# Einstein, Podolsky, Rosen & Bell
> R.1.4 proved that no atom carries a little arrow whose projections the analyzer reads off — and then
> deliberately left a door open, because a cleverer hidden mechanism might still work. This lesson
> shuts the door. Two atoms, three analyzer settings, one counting argument: any theory in which the
> atoms leave the source carrying complete instructions makes a prediction quantum mechanics violates,
> and nature sides with quantum mechanics.
## Learning Objectives
After this lesson you will be able to:
1. Write the two-atom **singlet state**, prove it has the same form in every basis, and derive its perfect anticorrelation along any shared axis.
2. State the EPR argument in Bohm's spin form — what "element of reality" means, and precisely which premise quantum mechanics denies.
3. Prove the **no-communication theorem** for this setup: show that one wing's statistics are independent of the other wing's setting and outcome, and explain why the frame-dependence of "first" is therefore harmless.
4. Explain why the classical coin analogy reproduces same-axis data and fails everywhere else, and state what a **local hidden variable** theory must supply.
5. Compute both predictions for the three-setting Bell experiment — the quantum $\tfrac12$ and the local-realist bound $\ge\tfrac59$ — and identify the four ingredients any Bell test needs.
6. Work the four-setting variant, show that both theories predict $\tfrac12$ there, and explain why the choice of settings is part of the experiment's design rather than a detail.
---
## Intuition
Everything so far has involved one atom at a time. Two atoms are where the trouble starts.
Build a source that emits atoms in pairs, flying apart in opposite directions, prepared so that their
spins are always found *opposite* — whenever the left analyzer says $+$, the right one says $-$. That
much is unremarkable; a machine that mails a left shoe to Tokyo and a right shoe to Toronto has the
same property, and nobody thinks the shoes are talking to each other.
Now the part that is not like shoes. The anticorrelation holds along *every* axis, provided the two
analyzers agree on which axis. Vertical and vertical: always opposite. Horizontal and horizontal:
always opposite. Both tilted $37°$: always opposite. And along each of these axes, each individual
result is 50–50 random.
Try to build that out of shoes. A shoe is a left shoe, period; there is no measurement you can make on
a shoe that comes out random. To reproduce our source with objects that carry definite properties, each
atom would have to carry an *answer for every possible axis* — a complete instruction set, prepared at
the source, with the right atom's instructions the exact complement of the left's. That is exactly the
patch Einstein, Podolsky, and Rosen proposed in 1935: quantum mechanics is not wrong, just incomplete;
the atoms carry hidden variables and the theory does not mention them.
It is an excellent idea. It is local, it is realistic, it explains all the data above, and it dissolves
the spookiness completely. And in 1964 John Bell found a way to test it — not test some particular
hidden-variable model, but test *every possible one at once* — by looking at what happens when the two
analyzers are set to *different* axes. The instruction-set hypothesis, it turns out, forces an
inequality on the statistics. Quantum mechanics violates it. Experiment agrees with quantum mechanics
[Fre, §2.3].
That is the whole lesson. The rest is making each step airtight.
---
## Theory
### Two atoms, and the notation for them
The state space of two spin-$\tfrac12$ atoms is the tensor product
$\mathbb{C}^2\otimes\mathbb{C}^2 = \mathbb{C}^4$, with product basis
$\lvert{+z}\rangle\otimes\lvert{+z}\rangle$ and its three companions. That is four symbols too many, so
we compact it: **the left slot is the left-moving atom, the right slot the right-moving one**,
$$
\lvert{+z,-z}\rangle \;\equiv\; \lvert{+z}\rangle\otimes\lvert{-z}\rangle ,
$$
and we use this comma notation for the rest of the lesson. The formal treatment of the tensor product
is [1.4.1 Multi-Qubit States](../../term-1-quantum-mechanics/c04-entanglement/01-multi-qubit-states.md);
all we need here is that it exists and that operators on one atom act on their own slot.
The source we need emits the **singlet state**
$$
\lvert{\Psi^-}\rangle = \frac{1}{\sqrt2}\Bigl(\lvert{+z,-z}\rangle - \lvert{-z,+z}\rangle\Bigr) ,
$$
the total-spin-zero combination of two spin-$\tfrac12$ particles. It is not exotic: it is the ground
state of the electron pair in a hydrogen molecule, and Bohm's 1951 proposal was simply to dissociate
such a molecule without disturbing the spins. This is one of the four Bell states of
[1.4.2 Bell States](../../term-1-quantum-mechanics/c04-entanglement/02-bell-states.md), where its
circuit construction is given; here we care only about its correlations.
