt{3}$.

Figure 6: Maximum quantum scores $\max_{\Delta}\mathbf{s}_{K,\Delta}^{(j)}$
against $j$ for (a) $j\leq 10$ and (b) $j\leq 200$. The score is zero whenever
$j<K/2$, always peaks at $j=K/2$, violates the classical bound for every
$j\geq K/2$, and converges to a limit as $j\to\infty$. We conjecture that the
limit is $\max_{\Delta}\mathbf{s}_{K,\Delta}^{\infty}$.

For other values of $K$, the maximum quantum scores
$\max_{\Delta}\mathbf{s}_{K,\Delta}^{(j)}$ are plotted against $j$ in Fig. 6.
Several features can be observed: the score is zero whenever $j<K/2$, the
maximum violation occurs at $j=K/2$, the classical bound is violated for every
$j\geq K/2$, and the scores appear to converge as $j\to\infty$, which is
consistent with the conjectured limit in Eq. (12).

A stronger form of the first observation is proved in Appendix E, where we
show that $\mathbf{s}_{K,\Delta}^{(j<K/2)}=0$ for every $K$ and $\Delta$. As
such, the precession protocol is also a dimension witness, as the violation of
the classical bound when performing the protocol on a uniformly-precessing
spin certifies that the dimension of the Hilbert space $d$ is lower bounded as
$d\geq K+1$.

Meanwhile, the maximum score for $j=K/2$ is analytically worked out in
Appendix F to be

$\displaystyle\mathbf{s}_{K,\Delta}^{(K/2)}$
$\displaystyle=2^{-(K-1)}\binom{K-1}{\lfloor\frac{K+\Delta}{2}\rfloor},$ (13)

which is achieved by the spin cat state

$\displaystyle\ket{\mathbf{s}_{K,\Delta}^{(K/2)}}$
$\displaystyle=\frac{1}{\sqrt{2}}\pqty{\ket{j,-j_{z}}+(-1)^{\lfloor(\Delta+K)/2\rfloor}\ket{j,j_{z}}}.$
(14)

The maximum value of $\mathbf{s}_{K,\Delta}^{(K/2)}$ for fixed $K$ is given by
the central binomial coefficient, which occurs when
$\lfloor(K+\Delta)/2\rfloor=\lceil(K-1)/2\rceil=K/2$. That is,

$\displaystyle\max_{\Delta}\mathbf{s}_{K,\Delta}^{(K/2)}$
$\displaystyle=\mathbf{s}_{K,\Delta=0}^{(j=K/2)}=2^{-(K-1)}\binom{K-1}{K/2},$
(15)
$\displaystyle\operatorname*{argmax}_{\Delta}\mathbf{s}_{K,\Delta}^{(K/2)}$
$\displaystyle\in[0,2).$

With the closed-form expression in Eq. (15), we show in Appendix F that
$\mathbf{s}_{K,\Delta=0}^{(j=K/2)}>\mathbf{s}_{K}^{c}$ for all even $K\geq 4$.
Therefore, the classical bound of the even-parity precession protocol can be
violated by all quantum systems with integral spins $j\geq 2$. This contrasts
with the odd-parity precession protocol, where no violation was found for
$j=2$ [7].

Finally, while we have taken the spin number $j$ to be fixed in the above
discussions, it is easily extendable the case where the protocol is performed
on $\vec{J}=\oplus_{j\in\mathcal{J}}\vec{J}^{(j)}$ for some set $\mathcal{J}$
of spins. Then, the maximum score will be given by
$\max_{j\in\mathcal{J}}\mathbf{s}_{K,\Delta}^{(j)}$, as the direct sum
structure of $\vec{J}$ would be imposed onto the operator $S_{K,\Delta}$.

##  IV Witnessing Entanglement

In this section, we apply the even-parity precession protocol to composite
systems, and demonstrate that it is an entanglement witness when performed on
certain collective coordinates.

###  IV.1 Witnessing Non-Gaussian Entanglement with Quadrature Measurements

A system of two identical linearly-coupled harmonic oscillators is governed by
the Hamiltonian

$H=\frac{P_{1}^{2}+P_{2}^{2}}{2m}+\frac{m\omega^{2}}{2}\pqty{X_{1}^{2}+X_{2}^{2}}-gX_{1}X_{2},$
(16)

where $X_{n}$ and $P_{n}$ are respectively the position and momentum of the
$n$th oscillator, $m$ their mass, $\omega$ the angular frequency of the trap,
and $g$ the coupling strength.

Equation (16) can also be rewritten as

$H=\frac{P_{+}^{2}}{2m}+\frac{m\omega_{+}^{2}}{2}X_{+}^{2}+\frac{P_{-}^{2}}{2m}+\frac{m\omega_{-}^{2}}{2}X_{-}^{2},$
(17)

where $X_{\pm}:=(X_{2}\pm X_{1})/\sqrt{2}$, $P_{\pm}:=(X_{2}\pm
X_{1})/\sqrt{2}$, and $\omega_{\pm}:=\omega\pm g/(2m)$. Here, $X_{+}$ is the
center-of-mass position of the two oscillators, and its evolution in time is
generated by the Hamiltonian in Eq. (17) is

$X_{+}(t)=\cos(\omega_{+}t)X_{+}+\sin(\omega_{+}t)P_{+}.$ (18)

As such, we can perform the precession protocol on the coupled harmonic
oscillators using $X_{+}(\theta/\omega_{+})$ as the uniformly-precessing
variable. For an initial state $\rho$ of the system, the expected score would
be $\mathbf{s}_{K,\Delta}^{\infty}={\tr}(\rho S_{K,\Delta}^{\infty})$, where
$S_{K,\Delta}^{\infty}$ is as defined in Eq. (6) with $X_{+}$ in place of $X$.
Now, let $\tr_{-}$ be the partial trace over the $(X_{-},P_{-})$ mode. Then,
since $S_{K,\Delta}^{\infty}$ is only defined on the center-of-mass mode,

$\mathbf{s}_{K,\Delta}^{\infty}=\tr(\rho
S_{K,\Delta}^{\infty})=\tr_{+}[(\tr_{-}\rho)S_{K,\Delta}^{\infty}].$ (19)

By the results of Appendix D, this means that
$s_{K,\Delta}^{\infty}\leq\mathbf{s}_{K}^{c}+2\mathcal{N}_{V}/K$, where
$\mathcal{N}_{V}$ is the Wigner negativity volume (8) of the state
$\tr_{-}\rho$.

