You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'We consider the bipartite spin entanglement between two identical fermions generated in spin-independent scattering. We show how the spatial degrees of freedom act as ancillas for the creation of entanglement to a degree that depends on the scattering angle, $\theta$. The number of Slater determinants generated in the process is greater than 1, corresponding to genuine quantum correlations between the identical fermions. The maximal entanglement attainable of 1 e-bit is reached at $\theta=\pi/2$. We also analyze a simple $\theta$ dependent Bell’s inequality, which is violated for $\pi/4<\theta\leq\pi/2$. This phenomenon is unrelated to the symmetrization postulate but does not appear for unequal particles.'
author:
- Lucas Lamata
- Juan León
title: 'Generation of bipartite spin entanglement via spin-independent scattering'
---

Introduction
============

Bipartite and multipartite entanglement is the main resource that gives rise to many of the applications of quantum information and computation, like for example quantum teleportation [@tele1; @tele2] and quantum cryptography [@crypto1; @Eke91], among others (see for instance Refs. [@nielsen; @GM02]). A compound system is entangled when it is impossible to attribute a complete set of properties to any of its parts. In this case, and for pure states, it is impossible to factor the state in a product of independent factors belonging to its parts. In this paper we will consider bipartite systems composed of two $s=\half$ fermions. Our aim is to uncover some specific features that apply when both particles are identical. They appear itemized in the next page.

States of two identical fermions have to obey the symmetrization postulate. This implies that they decompose into linear combinations of Slater determinants (SLs) of individual states. Naively, as these SLs cannot be factorized further, indistinguishability seems to imply entanglement. This is reinforced by the observation that the entropy of entanglement (EoE) is bounded from below by $S\geq 1$, well above the lower limit $S=0$ for a pair of non-entangled distinguishable particles. So, it looks like there is an inescapable amount of uncertainty, and hence of entanglement, in any state of two identical fermions. The above issue has been extensively examined in the literature [@entanglefermion1; @ESB+02; @entanglefermion2] with the following result: Part of the uncertainty (giving $S=1$) corresponds to the impossibility to individuate which one is the first or the second particle of the system. This explains why the lower limit for the EoE is 1. Consider for instance two identical $s=\half$ fermions in a singlet state $$\chi_S:=\frac{1}{\sqrt{2}}[\chi(1)^{\uparrow}\chi(2)^{\downarrow}-
\chi(1)^{\downarrow}\chi(2)^{\uparrow}].$$ The antisymmetrization does not preclude the assignment of properties to the particles, but only assigning them precisely to particle 1 or particle 2. The reduced density matrix of any of the particles is $\rho =
\frac{1}{2}\openone$ with an EoE $S(\rho) = 1$. The portion of $S$ above 1 (if any) is genuine entanglement as it corresponds to the impossibility of attributing precise properties to the particles of the system [@entanglefermion2]. Assume for instance that we endow the previous fermions with the capability of being outside ($\chi=\psi$) or inside ($\chi=\varphi$) the laboratory ($(\psi^i,\psi^j)=\delta^{ij}$, $(\varphi^i,\varphi^j)=\delta^{ij}$, $(\psi^i,\varphi^j)=0$, $i,j=\uparrow, \downarrow$). We now have two different possibilities: either the fermion outside has spin up ($\psi^{\uparrow}$) or spin down ($\psi^{\downarrow}$). Hence, there are two different SLs for a system built by a pair of particles with opposite spins, one outside, the other inside the laboratory $$\begin{aligned}
{\rm SL}(1,2)_1 &=&
\frac{1}{\sqrt{2}}[\psi(1)^{\uparrow}\,\varphi(2)^{\downarrow}-
\varphi(1)^{\downarrow}\, \psi(2)^{\uparrow}], \nonumber \\
{\rm SL}(1,2)_2 &=&
\frac{1}{\sqrt{2}}[\psi(1)^{\downarrow}\,\varphi(2)^{\uparrow}-
\varphi(1)^{\uparrow}\, \psi(2)^{\downarrow}]. \label{j1}\end{aligned}$$ They form two different biorthogonal states, the combination $[{\rm
SL}(1,2)_1-{\rm SL}(1,2)_2]/\sqrt{2}$ corresponding to the singlet and $[{\rm SL}(1,2)_1+{\rm SL}(1,2)_2]/\sqrt{2}$ to the triplet state (with respect to the total spin $\mathbf{s}=\mathbf{s}_1+\mathbf{s}_2$). An arbitrary state $\Phi(1,2)$ would then be a linear combination of these two SLs: $$\Phi(1,2)\,=\, c_1 {\rm SL}(1,2)_1\,+\,c_2 {\rm SL}(1,2)_2,\,\,
\sum_i|c_i|^2 = 1, \label{j2}$$ giving an EoE $$S\,=\,1-\sum_i |c_i|^2\, \log_2|c_i|^2 \,\geq \,1. \label{j3}$$ Clearly, when $c_1$ or $c_2$ vanish, we come back to $S=1$, as the only uncertainty left is the very identity of the particles. Summarizing, while indistinguishability is an issue to be solved by antisymmetrization within each SL, entanglement is an issue pertaining to the superposition of different SLs [@entanglefermion1; @ESB+02; @entanglefermion2]. At the end, we could even decide to call 1 to the variables of the outside particle, and forget about symmetrization $${\rm SL}(1,2) \rightarrow \, c_1
\psi(1)^{\uparrow}\,\varphi(2)^{\downarrow}\,+\,c_2
\psi(1)^{\downarrow}\,\varphi(2)^{\uparrow}, \label{j4}$$ as both particles are far away from each other. In this case, the EoE $S=\,-\sum_i |c_i|^2\, \log_2|c_i|^2\geq 0$ is lesser than the one corresponding to antisymmetrized states by a quantity of 1, which is just the uncertainty associated to antisymmetrization. From now on we will consider the latter definition of $S$, which gives the genuine amount of entanglement between the two particles. Notice that for half-odd $s$, the number $\#{\rm {\rm SL}}$ of Slater determinants is bounded by $\#{\rm {\rm SL}}\leq (2s+1)d/2$, where $d$ is the dimension of each Hilbert space of the configuration or momentum degrees of freedom for each of the two fermions.