### The singlet looks the same from every angle
The single most useful fact about $\lvert{\Psi^-}\rangle$ is that its form does not depend on the basis
you write it in. Take the rotated basis of R.1.2, $c = \cos(\theta/2)$, $s = \sin(\theta/2)$:
$$
\lvert{+\hat n}\rangle = c\lvert{+z}\rangle + s\lvert{-z}\rangle ,
\qquad
\lvert{-\hat n}\rangle = -s\lvert{+z}\rangle + c\lvert{-z}\rangle .
$$
Expand the two product terms:
$$
\lvert{+\hat n, -\hat n}\rangle = -cs\lvert{+z,+z}\rangle + c^2\lvert{+z,-z}\rangle - s^2\lvert{-z,+z}\rangle + sc\lvert{-z,-z}\rangle ,
$$
$$
\lvert{-\hat n, +\hat n}\rangle = -sc\lvert{+z,+z}\rangle - s^2\lvert{+z,-z}\rangle + c^2\lvert{-z,+z}\rangle + cs\lvert{-z,-z}\rangle .
$$
Subtract. The $\lvert{+z,+z}\rangle$ and $\lvert{-z,-z}\rangle$ terms cancel identically, and
$c^2 + s^2 = 1$ collapses the rest:
$$
\frac{1}{\sqrt2}\Bigl(\lvert{+\hat n,-\hat n}\rangle - \lvert{-\hat n,+\hat n}\rangle\Bigr)
= \frac{1}{\sqrt2}\Bigl(\lvert{+z,-z}\rangle - \lvert{-z,+z}\rangle\Bigr) = \lvert{\Psi^-}\rangle .
$$
The singlet is **rotationally invariant** — it is the unique two-qubit state with that property, and
the source therefore needs no preferred axis. Two consequences follow immediately.
*Perfect anticorrelation along any shared axis.* Written in the $\hat n$ basis, $\lvert{\Psi^-}\rangle$
contains no $\lvert{+\hat n,+\hat n}\rangle$ and no $\lvert{-\hat n,-\hat n}\rangle$ term, so if both
analyzers are set to $\hat n$ the outcomes are opposite with probability 1, whatever $\hat n$ is.
*Each wing alone is featureless.* The two branches have equal weight in every basis, so each analyzer
on its own reports 50–50 no matter how it is oriented. All the structure lives in the correlation.
### Correlations at a relative angle
Set the left analyzer to $\hat a$ and the right to $\hat b$, with relative angle $\theta$. Suppose the
left registers $+$; by the anticorrelation just proved, the right-hand atom then behaves exactly as
though it had been prepared in $\lvert{-\hat a}\rangle$. Applying the $\cos^2(\theta/2)$ law of R.1.2 to
that state and the analyzer $\hat b$:
$$
P(\text{right}=- \mid \text{left}=+) = \bigl|\langle{-\hat b}|{-\hat a}\rangle\bigr|^2 = \cos^2\frac{\theta}{2} .
$$
The same holds with all signs flipped, and each left outcome has probability $\tfrac12$, so overall
$$
P(\text{opposite}) = \cos^2\frac{\theta}{2},
\qquad
P(\text{same}) = \sin^2\frac{\theta}{2} .
$$
Sanity checks: $\theta = 0$ gives perfect anticorrelation, $\theta = 180°$ gives perfect *correlation*
(one analyzer reversed relabels its ports), and $\theta = 90°$ gives $\tfrac12$ — no correlation at all.
Those three cases are exactly the ones a hidden-variable theory finds easy, which is why Bell had to
look elsewhere.
```mermaid
flowchart LR
LA["left analyzer
setting a"] --> LD["+ or -"]
SRC["EPR source
singlet pairs"] --> LA
SRC --> RA["right analyzer
setting b"]
RA --> RD["+ or -"]
LD -.->|"compare later"| CMP["P(opposite)
= cos²(θ/2)"]
RD -.-> CMP
```
### The EPR argument
Einstein, Podolsky, and Rosen asked what this correlation implies about the world, and their argument
is tighter than its reputation [EPR35].
Their criterion: if, without in any way disturbing a system, we can predict with certainty the value of
a physical quantity, then there exists an **element of reality** corresponding to that quantity. It is a
modest-sounding principle, and deliberately so — it is meant to be the least a realist could ask for.
Now run it on the singlet. The two atoms are far apart. Measure the left atom along $z$; the right
atom's $z$ result is then certain. By the criterion, the right atom's $z$ projection is an element of
reality. But we could equally have chosen $x$, in which case the right atom's $x$ projection would be an
element of reality. And — this is the crucial step — *the two choices are made on the left, arbitrarily
far away*. If the left analyzer cannot influence the right atom (locality), then whatever is real about
the right atom was real before the choice was made. So the right atom must possess definite $z$ *and*
$x$ projections simultaneously.
Quantum mechanics says no such state exists: $[S_z, S_x] = i\hbar S_y \ne 0$ (R.1.4). EPR concluded that
quantum mechanics is therefore *incomplete* — an accurate statistical summary of a deeper theory whose
variables it does not name. Note what they did *not* claim: they never said quantum mechanics makes
wrong predictions. Their charge was that it is not the whole story.
The argument is valid. Its premises are locality and the reality criterion, and one of them has to go.
