Quantum Foundations · Chapter 13
What correlations do quantum composites allow, and why not more?
Starting point: Chapters 4, 7, and 8: composition, local tomography, and Hilbert space.
Anchor example: A Bell pair shared by two distant laboratories.
Principle at work: Local tomography realized as the tensor product of Hilbert spaces.
Foil: Local hidden-variable models below, and boxworld above.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
Early on we set up pairs of systems in any theory: each part has its own reduced state, and local tomography makes the joint state an array of local statistics. The reconstruction theorem then gave the quantum form of this rule. A pair is described on the tensor product \(\mathbb C^{N_A}\otimes\mathbb C^{N_B}\), with local effects \(E\otimes I\). We have used such pairs as examples, but not yet asked what they can do.
What is new about quantum pairs? We start with the state of one part. In matrix form it is a partial trace, and it can be mixed while the whole is pure, which no classical pair allows. That forces a distinction between product and entangled states, and a canonical form of any pure pair shows how the two parts share their information. Finally we return to the correlation game. Classical pairs are bounded by \(S\le2\) and boxworld reaches \(S=4\). Where does quantum theory lie, and what stops it?
The next chapter asks what changes when the two systems of a pair are identical.
Each part of a pair has a state of its own, because distant choices cannot change local statistics. What is that state as a matrix? We derive it from the local effects, which gives the partial trace. Then we compute it for a correlated pure pair, and find a part that is maximally mixed inside a whole that is pure.
Which density matrix describes A alone? By no-signaling, the statistics of measurements on A alone define a state \(\rho_A\). Its probabilities must match those of the joint state for every local effect:
\[\tr(\rho_AE)=\tr\big(\rho_{AB}(E\otimes I)\big)\quad\text{for every effect }E\text{ on A}.\]
Since the effects span all Hermitian matrices, this determines \(\rho_A\) uniquely. The resulting map \(\rho_{AB}\mapsto\rho_A\) is called the partial trace over B. In components, with bases \(\ket i\) of A and \(\ket j\) of B,
\[(\rho_A)_{ii'}=\sum_j\big(\rho_{AB}\big)_{ij,\,i'j}.\]
So the partial trace is not a separate postulate. It is what the reduced state must be once states are density operators and local effects are \(E\otimes I\).
What is the reduced state of a correlated pure pair? Take \(\ket{\Phi^+}=(\ket{00}+\ket{11})/\sqrt2\). Its density operator has entries \(\tfrac12\) in the four corners of the \(4\times4\) matrix indexed by \(00,01,10,11\), and zeros elsewhere. Summing over the second index gives \(\rho_A=\tfrac12\ketbra00+\tfrac12\ketbra11=I/2\). The whole is in a pure state, maximally specified, while each part is maximally uncertain.
Can this happen classically? No. A pure state of two classical systems is a single pair of configurations. Its marginals are single configurations, hence pure. In quantum theory the information in a pure joint state can reside entirely in correlations, with none in the parts. When we study open systems, this property becomes a principle of its own, purification.
Problem 13.1. Find \(\rho_A\) for \(\cos\alpha\ket{00}+\sin\alpha\ket{11}\). For which \(\alpha\) is it pure?
The cross terms \(\ket{00}\bra{11}\) disappear under the partial trace, leaving \(\rho_A=\mathrm{diag}(\cos^2\alpha,\sin^2\alpha)\). It is pure only when \(\cos\alpha\) or \(\sin\alpha\) is zero, that is, when the joint state is a product.
When does a pure whole have pure parts? We first separate the joint states that are built from independent parts from those that are not, for pure and for mixed states. Then a canonical form of any pure pair answers the question exactly, and gives a measure of how far a pure state is from a product.
Which pure pairs are built from independent parts? Call a pure joint state a product if it has the form \(\ket a\otimes\ket b\), and entangled otherwise. How can we tell from the amplitudes? Write \(\ket\psi=\sum_{ij}c_{ij}\ket{ij}\). A product has \(c_{ij}=a_ib_j\), so its array of coefficients is a column times a row, a matrix of rank one. Conversely, a rank-one array is a column times a row. For two qubits, rank one means \(c_{00}c_{11}=c_{01}c_{10}\).
What about mixed states? We follow the definition we gave for any theory: a state is separable if it is a mixture of products, and entangled otherwise. Separable states can be prepared by local operations and shared randomness, and entangled states cannot.