Much in the same way as above, we could consider one of the particles as right moving ($\chi=\psi_0$) the other as left moving ($\chi=\psi_\pi$), giving rise to two SLs in parallel with the above discussion. This is the first step towards the inclusion of the full set of commuting operators for the system. In addition to the spin components ($s_1, s_2$) or helicities, there are the total $\mathbf{P}$ and relative $\mathbf{p}$ momenta. In the center of mass (CoM) frame we could consider the system described by the continuum of SLs $$\begin{aligned}
{\rm SL}(1,2;\mathbf{p})_s &=&
\frac{1}{\sqrt{2}}[\psi(1)_0^{s}\,\psi(2)_\pi^{-s}-
\psi(1)_\pi^{-s}\, \psi(2)_0^{s}], \nonumber\\
{\rm SL}(1,2;\mathbf{p})_{-s} &=&
\frac{1}{\sqrt{2}}[\psi(1)_0^{-s}\,\psi(2)_\pi^{s}-
\psi(1)_\pi^{s}\, \psi(2)_0^{-s}], \label{j5}\end{aligned}$$ where $\psi(1)_0^s=\langle 1|\mathbf{p}\,s\rangle$ and $\psi(1)_\pi^s=\langle 1|-\mathbf{p}\,s\rangle$. The labels 0 and $\pi$ are the azimuthal angles when we laid the axes along $\mathbf{p}$. Finally, there is a pair of SLs for each $\mathbf{p}$, so that a general state made with two opposite spin particles with relative momentum $\mathbf{p}$ could be written in the form: $$\Phi(1,2)_\mathbf{p}^0\,=\, \sum_{s=\pm 1/2}\,c_s(\mathbf{p})\,{\rm
SL}(1,2;\mathbf{p})_s, \label{j6}$$ with $\sum_{s=\pm 1/2}\,|c_s(\mathbf{p})|^2 =1$. Again, we run into the impossibility to tell which is 1 and which is 2. In addition there may be some uncertainty about the total spin state, whether a singlet or a triplet, or conversely, about the spin component of any of the particles, $\psi_0$ or $\psi_\pi$.

After this discussion it should be clear to what extent entanglement and distinguishability belong to different realms [@entanglefermion1; @ESB+02; @entanglefermion2]. The only requirement to include identical particles is to symmetrize the expressions used for unlike particles. Until now, we have only considered the free case. We have to examine the case of two interacting particles, as interaction is expected to be the source of subsequent entanglement [@LM76; @Tor85; @PS03; @MY04; @AAM04; @SALW04; @H05; @TK05; @LLS05; @W05]. Obviously, the answer may depend on a tricky way on the detailed form of the interaction, of its spin dependence in particular . It also seems that the role of particles identity, if any, will be played through symmetrization.