Einstein was clear about which one he wanted to keep: he had spent a career removing action at a
distance from gravity and had no appetite for putting it back into mechanics. What he called
**spooky action at a distance** was the alternative he was arguing against, not a description of his own
view.
> **Historical note.** Podolsky, without Einstein's knowledge, gave the paper to *The New York Times*,
> which ran it on the front page as Einstein disproving quantum mechanics. Einstein was furious —
> physics is settled among physicists, not in the press — and never spoke to him again. It is a useful
> reminder that the paper's fame and its argument are two different things.
### Does the spookiness need delayed choice?
Freericks poses this as a problem [Fre, §2.3.1], and it is worth answering carefully because the
intuitive answer is wrong.
No, delayed choice is not required. Place the left analyzer slightly closer to the source, so the left
atom is measured first in the lab frame, and set both analyzers to $z$ from the start. The left result
is random; the right result is determined by it; and no signal could have travelled between them in
time. The correlation is already unexplainable by anything that happens after the pair separates. What
delayed choice adds is the closing of a specific loophole — the possibility that the atoms somehow knew
the analyzer settings in advance and coordinated accordingly. Rotating an analyzer while the atoms are
in flight removes that escape, which is why Aspect's group built rapidly switching analyzers in the
1980s. But it is a hardening of the experiment, not the source of the puzzle.
### The no-communication theorem
Before any of this can be called spooky, we must rule out the obvious worry: does the correlation let
you send a message faster than light? It does not, and the proof is short enough to do here.
Let the pair be in any joint state $\lvert{\Psi}\rangle$. Alice, on the left, chooses a measurement —
some complete set of orthogonal projectors $\{P_i\}$ on her factor, with $\sum_i P_i = \mathbb{1}$ — and
gets outcome $i$ with probability $p_i$. Bob, on the right, knows neither her choice nor her outcome.
His state is the average over her outcomes, weighted by their probabilities:
$$
\rho_B = \sum_i p_i \, \rho_B^{(i)}
= \sum_i \operatorname{Tr}_A\Bigl[(P_i\otimes\mathbb{1})\,\lvert{\Psi}\rangle\langle{\Psi}\rvert\,(P_i\otimes\mathbb{1})\Bigr] .
$$
Cyclicity of the trace over Alice's factor lets us combine the two projectors into $P_i^2 = P_i$, and
then completeness collapses the sum:
$$
\rho_B = \operatorname{Tr}_A\Bigl[\Bigl(\sum_i P_i\otimes\mathbb{1}\Bigr)\lvert{\Psi}\rangle\langle{\Psi}\rvert\Bigr]
= \operatorname{Tr}_A \lvert{\Psi}\rangle\langle{\Psi}\rvert .
$$
Bob's reduced state is the one he would have had if Alice had done nothing at all. It does not depend on
which observable she measured, or whether she measured. For the singlet it is $\rho_B = \tfrac12\mathbb{1}$
— maximally mixed, 50–50 along every axis, forever. This is the **no-communication theorem**, and it is
the reason entanglement coexists peacefully with relativity.
The physical reading is worth stating plainly. Alice does learn Bob's outcome instantly. But she cannot
*choose* it, because her own outcome is random; all she can send down her wing is a fair coin. A
correlation only becomes a message when the two halves are compared, and comparing requires a classical
channel running at or below the speed of light.
There is a bonus. Because no signal is needed, the fact that "which measurement happened first" is
frame-dependent causes no trouble. Take the left analyzer at $z$, the right at $x$. Analyzed left-first:
the left gives $\pm$ at random, the right atom is then $\mp z$, and measuring it along $x$ gives 50–50 —
uncorrelated. Analyzed right-first: the right gives $\pm$ at random, the left atom is then $\mp x$, and
measuring along $z$ gives 50–50 — uncorrelated. *Both orderings give identical predictions*, as they
must, because the joint probabilities are computed from one state and two commuting observables. A
theory that required a signal would have to say which way it went, and relativity refuses to tell it.
### The coin analogy, and where it breaks
The standard way to defuse the spookiness is a story about coins. A dealer in London posts a gold coin
to one customer and a silver coin to the other, choosing at random who gets which. When the customer in
Tokyo opens their parcel and sees gold, they know instantly that Toronto has silver. Nothing travelled;
the correlation was packed at the source.
For same-axis measurements this analogy is *exactly right*, and it is important to concede that. If we
only ever set both analyzers to $z$, the data are indistinguishable from coins, and anyone insisting on
spookiness is overreading them.
It breaks in two places [Fre, §2.3.3].
**First, the coins have values in transit and the atoms do not.** A gold coin is gold in the envelope.
The singlet, by contrast, has no fact of the matter about which side is $+$ until a measurement is made
— and R.1.4's conundrum of projections already showed that assigning such a fact to a single atom fails.
The coin story reproduces the correlations but posits definiteness that has independently been ruled out.