If $\rho$ is separable over the two oscillators, it is known that the Wigner
function of $\tr_{-}\rho$ must be nonnegative by Theorem 2 of Ref. [16], which
implies that $\mathcal{N}_{V}=0$. However, if $\mathcal{N}_{V}=0$, then
$s_{K,\Delta}^{\infty}\leq\mathbf{s}_{K}^{c}$. Taking the contrapositive
statement, if the even-parity precession protocol is performed on the center-
of-mass position $X_{+}$, then $s_{K,\Delta}^{\infty}>\mathbf{s}_{K}^{c}$
implies that the two oscillators are entangled. Furthermore, since
$2\mathcal{N}_{V}\geq K(s_{K,\Delta}^{\infty}-\mathbf{s}_{K}^{c})>0$, this
criteria only detects non-Gaussian entangled states of the system with
negative Wigner functions.

###  IV.2 Witnessing Genuine Multipartite Entanglement with Collective Spin
Measurements

Let us now consider an ensemble of $N$ particles, where the $n$th particle has
spin $j_{n}$ with angular momentum $\vec{J}^{(j_{n})}$, and the total spin
$\sum_{n=1}^{N}j_{n}$ is an integer. Then,
$\vec{J}:=\sum_{n}\vec{J}^{(j_{n})}$ rotates around the $-z$ axis as

$J_{x}(\theta)=e^{i\theta J_{z}/\hbar}J_{x}e^{-i\theta
J_{z}/\hbar}=\cos(\theta)J_{x}+\sin(\theta)J_{y}.$ (20)

As such, we can perform the precession protocol on the total angular momentum
$J_{x}(\theta)=\sum_{n=1}^{N}J_{x}^{(j_{n})}(\theta)$ for
$K=2\sum_{n=1}^{N}j_{n}$. For initial state $\rho$, the observed score will
then be $s_{K,\Delta}={\tr}(\rho S_{K,\Delta})$ with

$\displaystyle S_{K,\Delta}$
$\displaystyle:=\sum_{k=0}^{K-1}\frac{(-1)^{k}}{K}e^{-i\theta_{k}\frac{J_{z}}{\hbar}}\Pr(\frac{\absolutevalue\big{J_{x}}}{\hbar}\leq\frac{\Delta}{2})e^{i\theta_{k}\frac{J_{z}}{\hbar}}$
(21) $\displaystyle=\bigoplus_{\tilde{\jmath}}S_{K,\Delta}^{(\tilde{\jmath})}$
$\displaystyle=2^{-(K-1)}\binom{K-1}{\lfloor\frac{K}{2+\Delta}\rfloor}\Big{(}\outerproduct{\tfrac{K}{2},{\tfrac{K}{2}}_{z}}{\tfrac{K}{2},-{\tfrac{K}{2}_{z}}}$
$\displaystyle\qquad{}+{}\outerproduct{\tfrac{K}{2},-{\tfrac{K}{2}}_{z}}{\tfrac{K}{2},{\tfrac{K}{2}}_{z}}\Big{)},$

where $\tilde{\jmath}\leq\sum_{n}j_{n}=K/2$ indexes the irreducible blocks
that arise from the usual addition of angular momentum formula,
$S_{K,\Delta}^{(\tilde{\jmath})}$ is as defined in Eq. (10), and the explicit
form of $S_{K,\Delta}^{(\tilde{\jmath}\leq K/2)}$ is taken from Appendices E
and F.

For a state $\rho_{\lnot\text{GME}}$ that is not _genuine multipartite
entangled_ , that is, if $\rho_{\lnot\text{GME}}$ can be written as a convex
combination of states that are separable over some partition of the $N$
particles [17], we prove in Appendix G that

$\absolutevalue{\tr(\rho_{\lnot\text{GME}}S_{K,\Delta})}\leq\frac{\mathbf{s}_{K,\Delta}}{2}=:\mathbf{s}^{K\text{-sep}}_{K,\Delta}.$
(22)

As such, observing $\absolutevalue{\tr(\rho
S_{K,\Delta})}>\mathbf{s}^{K\text{-sep}}_{K,\Delta}$ implies that
$\rho\neq\rho_{\lnot\text{GME}}$. The even-parity precession protocol
performed on the total angular momentum of a spin ensemble is therefore a
witness of genuine multipartite entanglement.

In the case that all $N$ particles are qubits, the maximally-violating state
is

$\ket{\mathbf{s}_{K,\Delta}^{(K/2)}}=\frac{1}{\sqrt{2}}\pqty{\ket{\uparrow}^{\otimes
N}+(-1)^{\lfloor(\Delta+K)/2\rfloor}\ket{\downarrow}^{\otimes N}},$ (23)

where $\ket{\uparrow}:=\lvert{\frac{1}{2},{\frac{1}{2}}_{z}}\rangle$ and
$\ket{\downarrow}:=\lvert{\frac{1}{2},-{\frac{1}{2}}_{z}}\rangle$. This is
exactly the GHZ state of $N$ qubits, where $N$ is even.

Since our witness requires only measurements of the total angular momentum, it
belongs to the family of entanglement witnesses that utilize only collective
observables. Past witnesses in this family have been shown to detect Dicke and
many-body singlet states [18, 19], but the detection of GHZ states for more
than three qubits using such a witness was not partially solved until the odd-
parity precession protocol was introduced, which detects GHZ states with an
odd number of qubits [9]. Our witness therefore completely closes this gap.

