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Entangled states, singlet correlations $-\cos\theta$, the CHSH inequality $|S| \le 2$ and its quantum violation up to $2\sqrt{2}$, and the experiments that tested it.
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By the end of this lesson you will be able to compute correlations of entangled states, evaluate the CHSH quantity, explain Bell's theorem, and interpret the experiments that tested it.
You know the singlet and triplet states from adding angular momenta, spin probabilities along tilted axes, and the collapse of a state on measurement. You know that an overall state of two particles can be symmetric or antisymmetric. This last lesson asks what the singlet's correlations mean for our picture of reality, and how experiments have answered.
| Term | What it means |
|---|---|
| Product state | A two-particle state that can be written $\vert \alpha\rangle\vert \beta\rangle$, each particle with a state of its own. |
| Entangled state | A two-particle state that cannot be written as a product, such as the singlet. |
| Correlation | $E(a, b)$, the average of the product of two $\pm 1$ results measured along axes $a$ and $b$. |
| Local hidden variables | Preexisting values that fix every measurement's outcome, with no influence faster than light. |
| Bell's inequality | A bound, such as $\vert S\vert \le 2$, obeyed by every local hidden-variable theory. |
| CHSH quantity | $S = E(a, b) - E(a, b') + E(a', b) + E(a', b')$, named for Clauser, Horne, Shimony and Holt. |
| No-signaling | The fact that one side's choice of measurement does not change the other side's local probabilities. |
A product state $|\alpha\rangle|\beta\rangle$ gives each particle a state of its own. The singlet,
$$|\psi^-\rangle = \frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle\right),$$
cannot be written that way: it is entangled. Neither spin has a definite direction; each lab, measuring alone, gets up or down at random. Yet the results are correlated. If the axes are $\theta$ apart, the probability of equal results is $\sin^2(\theta/2)$, and the correlation, the average product of the $\pm 1$ outcomes, is
$$E(\theta) = -\cos\theta.$$
In 1935 Einstein, Podolsky and Rosen argued that since one lab's result lets it predict the other's with certainty, without disturbing the distant particle, the value must have been there all along — so quantum mechanics, which does not contain it, is incomplete. In 1964 John Bell turned that argument into a test. Suppose each pair carries hidden instructions fixing the result for every setting, and neither lab's choice affects the other's result. Then for two settings per side, the CHSH quantity obeys
$$S = E(a, b) - E(a, b') + E(a', b) + E(a', b'), \qquad |S| \le 2.$$
Quantum mechanics predicts up to $|S| = 2\sqrt{2} = 2.83$. The two cannot both be right, and experiments side with quantum mechanics.
Another way: picture
Picture two sealed envelopes prepared by a friend, one mailed to each of you, each containing answers to any question you might ask. Whatever the answers, your records can agree only so often: that is Bell's inequality, a limit on correlations that any set of pre-written answers must obey. Entangled particles beat that limit. No set of envelopes, however cleverly filled, could reproduce what they do — unless the envelopes could read which question the other person asked.
Another way: steps
Checks. Every correlation lies between $-1$ and $1$, so $|S|$ can never exceed $4$. With the optimal settings for photons, $0°$, $45°$, $22.5°$ and $67.5°$, each correlation is $\pm\cos 45° = \pm 0.707$, and $S = 2.83$. And the single-side probabilities must be $\tfrac{1}{2}$ whatever the other side does; if a calculation gives otherwise, something is wrong.
Suppose each pair carries values $A, A' = \pm 1$ for Alice's two settings and $B, B' = \pm 1$ for Bob's, fixed before measurement and unaffected by the other side's choice. For every single pair,
$$AB - AB' + A'B + A'B' = A(B - B') + A'(B + B').$$
Since $B$ and $B'$ are each $\pm 1$, either $B - B' = 0$ and $B + B' = \pm 2$, or the reverse. Either way the expression is $\pm 2$. Averaging over many pairs, with any distribution of hidden values, gives $|S| \le 2$.
