The definitive test of the non-locality of quantum physics is provided by a framework known as Bell’s inequality, developed by Bell (1964) (see Bell, 2004; Bell, 1966; Bell, 1981 and Clauser et al., 1969). The experimental tests of this logic have confirmed the quantum predictions, in ways that have also dealt with a number of possible loopholes in the logic (Aspect et al., 1982; Giustina et al., 2015; Hensen et al., 2015; Shalm et al., 2015).

Figure 1:

Illustration of the standard intuition about the local hidden variable version of the Bell’s inequality test, provided by J.S. Bell in his 1981 paper (Bell, 1981). Bell’s friend Bertlmann is eccentric and always wears different color socks. Thus, if you happen to see one sock, you know the other is a different color, without having to look at it. This is perfectly sensible. However, the challenging thing about Bell’s test is that your “measurement” of the color of the first sock actually determines what you see: it doesn’t actually have a specific color until this measurement, because the socks must be in a superposition (entangled) state to see the effect. Once you make that measurement, it “magically” influences the result that you get when you measure the other sock! This apparent non-local communication of the results of the first measurement, effectively changing the color of the sock on the other foot (“mid-step” as it were), is what boggles the mind. However, one could argue that this is merely revealing the fact that we have no idea what it means for something to be in a state of superposition, or how a measurement actually works to collapse this superposition into a distinct result.

The Bell’s inequality framework involves computing the statistics of measurements of an entangled quantum system of two particles, which means that the quantum wave is in a state of superposition. While we can easily represent this superposition in the mathematics of the Hilbert space, as we show below, it is unclear what the actual physical implication of such a superposition state is. People tend to think of it as an epistemic uncertainty, where somehow the system “really” is in some specific underlying state, but we just don’t yet know what that state is.

As illustrated in Figure 1 from Bell (1981), this epistemic interpretation corresponds to a local hidden variable theory, which was the point of the original thought experiment by Einstein et al. (1935) (EPR) that stimulated Bell’s work. Specifically, under the epistemic interpretation, the measurement process merely reveals what the unknown underlying state of the system was, thereby removing our lack of knowledge. The point of Bell’s framework is to show that if this was actually the case, then certain experimental results should follow. By contrast, the standard quantum math actually shows qualitatively different results.

As described in Figure 1, the contextual property of quantum mechanics predicts that the process of measurement is actually changing the underlying quantum state, effectively creating the resulting measurement outcome on the fly in the process of the measurement itself, rather than revealing some kind of unknown but fully pre-existing state. And because of the entangled nature of the two particles, the further prediction is that this creation process operating on one measurement of the overall quantum state should also affect the other measurement! In effect, measuring the color “pink” on one of Bertlmann’s feet actually creates, on the fly, the color of the other sock!

Figure 2:

Illustration of the standard Bell test setup from the pilot-wave perspective of particles interacting with wave packets. Somehow, the waves are in a state of superposition as a result of entanglement, and this causes the results of measurement at A to influence the outcome at B. Or is it the other way around? This is a fundamental ambiguity in the framework.

The standard language for describing the Bell test logic is that Alice (A) and Bob (B), who are situated in labs that could be rather far apart (i.e., spacelike separated in special relativity terms – further apart than light can travel within the relevant time frame of measurements), each make their own separate measurements on the entangled quantum state (Figure 2). When they later compare their data, they find that their results for events at the same point in time exhibit significant correlations, which reflect the influence of the measurement at A affecting the result at B. Or is it vice-versa?

From a standard Copenhagen perspective, the process of measurement involves instantaneously collapsing (projecting) the entire quantum wave state onto the basis vector (eigenvector) associated with the measurement. If you would then make an additional, subsequent measurement on the resulting post-collapse quantum state, it would just repeat the previous measurement outcome, because the underlying quantum state has been collapsed by the measurement process (this is contextuality).

Based on this general logic, it would seem that the precise order in which the measurements are made should matter somehow. If the A measurement happens first, and then B, the state should have collapsed to the A eigenvector, and then B operates on that collapsed state, and vice-versa for the opposite order. Furthermore, one could consider some kind of simultaneous collapse due to A and B happening at precisely the same time. As we’ll see below, each of these different scenarios indeed has a very different implication for the underlying quantum state in configuration space, but, because of the way that this space is constructed as a tensor product, the end measurement results all end up being identical!