**Second, and fatally, one pair of coins cannot answer questions about a different axis.** Set the left
analyzer to $z$ and the right to $x$ and the data come out uncorrelated; set both to $37°$ and they come
out perfectly anticorrelated. To reproduce that, each envelope must contain not one coin but an answer
for every axis anyone might ask about — which is no longer a coin analogy, it is the hidden-variable
hypothesis. So the analogy does not compete with quantum mechanics; it *is* the theory Bell will test,
in disguise.
### What a local hidden variable theory must supply
Let us state the hypothesis with no wiggle room, because the whole force of Bell's argument is that no
wiggle room remains [Fre, §2.3.4].
A **local hidden variable** theory says each atom leaves the source carrying an **instruction set**: a
function from analyzer setting to outcome, fixed at creation. Three words carry the weight.
- *Local:* the instructions are fixed at the source and cannot be altered afterwards by anything happening at the distant wing. No influence propagates faster than light.
- *Hidden:* we cannot read the instructions. They may be drawn from any probability distribution, and the distribution may be as baroque as you like.
- *Realistic:* the instructions cover *every* setting, whether or not it is chosen. This is not an extra assumption but a consequence of the first two: since the analyzer can be rotated after the pair is created — Wheeler's delayed choice again — an instruction set that only covered the actual setting would need to be updated non-locally.
The **instruction set** given to the right atom must be the complement of the left's, so that a
shared setting always yields opposite results. That single requirement is what makes the theory
testable, because it means one instruction set determines both wings.
Three warm-up cases show the hypothesis is not a straw man; it handles them effortlessly.
| Experiment | Quantum prediction | Instruction set that matches |
| --- | --- | --- |
| Both analyzers along $+z$ | always opposite, each wing 50–50 | left $+$ or $-$ at random; right the complement |
| Left $+z$, right $-z$ | always *the same*, each wing 50–50 | same instructions; reversing the right analyzer swaps its labels |
| Left $+z$, right $+x$ | uncorrelated, each wing 50–50 | independent fair coins for the two settings |
Bell's insight was that these three cases — $0°$, $180°$, $90°$ — are precisely the ones a hidden-variable
theory can absorb, and that they are therefore the wrong places to look.
### The three-setting Bell experiment
Here is the experiment that decides it [Fre, §2.3; Bell64]. Each analyzer can be set to one of three
orientations, coplanar and equally spaced:
$$
A = 0°, \qquad B = 120°, \qquad C = 240° ,
$$
chosen independently and uniformly at random on each side, on every shot. Package each analyzer into a
box with two indicator lights, $+$ and $-$; the setting is not recorded. One light flashes on each side
per shot, and the entire experiment consists of one question: **how often do the two boxes flash
opposite?**
```mermaid
flowchart LR
LB["left Bell analyzer
random A / B / C
lights + and -"] --- SRC["singlet source"]
SRC --- RB["right Bell analyzer
random A / B / C
lights + and -"]
LB --> Q{"how often
do they flash
opposite?"}
RB --> Q
Q --> QM["quantum: 1/2"]
Q --> HV["local realism: at least 5/9"]
```
**The quantum prediction.** Only the relative angle matters. Of the nine equally likely setting pairs,
three are aligned ($\theta = 0$) and six are misaligned by $\pm120°$. Using
$P(\text{opposite}) = \cos^2(\theta/2)$ with $\cos^2 0° = 1$ and $\cos^2 60° = \tfrac14$:
$$
P_{\text{QM}}(\text{opposite}) = \frac{1}{9}\left[3\times 1 + 6\times\frac14\right]
= \frac{1}{9}\cdot\frac{9}{2} = \frac12 .
$$
Exactly one half. (Freericks gets there by fixing the left result and averaging over the right
analyzer's three stops; the arithmetic is the same, $\tfrac13(1 + \tfrac14 + \tfrac14) = \tfrac12$.)
**The local-realist prediction.** Now the counting argument, which uses no quantum mechanics at all.
The left atom carries a triple of instructions $(L_A, L_B, L_C)$, each $\pm$; there are $2^3 = 8$ of
them. The right atom carries the complement, $R_X = \overline{L_X}$.
On a shot where the left analyzer is at setting $i$ and the right at setting $j$, the left light shows
$L_i$ and the right light shows $\overline{L_j}$. The lights are *opposite* exactly when
$L_i \ne \overline{L_j}$, that is, when
$$
L_i = L_j .
$$
So for a given instruction set, the fraction of the nine setting pairs that flash opposite is just the
fraction of pairs $(i,j)$ whose instructions agree. Count:
| Instruction set | Pattern | Agreeing pairs out of 9 | $P(\text{opposite})$ |
| --- | --- | --- | --- |
| $+\!+\!+$, $-\!-\!-$ | all alike | $3^2 = 9$ | $1$ |
| the other six | two alike, one different | $2^2 + 1^2 = 5$ | $5/9$ |
Every instruction set gives at least $\tfrac59$. And since the source's actual output is *some*
probability distribution over these eight sets — we are allowed to know nothing about it — the observed
frequency is a weighted average of numbers each of which is $\ge \tfrac59$. Therefore
$$
P_{\text{LHV}}(\text{opposite}) \;\ge\; \frac59 \approx 0.5556 .
$$
This is a **Bell inequality**: a bound on the statistics of *any* local hidden variable theory, derived
by counting, with no free parameters and no assumption about the distribution over instruction sets.
