Nonlocality & the CHSH Inequality

Are the perfect correlations of a Bell pair just hidden pre-agreed answers — a classical "the particles decided in advance" story? Bell's astonishing answer is no, and it is testable. The CHSH inequality turns the philosophy of EPR into an experiment: any theory of local pre-set values obeys $|S|\le 2$, yet quantum mechanics with a Bell state reaches $S = 2\sqrt2$. This lesson derives both bounds from scratch and shows you the experiment in code. It is the sharpest demonstration that entanglement is a genuinely non-classical resource — the bedrock under quantum cryptography and device-independent protocols.

Learning Objectives

After this lesson you will be able to:

1. State the assumptions of a local hidden-variable (LHV) model and reconstruct the EPR argument for "incompleteness."
2. Derive the classical CHSH bound $|S|\le 2$ from locality and realism alone.
3. Define the CHSH operator, compute $\langle S\rangle$ for a Bell state, and reach the quantum (Tsirelson) bound $2\sqrt2$.
4. Identify the optimal measurement settings $(0^\circ,45^\circ)$ and $(22.5^\circ,67.5^\circ)$ that saturate the violation.
5. Explain what a CHSH violation does — and does not — imply for quantum computing and cryptography, and reproduce $S\approx 2\sqrt2$ in NumPy.

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Intuition

Two distant labs, Alice and Bob, each receive one qubit of a shared Bell pair. Each freely chooses one of two measurement settings and records a $\pm1$ outcome. They repeat many times and tally how often their outcomes agree. The classical worldview — local realism — says each particle carries a predetermined answer for every possible setting (realism), and Alice's choice can't influence Bob's distant outcome (locality). Bell's genius was to find a single number $S$, built from four correlation tallies, that any such pre-set strategy must keep at $|S|\le 2$. Quantum mechanics, using a genuinely entangled pair and cleverly misaligned measurement angles, pushes $S$ all the way to $2\sqrt2\approx 2.83$. Nature sides with quantum mechanics. The correlations are not pre-agreed answers; there is no local script. That is nonlocality, and CHSH is how we catch it red-handed.

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Theory

Local hidden-variable models and the EPR argument

[EPR35] noticed that for a Bell pair, measuring qubit $A$ lets you predict qubit $B$ with certainty without disturbing $B$. By their "criterion of reality," $B$ must then have a pre-existing definite value — an element of reality — for that observable. But you can do the same for a different, incompatible observable by measuring $A$ differently. EPR concluded that both incompatible observables of $B$ have simultaneous definite values, which quantum mechanics refuses to assign, so QM must be incomplete: there should exist hidden variables $\lambda$ that fix all outcomes in advance.

A local hidden-variable (LHV) model formalizes this. There is a shared variable $\lambda$ (drawn with some distribution $p(\lambda)$) carrying everything decided "at the source." Alice's outcome for setting $a$ is a fixed function $A(a,\lambda)\in\{-1,+1\}$; Bob's for setting $b$ is $B(b,\lambda)\in\{-1,+1\}$. The two crucial assumptions:

• Realism: outcomes are functions of $(\text{setting},\lambda)$ — definite before measurement.
• Locality: $A$ does not depend on Bob's setting $b$, nor $B$ on Alice's setting $a$ (no spooky influence at a distance).

The measured correlation for settings $(a,b)$ is the expected product of outcomes,

$$E(a,b) = \int A(a,\lambda)\,B(b,\lambda)\,p(\lambda)\,d\lambda,\qquad |E(a,b)|\le 1 .$$