How can we test a mixed state for entanglement? Deciding whether a given mixed state is separable is in general a hard problem. But one necessary condition is easy to derive. Recall the partial transpose, which transposes the second factor of every product. On a product state it gives \(\rho_a\otimes\rho_b^T\). The transpose of a density matrix is again a density matrix, so this is a state. A mixture of products therefore has a partial transpose with no negative eigenvalue. We saw that \(\ket{\Phi^+}\) does not pass this test, since its partial transpose has eigenvalue \(-1/2\). Asher Peres proposed the test in 1996, and Michał, Paweł, and Ryszard Horodecki proved in the same year that for two qubits it is also sufficient: a two-qubit state is entangled exactly when its partial transpose has a negative eigenvalue.
Can every pure pair be written in a simple standard form? Write \(\ket\psi=\sum_{ij}c_{ij}\ket i\ket j\) and regard \(c\) as a matrix. Its singular value decomposition gives orthonormal bases \(\ket{a_k}\) of A and \(\ket{b_k}\) of B and nonnegative numbers \(s_k\) with
\[\ket\psi=\sum_ks_k\ket{a_k}\otimes\ket{b_k},\qquad\sum_ks_k^2=1 .\]
This is the Schmidt decomposition. What does it say about the parts? Taking the partial trace gives \(\rho_A=\sum_ks_k^2\ketbra{a_k}{a_k}\) and \(\rho_B=\sum_ks_k^2\ketbra{b_k}{b_k}\), which have the same nonzero eigenvalues. The number of nonzero \(s_k\) is the Schmidt rank, which is the rank of the array \(c\). So a pure whole has pure parts exactly when its Schmidt rank is one, which is exactly when it is a product.
How entangled is a pure state? The Shannon entropy of the distribution \(s_k^2\) measures it. It is the same whether computed from A or from B, it is zero for products, and for two qubits it reaches one bit when both \(s_k\) equal \(1/\sqrt2\).
Problem 13.2. Decide whether \((\ket{00}+\ket{01}+\ket{10}+\ket{11})/2\) and \((\ket{00}+\ket{01}+\ket{10}-\ket{11})/2\) are products.
For the first, \(c_{00}c_{11}=\tfrac14=c_{01}c_{10}\): it is \(\ket+\ket+\). For the second, \(c_{00}c_{11}=-\tfrac14\ne\tfrac14\): it is entangled. In fact its Schmidt coefficients are both \(1/\sqrt2\), so it is maximally entangled.
The correlation game separated shared plans (\(S\le2\)) from boxworld (\(S=4\)). Where do entangled quantum pairs fall? We first show that any model with local hidden variables obeys the classical limit, and that an entangled pair breaks it. Then we find the quantum limit, \(2\sqrt2\), from the Hilbert-space structure the principles produced, and see two other limits that structure imposes.
Could hidden variables explain the correlations of distant systems, provided each outcome depends only on local information? Make this precise. A local hidden-variable model assigns to each run a variable \(\lambda\), drawn from a distribution that does not depend on the measurement choices. Each laboratory’s outcome then depends only on its own choice and on \(\lambda\). For each \(\lambda\) the outputs are a shared plan, possibly randomized locally, and the bound for shared plans gives \(|S|\le2\). This is Bell’s theorem in the form of John Clauser, Michael Horne, Abner Shimony, and Richard Holt.
Does quantum theory obey the bound? Share \(\ket{\Phi^+}\), and measure \(\pm1\) observables \(\mathbf a\cdot\vec\sigma\) on the first qubit and \(\mathbf b\cdot\vec\sigma\) on the second, with \(\mathbf a\) and \(\mathbf b\) in the \(xz\)-plane. What is the correlation? First the basic cases. \(X\otimes X\) and \(Z\otimes Z\) both leave \(\ket{\Phi^+}\) unchanged, so their expectations are \(1\). \(X\otimes Z\) sends it to \((\ket{10}-\ket{01})/\sqrt2\), and \(Z\otimes X\) to \((\ket{01}-\ket{10})/\sqrt2\). Both are orthogonal to \(\ket{\Phi^+}\), so these expectations are \(0\). Expanding \(\mathbf a\cdot\vec\sigma\otimes\mathbf b\cdot\vec\sigma\) term by term then gives
\[\langle\mathbf a\cdot\vec\sigma\otimes\mathbf b\cdot\vec\sigma\rangle=a_xb_x+a_zb_z=\mathbf a\cdot\mathbf b .\]
Now choose \(A_0=Z\), \(A_1=X\) on the first qubit and \(B_0=(Z+X)/\sqrt2\), \(B_1=(Z-X)/\sqrt2\) on the second. The four correlations are \(\tfrac1{\sqrt2}\), \(\tfrac1{\sqrt2}\), \(\tfrac1{\sqrt2}\), and \(-\tfrac1{\sqrt2}\), so \(S=4/\sqrt2=2\sqrt2\approx2.83\). Experiments agree. In 2015, groups in Delft, Vienna, and Boulder observed violations of Bell inequalities while closing the main loopholes, the gaps through which a local model could escape, such as undetected particles or time for a signal between the laboratories.