##  V Conclusion

In this work, we have introduced an even-parity precession protocol that can
detect nonclassicality and entanglement by simply measuring a uniformly-
precessing observable along different angles, or equivalently, at one of
different times, in each round. A score is assigned based on whether or not it
falls within a region close to the origin, and the average score is calculated
after many rounds. If the observed score exceeds the maximum classical score,
the nonclassicality of the system is certified.

Violation of the classical bound is shown for states of the quantum harmonic
oscillators, and for all spin systems with $j\geq 2$. Unlike other
nonclassicality tests, the protocol does not require simultaneous or
sequential measurements. In contrast to previous works on the odd-parity
precession protocol, this protocol can also detect states that commute with
the parity operator.

Lastly, we applied the protocol to composite systems. We showed that the
precession protocol can detect non-Gaussian entangled states of two coupled
harmonic oscillators, and genuine multipartite entanglement of spin ensembles.
In particular, GHZ states of an even number of qubit can be detected, thereby
completely closing the long-standing problem of detecting GHZ states using
collective spin measurements.

###### Acknowledgements.

This work is supported by the National Research Foundation, Singapore, and
A*STAR under its CQT Bridging Grant.

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##  Appendix A Geometric Proof of Classical Bound

In this Appendix, we derive the classical bound for the even-parity precession
protocol for any $\Delta$ and $K$. We start by parameterizing the pure state
of a pair of uniformly-precessing classical variables as

$\pmqty{undef}=\pmqty{undef},$ (24)

where $A_{0}^{2}:=A_{x}^{2}+A_{y}^{2}$ and
$\phi_{0}:=\arg[A_{x}(0)+iA_{y}(0)]$.

For a given initial state, the score would be

$\displaystyle s_{K,\Delta}$
$\displaystyle=\frac{1}{K}\sum_{k=0}^{K-1}(-1)^{k}\Pr[\absolutevalue{A_{x}(\theta_{k})}\leq\frac{\Delta}{2}]$
(25)
$\displaystyle=\frac{1}{K}\sum_{k=0}^{K-1}(-1)^{k}\Pr[A_{0}\absolutevalue{\cos(\frac{\pi
k}{K}+\phi_{0})}\leq\frac{\Delta}{2}].$

It can be immediately seen that if $A_{0}\leq\Delta/2$, then
$A_{0}\absolutevalue{\cos(\frac{\pi k}{K}+\phi_{0})}\leq A_{0}\leq\Delta/2$.
Hence, since $s_{K,\Delta}\propto\sum_{k=0}^{K-1}(-1)^{k}=0$ for all initial
states with $A_{0}\leq\Delta/2$, we need only consider the cases when
$A_{0}>\Delta/2$.

It will be helpful to illustrate this geometrically. There are two main cases,
as illustrated in Fig. 7: the trajectory of the particle from $\theta=0$ to
$\theta=(K-1)/K$ intersects the $|A_{x}|\leq\Delta/2$ region (henceforth the
$\Delta$-region) only once like in Fig. 7(a–b), or twice like in Fig. 7(c–d).

Figure 7: Possible trajectories for the even-parity precession protocol when
$A_{0}>\Delta/2$, shown here for $K=12$.

Following the figure, we shall color the individual terms as follows:

$\displaystyle s_{K,\Delta}$
$\displaystyle=\pqty{\color[rgb]{0.9921875,0.64453125,0.25}\definecolor[named]{pgfstrokecolor}{rgb}{0.9921875,0.64453125,0.25}\frac{1}{K}}\Pr[\absolutevalue{\color[rgb]{0.9921875,0.64453125,0.25}\definecolor[named]{pgfstrokecolor}{rgb}{0.9921875,0.64453125,0.25}A_{x}(\theta_{0})}\leq\frac{\Delta}{2}]$
(26)
$\displaystyle\qquad{}+{}\pqty{\color[rgb]{0.296875,0.34765625,0.60546875}\definecolor[named]{pgfstrokecolor}{rgb}{0.296875,0.34765625,0.60546875}-\frac{1}{K}}\Pr[\absolutevalue{\color[rgb]{0.296875,0.34765625,0.60546875}\definecolor[named]{pgfstrokecolor}{rgb}{0.296875,0.34765625,0.60546875}A_{x}(\theta_{1})}\leq\frac{\Delta}{2}]$
$\displaystyle\qquad{}+{}\dots+\pqty{\color[rgb]{0.296875,0.34765625,0.60546875}\definecolor[named]{pgfstrokecolor}{rgb}{0.296875,0.34765625,0.60546875}-\frac{1}{K}}\Pr[\absolutevalue{\color[rgb]{0.296875,0.34765625,0.60546875}\definecolor[named]{pgfstrokecolor}{rgb}{0.296875,0.34765625,0.60546875}A_{x}(\theta_{K-1})}\leq\frac{\Delta}{2}].$

