Lecture 08 · Bell / Continuum

Bell's Theorem, Locality & Continuous States

Local hidden variable theories, the CHSH inequality, quantum violation of Bell's bound, what Bell's theorem actually proves, and the transition from discrete spins to continuous position and momentum.

Part I — The Local Hidden Variable Hypothesis

Einstein was deeply uncomfortable with quantum mechanics. His objection was not that it gave wrong predictions, but that it seemed incomplete. His view: quantum randomness is like classical statistical mechanics — an artifact of ignorance about underlying deterministic variables. There must be hidden variables $\lambda$ that, if known, would determine all outcomes with certainty.

A local hidden variable (LHV) theory formalizes this idea precisely. Each particle carries a hidden variable $\lambda$ drawn from some distribution $p(\lambda)$. When Alice measures spin along direction $\hat{n}$, the outcome is a deterministic function $A(\hat{n}, \lambda) \in \{+1, -1\}$. Similarly for Bob: $B(\hat{m}, \lambda) \in \{+1, -1\}$. The key locality assumption:

  • $A$ does not depend on Bob's detector setting $\hat{m}$
  • $B$ does not depend on Alice's detector setting $\hat{n}$

The two-particle correlation in a LHV theory is:

$$C_\mathrm{LHV}(\hat{n}, \hat{m}) = \int A(\hat{n},\lambda)\,B(\hat{m},\lambda)\,p(\lambda)\,d\lambda$$

The question Bell asked: is there any choice of $A$, $B$, and $p(\lambda)$ that reproduces all quantum mechanical predictions?

Part II — Bell's Inequality (CHSH)

Bell derived a constraint that any LHV theory must satisfy. The modern form, due to Clauser, Horne, Shimony, and Holt, uses four detector settings: Alice has directions $a, a'$ and Bob has directions $b, b'$. Define correlations $E(a,b) = \langle A(a)B(b)\rangle$. The CHSH inequality is:

CHSH Inequality (Bell)
$$\bigl|E(a,b) - E(a,b') + E(a',b) + E(a',b')\bigr| \leq 2$$

The proof is elementary. For fixed $\lambda$, define $A = A(a,\lambda)$, $A' = A(a',\lambda)$, $B = B(b,\lambda)$, $B' = B(b',\lambda)$, each $\in \{+1,-1\}$. Consider:

$$S(\lambda) = A(B - B') + A'(B + B')$$

Since $B, B' \in \{+1,-1\}$, exactly one of $(B-B')$ or $(B+B')$ equals zero and the other equals $\pm 2$. In both cases $|S(\lambda)| = 2$. After averaging over $\lambda$:

$$|\langle S\rangle| = |E(a,b) - E(a,b') + E(a',b) + E(a',b')| \leq \int |S(\lambda)|\,p(\lambda)\,d\lambda = 2$$

Part III — Quantum Violation of Bell's Inequality

For the singlet state, the quantum mechanical correlation between measurements along directions $\hat{n}$ and $\hat{m}$ is:

$$E_\mathrm{QM}(\hat{n},\hat{m}) = \langle\sigma_A\cdot\sigma_B\rangle = -\hat{n}\cdot\hat{m} = -\cos\theta_{nm}$$

where $\theta_{nm}$ is the angle between the two directions. Choosing the optimal angles $a = 0°$, $a' = 90°$, $b = 45°$, $b' = 135°$:

PairAngleCorrelation
$E(a,b)$$45°$$-\cos 45° = -1/\sqrt{2}$
$E(a,b')$$135°$$-\cos 135° = +1/\sqrt{2}$
$E(a',b)$$45°$$-\cos 45° = -1/\sqrt{2}$
$E(a',b')$$45°$$-\cos 45° = -1/\sqrt{2}$
$$S_\mathrm{QM} = \left|-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right| = \frac{4}{\sqrt{2}} = 2\sqrt{2} \approx 2.828$$
Tsirelson Bound
$$S_\mathrm{QM} = 2\sqrt{2} > 2 \quad\text{(violates CHSH by factor }\sqrt{2}\text{)}$$

The value $2\sqrt{2}$ is the quantum maximum — the Tsirelson bound. It is achieved by the singlet with this specific angle configuration. Quantum mechanics not only violates the LHV bound; it saturates its own maximum.