Can quantum pairs go higher than \(2\sqrt2\)? No.
Tsirelson’s bound. For any quantum state and any \(\pm1\) observables, \(|S|\le2\sqrt2\).
Proof. Let \(C=A_0\otimes(B_0+B_1)+A_1\otimes(B_0-B_1)\), so that \(S=\langle C\rangle\). Multiplying out with \(A_x^2=B_y^2=I\) gives
\[C^2=4\,I\otimes I-[A_0,A_1]\otimes[B_0,B_1].\]
The norm \(\|M\|\) of a matrix is the largest length \(\|M\ket v\|\) over unit vectors. Each observable has norm \(1\), so each commutator has norm at most \(2\), and their Kronecker product has norm at most \(4\). So \(C^2\) has eigenvalues at most \(8\). Since a variance is never negative, \(\langle C\rangle^2\le\langle C^2\rangle\le8\), and \(|S|\le2\sqrt2\). \(\square\)
What, then, excludes the PR box? Not a single principle used directly, but the structure the principles produced. Quantum correlations are expectations of operators whose squares are \(I\), on the tensor product of Hilbert spaces. That structure bounds the commutators, and so bounds \(S\). Boxworld has no such structure, and among our principles it fails continuous reversibility.
Does the Hilbert-space structure respect no-signaling, as the framework demands? Suppose B performs any measurement, with effects \(F_k\) summing to \(I\), and forgets the outcome. Then A’s statistics are \(\sum_k\tr\big(\rho(E\otimes F_k)\big)=\tr\big(\rho(E\otimes I)\big)\), the same whatever B measured.
Can an unknown state be copied? Suppose a unitary \(U\) satisfied \(U\ket\psi\ket0=\ket\psi\ket\psi\) and \(U\ket\phi\ket0=\ket\phi\ket\phi\). Unitaries preserve inner products, so \(\braket\phi\psi=\braket\phi\psi^2\), so \(\braket\phi\psi\) is \(0\) or \(1\). So only perfectly distinguishable or identical states can be cloned. This is the quantum form of the overlap of pure states that we first met on the disc.
Problem 13.3. With the same state and \(A_0=Z\), \(A_1=X\), use \(B_0=Z\), \(B_1=X\) instead. Compute \(S\).
The correlations are \(\mathbf a\cdot\mathbf b\): \(Z\)–\(Z\) gives \(1\), \(Z\)–\(X\) and \(X\)–\(Z\) give \(0\), \(X\)–\(X\) gives \(1\). So \(S=1+0+0-1=0\). The violation depends on choosing measurements at intermediate angles.
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
Find \(\rho_A\) for \((\ket{00}+\ket{01})/\sqrt2\). Is the state entangled?
Factor the state.
A product state has pure marginals.
The state is \(\ket0\otimes\ket+\), so \(\rho_A=\ketbra00\), pure, and the state is a product, not entangled.
What is the Schmidt rank of \((\ket{00}+\ket{11})/\sqrt2\), and what are its Schmidt coefficients?
It is already in Schmidt form.
Count nonzero terms.
Rank \(2\), with both coefficients \(1/\sqrt2\): maximally entangled for two qubits.
What CHSH winning probability corresponds to Tsirelson’s bound?
Winning probability is \(\tfrac12+S/8\).
\(S=2\sqrt2\).
\(\tfrac12+\tfrac{\sqrt2}4\approx0.854\), compared with \(0.75\) for local strategies and \(1\) for the PR box.
Compute \(\rho_A\) for \((\ket{00}+\ket{01}+\ket{10})/\sqrt3\), find its eigenvalues, and decide whether the state is entangled.