For every term where $k$ is even () and within the $\Delta$-region, there is a
contribution of $+1/K$ to $s_{K,\Delta}$. Similarly, terms where $k$ is odd ()
and within the $\Delta$-region contribute $-1/K$. Any pair of even and odd
points ($\leavevmode\hbox to5.38pt{\vbox
to5.38pt{\pgfpicture\makeatletter\hbox{\hskip 2.69096pt\lower-0.2723pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{{}{{}}{}{{{}} {}{}{}{}{}{}{}{}
}\pgfsys@beginscope\pgfsys@invoke{
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}\definecolor[named]{pgfstrokecolor}{rgb}{0.9921875,0.64453125,0.25}\pgfsys@color@rgb@stroke{0.9921875}{0.64453125}{0.25}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{2.41867pt}\pgfsys@moveto{2.15277pt}{2.41867pt}\pgfsys@curveto{2.15277pt}{3.60762pt}{1.18895pt}{4.57144pt}{0.0pt}{4.57144pt}\pgfsys@curveto{-1.18895pt}{4.57144pt}{-2.15277pt}{3.60762pt}{-2.15277pt}{2.41867pt}\pgfsys@curveto{-2.15277pt}{1.22972pt}{-1.18895pt}{0.2659pt}{0.0pt}{0.2659pt}\pgfsys@curveto{1.18895pt}{0.2659pt}{2.15277pt}{1.22972pt}{2.15277pt}{2.41867pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{2.41867pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}+\leavevmode\hbox
to5.38pt{\vbox to5.38pt{\pgfpicture\makeatletter\hbox{\hskip
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}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{{}{{}}{}{{{}} {}{}{}{}{}{}{}{}
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}{}\pgfsys@moveto{0.0pt}{2.41867pt}\pgfsys@moveto{2.15277pt}{2.41867pt}\pgfsys@curveto{2.15277pt}{3.60762pt}{1.18895pt}{4.57144pt}{0.0pt}{4.57144pt}\pgfsys@curveto{-1.18895pt}{4.57144pt}{-2.15277pt}{3.60762pt}{-2.15277pt}{2.41867pt}\pgfsys@curveto{-2.15277pt}{1.22972pt}{-1.18895pt}{0.2659pt}{0.0pt}{0.2659pt}\pgfsys@curveto{1.18895pt}{0.2659pt}{2.15277pt}{1.22972pt}{2.15277pt}{2.41867pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{2.41867pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}$ or $\leavevmode\hbox
to5.38pt{\vbox to5.38pt{\pgfpicture\makeatletter\hbox{\hskip
2.69096pt\lower-0.2723pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{{}{{}}{}{{{}} {}{}{}{}{}{}{}{}
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}\definecolor[named]{pgfstrokecolor}{rgb}{0.296875,0.34765625,0.60546875}\pgfsys@color@rgb@stroke{0.296875}{0.34765625}{0.60546875}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{2.41867pt}\pgfsys@moveto{2.15277pt}{2.41867pt}\pgfsys@curveto{2.15277pt}{3.60762pt}{1.18895pt}{4.57144pt}{0.0pt}{4.57144pt}\pgfsys@curveto{-1.18895pt}{4.57144pt}{-2.15277pt}{3.60762pt}{-2.15277pt}{2.41867pt}\pgfsys@curveto{-2.15277pt}{1.22972pt}{-1.18895pt}{0.2659pt}{0.0pt}{0.2659pt}\pgfsys@curveto{1.18895pt}{0.2659pt}{2.15277pt}{1.22972pt}{2.15277pt}{2.41867pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{2.41867pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}+\leavevmode\hbox
to5.38pt{\vbox to5.38pt{\pgfpicture\makeatletter\hbox{\hskip
2.69096pt\lower-0.2723pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{{}{{}}{}{{{}} {}{}{}{}{}{}{}{}
}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{1.07639pt}\pgfsys@invoke{
}\definecolor[named]{pgfstrokecolor}{rgb}{0.9921875,0.64453125,0.25}\pgfsys@color@rgb@stroke{0.9921875}{0.64453125}{0.25}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{2.41867pt}\pgfsys@moveto{2.15277pt}{2.41867pt}\pgfsys@curveto{2.15277pt}{3.60762pt}{1.18895pt}{4.57144pt}{0.0pt}{4.57144pt}\pgfsys@curveto{-1.18895pt}{4.57144pt}{-2.15277pt}{3.60762pt}{-2.15277pt}{2.41867pt}\pgfsys@curveto{-2.15277pt}{1.22972pt}{-1.18895pt}{0.2659pt}{0.0pt}{0.2659pt}\pgfsys@curveto{1.18895pt}{0.2659pt}{2.15277pt}{1.22972pt}{2.15277pt}{2.41867pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{2.41867pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}$) within the
$\Delta$-region contributes nothing to the score.

###  A.1 One Intersection

If there is only a single intersection, there are three possible cases: the
extreme points—the leftmost and rightmost points within the
$\Delta$-region—are both orange, both blue, or are different colours.

Keeping in mind that the colours of the points must alternate due to the
$(-1)^{k}$ factor, if the extreme points are both orange, then the
contribution of the points within the $\Delta$-region are

$\displaystyle\leavevmode\hbox to5.38pt{\vbox
to5.38pt{\pgfpicture\makeatletter\hbox{\hskip 2.69096pt\lower-0.2723pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{{}{{}}{}{{{}} {}{}{}{}{}{}{}{}
}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{1.07639pt}\pgfsys@invoke{
}\definecolor[named]{pgfstrokecolor}{rgb}{0.9921875,0.64453125,0.25}\pgfsys@color@rgb@stroke{0.9921875}{0.64453125}{0.25}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{2.41867pt}\pgfsys@moveto{2.15277pt}{2.41867pt}\pgfsys@curveto{2.15277pt}{3.60762pt}{1.18895pt}{4.57144pt}{0.0pt}{4.57144pt}\pgfsys@curveto{-1.18895pt}{4.57144pt}{-2.15277pt}{3.60762pt}{-2.15277pt}{2.41867pt}\pgfsys@curveto{-2.15277pt}{1.22972pt}{-1.18895pt}{0.2659pt}{0.0pt}{0.2659pt}\pgfsys@curveto{1.18895pt}{0.2659pt}{2.15277pt}{1.22972pt}{2.15277pt}{2.41867pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{2.41867pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
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}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
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} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}+\leavevmode\hbox
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} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}
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$\displaystyle=\underbrace{(\leavevmode\hbox to5.38pt{\vbox
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\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
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$0$}\\!\\!\\!\\!\\!\\!}{}+{}(\leavevmode\hbox to5.38pt{\vbox
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} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}})+\underbrace{\leavevmode\hbox
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} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}_{\text{contributes
$+1/K$}}$ $\displaystyle\implies$ $\displaystyle s_{K,\Delta}=1/K.$

This is also trivially true when the extreme points are the same, in the sense
that there is only one point within the $\Delta$-region.

A similar argument gives $s_{K,\Delta}=-1/K$ when the extreme points are both
blue, and $s_{K,\Delta}=0$ when the extreme points are different colours.
Therefore, the possible classical scores are $s_{K,\Delta}^{c}\in\\{0,\pm
1/K\\}$ when there is only one intersection.