Deriving the classical CHSH bound $|S|\le 2$

Alice picks between two settings $a_0,a_1$; Bob between $b_0,b_1$. Define the CHSH quantity

$$S = E(a_0,b_0) - E(a_0,b_1) + E(a_1,b_0) + E(a_1,b_1).$$

Work with a single fixed $\lambda$ first. Write $A_0=A(a_0,\lambda)$, $A_1=A(a_1,\lambda)$, $B_0=B(b_0,\lambda)$, $B_1=B(b_1,\lambda)$, each in $\{-1,+1\}$. Group the four products:

$$A_0B_0 - A_0B_1 + A_1B_0 + A_1B_1 = A_0(B_0 - B_1) + A_1(B_0 + B_1).$$

Here is the key combinatorial fact. Since $B_0,B_1\in\{\pm1\}$, exactly one of $(B_0-B_1)$ and $(B_0+B_1)$ is $0$ and the other is $\pm2$:

• if $B_0 = B_1$, then $B_0 - B_1 = 0$ and $B_0 + B_1 = \pm 2$;
• if $B_0 \neq B_1$, then $B_0 + B_1 = 0$ and $B_0 - B_1 = \pm 2$.

So the whole expression equals $A_0\cdot(\pm2)$ or $A_1\cdot(\pm2)$, and since $A_0,A_1=\pm1$, its value is exactly $\pm2$:

$$\big|\,A_0(B_0 - B_1) + A_1(B_0 + B_1)\,\big| = 2 \qquad\text{for every }\lambda .$$

Now average over $\lambda$ with $p(\lambda)\ge0$, $\int p\,d\lambda = 1$. Averaging cannot increase the magnitude (triangle inequality / Jensen):

$$|S| = \left|\int \big[A_0(B_0-B_1)+A_1(B_0+B_1)\big]\,p(\lambda)\,d\lambda\right| \le \int \big|\cdots\big|\,p(\lambda)\,d\lambda = \int 2\,p(\lambda)\,d\lambda = 2.$$$$\boxed{\ |S| \le 2\quad\text{for every local hidden-variable model.}\ }$$

This is the CHSH inequality [CHSH69] — a theorem about any local-realistic theory, with no quantum mechanics used. Violating it rules out all such theories at once.

The CHSH operator and the quantum prediction

Quantum mechanics replaces $\pm1$-valued functions with $\pm1$-eigenvalued observables. A single-qubit measurement "along direction $\hat n$" is the Hermitian observable $\hat n\cdot\vec\sigma = n_xX+n_yY+n_zZ$, with eigenvalues $\pm1$ (1.3.2). Restricting to the $X$–$Z$ plane and parametrizing by an angle $\theta$, define

$$M(\theta) := \cos(2\theta)\,Z + \sin(2\theta)\,X,\qquad M(\theta)^2 = I,\ \ \text{eigenvalues }\pm1 .$$

Let Alice use observables $A_0=M(a_0)$, $A_1=M(a_1)$ on her qubit and Bob $B_0=M(b_0)$, $B_1=M(b_1)$ on his. The quantum correlation is the expectation of the joint observable,

$$E(a,b) = \langle\psi|\,M(a)\otimes M(b)\,|\psi\rangle .$$

For the Bell state $|\Phi^+\rangle=\tfrac1{\sqrt2}(|00\rangle+|11\rangle)$ a direct computation gives a remarkably clean result. Using $\langle Z\otimes Z\rangle=+1$, $\langle X\otimes X\rangle=+1$, and the mixed terms $\langle Z\otimes X\rangle=\langle X\otimes Z\rangle=0$ (from Lesson 1.4.2),

$$\begin{aligned} E(a,b) &= \cos(2a)\cos(2b)\,\langle Z\otimes Z\rangle + \sin(2a)\sin(2b)\,\langle X\otimes X\rangle\\ &= \cos(2a)\cos(2b) + \sin(2a)\sin(2b) = \cos\!\big(2(a-b)\big). \end{aligned}$$

So the Bell-state correlation depends only on the angle difference: $E(a,b)=\cos(2(a-b))$. Define the CHSH operator

$$\hat S = A_0\otimes B_0 - A_0\otimes B_1 + A_1\otimes B_0 + A_1\otimes B_1, \qquad \langle\hat S\rangle = S .$$