In the following we will show that spin entanglement is generated for the case of two interacting spin-$\half$ identical particles, with the following features:

-   Spin-spin entanglement is generated even by spin independent interactions.

-   In this case, it is independent of any symmetrization procedure.

-   This phenomenon does not appear for unlike particles.

Spin entanglement via spin-independent scattering
=================================================

We first tackle the scattering of two unequal $s=\half$ particles $A$ and $B$ which run into each other with relative CoM momentum $\mathbf{p}$. We set the frame axes by the initial momentum $\mathbf{p}$ of particle $A$, and let the spin components be $s_a=s$ and $s_b=-s$ along an arbitrary but fixed axis. We will consider a spin independent Hamiltonian $H$, so the evolution conserves $\mathbf{s}_a$ and $\mathbf{s}_b$. We denote by $A_{\theta}^s$ $(B_{\theta}^s)$ the state of particle $A$ ($B$) that propagates along direction $\theta$ with spin $s$. In these conditions the scattering proceeds as: $$\Phi_{\mbox{in}}=\, A_0^s\, B_\pi^{-s}\longrightarrow\,
\Phi_{\mbox{out}}(\theta)=\,f_p(\theta) \,A_\theta^s \,
B_{\pi-\theta}^{-s},\label{j7}$$ where $\theta$ is the scattering angle and $f_p(\theta)$ the scattering amplitude. We will consider $\theta$ different from 0 or $\pi$ to avoid forward and backward directions. While the increase of uncertainty due to the interaction is clear, because a continuous manifold of final directions with probabilities $|f_p(\theta)|^2$ opened up from just one initial direction, spin remains untouched. The information about $s_a$ is the same before and after the scattering; as much as we knew the initial spin of $A$, we know its final spin whatever the final direction is. In other words, spin was not entangled by the interaction. We will now translate these well known facts to the case of identical particles, where they do not hold true.

Let particle $B$ be identical to $A$. Consider the same initial state as before: A particle $A$ with momentum $\mathbf{p}$ and spin $s$ runs into another $A$ with momentum $-\mathbf{p}$ and spin $-s$. Notice there is maximal information on the state. We could write $\Phi_{\mbox{in}}=\, A_0^s A_\pi^{-s}$, and eventually symmetrize. We now focus on the final state. It is no longer true that particle $A$ will come out with momentum $\mathbf{p'}$ and spin $s$ with amplitude $f_p(\theta)$ while the amplitude for coming out with momentum $\mathbf{p'}$ and spin $-s$ vanishes. Recalling that $B$ above did become $A$, the two cases $f_p(\theta) A_\theta^s B_{\pi-\theta}^{-s}$ and $f_p(\pi-\theta)A_{\pi-\theta}^{s} B_\theta^{-s}$ fuse into a unique state $$\Phi_{\mbox{out}}(\theta)=f_p(\theta) \,A_\theta^s\,
A_{\pi-\theta}^{-s}+f_p(\pi-\theta)\,A_{\pi-\theta}^{s} \,
A_\theta^{-s},\label{j8}$$ as shown in Figure \[figsse1bis\].

![(Color online) Schematic picture of the two channels that contribute to the spin-independent scattering of two identical fermions. The shaded regions denote an arbitrary spin-independent interaction between the two fermions. The vertical arrows $\uparrow$, $\downarrow$ indicate the corresponding third component of spin.[]{data-label="figsse1bis"}](scatteringbis.eps){width="8cm"}

Notice the uncertainty acquired by the spin: Now particle $A$ comes out from the interaction along $\theta$ either with spin $s$ or with spin $-s$, with relative amplitudes $f_p(\theta)$ and $f_p(\pi-\theta)$ respectively. In other words, spin was entangled during the spin independent evolution. Here, it is not the spin dependence of the interaction, but the existence of additional degrees of freedom which generate spin-spin entanglement. These, act as ancillas creating an effective spin-spin interaction that entangles the two fermions. The ancilla and the degree of entanglement depend on the scattering angle $\theta$. Notice that for $\theta=\pi/2$ both amplitudes $f_p(\theta)$ and $f_p(\pi-\theta)$ become equal, so that the degree of generated entanglement is maximal, 1 e-bit. On the other hand, for $\theta\simeq 0$, it generally holds $f_p(\theta)\gg
f_p(\pi-\theta)$, so that in the forward and backward scattering almost no entanglement would be generated. However, this depends on the specific interaction. In Sec. \[asecslo\] we will clarify this point with Coulomb interaction.