###  A.2 Two Intersections

With two intersections, one at $A_{y}>0$ and another at $A_{y}<0$, we can
consider the contributions of each intersection individually using the
previous technique, then sum them up to obtain the total score. Doing so, it
seems that the scores $s_{K,\Delta}=\pm 2/K$ should be possible when the
extreme points of both intersections are the same colour. In this section, we
show that this can never be the case.

Figure 8: Geometric argument that the initial point must always be within the
$\Delta$-region when there are two intersections.

We shall again tackle this geometrically. For a trajectory to intersect the
$\Delta$-region twice, it has to exit and reenter the region. Since the
trajectory is along a circular path while the rectangular region extends from
$A_{y}=-\infty$ to $A_{y}=\infty$, it must exit and reenter at one of the two
edges of the region (the lines $A_{x}=\Delta/2$ and $A_{x}=-\Delta/2$). Let us
take that the trajectory crosses $A_{x}=\Delta/2$, as shown in Fig. 8(a) (the
argument proceeds similarly with $A_{x}=-\Delta/2$). As the precession is
clockwise and the center of the circular arc is at the origin, the exit occurs
at $A_{y}>0$ and the entrance at $A_{y}<0$ (conversely, exit at $A_{y}<0$ and
entrance at $A_{y}>0$ if trajectory crosses $A_{x}=-\Delta/2$).

Now, this means that the circular arc that goes from the exit point to the
entrance point must be included in the trajectory, and the initial point
should occur somewhere before the exit point, as marked out in Fig. 8(b).

However, we also know that the full trajectory must be a circular arc that
subtends an angle of $\theta_{K-1}-\theta_{0}<\pi$. Hence, as shown in Fig.
8(c), the initial point can be _at most_ an angle $\pi$ away from the entrance
point (initial point precedes the exit, which precedes the entrance).

As such, we conclude that the initial point is between an angle $\pi$ away
from the entrance, and the exit point. We can see in Fig. 8(d) that this
implies that (1) the initial point must be in the $\Delta$-region and (2) the
initial point occurs in the $A_{y}>0$ plane. Again invoking that the full
circular arc must be smaller than a semicircle, the trajectory will not
reenter the $\Delta$-region from the left, which means that (3) the initial
point is an extreme point of the $\Delta$-region for the intersection at
$A_{y}>0$.

This considerations lead us to the conclusion that when the trajectory crosses
$A_{x}=1/K$, the points within $\Delta$-region at $A_{y}>0$, since one of its
extreme points is the orange initial point, can only contribute $0$ or $+1/K$
to the total score.

Repeating this argument to find the possible position of the final ($k=K-1$)
point, we will find that the blue final point (1) must be in the
$\Delta$-region, (2) occurs in the $A_{y}<0$ plane, and (3) is an extreme
point for the intersection at $A_{y}<0$. Therefore, when the trajectory
crosses $A_{x}=1/K$, the points within $\Delta$-region at $A_{y}<0$ can only
contribute $0$ or $-1/K$ to the total score.

As such, when the trajectory crosses $A_{x}=1/K$, the total score can be at
most $s_{K,\Delta}^{c}=+1/K+0=1/K$ and at least
$s_{K,\Delta}^{c}=0+(-1/K)=-1/K$. A similar argument gives the same conclusion
when the trajectory crosses $A_{y}=-1/K$, and therefore we have that the
classical bound is

$\absolutevalue{s_{K,\Delta}^{c}}\leq\frac{1}{K}=:\mathbf{s}_{K,\Delta}^{c}.$
(28)

##  Appendix B Matrix Elements of $S_{K,\Delta}^{\infty}$ for the Quantum
Harmonic Oscillator

In this appendix, we derive the matrix elements of $S_{K,\Delta}^{\infty}$. We
shall first show that $\bra{n}S_{K,\Delta}^{\infty}\ket{n^{\prime}}$ is
nonzero only when $n-n^{\prime}=(2l+1)K$, where $n$ and $n^{\prime}$ are the
quantum numbers of the corresponding eigenstates with
$H\ket{n}=(n+\frac{1}{2})\hbar\omega$, and $l$ is an integer.

We begin by defining the averaging operation $E_{K}^{\infty}[\bullet]$ as

$E_{K}^{\infty}[\bullet]:=\frac{1}{K}\sum_{k=0}^{K-1}(-1)^{k}e^{i\theta_{k}\frac{H}{\hbar\omega}}[\bullet]e^{-i\theta_{k}\frac{H}{\hbar\omega}}.$
(29)

In terms of this,

$S_{K,\Delta}^{\infty}=E_{K}^{\infty}\bqty{\int_{-\sqrt{\frac{\hbar\Delta^{2}}{2m\omega}}}^{\sqrt{\frac{\hbar\Delta^{2}}{2m\omega}}}\differential{x}\outerproduct{x}{x}}=:E_{K}^{\infty}\bqty{P^{x}_{\Delta}},$
(30)

where we defined the projector $P^{x}_{\Delta}$. Expanding $P^{x}_{\Delta}$ in
the number basis, we have
$P^{x}_{\Delta}=\sum_{n,n^{\prime}=0}^{\infty}P^{x}_{\Delta|n,n^{\prime}}\outerproduct{n}{n^{\prime}}$
with matrix elements
$P^{x}_{\Delta|n,n^{\prime}}:=\bra{n}P^{x}_{\Delta}\ket{n^{\prime}}$. Plugging
this into the expression of $S_{K,\Delta}^{\infty}$,

$E_{K}^{\infty}\bqty{P^{x}_{\Delta}}=\frac{1}{K}\sum_{n,n^{\prime}}P^{x}_{\Delta|n,n^{\prime}}\outerproduct{n}{n^{\prime}}\sum_{k=0}^{K-1}(-1)^{k}e^{-i\theta_{k}(n-n^{\prime})}$
(31)

With $\theta_{k}=\pi/K$, we get
$\sum_{k=0}^{K-1}(-1)^{k}e^{-i\theta_{k}n-n^{\prime}}=0$ when
$n-n^{\prime}=2lK$ for integer $l$, and
$\sum_{k=0}^{K-1}(-1)^{k}e^{-i\theta_{k}\frac{n-n^{\prime}}{\hbar\omega}}=K$
for $n-n^{\prime}=(2l+1)K$ with an integer $l$. These cover the cases for
$n-n^{\prime}$ even.