This situation reflects the deeper problem that the quantum state is fundamentally incompatible with special relativity, even though the overall resulting phenomena (i.e., the measurements) are not (Bell, 1975; Aharonov & Albert, 1981; Peres, 2000; Peres & Terno, 2004; Norsen, 2011). This incompatibility is intuitively obvious, similar to the Coulomb gauge for the electric field, given that the quantum configuration space is manifestly non-local, and yet the theory says it must collapse instantaneously everywhere.

At the phenomenological level, it has been shown that the non-local quantum correlations cannot be used to actually send signals in a way that would violate special relativity (Bell, 1975; Ballentine & Jarrett, 1987), but this does not mean that the underlying mathematical framework that is used to explain these results is not fundamentally incompatible. Indeed, the authors discussing this conflict invoke the idea that the quantum state is just a calculational tool, not a “beable” (a physically real thing), so we shouldn’t be too disturbed by this situation (Bell, 1975; Peres & Terno, 2004).

But the obvious problem with such an approach is that there is at this point no viable alternative framework that can otherwise explain how these non-local correlations in the measurements might arise in a physical system whose mechanisms obey the constraints of special relativity. Thus, we appear to have a physically real, replicable phenomenon that is otherwise inexplicable, except by using a calculational tool that is manifestly non-local.

The non-locality of the configuration space does create very “real” problems for the Bohmian version of the pilot wave framework, because it postulates that the particle positions are continuously well-defined in 3D space, and yet they are being guided around by a non-local, exponentially-large configuration space wave state. Thus, it is not possible to dodge the conflict with special relativity in this framework by hiding behind the calculational tool argument, and this conflict thus remains one of the most significant difficulties with the pilot wave framework (Norsen, 2015; Norsen, 2011).

The concept of superposition in the position basis is particularly interesting in the pilot-wave framework, because the positions of the discrete particles are continuously defined: the particles themselves are always in a specific, well-defined location. Thus, how can there be a superposition state in such a case? The answer is that the corresponding quantum wave can have the superposition state, and this ends up having the appropriate effects on the discrete particles such that they reflect the probabilities that you would otherwise get from the standard quantum formalism.

Here are a few other relevant high-level points.

• Superposition is entirely a wave-specific phenomenon. Any kind of classical wave automatically exhibits superposition. And yet it still looks a bit like magic to see the superposition in effect in a classical wave.

• Extensive work has shown that classical waves, and corresponding optical wave phenomena, exhibit entanglement-like phenomena when put into a state of superposition (Spreeuw, 1998; Qian & Eberly, 2011; Qian et al., 2015; Karimi & Boyd, 2015; Khrennikov, 2006).

• However, superposition alone is insufficient. You also need the outer-product nature of configuration space, which keeps the wave state of the two different particles separated, even as they are in a state of superposition. So the superposition is fundamentally about the possible state of each particle separately, not about both particles being somehow superposed directly on top of each other. The space-like separation of A and B is indeed built into the tensor-product state representation, based on the requirement that such separated entities should be in different commuting parts of the state space (Peierls, 1952).

We return to these issues after exploring a fully worked-out example using the standard Hilbert space formalism, to get a better sense of what is happening at a mathematical level.

Example

We’ll use the modern terminology from quantum computing, in terms of a qubit, which has a quantum binary state. In the actual experiments, this is typically either the polarization direction of a photon, or the spin direction of a fermion.

The ket states for each qubit are denoted:

\[ |0\rangle = \begin{pmatrix}1\\0\end{pmatrix}, \qquad |1\rangle = \begin{pmatrix}0\\1\end{pmatrix} \]

The full tensor product state space for the two-qubit space is thus a 4 dimensional vector, with the subscripts representing the 0 or 1 basis for each qubit of a and b, which is a 2x2 matrix that is then stretched out into a 1d vector using the Kronecker convention:

\[ |a \rangle \langle b | = \begin{pmatrix} a_0 b_0 & a_0 b_1 \\ a_1 b_0 & a_1 b_1 \end{pmatrix} = \begin{pmatrix} a_0 b_0 \\ a_0 b_1 \\ a_1 b_0 \\ a_1 b_1 \end{pmatrix} \]

Thus, the four distinct pure outcome states in this outer-product state are:

\[ |00\rangle = \begin{pmatrix}1\\0\\0\\0\end{pmatrix}\quad |01\rangle = \begin{pmatrix}0\\1\\0\\0\end{pmatrix}\quad |10\rangle = \begin{pmatrix}0\\0\\1\\0\end{pmatrix}\quad |11\rangle = \begin{pmatrix}0\\0\\0\\1\end{pmatrix} \]