The proof uses only two assumptions: that outcomes are determined by values present in advance (realism), and that one side's setting cannot influence the other side's outcome (locality). Quantum mechanics violates the inequality, so at least one assumption must fail. This does not depend on quantum mechanics being correct: it is a constraint on nature, tested directly by experiment.
The first tests, by John Clauser and Stuart Freedman at Berkeley in 1972, used photon pairs from calcium atoms and found violations. Alain Aspect's group near Paris in 1982 switched the polarizer settings while the photons were in flight, closing a gap through which the settings might have been communicated, and measured $S = 2.70$. Anton Zeilinger's group in Innsbruck in 1998 used truly random, fast switching across $400$ m.
Two loopholes remained: detectors that miss many photons might select an unrepresentative sample, and settings chosen too slowly might leak. In 2015 three experiments closed both at once — with electron spins in diamond separated by $1.3$ km in Delft, and with photons at NIST in Boulder and in Vienna. All found violations. Clauser, Aspect and Zeilinger shared the 2022 Nobel Prize in Physics for this work.
It is tempting to conclude that measuring one particle instantly changes the other, and to use this to send messages. It cannot be done. Whatever Alice measures, Bob's results alone are up or down with probability $\tfrac{1}{2}$ each; his local statistics do not depend on her choice of axis. Only when the two records are compared, by a telephone call or an email traveling no faster than light, do the correlations appear.
This no-signaling property follows from the formalism: Bob's reduced state, obtained by averaging over Alice's outcomes, is the same whatever she measures. So entanglement coexists with relativity. What Bell's theorem rules out is not relativity but the comfortable picture of particles carrying definite local properties. The correlations are real, stronger than any classical ones, and unusable for sending messages — a combination with no classical counterpart.
What began as a debate about interpretation has become a technology. Entanglement lets two parties share a secret key whose security is certified by a Bell violation itself (the Ekert protocol of 1991): if $S$ is close to $2\sqrt{2}$, no eavesdropper can have much information about the results. Quantum teleportation uses a shared entangled pair plus two classical bits to transfer an unknown quantum state from one place to another, first demonstrated in 1997.
Quantum computers rely on entanglement among many qubits; without it, their states could be simulated efficiently on ordinary computers. The no-cloning theorem — an unknown quantum state cannot be copied, because copying would be a nonlinear operation — underpins the security of these protocols. Networks that distribute entanglement between cities are being built in the United States, Europe and China as the first steps toward a quantum internet.
The singlet has a definite total state while its parts have none: the whole is more definite than the parts. This is the heart of entanglement, and it is the generic situation, not an exception. Almost every state of two or more interacting particles is entangled; product states are the special case.
Entanglement also explains why the quantum world looks classical from outside. When a system interacts with its environment — air molecules, stray photons — it becomes entangled with it, and the superpositions of the system alone lose their phase relationships, a process called decoherence. That is why cats are never seen in superposition and why quantum computers must be isolated so carefully. The same physics that Bell tests certify is what makes quantum technology hard: entanglement with the outside world destroys the delicate entanglement inside.
In 2015 a team led by Ronald Hanson at Delft University of Technology placed two electron spins in diamond crystals $1.3$ km apart on the university campus. Each spin emitted a photon entangled with it; when the photons met midway and were detected together, the spins were left entangled. Random number generators then chose the measurement settings at each lab, and the spins were read out in less time than light takes to cross between them, closing the locality loophole. Every trial produced a result, closing the detection loophole.
Over $245$ trials the team found $S = 2.42 \pm 0.20$, a violation of Bell's inequality. Within months, photon experiments at NIST in Boulder and in Vienna confirmed it with far larger statistics. In 2025 NIST and the University of Colorado used a loophole-free Bell test to build a public randomness beacon, CURBy, whose numbers are certified unpredictable by the violation itself — because local hidden values would have been needed to predict them, and Bell tests have ruled those out.