**The verdict.** $\tfrac12 < \tfrac59$. Quantum mechanics predicts a value the hypothesis forbids, and
the gap is $0.0556$ — small, but a probability gap of that size is settled by a few thousand shots
(exercise E4). Experiments say $\tfrac12$.
Notice how little went into the argument. No wavefunctions, no operators, no commutators — just the
observation that a triple of $\pm$ symbols cannot have fewer than five of its nine self-comparisons
agree. Mermin's 1981 exposition made exactly this point: the machinery of quantum mechanics is not
needed to see that something is wrong with the classical picture, only the ability to count to nine.
### The four-setting variant, and why the design matters
Freericks closes the section with a variant that looks like an obvious improvement and is not
[Fre, §§2.3.7–2.3.8]. Use four settings at $90°$ intervals: $+z\,(0°)$, $+x\,(90°)$, $-z\,(180°)$,
$-x\,(270°)$, again chosen independently and uniformly.
*Quantum.* Of the sixteen setting pairs, four are aligned, eight differ by $90°$, and four differ by
$180°$:
$$
P_{\text{QM}}(\text{opposite}) = \frac{1}{16}\left[4\cos^2 0° + 8\cos^2 45° + 4\cos^2 90°\right]
= \frac{1}{16}\left[4 + 4 + 0\right] = \frac12 .
$$
*Local realism.* Instruction sets are now quadruples, $2^4 = 16$ of them, and opposite flashes again
correspond to agreeing pairs $(i,j)$:
| Pattern of the quadruple | Number of such sets | Agreeing pairs out of 16 | $P(\text{opposite})$ |
| --- | --- | --- | --- |
| $4$–$0$ (e.g. $+\!+\!+\!+$) | 2 | $4^2 = 16$ | $1$ |
| $3$–$1$ (e.g. $+\!+\!+\!-$) | 8 | $3^2 + 1^2 = 10$ | $5/8$ |
| $2$–$2$ (e.g. $+\!+\!-\!-$) | 6 | $2^2 + 2^2 = 8$ | $1/2$ |
The minimum is now exactly $\tfrac12$ — the quantum value. A source that emitted only *balanced*
instruction sets, two pluses and two minuses, would reproduce the quantum statistics of this experiment
perfectly.
So the four-setting experiment *cannot* distinguish the theories. This is not a defect of local
realism; it is a defect of the experiment, and the lesson is important enough to state in bold: **a Bell
test is a design problem, and most designs fail.** The three equally spaced settings work because $120°$
puts $\cos^2 60° = \tfrac14$ where a counting argument needs $\tfrac{1}{3}$-ish; the four perpendicular
settings fail because $90°$ and $180°$ are exactly the angles a hidden-variable theory already handles
(see the warm-up table). Nothing here rehabilitates local realism — the three-setting experiment has
been done, repeatedly, and it comes out quantum. It only shows that "we measured entangled particles and
saw correlations" is not by itself evidence of anything.
### What a Bell experiment needs
Gathering the requirements [Fre, §2.3.5], a test of local realism against quantum mechanics must:
1. **Create entangled pairs.** Product states have local descriptions by construction.
2. **Use settings that differ, at least some of the time.** With aligned analyzers the coin story is exact.
3. **Use settings away from $0°$, $90°$, and $180°$, at least some of the time.** Those three are the angles a hidden-variable theory absorbs.
4. **Predict different numbers.** The whole point is a gap between the two predictions that data can resolve.
Two things are *not* on the list, though both make a real experiment more convincing. Changing the
settings while the particles are in flight is not required for the puzzle, only for closing the locality
loophole. And the particles need not be atoms — every actual Bell test to date has used photons, or
photon-mediated entanglement between distant electron spins, because they are enormously easier to make
and separate.
### The experimental record
The chain of experiments is worth knowing because it is a case study in taking loopholes seriously.
Freedman and Clauser (1972) and Aspect and collaborators (1981–82) saw the violation, with Aspect's
group adding fast switching to address locality. Weihs et al. (1998) enforced spacelike separation
between the setting choices and the distant detections. Rowe et al. (2001) closed the detection loophole
using trapped ions, where essentially every event is registered. In 2015 three groups — Hensen et al.
with entangled electron spins in diamond, and Giustina et al. and Shalm et al. with photons — closed all
the major loopholes simultaneously. The 2022 Nobel Prize in Physics went to Aspect, Clauser, and
Zeilinger for this line of work.