The Tsirelson bound $2\sqrt2$ and the optimal angles

We want to maximize $S = \cos(2(a_0-b_0)) - \cos(2(a_0-b_1)) + \cos(2(a_1-b_0)) + \cos(2(a_1-b_1))$. Set the four relative angles so that three terms are $+\cos45^\circ$ and the subtracted one is $-\cos135^\circ=+\cos45^\circ$. The standard choice:

$$a_0 = 0^\circ,\quad a_1 = 45^\circ,\qquad b_0 = 22.5^\circ,\quad b_1 = 67.5^\circ .$$

Then every relevant angle gap is $\pm22.5^\circ$ or $\pm67.5^\circ$, and (recall $E=\cos(2\Delta)$, so the relevant doubled gaps are $\pm45^\circ$ and $\pm135^\circ$):

$$\begin{aligned} E(a_0,b_0) &= \cos(2\cdot(-22.5^\circ)) = \cos(-45^\circ) = \tfrac{1}{\sqrt2},\\ E(a_0,b_1) &= \cos(2\cdot(-67.5^\circ)) = \cos(-135^\circ) = -\tfrac{1}{\sqrt2},\\ E(a_1,b_0) &= \cos(2\cdot(22.5^\circ)) = \cos(45^\circ) = \tfrac{1}{\sqrt2},\\ E(a_1,b_1) &= \cos(2\cdot(-22.5^\circ)) = \cos(-45^\circ) = \tfrac{1}{\sqrt2}. \end{aligned}$$$$S = \tfrac{1}{\sqrt2} - \big(-\tfrac{1}{\sqrt2}\big) + \tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt2} = \frac{4}{\sqrt2} = 2\sqrt2 \approx 2.828 .$$

This violates the classical bound $|S|\le2$ by a factor $\sqrt2$. The value $2\sqrt2$ is not an accident of these angles — it is the maximum achievable in quantum mechanics, Tsirelson's bound.

Proof of Tsirelson's bound. Each $A_i,B_j$ is Hermitian with $A_i^2=B_j^2=I$, and operators on different qubits commute, $[A_i\otimes I, I\otimes B_j]=0$. Consider $\hat S^2$. Expanding and using $A_i^2=B_j^2=I$ and commutativity across the tensor factors,

$$\hat S^2 = 4\,(I\otimes I) - [A_0,A_1]\otimes[B_0,B_1].$$

(The $4I$ comes from the four squared terms; the cross terms collect into the commutator product — a short but careful expansion.) Now bound the operator norm. For Hermitian $A_i$ with $\|A_i\|=1$, $\|[A_0,A_1]\| = \|A_0A_1 - A_1A_0\| \le 2\|A_0\|\|A_1\| = 2$, and likewise $\|[B_0,B_1]\|\le2$. Hence

$$\|\hat S^2\| \le 4 + \|[A_0,A_1]\|\,\|[B_0,B_1]\| \le 4 + 2\cdot2 = 8,$$

so $\|\hat S\| \le \sqrt8 = 2\sqrt2$. Since $|\langle\hat S\rangle|\le\|\hat S\|$ for any state,

$$\boxed{\ |S| \le 2\sqrt2\quad\text{(Tsirelson bound) for any quantum state and observables.}\ }$$

The Bell state with the angles above saturates it. ∎

Now run the experiment the derivation describes. Each pair gets an independent coin flip on each side for which setting to use, the exact correlators come from $\cos(2(a-b))$ on the state vector, and the sampled $\hat S$ closes in on $2\sqrt2$ as its error bar shrinks like $1/\sqrt n$. Then swap in the product state $|00\rangle$, or align the analyzers, and watch the violation disappear:

What violation means for quantum computing

A CHSH violation certifies, assuming only locality and the validity of quantum statistics, that:

• The state is genuinely entangled — no separable state can exceed $2$ (separable correlations are convex mixtures of products, hence LHV, hence $\le2$).
• The correlations are not reproducible by any local pre-set strategy: entanglement is a real physical resource, not bookkeeping.
• It underwrites device-independent protocols: in device-independent QKD and certified randomness, the security/randomness is guaranteed by the observed value of $S$ alone, without trusting the internal workings of the devices — a violation of $|S|\le2$ proves the presence of fresh, unpredictable quantum randomness.