This innocent-looking configuration space organization already contains critical properties that determine several features of the resulting calculations. First, note that there is no position space represented, and a given state instance contains the full potentially non-local information about each particle, even if they might be spacelike separated. Furthermore, each particle that will end up going to the A or B detector has its own separate set of states (which then get integrated in the tensor product), which automatically makes any measurement operating on each such particle a commuting operator, which is thus not subject to the collapse effects of a measurement on the other particle. This is what allows the measurements at A and B to happen in any order and still produce the same overall result, even though the effects of each such measurement on the underlying state does depend on the order.

The maximally entangled state of this system \(\psi_\otimes\) is a superposition of the case where either qubit is either on or off:

\[ |\psi_\otimes\rangle = \frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle\right) = \frac{1}{\sqrt 2}\begin{pmatrix}1\\0\\0\\1\end{pmatrix} \]

You can verify it is entangled by computing the determinant of the 2x2 matrix version of the state – if this is non-zero, then it is entangled. The determinant \(\det\) is the product of the diagonal elements minus the off-diagonal ones:

\[ C = \begin{pmatrix}a & b\\ c & d\end{pmatrix}, \qquad \det C = ad - bc \]

\[ C_\otimes = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}, \qquad \det C_\otimes = 1 \neq 0 \]

As explained in measurement, the density operator \(\rho\) contains all the information about all possible measurements on the given quantum state, providing the correlation matrix for different possible outcomes. For this entangled case, it is:

\[ \rho_\otimes = |\psi_\otimes\rangle \langle \psi_\otimes | = \begin{pmatrix} 0.5 & 0 & 0 & 0.5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0.5 & 0 & 0 & 0.5 \end{pmatrix} \]

The critical point is that there are off-diagonal elements here, which directly capture the presence of a correlation between the probability of a \(|00\rangle\) outcome and a \(|11\rangle\) outcome. Specifically, in the first column, the bottom-left cell represents the probability for a \(|00\rangle\) and \(|11\rangle\) outcome, and the upper-right cell represents the probability of a \(|11\rangle\) and \(|00\rangle\) outcome. You can see how this turns into the measured correlations by applying the projection operators, as described below.

The essential point is that these off-diagonal elements are only possible in a state created from a superposition, which happens via the process of entanglement in preparing the initial state of the system.

There are two key contrasting cases which do not produce the uniquely quantum non-locality result. The first one is a pure product state where for example the two qubits are both in the pure 0 state:

\[ |\psi_*\rangle = |0\rangle_A \otimes |0\rangle_B = \begin{pmatrix}1\\0\end{pmatrix}\otimes\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\\0\\0\end{pmatrix} \]

The determinant of the corresponding 2x2 matrix is zero:

\[ C_* = \begin{pmatrix}1 & 0\\0&0\end{pmatrix}, \qquad \det C_* = 0 \]

and the density matrix has no off-diagonal elements:

\[ \rho_* = |\psi_*\rangle \langle \psi_* | = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

The more interesting contrast corresponds to a local hidden variable case, where half the time a \(|00\rangle\) state is prepared, and the other half it is \(|11\rangle\), as compared to the single state that is in a state of superposition. This contrast highlights precisely what it is about the superposition state that differs from a standard probabilistic situation. This then provides the essential insight about entanglement.

\[ \rho_+ = \tfrac12 |00\rangle \langle00| + \tfrac12|11\rangle\langle11| \]

\[ \rho_+ = \begin{pmatrix} 0.5 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0.5 \end{pmatrix} \]

Critically, this also has no off-diagonal elements, and thus only produces classical-like probability results.

Thus, there is an obvious, absolute conclusion from this comparison:

• • superposition (entanglement) is an essential ingredient in producing non-local quantum effects.

Any local hidden variable strategy that does not result in a wave state that is in a state of superposition is not going to capture the quantum effects, because, as you can see in the above math, the resulting probabilities for mixing the different states are purely classical, acting as simple weighted-average multipliers on the resulting density operator.