Part IV — What Bell's Theorem Actually Proves

Bell's theorem is a mathematical theorem, not an experimental result. The structure of the argument is a logical chain:

  1. Assume LHV. Then necessarily $|S| \leq 2$ — this is the proven mathematical bound.
  2. Experiment measures $|S| \approx 2\sqrt{2}$. Therefore LHV is false.

What must be given up? The LHV framework has two key assumptions:

  • Realism: outcomes are determined by pre-existing values of $\lambda$.
  • Locality: Alice's outcome doesn't depend on Bob's setting, and vice versa.

Bell's theorem says you can't have both. Most interpretations preserve locality (no faster-than-light signals are possible with entangled particles — you cannot use them to communicate) and abandon hidden variable realism. Bohr's response: the question "what is the spin before measurement?" is simply meaningless — there is no spin value before it is measured.

Part V — Continuum States: From Discrete to Continuous

All our quantum mechanics so far has involved finite-dimensional Hilbert spaces — the qubit, multi-spin systems. Now we turn to a particle on a line. The new observable is position $\hat{x}$ with a continuous spectrum $x \in \mathbb{R}$.

The position eigenstates $|x\rangle$ satisfy $\hat{x}|x\rangle = x|x\rangle$ and are labeled by a continuous parameter. The normalization and completeness relations replace their discrete counterparts:

PropertyDiscreteContinuous
Normalization$\langle n|m\rangle = \delta_{nm}$$\langle x|x'\rangle = \delta(x-x')$
Completeness$\sum_n |n\rangle\langle n| = 1$$\int |x\rangle\langle x|\,dx = 1$

The wave function is the projection of the state onto position eigenstates:

$$\psi(x) = \langle x|\psi\rangle$$

The physical interpretation: $|\psi(x)|^2\,dx$ is the probability of finding the particle between $x$ and $x + dx$. The normalization condition becomes $\int_{-\infty}^\infty |\psi(x)|^2\,dx = 1$.

Part VI — The Dirac Delta Function

The Dirac delta $\delta(x - x_0)$ appears everywhere in continuous quantum mechanics. It is not an ordinary function — it is a distribution (generalized function) defined by its action on test functions:

$$\int_{-\infty}^\infty f(x)\,\delta(x - x_0)\,dx = f(x_0)$$

Key properties:

  • $\delta(-x) = \delta(x)$ — even function
  • $x\,\delta(x) = 0$
  • $\delta(ax) = \delta(x)/|a|$ for $a \neq 0$
  • Fourier representation: $\delta(x) = \dfrac{1}{2\pi}\displaystyle\int_{-\infty}^\infty e^{ikx}\,dk$

The position eigenstate in position space is: $\langle x|x_0\rangle = \delta(x - x_0)$ — an infinitely sharp spike at $x_0$. This encodes "perfect localization" but, by the Heisenberg uncertainty principle, requires completely indefinite momentum.

Part VII — Momentum Eigenstates

The momentum operator in position space is $\hat{p} = -i\hbar\,d/dx$. Its eigenstates satisfy $\hat{p}|p\rangle = p|p\rangle$, which in position space becomes the differential equation:

$$-i\hbar\frac{d}{dx}\langle x|p\rangle = p\langle x|p\rangle \qquad\Rightarrow\qquad \langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}}\,e^{ipx/\hbar}$$
Momentum Eigenstate in Position Space
$$\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}}\,e^{ipx/\hbar}$$

This is a plane wave — completely delocalized in position. Normalization gives $\langle p|p'\rangle = \delta(p - p')$. The momentum-space wave function:

$$\tilde{\psi}(p) = \langle p|\psi\rangle = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty \psi(x)\,e^{-ipx/\hbar}\,dx$$

This is the Fourier transform of $\psi(x)$! The intimate connection between position and momentum is precisely the Fourier duality — and this is why they obey the Heisenberg uncertainty relation $\Delta x\,\Delta p \geq \hbar/2$.

Exercises — Lecture 8

Eight exercises spanning Bell inequalities, hidden variable models, the Dirac delta, and momentum eigenstates. Click any exercise to reveal the full worked solution.

1
CHSH Bound Verification
EasyBell Inequality

The CHSH bound states that any local hidden variable model satisfies $|S_\mathrm{CHSH}| \leq 2$. The quantum maximum is $2\sqrt{2} \approx 2.83$.