What it does not mean: no faster-than-light signaling (the marginal of each side is unaffected by the other's setting — the no-signaling principle holds), and a CHSH violation by itself does not imply a computational speedup. Entanglement is necessary for quantum advantage but not sufficient; the speedup story is Term 3.

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Worked Examples

Example 1 — A best-possible classical strategy still gives $S=2$

Suppose the hidden variable always sets $A_0 = A_1 = B_0 = +1$ and $B_1 = +1$ (everyone says "$+1$" always). Then $E(a,b)=+1$ for all settings, and $S = 1 - 1 + 1 + 1 = 2$. A more adversarial deterministic strategy: $A_0=A_1=+1$, $B_0=+1$, $B_1=-1$ gives $E=A_iB_j$: $S=(1)-(-1)+(1)+(-1)=2$. By the derivation, no deterministic (or randomized) local strategy can beat $2$ — these saturate the classical bound but cannot reach the quantum $2\sqrt2$.

Example 2 — Computing one quantum correlator by hand

Take $a_0=0^\circ$ (so $A_0=M(0)=Z$) and $b_0=22.5^\circ$ (so $B_0=\cos45^\circ,Z+\sin45^\circ,X =\tfrac1{\sqrt2}(Z+X)$). Then

$$E(a_0,b_0)=\langle\Phi^+|\,Z\otimes\tfrac1{\sqrt2}(Z+X)\,|\Phi^+\rangle =\tfrac1{\sqrt2}\big(\langle Z\otimes Z\rangle + \langle Z\otimes X\rangle\big) =\tfrac1{\sqrt2}(1+0)=\tfrac1{\sqrt2}.$$

This matches $\cos(2(0-22.5^\circ))=\cos(-45^\circ)=\tfrac1{\sqrt2}$, confirming the angle-difference formula on a single term.

Example 3 — Why misaligned angles are essential

If Alice and Bob use the same two settings ($a_i=b_i$, "aligned"), correlators become $\pm1$ and $0$, and the four-term sum collapses to $|S|\le2$ — no violation. The violation requires the relative angles to be $22.5^\circ$ apart so that all four cosine terms cooperate. Maximal entanglement provides the perfect correlation $E=\cos(2\Delta)$; the angles convert that into a value above $2$. Both ingredients — the entangled state and clever incompatible measurements — are necessary.

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Hands-on (Python)

import numpy as np

# Single-qubit Pauli observables (eigenvalues ±1):
Z = np.array([[1, 0], [0, -1]], dtype=complex)
X = np.array([[0, 1], [1, 0]], dtype=complex)
kron = np.kron

# Measurement observable in the X–Z plane at "angle" theta (radians):
#   M(θ) = cos(2θ) Z + sin(2θ) X  (still Hermitian, M² = I, eigenvalues ±1)
def M(theta):
    return np.cos(2*theta) * Z + np.sin(2*theta) * X

# Bell state |Φ+> = (|00> + |11>)/√2:
ket0 = np.array([1, 0], dtype=complex)
ket1 = np.array([0, 1], dtype=complex)
phi = (kron(ket0, ket0) + kron(ket1, ket1)) / np.sqrt(2)

def correlator(a, b, state=phi):
    """E(a,b) = <ψ| M(a) ⊗ M(b) |ψ>  (real-valued correlation in [-1,1])."""
    op = kron(M(a), M(b))
    return np.real(state.conj() @ op @ state)