When $n-n^{\prime}$ is odd, we have
$e^{-i\pi\frac{H}{\hbar\omega}}\ket{x}=\ket{-x}$ and
$\int_{-x_{0}}^{x_{0}}\differential{x}\outerproduct{x}{x}=\int_{-x_{0}}^{x_{0}}\differential{x}\outerproduct{-x}{-x}$,
which tells us that

$\displaystyle P^{x}_{\Delta|n,n^{\prime}}$ $\displaystyle=$
$\displaystyle\bra{n}\bqty{\int_{-\sqrt{\frac{\hbar\Delta^{2}}{2m\omega}}}^{\sqrt{\frac{\hbar\Delta^{2}}{2m\omega}}}\differential{x}e^{-i\pi\frac{H}{\hbar\omega}}\outerproduct{-x}{-x}e^{i\pi\frac{H}{\hbar\omega}}}\ket{n^{\prime}}$
(32) $\displaystyle=$ $\displaystyle
e^{-i\pi(n-n^{\prime})}P^{x}_{\Delta|n,n^{\prime}}=-P^{x}_{\Delta|n,n^{\prime}}$

As such, $P^{x}_{\Delta|n,n^{\prime}}=0$ for $n-n^{\prime}$ odd. From the
above argument, we conclude that the matrix elements of the operator
$S_{K,\Delta}^{\infty}$ are nonzero only when $n-n^{\prime}=(2l+1)K$ with an
integer $l$.

Using the technique by Mawby and Halliwell [20], we can also work out
$P^{x}_{\Delta|n,n^{\prime}}$, which gives us the nonzero matrix elements of
$S_{K,\Delta}^{\infty}$. Therefore,

$\bra{n}S_{K,\Delta}^{\infty}\ket{n^{\prime}}=\begin{cases}\frac{2e^{-\frac{\Delta^{2}}{4}}\bqty{n^{\prime}H_{n}(\frac{\Delta}{2})H_{n^{\prime}-1}(\frac{\Delta}{2})-nH_{n^{\prime}}(\frac{\Delta}{2})H_{n-1}(\frac{\Delta}{2})}}{(n-n^{\prime})\sqrt{2^{n+n^{\prime}}n!n^{\prime}!}}&\text{if
$(n-n^{\prime})\bmod K=0$ and $(n-n^{\prime})\bmod(2K)\neq 0$,}\\\
0&\text{otherwise,}\end{cases}$ (33)

where $H_{n}(x)=(-1)^{n}e^{x^{2}}\frac{d^{n}e^{-x^{2}}}{dx^{n}}$ are the
Hermite polynomials.

##  Appendix C Quantum Score for the Harmonic Oscillator Matches or Violates
Classical Bound

With a similar geometric proof as Appendix A, it can be shown that initial
points $(x,p)$ with
$\absolutevalue{x}\leq\Delta/2,p>\Delta/[2\sin(\pi/K)]=:p_{0}$ obtain the
score $1/K$. As such, we can lower bound the quantum score in terms of the
Wigner function as

$\displaystyle s_{K,\Delta}^{\infty}$
$\displaystyle\geq\frac{1}{K}\int_{-\frac{\Delta}{2}}^{\frac{\Delta}{2}}\differential{x}\int_{p_{0}}^{\infty}\differential{p}W(x,p)$
(34)
$\displaystyle\qquad{}-{}\frac{1}{K}\bqty{1-\int_{-\frac{\Delta}{2}}^{\frac{\Delta}{2}}\differential{x}\int_{p_{0}}^{\infty}\differential{p}W(x,p)}$
$\displaystyle=\frac{1}{K}\bqty{2\int_{-\frac{\Delta}{2}}^{\frac{\Delta}{2}}\differential{x}\int_{p_{0}}^{\infty}\differential{p}W(x,p)-1}$

Given the state $W(x,p)\propto
e^{-\sigma^{2}x^{2}}e^{-(p-p_{0}-\Delta\sigma^{2}/2)^{2}/\sigma^{2}}$, the
lower bound approaches $1/K$ as $\sigma\to 0$, which implies that the maximum
score of quantum harmonic oscillator is at least the classical bound.

##  Appendix D Relationship Between Quantum Violation and Wigner negativity

The score achieved by a state given by $W(x,p)$ can be calculated as

$\displaystyle s_{K,\Delta}^{\infty}$
$\displaystyle=\frac{1}{K}\sum_{k=1}^{\infty}(-1)^{k}\int\differential{x}\int\differential{p}\Theta(|x|\leq\Delta/2)W(x,p;t_{k})$
(35)
$\displaystyle=\frac{1}{K}\sum_{k=1}^{\infty}(-1)^{k}\int\differential{x}\int\differential{p}\Theta(|x|\leq\Delta/2)$
$\displaystyle\quad
W(\cos(t_{k})x-\sin(t_{k})p,\cos(t_{k})p+\sin(t_{k})x;t=0)$
$\displaystyle=\int\differential{x}\int\differential{p}\frac{1}{K}\sum_{k=1}^{\infty}(-1)^{k}$
$\displaystyle\quad\Theta(|\cos(t_{k})x+\sin(t_{k})p|\leq\Delta/2)W(x,p)$
$\displaystyle=\mathbf{s}_{K,\Delta}^{c}\int\differential{x}\int\differential{p}\bqty{\Theta_{+}(x,p)-\Theta_{-}(x,p)}W(x,p)$

where $\Theta_{\pm}(x,p)$ take the value $1$ in the regions where the
classical score is $\pm\mathbf{s}_{K,\Delta}^{c}$ and $0$ otherwise. The
regions are mutually exclusive, so $|\Theta_{+}(x,p)-\Theta_{-}(x,p)|\leq 1$.
Writing $W(x,p)=W^{+}(x,p)-W^{-}(x,p)$ with $W^{\pm}(x,p):=[|W(x,p)|\pm
W(x,p)]/2\geq 0$,