Interestingly, there is indeed a literature on the effects of superposition in classical waves, i.e., in electromagnetic waves, which reproduce many of the quantum-like effects of superposition (Spreeuw, 1998; Qian & Eberly, 2011; Qian et al., 2015; Karimi & Boyd, 2015; Khrennikov, 2006). However, these experiments do not capture the non-locality present in the Bell’s test experiments, which follows naturally from the tensor-product nature of the configuration space used in the above Hilbert-space formulation.

The main question in this context is thus, how could Nature actually exhibit the corresponding non-locality in terms of a universal basis space representation that doesn’t just build it in as an assumption by using a tensor space?

The measurement observables

Because you cannot directly observe the density matrix values shown above, actual experiments involve setting the parameters on the A and B detectors that each measure a spin component in the \(x\)–\(z\) plane, at angles \(\alpha\) and \(\beta\). The outcome for A when set at an angle of \(\alpha\) is:

\[ A(\alpha) = \sin\alpha\,\sigma_x + \cos\alpha\,\sigma_z = \begin{pmatrix}\cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha\end{pmatrix} \]

and likewise for \(B(\beta)\). The \(\sigma_x\) and \(\sigma_z\) are the Pauli matrices. Each has eigenvalues \(\pm 1\), with eigenvectors

\[ |{+}_\alpha\rangle = \begin{pmatrix}\cos\frac{\alpha}{2}\\[2pt] \sin\frac{\alpha}{2}\end{pmatrix},\qquad |{-}_\alpha\rangle = \begin{pmatrix}-\sin\frac{\alpha}{2}\\[2pt] \cos\frac{\alpha}{2}\end{pmatrix} \]

The spectral projectors are:

\[ P^A_\pm(\alpha) = |{\pm}_\alpha\rangle\langle{\pm}_\alpha| = \tfrac12\begin{pmatrix}1\pm\cos\alpha & \pm\sin\alpha\\ \pm\sin\alpha & 1\mp\cos\alpha\end{pmatrix}, \qquad A(\alpha) = P^A_+ - P^A_- \]

Alice measuring \(\alpha\) and Bob measuring \(\beta\) is a single four-outcome projective measurement on the joint space, with rank-1 projectors

\[ \Pi_{s_A s_B} = P^A_{s_A}(\alpha) \otimes P^B_{s_B}(\beta), \qquad s_A, s_B \in \{+,-\} \]

These are four orthogonal rank-1 projectors summing to \(I_4\). Outcome probabilities by the Born rule:

\[ P(s_A, s_B \mid \alpha,\beta) = \langle \psi_\otimes|\,\Pi_{s_A s_B}\,|\psi_\otimes\rangle = \left|\left(\langle s_A{}_\alpha| \otimes \langle s_B{}_\beta|\right)|\psi_\otimes\rangle\right|^2 \]

Using \(\langle {+}_\alpha|0\rangle = \cos\frac\alpha2\), \(\langle{+}_\alpha|1\rangle = \sin\frac\alpha2\), \(\langle{-}_\alpha|0\rangle = -\sin\frac\alpha2\), \(\langle{-}_\alpha|1\rangle = \cos\frac\alpha2\), and \(|\psi_\otimes\rangle = \frac{1}{\sqrt2}(|00\rangle + |11\rangle)\):

\[ \langle {+}_\alpha {+}_\beta|\psi_\otimes\rangle = \tfrac{1}{\sqrt2}\left[\cos\tfrac\alpha2\cos\tfrac\beta2 + \sin\tfrac\alpha2\sin\tfrac\beta2\right] = \tfrac{1}{\sqrt2}\cos\tfrac{\alpha-\beta}{2} \]

\[ \langle {+}_\alpha {-}_\beta|\psi_\otimes\rangle = \tfrac{1}{\sqrt2}\left[-\cos\tfrac\alpha2\sin\tfrac\beta2 + \sin\tfrac\alpha2\cos\tfrac\beta2\right] = \tfrac{1}{\sqrt2}\sin\tfrac{\alpha-\beta}{2} \]

\[ \langle {-}_\alpha {+}_\beta|\psi_\otimes\rangle = \tfrac{1}{\sqrt2}\left[-\sin\tfrac\alpha2\cos\tfrac\beta2 + \cos\tfrac\alpha2\sin\tfrac\beta2\right] = -\tfrac{1}{\sqrt2}\sin\tfrac{\alpha-\beta}{2} \]

\[ \langle {-}_\alpha {-}_\beta|\psi_\otimes\rangle = \tfrac{1}{\sqrt2}\left[\sin\tfrac\alpha2\sin\tfrac\beta2 + \cos\tfrac\alpha2\cos\tfrac\beta2\right] = \tfrac{1}{\sqrt2}\cos\tfrac{\alpha-\beta}{2} \]