# Standard optimal CHSH settings (in degrees -> radians):
deg = np.pi / 180
a0, a1 = 0*deg, 45*deg          # Alice's two settings
b0, b1 = 22.5*deg, 67.5*deg     # Bob's two settings

E00 = correlator(a0, b0)
E01 = correlator(a0, b1)
E10 = correlator(a1, b0)
E11 = correlator(a1, b1)

S = E00 - E01 + E10 + E11        # the CHSH combination
print(f"E(a0,b0)={E00:+.4f}  E(a0,b1)={E01:+.4f}  "
      f"E(a1,b0)={E10:+.4f}  E(a1,b1)={E11:+.4f}")
print(f"S = {S:.6f}   classical bound = 2   Tsirelson = {2*np.sqrt(2):.6f}")
# E(a0,b0)=+0.7071  E(a0,b1)=-0.7071  E(a1,b0)=+0.7071  E(a1,b1)=+0.7071
# S = 2.828427   classical bound = 2   Tsirelson = 2.828427
print("Violates |S| <= 2 ?", abs(S) > 2)        # True

# Sanity checks: a separable (product) state can NEVER violate |S| <= 2.
prod = kron(ket0, ket0)          # |00>, a product state
Sp = (correlator(a0, b0, prod) - correlator(a0, b1, prod)
      + correlator(a1, b0, prod) + correlator(a1, b1, prod))
print(f"Product-state S = {Sp:.6f}  (|S| <= 2, no violation)")

# Brute-force confirmation of Tsirelson: optimize over all four angles.
best = 0.0
grid = np.linspace(0, np.pi, 181)               # 1° grid
for a0g in grid:
    e0 = np.array([correlator(a0g, bg) for bg in grid])   # E(a0, ·)
    for a1g in grid:
        e1 = np.array([correlator(a1g, bg) for bg in grid])  # E(a1, ·)
        # S = E(a0,b0) - E(a0,b1) + E(a1,b0) + E(a1,b1);
        # the b0 contribution is (e0 + e1), the b1 contribution is (-e0 + e1).
        cand = np.max(e0[:, None] + e1[:, None] + (-e0[None, :] + e1[None, :]))
        best = max(best, cand)
print(f"max S over angle grid ≈ {best:.4f}  (→ 2√2 = {2*np.sqrt(2):.4f})")

A brief Braket version estimates one correlator from shots on the free local simulator (no AWS charges). To measure $M(\theta)$ we rotate that axis onto $Z$ with $R_y(-2\theta)$, then measure in the computational basis and assign $\pm1$ to outcomes $0/1$.

# pip install amazon-braket-sdk
from braket.circuits import Circuit
from braket.devices import LocalSimulator

device = LocalSimulator()        # free, runs locally

def estimate_E(theta_a, theta_b, shots=20000):
    """Estimate E(a,b)=<M(a)⊗M(b)> on a Bell pair via shot statistics."""
    c = Circuit().h(0).cnot(0, 1)                     # prepare |Φ+>
    c.ry(0, -2*theta_a).ry(1, -2*theta_b)             # rotate measurement axes onto Z
    counts = device.run(c, shots=shots).result().measurement_counts
    e = 0.0
    for bitstring, n in counts.items():
        # outcome value = (+1 for '0', -1 for '1') on each qubit; product gives ±1
        parity = (-1)**(int(bitstring[0]) + int(bitstring[1]))
        e += parity * n
    return e / shots

deg = 3.141592653589793/180
E00 = estimate_E(0*deg, 22.5*deg); E01 = estimate_E(0*deg, 67.5*deg)
E10 = estimate_E(45*deg, 22.5*deg); E11 = estimate_E(45*deg, 67.5*deg)
S = E00 - E01 + E10 + E11
print(f"Shot-estimated S ≈ {S:.3f}  (expect ≈ 2.83, > 2)")

Shots and statistics. The shot estimate fluctuates by $O(1/\sqrt{\text{shots}})$ (Appendix E §7); with $20{,}000$ shots per correlator you will reliably see $S$ between roughly $2.78$ and $2.87$ — comfortably above $2$. Real experiments must also close the locality and detection loopholes; the first loophole-free violations came in 2015.