The result, restated as carefully as it deserves: **local realism is inconsistent with experiment.** At
least one of "the outcomes were determined before measurement" and "no influence travels faster than
light" is false. Quantum mechanics keeps the second and abandons the first. It is possible to construct
theories that do the reverse — Bohmian mechanics is explicitly nonlocal — but nobody gets to keep both.
### Closing the loop from R.1.4
R.1.4 ended with an explicit promissory note. The conundrum of projections proved that an atom cannot
carry a classical *vector* whose components the analyzer reads off, and then said, in so many words,
that this does not by itself rule out every hidden-variable theory: a contrived enough one, with extra
machinery keyed to the apparatus, might survive.
That note is now paid. The instruction-set hypothesis is exactly the general form of the contrivance —
arbitrary machinery, arbitrary distribution, an answer prepared for every setting — and the three-setting
Bell experiment refutes it by counting. What survived R.1.4 does not survive this.
Two things are worth keeping straight about the scope. The refutation is of *local* hidden variables;
nonlocal ones remain logically available and unattractive. And this lesson has been deliberately
experiment-first: the same result stated in operator language, with correlation functions
$E(a,b) = -\hat a\cdot\hat b$, the CHSH combination, and the Tsirelson bound $2\sqrt2$, is
[1.4.3 Nonlocality & CHSH](../../term-1-quantum-mechanics/c04-entanglement/03-nonlocality-chsh.md). The
two routes reach the same place; go read that one next if you want the algebra that generalizes.
---
## Hands-on (Python)
```python
import numpy as np
from itertools import product
# ---- Singlet, in any basis ----------------------------------------------
pz, mz = np.array([1, 0], complex), np.array([0, 1], complex)
def n_basis(theta):
c, s = np.cos(theta / 2), np.sin(theta / 2)
return c * pz + s * mz, -s * pz + c * mz
def singlet(theta=0.0):
p, m = n_basis(theta)
return (np.kron(p, m) - np.kron(m, p)) / np.sqrt(2)
ref = singlet(0.0)
print("singlet is basis independent:",
all(abs(abs(np.vdot(ref, singlet(t))) - 1) < 1e-12
for t in np.deg2rad([37, 90, 120, 251])))
# ---- Joint probabilities at settings (a, b) -----------------------------
def joint(a, b):
"""P[(left sign, right sign)] for analyzer angles a, b (radians)."""
psi, out = singlet(), {}
for sl, vl in zip((+1, -1), n_basis(a)):
for sr, vr in zip((+1, -1), n_basis(b)):
out[(sl, sr)] = abs(np.vdot(np.kron(vl, vr), psi)) ** 2
return out
def p_opposite(a, b):
j = joint(a, b)
return j[(+1, -1)] + j[(-1, +1)]
print("\n theta P(opposite) cos^2(theta/2)")
for deg in (0, 60, 90, 120, 180):
th = np.deg2rad(deg)
print(f" {deg:4d} {p_opposite(0.0, th):.4f} {np.cos(th/2)**2:.4f}")
# ---- No communication: Bob's marginal never moves -----------------------
print("\nBob's P(+) as Alice changes her setting:")
for deg in (0, 45, 90, 137):
j = joint(np.deg2rad(deg), 0.0)
print(f" Alice at {deg:3d} deg -> P(Bob +) = {j[(+1, +1)] + j[(-1, +1)]:.4f}")
# ---- Three-setting Bell test -------------------------------------------
def quantum_opposite(settings):
s = np.deg2rad(settings)
return np.mean([[p_opposite(a, b) for b in s] for a in s])
def lhv_opposite(instr):
"""Lights flash opposite iff the left instructions at the two stops agree."""
n = len(instr)
return sum(instr[i] == instr[j] for i in range(n) for j in range(n)) / n**2
three = [0, 120, 240]
print(f"\nTHREE stops {three}")
print(f" quantum P(opposite) = {quantum_opposite(three):.4f}")
worst = min(lhv_opposite(s) for s in product([+1, -1], repeat=3))
for s in product([+1, -1], repeat=3):
tag = "".join("+" if x > 0 else "-" for x in s)
print(f" instructions {tag} P(opposite) = {lhv_opposite(s):.4f}")
print(f" local-realist minimum = {worst:.4f} ( = 5/9 )")
# ---- Four-setting variant: the gap closes ------------------------------
four = [0, 90, 180, 270]
print(f"\nFOUR stops {four}")
print(f" quantum P(opposite) = {quantum_opposite(four):.4f}")
seen = {}
for s in product([+1, -1], repeat=4):
seen.setdefault(round(lhv_opposite(s), 4), 0)
seen[round(lhv_opposite(s), 4)] += 1
for p, count in sorted(seen.items()):
print(f" {count:2d} instruction sets give P(opposite) = {p:.4f}")
print(" local-realist minimum = 0.5000 -> no gap, the test is blind")
```
The three-setting block prints `0.5000` for quantum against a local-realist minimum of `0.5556`; the
four-setting block prints `0.5000` for *both*, which is the whole point of §2.3.8.