$\displaystyle\frac{s_{K,\Delta}^{\infty}}{\mathbf{s}_{K,\Delta}^{c}}$
$\displaystyle=\int\differential{x}\int\differential{p}\bqty{\Theta_{+}(x,p)-\Theta_{-}(x,p)}W(x,p)$
(36) $\displaystyle s_{K,\Delta}^{\infty}K$
$\displaystyle=\int\differential{x}\int\differential{p}W(x,p)$
$\displaystyle+\int\differential{x}\int\differential{p}\underbrace{\bqty{1+\Theta_{-}(x,p)-\Theta_{+}(x,p)}}_{\geq
0,\leq 2}\bqty{-W(x,p)}$
$\displaystyle\leq\int\differential{x}\int\differential{p}W(x,p)+2\int\differential{x}\int\differential{p}W^{-}(x,p)$
$\displaystyle=2\mathcal{N}_{V}+1,$

where $\mathcal{N}_{V}:=\iint\differential{x}\differential{p}W^{-}(x,p)$ is
the Wigner negativity volume.

Rearranging (36) and noting that $\mathbf{s}_{K,\Delta}^{c}=1/K$, we get

$K(|s_{K,\Delta}^{\infty}|-\mathbf{s}_{K,\Delta}^{c})\leq 2\mathcal{N}_{V}.$
(37)

##  Appendix E Scores when $j<K/2$

In this appendix, we derive the matrix elements of $S_{K,\Delta}^{(j)}$. As
the steps are similar to the quantum harmonic oscillator, we provide only a
sketch of the proofs: one can refer to Appendix B for the details.

We again define

$E_{K}^{(j)}[\bullet]:=\frac{1}{K}\sum_{k=0}^{K-1}(-1)^{k}e^{-i\theta_{k}\frac{J_{z}^{(j)}}{\hbar}}[\bullet]e^{i\theta_{k}\frac{J_{z}^{(j)}}{\hbar}},$
(38)

with which we have $S_{K,\Delta}^{(j)}=E_{K}^{(j)}[P^{(j)}_{\Delta}]$ in terms
of the projector

$\displaystyle P^{(j)}_{\Delta}$
$\displaystyle:=\Pr(\frac{|J_{x}^{(j)}|}{\hbar}\leq\frac{\Delta}{2})$ (39)
$\displaystyle=\sum_{m,m^{\prime}=-j}^{j}P^{(j)}_{\Delta|m,m^{\prime}}\outerproduct{j,m_{z}}{j,m_{z}^{\prime}}.$

As before, $\bra{j,m_{z}}S_{K,\Delta}^{(j)}\ket{j,m_{z}^{\prime}}=0$ when
$m-m^{\prime}=2lK$ for integer $l$, and
$\bra{j,m_{z}}S_{K,\Delta}^{(j)}\ket{j,m_{z}^{\prime}}=P^{(j)}_{\Delta|m,m^{\prime}}$
when $m-m^{\prime}=(2l+1)K$ for integer $l$. These cover the cases for
$m-m^{\prime}$ even.

When $m-m^{\prime}$ is odd, $e^{-i\pi J_{z}/\hbar}J_{x}e^{i\pi
J_{z}/\hbar}=-J_{x}$ implies that $e^{-i\pi
J_{z}/\hbar}P_{\Delta}^{(j)}e^{i\pi J_{z}/\hbar}=P_{\Delta}^{(j)}$, with which

$\displaystyle P_{\Delta|m,m^{\prime}}^{(j)}$
$\displaystyle=\bra{j,m_{z}}e^{-i\pi\frac{J_{z}}{\hbar}}P_{\Delta}^{(j)}e^{i\pi\frac{J_{z}}{\hbar}}\ket{j,m^{\prime}_{z}}$
(40)
$\displaystyle=e^{-i\pi(m-m^{\prime})}P^{(j)}_{\Delta|m,m^{\prime}}=-P^{(j)}_{\Delta|m,m^{\prime}}.$

As such, $P_{\Delta|m,m^{\prime}}^{(j)}=0$ for $m-m^{\prime}$ odd.

From the derivation above, the matrix elements of $S_{K,\Delta}^{(j)}$,
written in the basis of $J_{z}$, are nonzero only when $m-m^{\prime}=(2l+1)K$
for integer $l$. When $j<K/2$, $|m-m^{\prime}|\leq K$ implies that
$m-m^{\prime}\neq(2l+1)K$ for all $m,m^{\prime}$, which implies that the
matrix is a zero matrix. Therefore, $s_{K,\Delta}^{(j<K/2)}=0$.

##  Appendix F Scores when $j=K/2$

When $j=K/2$, the only nonzero matrix elements of $S_{K,\Delta}^{(j)}$ are
$\langle{\frac{K}{2},\pm\frac{K}{2}_{z}}\rvert
S_{K,\Delta}^{(j)}\lvert{\frac{K}{2},\mp\frac{K}{2}_{z}}\rangle$. The
eigenvalue of $S_{K,\Delta}^{(j=K/2)}$ can be solved with standard techniques,
which gives the maximum score