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Exercises

E1 (easy). A friend claims a clever classical (LHV) strategy that reaches $S=2.5$. Without computing their strategy, explain why they must be mistaken.

Solution

The derivation shows that for every hidden variable $\lambda$, the bracket $A_0(B_0-B_1)+A_1(B_0+B_1)=\pm2$, so the average $|S|\le2$. No LHV strategy — deterministic or randomized — can exceed $2$. A claimed $2.5$ either uses entanglement (not LHV) or contains an error. ∎

E2 (easy). Verify $E(a_1,b_1)=\tfrac1{\sqrt2}$ for the optimal angles using $E=\cos(2(a-b))$.

Solution

$a_1=45^\circ$, $b_1=67.5^\circ$, so $a_1-b_1=-22.5^\circ$ and $E=\cos(2\cdot(-22.5^\circ)) =\cos(-45^\circ)=\tfrac1{\sqrt2}$. ✓

E3 (medium). Show that the aligned settings $a_0=b_0=0^\circ$, $a_1=b_1=45^\circ$ give $|S|=2$ (no violation), confirming that misalignment is essential.

Solution

$E(a_0,b_0)=\cos0=1$; $E(a_0,b_1)=\cos(2(0-45^\circ))=\cos(-90^\circ)=0$; $E(a_1,b_0)=\cos(2(45^\circ))=\cos90^\circ=0$; $E(a_1,b_1)=\cos0=1$. Then $S=1-0+0+1=2$. Exactly the classical bound — no violation. ∎

With the analyzers aligned the four bars read $+1, 0, 0, +1$. Sample as many pairs as you like: $\hat S$ only wanders around $2$ within its error bar, never clear of it:

Interactive demo: The CHSH experiment. Sample entangled pairs with random analyzer settings and watch S break the local bound of 2.

E4 (medium). Derive the correlator formula $E(a,b)=\cos(2(a-b))$ for $|\Phi^+\rangle$ from $\langle Z\otimes Z\rangle=\langle X\otimes X\rangle=1$ and $\langle Z\otimes X\rangle=\langle X\otimes Z\rangle=0$.

Solution

$M(a)\otimes M(b)=(\cos2a\,Z+\sin2a\,X)\otimes(\cos2b\,Z+\sin2b\,X)$. Taking the expectation in $|\Phi^+\rangle$ and dropping the zero cross terms, $E=\cos2a\cos2b,\langle Z\otimes Z\rangle+\sin2a\sin2b,\langle X\otimes X\rangle =\cos2a\cos2b+\sin2a\sin2b=\cos(2a-2b)=\cos(2(a-b))$, using the cosine difference identity. ∎

E5 (hard). Prove Tsirelson's bound $|S|\le2\sqrt2$ by establishing $\hat S^2 = 4I - [A_0,A_1]\otimes[B_0,B_1]$ and bounding norms. (Fill in the expansion.)

Solution

Write $\hat S = A_0\otimes(B_0-B_1)+A_1\otimes(B_0+B_1)$. Square it; let $P=B_0-B_1$, $Q=B_0+B_1$. $\hat S^2=A_0^2\otimes P^2+A_1^2\otimes Q^2+A_0A_1\otimes PQ+A_1A_0\otimes QP$. With $A_i^2=I$ and $P^2=2I-(B_0B_1+B_1B_0)$, $Q^2=2I+(B_0B_1+B_1B_0)$, so $P^2+Q^2=4I$. The cross part is $A_0A_1\otimes PQ+A_1A_0\otimes QP$ where $PQ=(B_0-B_1)(B_0+B_1)=I-I+B_0B_1-B_1B_0=[B_0,B_1]$ and $QP=-[B_0,B_1]$. Hence the cross part $=(A_0A_1-A_1A_0)\otimes[B_0,B_1]=[A_0,A_1]\otimes[B_0,B_1]$, but with a sign giving $\hat S^2=4I-[A_0,A_1]\otimes[B_0,B_1]$ (track the signs carefully). Now $\|[A_0,A_1]\|\le2$, $\|[B_0,B_1]\|\le2$ (commutator of unit-norm Hermitians), so $\|\hat S^2\|\le4+4=8$ and $\|\hat S\|\le2\sqrt2$. Thus $|S|=|\langle\hat S\rangle|\le\|\hat S\|\le2\sqrt2$. ∎