---
## Exercises
**E1 (easy).** Both analyzers are set to $\hat n$ at $37°$ from vertical. What is the probability that
they read opposite? Now set the left to $37°$ and the right to $127°$. Answer both from the general
formula and say which experimental fact each case illustrates.
Solution
*Both at $37°$.* Relative angle $\theta = 0$, so $P(\text{opposite}) = \cos^2 0 = 1$ — always opposite.
This is the rotational invariance of the singlet: the perfect anticorrelation holds along *every* shared
axis, not just the vertical, which is what makes the source impossible to imitate with a single pair of
labelled objects.
*Left $37°$, right $127°$.* Relative angle $\theta = 90°$, so
$P(\text{opposite}) = \cos^2 45° = \tfrac12$ — no correlation whatsoever. Each wing is still 50–50
individually. This is one of the three "easy" angles that any hidden-variable theory reproduces with
independent fair coins, which is exactly why a Bell test must avoid relying on it.
Solution
Nothing changes for Bob in any case. Averaged over Alice's outcomes — which he does not know — his
reduced state is $\rho_B = \tfrac12\mathbb{1}$, so he finds $+$ with probability $\tfrac12$ along $z$,
along $x$, and along every other axis.
It is worth seeing it concretely for the $z$ case: with probability $\tfrac12$ Alice got $+$ and Bob's
atom is $\lvert{-z}\rangle$; with probability $\tfrac12$ she got $-$ and it is $\lvert{+z}\rangle$. The
mixture is $\tfrac12(\lvert{+z}\rangle\langle{+z}\rvert + \lvert{-z}\rangle\langle{-z}\rvert) = \tfrac12\mathbb{1}$.
Doing the same with $x$ gives $\tfrac12(\lvert{+x}\rangle\langle{+x}\rvert + \lvert{-x}\rangle\langle{-x}\rvert)$,
which is the *same* matrix. Bob cannot tell which observable Alice measured, or whether she measured at
all — the no-communication theorem in its most concrete form.
Solution
Quantum first: perpendicular analyzers means $\theta = 90°$, so
$P(\text{opposite}) = \cos^2 45° = \tfrac12$ and each wing reads $\pm$ half the time.
Now the instruction sets. Say the left analyzer is at $z$ and the right at $x$: the left light shows
$L_z$ and the right light shows $\overline{L_x}$, so they flash opposite iff $L_z = L_x$ — the two
symbols of the left instruction agree.
- **(2)** $+\!+$ and $-\!-$ both have agreeing symbols, so the lights are *always* opposite: $P = 1$. ✗
- **(3)** $+\!-$ and $-\!+$ both disagree, so the lights are always the same: $P = 0$. ✗
- **(4)** $+\!+$ always: $P = 1$, and worse, the left light always shows $+$ when the left analyzer is at $z$, so the wings are not even individually 50–50. ✗
- **(1)** Two of the four sets agree and two disagree, so $P(\text{opposite}) = \tfrac12$. ✓ Checking the other configuration (left at $x$, right at $z$) gives the same count by symmetry, and each wing is 50–50 because each symbol is $\pm$ equally often.
So option (1) works. Note what this shows: perpendicular settings are *easy* for hidden variables, and a
theory that survives here is not thereby in any trouble. The three-setting experiment is needed precisely
because the easy angles prove nothing.
Solution
**(a)** The gap is $\Delta = \tfrac59 - \tfrac12 = \tfrac1{18} = 0.0556$. Estimating a probability near
$\tfrac12$ from $N$ shots has standard error $\sigma_{\hat p} = \sqrt{p(1-p)/N} \approx 0.5/\sqrt N$
(R.1.3). Requiring $\Delta \ge 3\sigma_{\hat p}$:
$$
0.0556 \ge \frac{1.5}{\sqrt N}
\quad\Longrightarrow\quad
\sqrt N \ge 27
\quad\Longrightarrow\quad
N \gtrsim 730 .
$$
Under a thousand pairs for a $3\sigma$ result — and note this is the *worst case*, since the bound
$\tfrac59$ is attained only if the source never emits an all-alike instruction set; any admixture of
those pushes the local-realist prediction higher and widens the gap.
**(b)** With settings $0°$, $90°$, $180°$ the relative angles among the nine pairs are: $0°$ three times,
$90°$ four times, $180°$ twice. Quantum:
$$
P_{\text{QM}} = \tfrac19\left[3\cos^2 0° + 4\cos^2 45° + 2\cos^2 90°\right]
= \tfrac19\left[3 + 2 + 0\right] = \tfrac59 .
$$
Which is *exactly* the local-realist bound. The two predictions coincide and the experiment is blind —
the same failure as the four-setting design, for the same reason. Easy alignment is worth nothing if the
angles are the ones both theories agree on. Choosing $120°$ spacing is not aesthetic; it is the
experiment.