Writing \(\delta \equiv \alpha - \beta\), the joint probabilities are

\[ P(\pm,\pm) = \tfrac12\cos^2\tfrac{\delta}{2}, \qquad P(\pm,\mp) = \tfrac12\sin^2\tfrac\delta2 \]

Normalization: \(\cos^2\frac\delta2 + \sin^2\frac\delta2 = 1\) ✓

Marginals (no-signalling):

\[ P(A = +) = P(+,+) + P(+,-) = \tfrac12\cos^2\tfrac\delta2 + \tfrac12\sin^2\tfrac\delta2 = \tfrac12 \]

Alice’s marginal is \(\tfrac12\) regardless of \(\beta\) — Bob’s setting choice is invisible to Alice. All the structure is in the correlations, none in the marginals. This is why entanglement doesn’t permit signalling.

The correlation function

\[ E(\alpha,\beta) = \langle \psi_\otimes|A(\alpha)\otimes B(\beta)|\psi_\otimes\rangle = P(+,+) + P(-,-) - P(+,-) - P(-,+) \]

\[ = \cos^2\tfrac\delta2 - \sin^2\tfrac\delta2 = \boxed{\cos(\alpha - \beta)} \]

A cross-check without projectors: for \(|\psi_\otimes\rangle = \frac{1}{\sqrt2}\sum_i|ii\rangle\) one has \(\langle\psi_\otimes|M\otimes N|\psi_\otimes\rangle = \frac12\operatorname{tr}(MN^T)\), and \(\operatorname{tr}\!\big(A(\alpha)B(\beta)\big) = 2\cos(\alpha-\beta)\). ✓

6. CHSH and the classical bound

With two settings each (\(a, a'\) for Alice; \(b, b'\) for Bob):

\[ S = E(a,b) - E(a,b') + E(a',b) + E(a',b') \]

Local hidden variable bound. If each run carries predetermined values \(A_a, A_{a'}, B_b, B_{b'} \in \{-1,+1\}\) fixed by some \(\lambda\), then for each \(\lambda\):

\[ S(\lambda) = A_a(B_b - B_{b'}) + A_{a'}(B_b + B_{b'}) \]

Since \(B_b, B_{b'} = \pm1\): either \(B_b = B_{b'}\), making the first bracket \(0\) and the second \(\pm2\); or \(B_b = -B_{b'}\), making the second \(0\) and the first \(\pm2\). Either way \(|S(\lambda)| \le 2\), and averaging over \(\lambda\) preserves it:

\[ \boxed{|S_{\text{LHV}}| \le 2} \]

Quantum prediction. With \(E = \cos(\alpha-\beta)\):

\[ S = \cos(a-b) - \cos(a-b') + \cos(a'-b) + \cos(a'-b') \]

Choosing \(a = 0,\ a' = \frac\pi2,\ b = \frac\pi4,\ b' = \frac{3\pi}{4}\) makes all four terms contribute \(+\frac{\sqrt2}{2}\):

\[ S = 4\cdot\tfrac{\sqrt2}{2} = 2\sqrt2 \approx 2.8284 \]

7. Now with explicit numbers

The four observables

Each has eigenvalues exactly \(\{+1, -1\}\).

The four 4×4 tensor products

Shown multiplied by \(\sqrt2\) so the entries are integers. Acting on \(|\psi_\otimes\rangle = \frac{1}{\sqrt2}(1,0,0,1)^T\):

Worked example for \(A \otimes B\), by hand. Let \(M = \frac{1}{\sqrt2}\begin{psmallmatrix}1&1&0&0\\1&-1&0&0\\0&0&-1&-1\\0&0&-1&1\end{psmallmatrix}\) and \(v = \frac{1}{\sqrt2}(1,0,0,1)^T\):

\[ Mv = \tfrac{1}{\sqrt2}\cdot\tfrac{1}{\sqrt2}\begin{pmatrix}1\cdot1 + 1\cdot 0\\ 1\cdot1 - 1\cdot0 \\ -1\cdot 0 - 1\cdot 1\\ -1\cdot0 + 1\cdot1\end{pmatrix} = \tfrac12\begin{pmatrix}1\\1\\-1\\1\end{pmatrix} \]

\[ v^T(Mv) = \tfrac{1}{\sqrt2}(1,0,0,1)\cdot\tfrac12(1,1,-1,1)^T = \tfrac{1}{2\sqrt2}(1 + 1) = \tfrac{1}{\sqrt2} = 0.707107 \;\checkmark \]

All joint probabilities

Every marginal is exactly \(0.500000\) for both parties in all four rows.