Symmetrization does not change this, it only expresses that we can not tell which one is 1 and which one is 2. The properly symmetrized initial state is $$\begin{aligned}
\Phi_{\mbox{in}}&=&{\rm SL}(1,2;\mathbf{p})_s\nonumber\\
&=&\frac{1}{\sqrt{2}}[A(1)_0^s A(2)_\pi^{-s}-A(1)_\pi^{-s}A(2)_0^s].
\label{j9}\end{aligned}$$ The scattering process could be written in terms of SLs as $$\begin{aligned}
&{\rm SL}(1,2;\mathbf{p})_s\longrightarrow &\nonumber\\
&f_p(\theta)\, {\rm SL}(1,2;\mathbf{p}')_s - f_p(\pi-\theta)\, {\rm
SL}(1,2;\mathbf{p}')_{-s},& \label{j10}\end{aligned}$$ where $\mathbf{p}'$ is the final momentum and the Slater determinants are given in (\[j5\]). Both, this expression and Eq.(\[j8\]), describe the same physical situation and lead to the same entanglement generation.

The bosonic case may be analyzed in an analogous way. The modification for two-dimensional spin Hilbert spaces (i.e. photons) would be a sign change in Eqs. (\[j5\]), (\[j9\]) and (\[j10\]), as bosonic statistics has associated symmetric states. The equivalent of Eq. (\[j10\]) for bosons is a genuine entangled state for $\theta\neq0,\pi$, much as in the fermionic case.

A specific example: Coulomb scattering at lowest order\[asecslo\]
=================================================================

We now consider Coulomb interaction at lowest order to illustrate the reasonings presented above. In this case $$\begin{aligned}
f_p(\theta) & = & \frac{N(e)}{t(\theta)},\nonumber\\
f_p(\pi-\theta) & = & \frac{N(e)}{u(\theta)},\label{eqsse11}\end{aligned}$$ where $N(e)$ is a numerical factor depending on the charge $e$. $t(\theta)$ and $u(\theta)$ are two of the Mandelstam variables, associated to $t$ and $u$ channels respectively, and depending on the scattering angle $\theta$. for initial $p$ and final $p'$ relative 4-momenta of the scattering fermions, they are given by $t=(p-p')^2$, $u=(p+p')^2$. In the CoM frame, $$\begin{aligned}
 t(\theta)\!\! & := & 2(m^2-E^2)(1-\cos\theta),\nonumber\\
 u(\theta)\!\! & := & 2(m^2-E^2)(1+\cos\theta),\label{eqsse12}\end{aligned}$$ where $m$ is the mass of each fermion and $2E$ is the available energy.

According to this, the spin part of the state (\[j8\]) for this case, properly normalized, is $$\begin{aligned}
|\chi_{\theta}\rangle=f_+(\theta)|\!\!\uparrow\downarrow\rangle-
f_-(\theta)|\!\!\downarrow\uparrow\rangle,\label{eqsse13}\end{aligned}$$ being $$\begin{aligned}
f_{\pm}(\theta):=\frac{1\pm\cos\theta}{\sqrt{2(1+\cos^2\theta)}}.\label{eqsse14}\end{aligned}$$ The two amplitudes $f_+$ and $f_-$ vary monotonously as $\theta$ grows, becoming equal for $\theta=\pi/2$. The physical meaning for this is that for $\theta\rightarrow0$, the knowledge about the system is maximal and the entanglement minimal (zero), and for increasing $\theta$ the knowledge of the system decreases continuously until reaching its minimum value at $\theta=\pi/2$. Accordingly, the entanglement grows with $\theta$ until reaching its maximum value for $\theta=\pi/2$.

We plot in Figure \[figsse2\] the EoE [@TK05] $S(\theta)=-f_+(\theta)^2\log_2f_+(\theta)^2-f_-(\theta)^2\log_2f_-(\theta)^2$ of state (\[eqsse13\]) as a function of $\theta$, for $0<\theta\leq\pi/2$. The entanglement grows monotonically until $\theta=\pi/2$, where it becomes maximal (1 e-bit).