Solution
**(a)** Among the sixteen ordered setting pairs the relative angles are $0°$ (4 pairs: each setting with
itself), $90°$ (8 pairs), and $180°$ (4 pairs: $+z$ with $-z$ both ways, $+x$ with $-x$ both ways). Then
$$
P_{\text{QM}} = \tfrac1{16}\left[4(1) + 8\left(\tfrac12\right) + 4(0)\right] = \tfrac{8}{16} = \tfrac12 .
$$
**(b)** Opposite flashes occur iff $L_i = L_j$, so the count is the number of agreeing ordered pairs. If
the quadruple has $k$ pluses and $4-k$ minuses, that number is $k^2 + (4-k)^2$.
- $+\!+\!+\!+$: $k = 4$, count $= 16 + 0 = \mathbf{16}$, so $P = 1$.
- $+\!+\!+\!-$: $k = 3$, count $= 9 + 1 = \mathbf{10}$, so $P = \tfrac58$.
- $+\!+\!-\!-$: $k = 2$, count $= 4 + 4 = \mathbf{8}$, so $P = \tfrac12$.
**(c)** $f(k) = k^2 + (4-k)^2$ is a convex parabola in $k$, minimized at $k = 2$ with $f(2) = 8$. So
$P(\text{opposite}) \ge \tfrac8{16} = \tfrac12$ for every instruction set, with equality exactly for the
six **balanced** quadruples ($+\!+\!-\!-$, $+\!-\!+\!-$, $+\!-\!-\!+$, $-\!+\!+\!-$, $-\!+\!-\!+$,
$-\!-\!+\!+$). A source emitting only those reproduces the quantum statistics of this experiment exactly.
**(d)** It does not rescue anything, because a theory is refuted by *any* experiment it gets wrong, and
the three-setting experiment gets it wrong. Showing that some other experiment cannot tell two theories
apart is a statement about that experiment. (Compare: a scale that reads to the nearest kilogram cannot
distinguish two people whose masses differ by a gram; that is not evidence they weigh the same.)
The general principle. For $M$ settings the local-realist bound is
$P(\text{opposite}) \ge \min_k \left[k^2 + (M-k)^2\right]/M^2$, which is $\tfrac12$ whenever $M$ is even
(take $k = M/2$) and $\tfrac12 + \tfrac{1}{2M^2}$ when $M$ is odd. Meanwhile the quantum prediction for
$M$ settings spread evenly over a full turn, spacing $360°/M$, is $\tfrac12$ for every $M$ — the
$\cos^2$ values average to $\tfrac12$ over equally spaced angles. So an even number of settings leaves
no gap at all, while an odd number gives a gap of $1/2M^2$: largest at the smallest useful value,
$M = 3$, where it is $\tfrac1{18}$. Bell's three-setting design is not one choice among many; within
this family it is the only one that both works and is optimal.
Answers
1. $\lvert{\Psi^-}\rangle = \tfrac{1}{\sqrt2}(\lvert{+z,-z}\rangle - \lvert{-z,+z}\rangle)$. Expanding in a rotated basis, the $\lvert{+\hat n,+\hat n}\rangle$ and $\lvert{-\hat n,-\hat n}\rangle$ terms cancel and $c^2+s^2 = 1$ restores the original form. It follows that the outcomes are perfectly opposite along *any* shared axis, and that each wing on its own is 50–50 along every axis.
2. (i) Measuring the left atom lets us predict the right atom's result with certainty without disturbing it, so by the reality criterion that quantity is an element of reality. (ii) The left axis can be chosen freely and far away, so under locality the right atom must possess definite values for *all* axes at once. (iii) Quantum mechanics has no such state, so it is incomplete. The premises are locality and the reality criterion; quantum mechanics abandons the second — the values are not there before measurement.
3. Bob's state averaged over Alice's unknown outcomes is $\sum_i \operatorname{Tr}_A[(P_i\otimes\mathbb{1})\lvert{\Psi}\rangle\langle{\Psi}\rvert(P_i\otimes\mathbb{1})] = \operatorname{Tr}_A\lvert{\Psi}\rangle\langle{\Psi}\rvert$, using $P_i^2 = P_i$ under the cyclic trace and $\sum_i P_i = \mathbb{1}$. Since no signal is needed to produce the correlations, it does not matter that different frames disagree about the ordering; analyzing the experiment in either order gives identical joint probabilities.
4. It fails because the coins have definite values in transit (which R.1.4 ruled out for atoms) and, decisively, because one pair of coins cannot answer questions about a *different* axis. Repairing it means giving each atom an answer for every possible axis — which is precisely the local hidden variable hypothesis Bell refutes.
5. Quantum: $P(\text{opposite}) = \tfrac19[3(1) + 6(\tfrac14)] = \tfrac12$. Local realism: every instruction triple gives $\ge \tfrac59$, so any mixture does too. The four-setting version is blind because its local-realist minimum is also exactly $\tfrac12$ — the balanced instruction sets reproduce the quantum statistics — so the two predictions coincide.