Sample check of row 1: \(\frac12\cos^2(-22.5°) = \frac12(0.923880)^2 = \frac12(0.853553) = 0.426777\) ✓, and \(\frac12\sin^2(-22.5°) = \frac12(0.382683)^2 = 0.073223\) ✓.

Assembling \(S\)

\[ S = (+0.707107) - (-0.707107) + (0.707107) + (0.707107) \]

\[ = 0.707107 \times 4 = \mathbf{2.828427} \]

\[ 2\sqrt2 = 2.828427\ \checkmark \qquad\text{versus the classical bound } 2 \]

Violation ratio: \(2\sqrt2 / 2 = \sqrt2 \approx 1.414\).

Form the CHSH operator on \(\mathbb{C}^4\):

\[ \hat S = A\otimes B - A\otimes B' + A'\otimes B + A'\otimes B' \]

Applying it to the Bell state numerically gives \(\hat S|\psi_\otimes\rangle = 2\sqrt2\,|\psi_\otimes\rangle\) — the Bell state is an exact eigenvector of the CHSH operator at these settings, with eigenvalue \(2\sqrt2\). That’s why the violation is maximal: you can’t do better than an eigenvalue, and Tsirelson’s bound says no eigenvalue is larger. The one-line reason (Landau’s identity):

\[ \hat S^2 = 4\,I \;+\; [A,A']\otimes[B,B'] \]

Since \(\|A\| = \|A'\| = 1\), \(\|[A,A']\| \le 2\), so \(\|\hat S^2\| \le 4 + 4 = 8\) and \(\|\hat S\| \le 2\sqrt2\). If \(A\) and \(A'\) commuted — the classical case, where both have simultaneous definite values — the commutator term vanishes and you recover \(|S| \le 2\) exactly. The violation is precisely the non-commutativity of the two measurement settings, amplified through the tensor product structure.

The correlation factorizes — symbolically

This is the whole difference, in one line. Because the state is a product, the expectation of any product observable splits:

\[ E(\alpha,\beta) = \langle 0|\otimes\langle 0|\;A(\alpha)\otimes B(\beta)\;|0\rangle\otimes|0\rangle = \underbrace{\langle 0|A(\alpha)|0\rangle}_{\text{Alice alone}}\cdot\underbrace{\langle 0|B(\beta)|0\rangle}_{\text{Bob alone}} \]

With \(\langle 0|A(\alpha)|0\rangle = A(\alpha)_{00} = \cos\alpha\):

\[ \boxed{E_{\text{prod}}(\alpha,\beta) = \cos\alpha\,\cos\beta} \qquad\text{versus}\qquad E_{\psi_\otimes}(\alpha,\beta) = \cos(\alpha-\beta) \]

This is the structural difference. \(\cos\alpha\cos\beta\) is a separable function — a product of a function of \(\alpha\) alone and a function of \(\beta\) alone. \(\cos(\alpha-\beta)\) is not; it cannot be factored, because \(\cos(\alpha-\beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta\) has two terms. That second term \(\sin\alpha\sin\beta\) is the entire Bell violation.

The joint probabilities factorize too

\[ P(s_A,s_B) = \left|\langle s_A{}_\alpha|0\rangle\right|^2 \cdot \left|\langle s_B{}_\beta|0\rangle\right|^2 = P(s_A)\cdot P(s_B) \]

with \(P(A{=}+) = \cos^2\frac\alpha2\), \(P(B{=}+) = \cos^2\frac\beta2\). The outcomes are statistically independent — the joint distribution is literally the product of the marginals, for every setting pair. There is nothing to correlate.

Numbers at the Bell-optimal angles

Factorization check on row 3: \(P(+,+) = 0.5 \times 0.853553 = 0.426777\) ✓ exactly.

\[ S = 0.707107 - (-0.707107) + 0 + 0 = \mathbf{1.414214} = \sqrt2 \]

Note that Alice’s marginal here does depend on her own setting (\(1.0\) vs \(0.5\)) — unlike the Bell state, where it was always \(0.5\). That’s a giveaway that this isn’t a fair comparison. Hence:

So no local measurement on either qubit alone can tell them apart. Only the correlations across settings can.