In 2017 the Chinese satellite Micius distributed entangled photon pairs to two ground stations $1200$ km apart, and the stations measured a Bell violation. The photons traveled through the thin upper atmosphere and vacuum, suffering far less loss than they would have in that much optical fiber. The same satellite then carried out entanglement-based key distribution between the stations.
In the United States, the Department of Energy's national laboratories and universities in the Chicago area operate a quantum network testbed of more than a hundred miles of fiber, and New York and Boston groups have demonstrated entanglement distribution over installed telecommunications fiber. The goal is a quantum internet, in which entangled pairs shared between distant nodes let users generate provably secure keys, link quantum computers and synchronize clocks beyond classical limits — all resting on correlations that no local theory can explain.
Because a measurement on one particle seems to settle the other's result instantly, it is natural to imagine sending a message by choosing what to measure. But Bob's results alone are always a fair coin toss, whatever Alice does. The correlations exist only in the comparison of both records, which requires ordinary communication. Entanglement is a correlation, not a channel.
A second misconception is that Bell's theorem proves quantum mechanics is "nonlocal" in a way that conflicts with relativity. What it proves is that no theory of local, predetermined values can match the observed correlations. Quantum mechanics does match them while respecting no-signaling. Physicists still debate whether to give up locality, realism or something subtler, but all agree on the experimental facts.
Write the singlet along any axis $\hat{n}$.
$|\psi^-\rangle = \tfrac{1}{\sqrt{2}}(|\uparrow_n\downarrow_n\rangle - |\downarrow_n\uparrow_n\rangle)$
Spin zero has the same form in every basis.
Let Alice find up along $\hat{a}$.
$\text{Bob's spin} \to |\downarrow_a\rangle$
Collapse of the joint state.
Find the chance Bob gets up along $\hat{b}$.
$|\langle\uparrow_b|\downarrow_a\rangle|^2 = \sin^2\dfrac{\theta}{2}$
Tilted-axis probabilities.
Find the probability of equal results.
$P_{\text{same}} = \sin^2\dfrac{\theta}{2}, \qquad P_{\text{diff}} = \cos^2\dfrac{\theta}{2}$
By symmetry, the same if Alice finds down.
Find the correlation.
$E(\theta) = \sin^2\dfrac{\theta}{2} - \cos^2\dfrac{\theta}{2} = -\cos\theta$
A double-angle identity.
Write the correlation for $(|HH\rangle + |VV\rangle)/\sqrt{2}$.
$E(\alpha, \beta) = \cos 2(\alpha - \beta)$
Polarizers at equal angles always agree.
Choose the settings.
$a = 0°, \ a' = 45°, \ b = 22.5°, \ b' = 67.5°$
The optimal choice.
Evaluate $E(a, b)$ and $E(a, b')$.
$\cos(-45°) = 0.707, \qquad \cos(-135°) = -0.707$
Angle differences $22.5°$ and $67.5°$, doubled.
Evaluate $E(a', b)$ and $E(a', b')$.
$\cos 45° = 0.707, \qquad \cos(-45°) = 0.707$
Angle differences $22.5°$ and $-22.5°$, doubled.
Combine them into the CHSH quantity.
$S = 0.707 - (-0.707) + 0.707 + 0.707 = 2.828$
The minus sign on the second term makes all four add.
Compare with the bounds.
$2 < 2.828 = 2\sqrt{2}$
The largest violation quantum mechanics permits.
Assign hidden values to one pair.
$A, A', B, B' \in \{+1, -1\}$
Predetermined results for each setting.
Form the CHSH combination for that pair.
$AB - AB' + A'B + A'B' = A(B - B') + A'(B + B')$
Factor by Alice's values.
Consider $B = B'$.
$B - B' = 0, \quad B + B' = \pm 2 \quad\Rightarrow\quad \pm 2$
Only the second bracket survives.
Consider $B = -B'$.
$B - B' = \pm 2, \quad B + B' = 0 \quad\Rightarrow\quad \pm 2$
Only the first bracket survives.
