Chapter 08

Bell correlations and the CHSH inequality

Alice and Bob each choose one of two measurements and record $+1$ or $-1$. If each pair of systems carries instructions fixed at the source, a certain combination of their four correlations can never exceed 2. For a Bell pair, quantum mechanics predicts $2\sqrt2$. Which assumptions does that disagreement rule out, and which does it leave standing?

Before you start: This chapter needs tensor products, two-outcome measurements with outcomes $\pm1$, and expectation values: Pauli means (Chapter 1), the Bell pair (Chapter 4), and reduced states for the last section (Chapter 3). The commutator appears in an optional derivation (Chapter 4). Notation guide.

State the experiment before interpreting it

A source sends one system to Alice and one to Bob. Alice chooses setting $x\in\{0,1\}$ and records $a\in\{+1,-1\}$. Bob independently chooses $y\in\{0,1\}$ and records $b\in\{+1,-1\}$. Repeating the preparation estimates four correlators $E_{xy}=\sum_{a,b}ab\,p(a,b|x,y)$.

A local hidden-variable model assumes a shared variable $\lambda$ distributed according to $q(\lambda)$, independently of the chosen settings, and factorised responses conditional on it:

$$p(a,b|x,y)=\int d\lambda\,q(\lambda)\,p(a|x,\lambda)p(b|y,\lambda).$$

The factorisation expresses locality within this model. Alice's response does not depend on Bob's setting once $\lambda$ is specified, and conversely. The setting-independence assumption is separate: allowing the source variable to depend on the settings changes the class of models being tested.

The four correlators come from four groups of trials, one per setting pair. We never measure both $A_0$ and $A_1$ on the same member of a pair. A hidden-variable model nevertheless assigns responses for either possible setting, using the same distribution of $\lambda$. That shared assignment lets us combine the four averages into one bound. If the source instead used a different distribution for each chosen setting pair, the next averaging step would not follow from this model.

Derive the classical bound

First take deterministic responses $A_0,A_1,B_0,B_1\in\{+1,-1\}$ for a fixed $\lambda$. (In this section $A_x$ and $B_y$ are the $\pm1$ numbers the model assigns; in the quantum calculation below the same letters denote the observables, whose measured values are again $\pm1$.) Define

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

If $B_0=B_1$, the second bracket vanishes and the first is $\pm2$. If $B_0=-B_1$, the first vanishes and the second is $\pm2$. In either case $C=\pm2$. In this model each correlator is the $q$-average of the product of responses, $E_{xy}=\int d\lambda\,q(\lambda)\,A_x(\lambda)B_y(\lambda)$, so the combination $S$ is the $q$-average of $C(\lambda)$ and $|S|\le\int d\lambda\,q(\lambda)\,|C(\lambda)|=2$. Averaging over $\lambda$ therefore gives the CHSH bound

$$|S|\le2,\qquad S=E_{00}+E_{01}+E_{10}-E_{11}.$$

Stochastic local responses do not raise the bound. Their local random seeds may be included in $\lambda$, making the enlarged model a mixture of deterministic assignments. An observed violation excludes this class of models under the stated assumptions; it is not a conclusion obtained merely from correlated outcomes.

Calculate a quantum violation

Use $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$. Direct action on the two terms gives

$$\langle Z\otimes Z\rangle=1,\quad\langle X\otimes X\rangle=1,\quad\langle Z\otimes X\rangle=\langle X\otimes Z\rangle=0.$$

For example, $Z\otimes X$ takes the state into $(|01\rangle-|10\rangle)/\sqrt2$, orthogonal to the original. Choose Alice's observables $A_0=Z,A_1=X$, and Bob's $B_0=(Z+X)/\sqrt2,B_1=(Z-X)/\sqrt2$. The latter are valid binary observables: each is Hermitian and squares to $I$ because $XZ+ZX=0$.

Expanding each tensor product and using the four means above gives $E_{00}=E_{01}=E_{10}=1/\sqrt2$ and $E_{11}=-1/\sqrt2$. Hence $S=2\sqrt2$, greater than two. Each correlator averages the product of two measured signs. Three of these averages are positive and the fourth negative; the minus sign in the CHSH combination makes all four contributions add.

A geometric reading helps. Write each observable as a direction in the $x$–$z$ plane of the Bloch sphere: $\cos t\,Z+\sin t\,X$ points at angle $t$ from $z$. For $|\Phi^+\rangle$ the correlator of two such directions is the cosine of the angle between them, since $\langle Z\otimes Z\rangle=\langle X\otimes X\rangle=1$ and the mixed terms vanish. Our settings put $A_0$ at $0^\circ$, $A_1$ at $90^\circ$, $B_0$ at $45^\circ$ and $B_1$ at $-45^\circ$. Three pairs are $45^\circ$ apart and one, $A_1$ and $B_1$, is $135^\circ$ apart, which gives the three values $\cos45^\circ$ and the one $\cos135^\circ$.