$\displaystyle\mathbf{s}_{K,\Delta}^{(j)}$
$\displaystyle=\absolutevalue{\bra{\tfrac{K}{2},\tfrac{K}{2}_{z}}\Pr(\frac{\absolutevalue\big{J_{x}^{(j)}}}{\hbar}\leq\frac{\Delta}{2})\ket{\tfrac{K}{2},-\tfrac{K}{2}_{z}}}$
(41)
$\displaystyle=\absolutevalue{\sum_{m=-(\lfloor\tfrac{\Delta}{2}+\tfrac{K}{2}\rfloor-\frac{K}{2})}^{\lfloor\tfrac{\Delta}{2}+\tfrac{K}{2}\rfloor-\frac{K}{2}}\innerproduct{\tfrac{K}{2},\tfrac{K}{2}_{z}}{\tfrac{K}{2},m_{x}}\\!\innerproduct{\tfrac{K}{2},m_{x}}{\tfrac{K}{2},-\tfrac{K}{2}_{z}}}$

The quantities $\innerproduct{j,m_{z}}{j,m_{x}^{\prime}}$, which are elements
of the so-called “Wigner $d$-matrix”, which has the known relations [15]

$\displaystyle\innerproduct{j,m_{z}^{\prime}}{j,m_{x}}$
$\displaystyle=(-1)^{m-m^{\prime}}\innerproduct{j,m_{z}}{j,m_{x}^{\prime}}$
(42) $\displaystyle=(-1)^{j+m}\innerproduct{j,-m_{z}^{\prime}}{j,m_{x}},$

and

$\innerproduct{j,j_{z}}{j,m_{x}}=2^{-j}\sqrt{\binom{2j}{j-m}}.$ (43)

With these relations, we obtain

$\bra{\tfrac{K}{2},\tfrac{K}{2}_{z}}\ket{\tfrac{K}{2},m_{x}}=(-1)^{\frac{K}{2}-m}\,2^{-\frac{K}{2}}\sqrt{\binom{K}{\frac{K}{2}-m}}$
(44)
$\bra{\tfrac{K}{2},m_{x}}\ket{\tfrac{K}{2},-\tfrac{K}{2}_{z}}=2^{-\frac{K}{2}}\sqrt{\binom{K}{\frac{K}{2}-m}}$
(45)

we can calculate the maximum score to be

$\displaystyle\mathbf{s}_{K,\Delta}^{(j)}$
$\displaystyle=\absolutevalue{\sum_{m=-(\lfloor\tfrac{\Delta}{2}+\tfrac{K}{2}\rfloor-\frac{K}{2})}^{\lfloor\tfrac{\Delta}{2}+\tfrac{K}{2}\rfloor-\frac{K}{2}}(-1)^{\frac{K}{2}-m}\,2^{-K}\binom{K}{\frac{K}{2}-m}}$
(46) $\displaystyle=2^{-(K-1)}\binom{K-1}{\lfloor\frac{K+\Delta}{2}\rfloor}.$

This quantity is maximized for the central binomial coefficient
$\max_{\Delta}\binom{K-1}{\lfloor\frac{K+\Delta}{2}\rfloor}=\binom{K-1}{\frac{K}{2}}$.
Using mathematical induction, we can also prove that
$\max_{\Delta}\mathbf{s}_{K,\Delta}^{(j=K/2)}=2^{-K+1}\binom{K-1}{\frac{K}{2}}>1/K=\mathbf{s}_{K}^{c}$
when $K\geq 4$, which means that spin angular momentum systems can violate the
classical bound for all $K\geq 4$.

##  Appendix G Proof of Separable Bound for Spin Ensembles

Partition a system into two, and write a pure separable state with respect to
this bipartition as $\ket{\psi}=\ket{\psi_{1}}\otimes\ket{\psi_{2}}$, where
$\ket{\psi_{1,2}}$ denote the states corresponding to each subsystem. For such
a state, the observed score will be

$\displaystyle S_{\mathrm{sep}}$ $\displaystyle=$
$\displaystyle\bra{\psi_{1}}\otimes\bra{\psi_{2}}S_{K,\Delta}\ket{\psi_{1}}\otimes\ket{\psi_{2}}$
(47) $\displaystyle=$
$\displaystyle\bra{\psi_{1}}\Tr_{2}[S_{K,\Delta}(I_{1}\otimes\ket{\psi_{2}}\bra{\psi_{2}})]\ket{\psi_{1}}.$

Thus, the task at hand is to determine the maximum eigenvalue of the operator
$\Tr_{2}[S_{K,\Delta}(I_{1}\otimes\ket{\psi_{2}}\bra{\psi_{2}})]$.

Here we can define $\ket{0}=\otimes\ket{j_{n},-j_{n}}$ and
$\ket{1}=\otimes\ket{j_{n},j_{n}}$, where $\ket{j_{n},\pm j_{n}}$ are
eigenstates of $J_{z}$. Then we can write
$I_{1}=\ket{0}_{1}\bra{0}_{1}+\ket{1}_{1}\bra{1}_{1}+\sum_{k}\ket{k}_{1}\bra{k}_{1}$
and $\ket{\psi_{2}}=C_{1}\ket{0}_{2}+\ket{1}_{2}+\sum_{l}C_{l}\ket{l}_{2}$.
From the expression of $S_{K,\Delta}$, it is not difficult to get

$S_{K,\Delta}(\ket{k}_{1}\bra{k}_{1}\otimes\ket{l}_{2}\bra{l}_{2})=0$ (48)

So finally we can simplify the expression as

$\begin{split}&\Tr_{2}[S_{K,\Delta}(I_{1}\otimes\ket{\psi_{2}}\bra{\psi_{2}})]\\\
&=2^{-K+1}\binom{K-1}{\lfloor\frac{K+\Delta}{2}\rfloor}(C_{1}C_{2}^{*}\ket{1}_{1}\bra{0}_{1}+C_{1}^{*}C_{2}\ket{0}_{1}\bra{1}_{1})\end{split}$
(49)

We get eigenvalue
$\lambda_{\pm}=\pm\absolutevalue{C_{1}}\absolutevalue{C_{2}}$. With
$\absolutevalue{C_{1}}^{2}+\absolutevalue{C_{2}}^{2}\leq 1$, the maximum score
is
$\textbf{s}_{K,\Delta}=2^{-K}\binom{K-1}{\lfloor\frac{K+\Delta}{2}\rfloor}$.