Average over many pairs.
$|S| = |\langle AB\rangle - \langle AB'\rangle + \langle A'B\rangle + \langle A'B'\rangle| \le 2$
An average of numbers each equal to $\pm 2$.
Identify the assumptions.
$\text{realism: values exist}; \quad \text{locality: } B \text{ does not depend on Alice's setting}$
Both are used.
Compare with experiment.
$S_{\text{measured}} = 2.70 \ (1982), \ 2.42 \ (2015)$
Both exceed $2$ by many standard deviations.
Draw the conclusion.
$\text{local realism fails}$
Nature is not described by local predetermined values.
Write the correlation formula.
$E(\theta) = -\cos\theta$
For spin-½ singlets.
Substitute the angle.
$E = -\cos 60°$
$\theta = 60°$.
Evaluate the correlation.
Two electrons are prepared in the singlet state and sent to distant labs. Both labs measure spin along the same axis. What do they find?
Complete the worked solution: $520$ singlet pairs are measured along axes $120°$ apart. Find the expected number of pairs with equal results, with opposite results, and the difference between the two.
Multiply by the probability of equal results.
$N_{\text{same}} = \tfrac{3}{4} \times 520 =$ a
$\sin^2(120°/2) = \tfrac{3}{4}$.
Multiply by the probability of opposite results.
$N_{\text{diff}} = \tfrac{1}{4} \times 520 =$ b
The remaining quarter.
Subtract the opposite count from the equal count.
$N_{\text{same}} - N_{\text{diff}} =$ c
Divided by the total, this is $E = 0.5$.
Match each idea about entanglement to its expression.
| $\frac{1}{\sqrt{2}}(\uparrow\downarrow - \downarrow\uparrow)$ | $-\cos\theta$ | $|S| \le 2$ | $|S| \le 2\sqrt{2}$ | |
|---|---|---|---|---|
| the singlet state | ||||
| the singlet correlation | ||||
| the local hidden-variable bound | ||||
| the quantum bound |
For the singlet, fill in the correlation $E(\theta)$ and the probability that both labs get the same result, for axes $\theta$ apart.
| $E(\theta)$ | $P_{\text{same}}$ | |
|---|---|---|
| $\theta = 0°$ | ||
| $\theta = 60°$ | ||
| $\theta = 90°$ | ||
| $\theta = 120°$ |
Two spins in the singlet are measured along axes $90°$ apart. What is the probability that both results are up?
Answer:
Photon pairs are prepared in $(|HH\rangle + |VV\rangle)/\sqrt{2}$. Alice measures polarization at $a = 0°$ or $a' = 45°$, Bob at $b = 30°$ or $b' = 60°$. Find the CHSH quantity $S = E(a, b) - E(a, b') + E(a', b) + E(a', b')$.
Answer:
In a 1998 photon experiment in Innsbruck with fast random switching, the measured correlations were $E(a, b) = 0.683$, $E(a, b') = -0.682$, $E(a', b) = 0.682$ and $E(a', b') = 0.683$. What is the CHSH value $S$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Complete the worked solution: $384$ singlet pairs are measured along axes $120°$ apart. Find the expected number of pairs with equal results, with opposite results, and the difference between the two.
Multiply by the probability of equal results.
$N_{\text{same}} = \tfrac{3}{4} \times 384 =$ a
$\sin^2(120°/2) = \tfrac{3}{4}$.
Multiply by the probability of opposite results.
$N_{\text{diff}} = \tfrac{1}{4} \times 384 =$ b
The remaining quarter.
Subtract the opposite count from the equal count.
$N_{\text{same}} - N_{\text{diff}} =$ c
Divided by the total, this is $E = 0.5$.
You can explain and compute Bell tests. Explain to someone why entanglement cannot be used to send a message faster than light.
17. Your turn: for the singlet, what is the correlation between results along axes $60°$ apart?, step 3
$E = -0.5$
Opposite results three times out of four.