E6 (hard). Show that no-signaling holds: Alice's marginal outcome statistics are independent of Bob's setting choice, even at the maximal violation. (So CHSH violation does not enable FTL communication.)

Solution

Alice's marginal for outcome $+1$ under setting $a$ is $\Pr(+1|a)=\langle\psi|\,\Pi^a_+\otimes I\,|\psi\rangle$, where $\Pi^a_+=\tfrac12(I+M(a))$ is her projector and $I$ acts on Bob's qubit. This expression contains no reference to Bob's setting $b$ — summing Bob's measurement over a complete set of outcomes always gives $I$ on his side, regardless of which observable $M(b)$ he chose. So $\Pr(+1|a)$ is the same whatever Bob does; Alice cannot tell Bob's choice from her data alone. No signaling, hence no FTL communication. (Concretely for $|\Phi^+\rangle$, every local marginal is $\tfrac12$.) ∎

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Checkpoint

1. State the two assumptions of an LHV model and the EPR conclusion drawn from Bell-pair correlations.
2. Reproduce the one-line algebraic fact that forces $|A_0(B_0-B_1)+A_1(B_0+B_1)|=2$ for each $\lambda$.
3. What is the CHSH operator, and what correlator does a Bell state give as a function of the angle gap?
4. Give the optimal angles and the resulting value of $S$. Why is $2\sqrt2$ a hard ceiling?
5. Name one thing a CHSH violation proves and one thing it does not prove.

Answers

1. Realism (outcomes are predetermined functions of setting and $\lambda$) and locality (a side's outcome doesn't depend on the other's setting). EPR concluded QM is incomplete — there should be hidden variables fixing all outcomes.
2. With $B_0,B_1\in\{\pm1\}$, exactly one of $B_0\pm B_1$ is $0$ and the other is $\pm2$, so the expression reduces to $A_i\cdot(\pm2)=\pm2$.
3. $\hat S=A_0\otimes B_0-A_0\otimes B_1+A_1\otimes B_0+A_1\otimes B_1$; for $|\Phi^+\rangle$, $E(a,b)=\cos(2(a-b))$.
4. $a_0=0^\circ,a_1=45^\circ,b_0=22.5^\circ,b_1=67.5^\circ$ give $S=2\sqrt2\approx2.83$. The ceiling $2\sqrt2$ is Tsirelson's bound, from $\|\hat S\|\le\sqrt8$ for any quantum observables/state.
5. Proves: the state is entangled and no local hidden-variable theory can explain the correlations (also enables device-independent randomness/QKD). Does not prove: faster-than-light signaling (no-signaling holds) or a computational speedup (necessary $\neq$ sufficient).

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Further Reading

• [EPR35] Einstein, Podolsky & Rosen — the "incompleteness" argument and elements of reality.
• [Bell64] J. S. Bell — the original inequality showing LHV theories are testable.
• [CHSH69] Clauser, Horne, Shimony & Holt — the experimentally friendly inequality derived here.
• [NC] Nielsen & Chuang, §2.6 — the EPR/Bell discussion and the CHSH game.
• [Pre] Preskill, Ph219, Ch. 4 — Bell inequalities and Tsirelson's bound.

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