![(Color online) EoE $S(\theta)$ as a function of $\theta$.[]{data-label="figsse2"}](sdef.eps)

$\theta$-dependence of Bell’s inequality violation
==================================================

In order to analyze the role the $\theta$ scattering angle plays in the generation of these genuine quantum correlations, we consider now the degree of violation of Bell’s inequality as a function of $\theta$. To this purpose, we define [@Bell64; @CS78] the observable $$\begin{aligned}
E(\mathbf{\hat{a}},\mathbf{\hat{b}})&:=&\langle\Phi|(\mathbf{\sigma}^{(1)}\cdot
\mathbf{\hat{a}}\otimes\mathbf{\sigma}^{(2)}\cdot
\mathbf{\hat{b}})|\Phi\rangle\label{eqsse15}\\&=&-[\hat{a}_z\hat{b}_z+2f_p(\theta)f_p(\pi-\theta)(\hat{a}_x\hat{b}_x+\hat{a}_y\hat{b}_y)],\nonumber\end{aligned}$$ where $|\Phi\rangle:=|\Phi_{\rm out}(\theta)\rangle$ is the (normalized) state (\[j8\]) and $\mathbf{\hat{a}}$, $\mathbf{\hat{b}}$ are arbitrary unit vectors. In Eq. (\[eqsse15\]) we consider the amplitudes $f_p(\theta)$ and $f_p(\pi-\theta)$ normalized for each $\theta$, in the form $|f_p(\theta)|^2+|f_p(\pi-\theta)|^2=1$. We consider three coplanar unit vectors, $\mathbf{\hat{a}}$, $\mathbf{\hat{b}}$ and $\mathbf{\hat{c}}$. $(\widehat{\mathbf{\hat{a}},\mathbf{\hat{b}}})=\pi/3$, $(\widehat{\mathbf{\hat{a}},\mathbf{\hat{c}}})=2\pi/3$ and $(\widehat{\mathbf{\hat{b}},\mathbf{\hat{c}}})=\pi/3$. We have $$\begin{aligned}
&&|E(\mathbf{\hat{a}},\mathbf{\hat{b}})-E(\mathbf{\hat{a}},\mathbf{\hat{c}})|=1,\nonumber\\
&&F(\theta):=1+E(\mathbf{\hat{b}},\mathbf{\hat{c}})=\frac{5}{4}-\frac{3}{2}f_p(\theta)f_p(\pi-\theta).\end{aligned}$$ The Bell’s inequality, given by [@Bell64; @CS78] $$\begin{aligned}
|E(\mathbf{\hat{a}},\mathbf{\hat{b}})-E(\mathbf{\hat{a}},\mathbf{\hat{c}})|\leq
1+E(\mathbf{\hat{b}},\mathbf{\hat{c}}),\end{aligned}$$ will then be $$\begin{aligned}
F(\theta)\geq 1.\end{aligned}$$

For the particular case of Coulomb interaction at lowest order here considered, $2f_p(\theta)f_p(\pi-\theta)=2f_+(\theta)f_-(\theta)=(1-\cos^2\theta)/(1+\cos^2\theta)$ and thus the critical angle for which the inequality becomes violated is $\theta_c=\pi/4$ for $F(\theta_c)=1$. For $\theta_c<\theta\leq\pi/2$ the Bell’s inequality does not hold. We show in Figure \[figsse3\] the $\theta$ dependence of $F(\theta)$ together with the classical-quantum border, $F=1$, at $\theta_c=\pi/4$. Thus, for experiments with $\theta>\pi/4$ one could be able in principle to discriminate between local realism and quantum mechanics.

![(Color online) $F(\theta)$ as a function of $\theta$. The classical-quantum border corresponds to $F(\theta_c)=1$, with $\theta_c=\pi/4$.[]{data-label="figsse3"}](belldef.eps)

This is in contrast with recent analysis of Bell’s inequalities violation in elementary particle systems [@Go04; @Bra05; @Tor86], where the emphasis was placed on flavor entanglement, $K^0\bar{K}^0$, $B^0\bar{B}^0$, and the like. These analysis presented [@Bra04] two kinds of drawbacks coming from the lack of experimenter’s free will and from the unitary evolution with decaying states. These issues reduce the significance of the experiments up to the point of preventing their use as tests of quantum mechanics versus local realistic theories. The spin-spin entanglement analyzed in this paper does not have this kind of problems and could be used in principle for that purpose.

Conclusions
===========

In summary, we analyzed the relation between entanglement and antisymmetrization for identical particles, in the context of spin-independent particle scattering. We showed that, in order to create genuine spin-spin quantum correlations between two $s=\half$ fermions, spin-dependent interactions are not compulsory. The identity of the particles along with an interaction between degrees of freedom different from the spin, suffice for this purpose. The entanglement generated this way is not a fictitious one due to antisymmetrization, but a real one, and violates a certain Bell’s inequality for $\theta>\theta_c=\pi/4$.

ACKNOWLEDGMENTS {#acknowledgments .unnumbered}
===============

This work was partially supported by the Spanish MEC project No. FIS2005-05304. L.L. acknowledges support from the FPU grant No. AP2003-0014.

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