The correlation function

\[ E_{\text{mix}}(\alpha,\beta) = \operatorname{Tr}\!\left[\rho_{\text{mix}}\,A(\alpha)\otimes B(\beta)\right] = \tfrac12\langle00|A\otimes B|00\rangle + \tfrac12\langle11|A\otimes B|11\rangle \]

\[ = \tfrac12\big[\cos\alpha\cos\beta\big] + \tfrac12\big[(-\cos\alpha)(-\cos\beta)\big] = \boxed{\cos\alpha\,\cos\beta} \]

Same separable form as the pure product state.

Numbers

\[ S = 0.707107 - (-0.707107) + 0 + 0 = \mathbf{1.414214} = \sqrt2 \]

The diagnostic: where exactly they diverge

Put the two states’ correlations side by side:

| settings | \(E\) for \(|\psi_\otimes\rangle\) | \(E\) for \(\rho_{\text{mix}}\) | same? |

Look at the first two rows: for Alice’s setting \(a = 0°\) (i.e. \(A = \sigma_z\)), the mixture reproduces the Bell state’s joint probabilities exactly, to every decimal place — not just the correlation but all four \(P(s_A,s_B)\). The classical mixture is a perfect impostor as long as Alice measures \(\sigma_z\).

It fails the moment Alice switches to \(a' = 90°\), where \(A' = \sigma_x\). \(\sigma_x\) is off-diagonal, so its expectation reads the coherences — precisely the matrix elements that \(\rho_{\text{mix}}\) lacks. Both of those rows collapse to \(E = 0\), killing half the CHSH sum.

The single cleanest statement of the difference. Measure both parties along the same in-plane angle \(\theta\):

\[ E_{\psi_\otimes}(\theta,\theta) = \cos(\theta-\theta) = 1 \quad\text{for every }\theta\]

\[E_{\text{mix}}(\theta,\theta) = \cos^2\theta \quad\Rightarrow\quad 1 \text{ at } 0°,\; 0.5 \text{ at } 45°,\; 0 \text{ at } 90° \]

The Bell state is perfectly correlated in every basis. The mixture is perfectly correlated in one basis and uncorrelated in the conjugate one. That basis-independence is what has no classical analogue, and CHSH is the device that detects it by querying two incompatible bases at once.

Why no separable state can ever exceed 2

For \(\rho_{\text{mix}}\) the local hidden variable model is not hypothetical — write it down:

• • \(\lambda \in \{0, 1\}\), each with probability \(\tfrac12\) (which pair was emitted).
• • Given \(\lambda\), Alice outputs \(+1\) with probability \(|\langle{+}_\alpha|\lambda\rangle|^2\), using only \(\alpha\) and \(\lambda\).
• • Given \(\lambda\), Bob independently outputs \(+1\) with probability \(|\langle{+}_\beta|\lambda\rangle|^2\), using only \(\beta\) and \(\lambda\).

Then \(E(\alpha,\beta) = \frac12[\cos\alpha\cos\beta] + \frac12[(-\cos\alpha)(-\cos\beta)] = \cos\alpha\cos\beta\) ✓ — reproduced exactly, with no communication. Since an LHV model exists, the \(|S| \le 2\) proof from before applies verbatim.

Maximizing over all settings for \(E = \cos\alpha\cos\beta\):

\[ S = \cos a\,(\cos b - \cos b') + \cos a'\,(\cos b + \cos b') \]

\[ \max S = |\cos b - \cos b'| + |\cos b + \cos b'| = 2\max(|\cos b|, |\cos b'|) \le 2 \]

The bound \(2\) is reachable (e.g. \(a = 0°, b = 0°, b' = 180°\)) — but only by the degenerate choice \(B' = -B\), where Bob isn’t really using two independent settings. Saturated, never exceeded.

Summary

| | \(|\psi_\otimes\rangle\) | \(|0\rangle|0\rangle\) | \(\rho_{\text{mix}}\) |

The one-sentence version: entanglement shows up as a correlation function that cannot be factored into a function of Alice’s setting times a function of Bob’s, and CHSH is exactly the linear combination that is bounded by 2 for every factorizable correlation and reaches \(2\sqrt2\) for the non-factorizable one.