(a) What is the maximum possible value of $|S|$ for a LHV theory?

(b) Can a LHV model reach $2\sqrt{2}$? Why or why not?

Complete Solution
a

For any fixed $\lambda$, we showed $|S(\lambda)| = 2$. After averaging over the hidden variable distribution $p(\lambda)$:

$$|S_\mathrm{LHV}| = \left|\int S(\lambda)p(\lambda)\,d\lambda\right| \leq \int|S(\lambda)|p(\lambda)\,d\lambda = 2\int p(\lambda)\,d\lambda = 2$$

The maximum is exactly $2$, achieved when all $\lambda$ yield the same sign of $S(\lambda)$.

b
Answer
No LHV model can reach $2\sqrt{2}$. The CHSH bound $|S| \leq 2$ is a hard mathematical limit for any theory satisfying locality and realism. The value $2\sqrt{2}$ is the Tsirelson bound — the maximum achievable by quantum mechanics, which is itself bounded from above. The gap between $2$ and $2\sqrt{2}$ is experimentally measurable and has been confirmed in numerous Bell test experiments.
2
Dirac Delta Property
EasyDelta Function

Evaluate using the defining property $\int f(x)\delta(x-x_0)\,dx = f(x_0)$:

(a) $\displaystyle\int_{-\infty}^\infty e^{2x}\,\delta(x-3)\,dx$

(b) $\displaystyle\int_{-\infty}^\infty \cos(x)\,\delta(x - \pi/2)\,dx$

Complete Solution
a

Here $f(x) = e^{2x}$ and $x_0 = 3$. The delta function evaluates $f$ at $x_0$:

$$\int_{-\infty}^\infty e^{2x}\,\delta(x-3)\,dx = e^{2\cdot 3} = e^6$$
b

Here $f(x) = \cos(x)$ and $x_0 = \pi/2$:

$$\int_{-\infty}^\infty \cos(x)\,\delta\!\left(x - \frac{\pi}{2}\right)\,dx = \cos\!\left(\frac{\pi}{2}\right) = 0$$
Answers
(a) $e^6 \approx 403.4$. (b) $0$. The delta function simply evaluates the test function at its center — no integration is actually performed.
3
Optimal Bell Angles
MediumCHSH Violation

For the singlet with correlation $C(\theta) = -\cos\theta$ (where $\theta$ is the angle between detector directions), use angles $a = 0°$, $a' = 90°$, $b = 45°$, $b' = 135°$.

(a) Compute the four correlations $E(a,b)$, $E(a,b')$, $E(a',b)$, $E(a',b')$.

(b) Compute $S_\mathrm{CHSH} = |E(a,b) - E(a,b') + E(a',b) + E(a',b')|$.

(c) Confirm $S = 2\sqrt{2}$.

Complete Solution
a

Angle between detectors equals $|b - a|$ (taking the formula $C = -\cos\theta$):

$$E(a,b) = -\cos(45°) = -\frac{1}{\sqrt{2}}$$
$$E(a,b') = -\cos(135°) = -(-\tfrac{1}{\sqrt{2}}) = +\frac{1}{\sqrt{2}}$$
$$E(a',b) = -\cos(|90°-45°|) = -\cos(45°) = -\frac{1}{\sqrt{2}}$$
$$E(a',b') = -\cos(|90°-135°|) = -\cos(45°) = -\frac{1}{\sqrt{2}}$$
b–c
$$S = \left|-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right| = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$$
Confirmed
$S = 2\sqrt{2} \approx 2.828 > 2$. This exceeds the CHSH bound by a factor of $\sqrt{2}$, conclusively ruling out local hidden variable theories.
4
Hidden Variable Model for Single Particle
MediumConceptual

Let $\lambda \in [0, 2\pi)$ be uniform. Define $A(\hat{n}, \lambda) = +1$ if $\lambda$ lies in the hemisphere "above" $\hat{n}$, $-1$ otherwise.

(a) Show $P(A = +1) = 1/2$ for any direction $\hat{n}$.

(b) Show $\langle A(\hat{n})\rangle = 0$.

(c) Explain why this model cannot work for two correlated spins (Bell's theorem).