Why the quantum value is bounded too

No choice of quantum settings or state does better than $2\sqrt2$, a result known as Tsirelson’s bound, so the settings above are optimal.

Why no quantum state exceeds $2\sqrt2$

Recall the commutator $[A,B]=AB-BA$. It measures the difference between the two multiplication orders. Because Alice's and Bob's operators act on separate tensor factors, they commute with each other; the commutators below compare the two choices on the same side.

For arbitrary Hermitian binary observables with square $I$, define the CHSH operator $\mathcal B=A_0\otimes(B_0+B_1)+A_1\otimes(B_0-B_1)$. Expanding its square yields

$$\mathcal B^2=4I-[A_0,A_1]\otimes[B_0,B_1].$$

Each observable has operator norm one, so each commutator has norm at most two. Therefore $\|\mathcal B^2\|\le8$, implying $|\langle\mathcal B\rangle|\le2\sqrt2$. Our settings attain this bound. If either pair of local observables commutes, the commutator term vanishes and this construction cannot exceed two.

Correlations do not provide a signalling channel

Alice's reduced state is $I/2$. If Bob performs any local measurement with operators $M_b$ and keeps its outcome private, Alice's new state is

$$\rho'_A=\operatorname{Tr}_B\sum_b(I\otimes M_b)\rho_{AB}(I\otimes M_b^\dagger)=\operatorname{Tr}_B\rho_{AB}=\rho_A.$$

The equality uses $\sum_bM_b^\dagger M_b=I$ and the cyclic trace over $B$. Alice's unconditioned outcome probabilities cannot depend on Bob's measurement choice. Conditional states can differ if Bob later reports his result; obtaining those selected statistics requires communication.

One more correlator is worth recording for later chapters: for this Bell state $\langle Y\otimes Y\rangle=-1$. Identical measurements on the two halves agree for $X$ and for $Z$ and disagree for $Y$; the sign depends on the axis.

Where this leaves us

With the settings above, a Bell pair gives $S=2\sqrt2$, and no model of the stated kind can give more than 2. What that rules out is the class of models with a shared variable whose distribution does not depend on the settings, and responses at each side that depend only on the local setting and that variable. It does not rule out models that drop one of those assumptions, and it does not allow signalling: Alice's own statistics are unchanged by anything Bob chooses. An experiment must also close practical gaps, such as detector efficiency and the timing of the setting choices, before its data test the ideal prediction computed here.

Correlations alone carry no message. Combined with a classical message, though, a shared Bell pair can move an unknown state from one lab to another, which no-cloning seemed to make hard. The next chapter tracks exactly where the state is at each step.

Problems

1. For $\rho_v=v|\Phi^+\rangle\langle\Phi^+|+(1-v)I/4$, evaluate $S$ with the settings above. When is CHSH violated?

Worked answer

Expectation values are linear in the state. For any of the settings,

$$E_{xy}(\rho_v)=vE_{xy}(\Phi^+)+(1-v)\operatorname{Tr}[(I/4)(A_x\otimes B_y)].$$

The second trace equals $\operatorname{Tr}(A_x)\operatorname{Tr}(B_y)/4=0$, because both factors are traceless. Each correlator is therefore multiplied by $v$, and $S(\rho_v)=2\sqrt2v$. Solving $2\sqrt2v>2$ gives $v>1/\sqrt2$. At equality the classical bound is just reached; below it this particular test does not certify entanglement.

The tempting wrong turn is to read $v=1/\sqrt2$ as the point where entanglement disappears. This family is entangled for every $v>1/3$. Between $1/3$ and $1/\sqrt2$ the state is entangled, but this test cannot show it.

2. Find $p(a,b|x,y)$ for our Bell-state settings from $E_{xy}$.

Worked answer

The projector for Alice’s sign $a$ is $(I+aA_x)/2$, and Bob’s is $(I+bB_y)/2$. Expand their tensor product before taking its expectation:

$$p(a,b|x,y)=\frac14[1+a\langle A_x\otimes I\rangle+b\langle I\otimes B_y\rangle+abE_{xy}].$$

The two single-party means vanish because each reduced state is $I/2$ and the observables are traceless. Hence $p(a,b|x,y)=(1+abE_{xy})/4$. Summing over $b=+1,-1$ gives $[1+aE_{xy}+1-aE_{xy}]/4=1/2$. The marginal cannot reveal Bob’s setting even though the joint distribution depends on it.

References and further reading

Nielsen & Chuang, §2.6 (EPR and Bell inequalities). Sakurai & Napolitano, spin correlation measurements and Bell’s inequality (3rd edition, §3.10).