With the coefficient array \(c\), \(\rho_A=cc^\dagger\).
\(c=\tfrac1{\sqrt3}\begin{pmatrix}1&1\\1&0\end{pmatrix}\).
\(\rho_A=\tfrac13\begin{pmatrix}2&1\\1&1\end{pmatrix}\), with eigenvalues \((3\pm\sqrt5)/6\approx0.873\) and \(0.127\). The reduced state is mixed, so the pure joint state is entangled, though not maximally.
Find the Schmidt coefficients of \((\ket{00}+\ket{01}+\ket{10})/\sqrt3\) from the singular values of its coefficient array, and check them against the previous exercise.
Singular values are square roots of the eigenvalues of \(cc^\dagger\).
Their squares sum to one.
The singular values are \(\sqrt{(3\pm\sqrt5)/6}\approx0.934\) and \(0.357\), whose squares are the eigenvalues of \(\rho_A\) found before. Both reduced states share these eigenvalues, as the Schmidt decomposition guarantees.
The Werner state is \(v\ketbra{\Phi^+}{\Phi^+}+(1-v)I/4\). (a) Find its CHSH value with the optimal measurements found for \(\ket{\Phi^+}\). (b) For which \(v\) does it violate the local bound?
The noise term contributes zero to every correlation.
Scale \(2\sqrt2\) by \(v\).
(a) \(S=2\sqrt2\,v\). (b) Violation requires \(v>1/\sqrt2\approx0.707\). Noise degrades the correlations linearly; experiments must keep it below this threshold.
Show directly that no unitary can copy both \(\ket0\) and \(\ket+\).
Compare inner products before and after.
\(\braket0+=1/\sqrt2\).
Copying requires \(U\ket0\ket0=\ket0\ket0\) and \(U\ket+\ket0=\ket+\ket+\). Inner products before: \(\braket0+=1/\sqrt2\); after: \(\braket0+^2=\tfrac12\). Since \(1/\sqrt2\ne\tfrac12\), no such \(U\) exists. Copying is possible only for states whose overlap is \(0\) or \(1\).
A student says: “When Alice measures her half of a Bell pair, Bob’s qubit changes instantly, so they can communicate faster than light.” Explain why not.
What is Bob’s reduced state before and after Alice’s measurement?
Bob does not know Alice’s outcome.
Bob’s reduced state is \(I/2\) whatever Alice does: averaged over her outcomes, which he does not know, his statistics are unchanged, as no-signaling requires. The correlation becomes visible only when outcomes are compared through an ordinary channel. Entanglement creates correlations, not a channel.
Find the eigenvalues of the partial transpose of the Werner state, determine for which \(v\) it is entangled, and compare with the CHSH threshold.
The partial transpose of \(\ketbra{\Phi^+}{\Phi^+}\) is \(\tfrac12S\).
\(S\) has eigenvalues \(1,1,1,-1\).
The partial transpose is \(\tfrac v2S+\tfrac{1-v}4I\), with eigenvalues \(\tfrac{1+v}4\) (three times) and \(\tfrac{1-3v}4\). It is negative, so the state is entangled, exactly when \(v>\tfrac13\). For \(\tfrac13<v\le1/\sqrt2\) the state is entangled yet does not violate CHSH: entanglement and Bell nonlocality are different properties.
For \(\pm1\) observables (\(A_x^2=B_y^2=I\)), verify \(C^2=4I-[A_0,A_1]\otimes[B_0,B_1]\) where \(C=A_0\otimes(B_0+B_1)+A_1\otimes(B_0-B_1)\), and deduce Tsirelson’s bound.
Expand and use \(A_x^2=I\), \(B_y^2=I\).
\(\|[A_0,A_1]\|\le2\).
Expanding, the squared terms give \(I\otimes(B_0+B_1)^2+I\otimes(B_0-B_1)^2=I\otimes4I\), and the cross terms combine into \(A_0A_1\otimes(B_0+B_1)(B_0-B_1)+A_1A_0\otimes(B_0-B_1)(B_0+B_1)\), which simplifies to \(-[A_0,A_1]\otimes[B_0,B_1]\). Each commutator has norm at most \(2\), so \(C^2\le8I\) and \(|\langle C\rangle|\le2\sqrt2\). For commuting observables the commutator vanishes and \(|S|\le2\), the classical bound.