Complete Solution
a

For any direction $\hat{n}$, exactly half the interval $[0, 2\pi)$ lies in the hemisphere above $\hat{n}$.

$$P(A=+1) = \frac{\pi}{2\pi} = \frac{1}{2} \checkmark$$
b
$$\langle A\rangle = \frac{1}{2}(+1) + \frac{1}{2}(-1) = 0 \checkmark$$
c
The Core Problem
For two correlated spins, we need to reproduce the quantum correlation $E(\hat{n},\hat{m}) = -\cos\theta$. While this model gives the correct single-particle statistics, no assignment of functions $A(\hat{n},\lambda)$ and $B(\hat{m},\lambda)$ can simultaneously satisfy $E = -\cos\theta$ for all angles AND the CHSH bound $|S| \leq 2$. Bell's theorem proves this is impossible — the cosine function gives $S = 2\sqrt{2}$ for optimal angles, violating the LHV constraint.
5
Momentum Operator Eigenvalue
MediumMomentum

Set $\hbar = 1$. Show that $\psi_p(x) = \dfrac{1}{\sqrt{2\pi}}\,e^{ipx}$ is an eigenfunction of $\hat{p} = -i\dfrac{d}{dx}$ with eigenvalue $p$.

Then show that $\langle\psi_p|\psi_{p'}\rangle = \delta(p-p')$ using the Fourier representation of the delta function.

Complete Solution
eigenvalue
$$\hat{p}\,\psi_p(x) = -i\frac{d}{dx}\left(\frac{1}{\sqrt{2\pi}}\,e^{ipx}\right) = -i\cdot(ip)\cdot\frac{e^{ipx}}{\sqrt{2\pi}} = p\cdot\psi_p(x) \checkmark$$

Eigenvalue is $p$.

orthogonality
$$\langle\psi_p|\psi_{p'}\rangle = \int_{-\infty}^\infty \psi_p^*(x)\,\psi_{p'}(x)\,dx = \frac{1}{2\pi}\int_{-\infty}^\infty e^{-ipx}\,e^{ip'x}\,dx = \frac{1}{2\pi}\int_{-\infty}^\infty e^{i(p'-p)x}\,dx$$
Result
Using the Fourier representation $\delta(k) = \tfrac{1}{2\pi}\int e^{ikx}\,dx$: $$\langle\psi_p|\psi_{p'}\rangle = \delta(p'-p) = \delta(p-p') \checkmark$$ The momentum eigenstates form a continuous orthonormal family — orthogonal in the delta-function sense.
6
Bell Inequality Proof
HardProof

Prove the CHSH inequality $|E(a,b) - E(a,b') + E(a',b) + E(a',b')| \leq 2$ for any LHV model where $A(a,\lambda) = \pm 1$ and $B(b,\lambda) = \pm 1$.

Complete Solution
setup

For each fixed $\lambda$, let $A = A(a,\lambda)$, $A' = A(a',\lambda)$, $B = B(b,\lambda)$, $B' = B(b',\lambda) \in \{+1,-1\}$. Define:

$$S(\lambda) = AB - AB' + A'B + A'B' = A(B-B') + A'(B+B')$$
case analysis

Since $B, B' \in \{+1,-1\}$, only two cases:

Case 1: $B = B'$. Then $B - B' = 0$ and $B + B' = \pm 2$. So $S(\lambda) = A'(\pm 2) = \pm 2$.

Case 2: $B = -B'$. Then $B + B' = 0$ and $B - B' = \pm 2$. So $S(\lambda) = A(\pm 2) = \pm 2$.

In all cases: $|S(\lambda)| = 2$.

average
$$|E(a,b) - E(a,b') + E(a',b) + E(a',b')| = \left|\int S(\lambda)\,p(\lambda)\,d\lambda\right| \leq \int |S(\lambda)|\,p(\lambda)\,d\lambda = 2 \cdot 1 = 2$$
QED
The bound $|S| \leq 2$ follows from the algebraic identity $|S(\lambda)| = 2$ for all $\lambda$ and the fact that $p(\lambda)$ is a normalized distribution. No assumptions about the specific form of $A$, $B$, or $p(\lambda)$ are needed beyond $A, B \in \{+1,-1\}$ and locality.
7
Fourier Transform of Gaussian
HardFourier Analysis

Compute $\tilde{\psi}(p) = \dfrac{1}{\sqrt{2\pi}}\displaystyle\int \psi(x)\,e^{-ipx}\,dx$ for $\psi(x) = \left(\dfrac{2}{\pi}\right)^{1/4}\!e^{-x^2}$.

Use the Gaussian integral: $\displaystyle\int e^{-ax^2 + bx}\,dx = \sqrt{\dfrac{\pi}{a}}\,e^{b^2/4a}$.

Complete Solution
integral

Substitute the definitions:

$$\tilde{\psi}(p) = \frac{1}{\sqrt{2\pi}}\left(\frac{2}{\pi}\right)^{1/4}\int_{-\infty}^\infty e^{-x^2}\,e^{-ipx}\,dx = \frac{1}{\sqrt{2\pi}}\left(\frac{2}{\pi}\right)^{1/4}\int e^{-x^2 - ipx}\,dx$$

This matches the Gaussian formula with $a = 1$ and $b = -ip$:

$$\int e^{-x^2 - ipx}\,dx = \sqrt{\pi}\,e^{(-ip)^2/4} = \sqrt{\pi}\,e^{-p^2/4}$$
simplify
$$\tilde{\psi}(p) = \frac{1}{\sqrt{2\pi}}\cdot\left(\frac{2}{\pi}\right)^{1/4}\cdot\sqrt{\pi}\cdot e^{-p^2/4} = \frac{\pi^{1/2}}{\sqrt{2\pi}}\cdot\left(\frac{2}{\pi}\right)^{1/4}\cdot e^{-p^2/4} = \frac{2^{1/4}}{\pi^{1/4}}\cdot\frac{1}{\sqrt{2}}\cdot e^{-p^2/4}$$
$$= \left(\frac{2}{\pi}\right)^{1/4}\cdot\frac{1}{\sqrt{2}}\cdot e^{-p^2/4}$$
Key Result
The Fourier transform of a Gaussian is a Gaussian with reciprocal width. The input has width $\sigma_x = 1/\sqrt{2}$ (from $e^{-x^2} = e^{-x^2/(2\sigma_x^2)}$ with $\sigma_x^2 = 1/2$). The output has width $\sigma_p = \sqrt{2}$ in $p$. The product $\sigma_x\cdot\sigma_p = \tfrac{1}{\sqrt{2}}\cdot\sqrt{2} = 1 = 2\cdot(\hbar/2)$, saturating the Heisenberg bound. A Gaussian wave packet is the minimum uncertainty state.
8
Completeness Relation for Position States
HardDirac Notation

The completeness relation $\int |x\rangle\langle x|\,dx = \mathbf{1}$ means: for any state $|\psi\rangle$, $\int |x\rangle\langle x|\psi\rangle\,dx = |\psi\rangle$.

(a) Apply $\langle y|$ to both sides and use $\langle y|x\rangle = \delta(y-x)$ to derive $\psi(y) = \int\delta(y-x)\,\psi(x)\,dx$. What property does this verify?

(b) Insert completeness between $\langle p|$ and $|\psi\rangle$ to show that $\tilde{\psi}(p) = \langle p|\psi\rangle$ is the Fourier transform of $\psi(x)$.

Complete Solution
a

Act with $\langle y|$ from the left on the completeness relation $\int|x\rangle\langle x|\,dx\,|\psi\rangle = |\psi\rangle$:

$$\langle y|\psi\rangle = \int\langle y|x\rangle\langle x|\psi\rangle\,dx = \int\delta(y-x)\,\psi(x)\,dx = \psi(y)$$

This is the reproducing property of the Dirac delta: $\int\delta(y-x)f(x)\,dx = f(y)$. The delta function acts as the identity kernel in position space.

b

Insert the completeness relation $\mathbf{1} = \int|x\rangle\langle x|\,dx$ into $\langle p|\psi\rangle$:

$$\tilde{\psi}(p) = \langle p|\psi\rangle = \int\langle p|x\rangle\langle x|\psi\rangle\,dx = \int\frac{1}{\sqrt{2\pi\hbar}}\,e^{-ipx/\hbar}\cdot\psi(x)\,dx$$
Key Insight
This is exactly the Fourier transform of $\psi(x)$ evaluated at momentum $p$. The momentum-space wave function IS the Fourier transform of the position-space wave function. Position and momentum are Fourier conjugates — this is the deep origin of the Heisenberg uncertainty relation $\Delta x\,\Delta p \geq \hbar/2$.