Composite Systems and Entanglement

Composite systems and entanglement¶

Learning objectives¶

After this chapter, you should be able to:

• construct product states and operators with tensor products;

• calculate joint, marginal, and conditional probabilities;

• distinguish separable states from entangled states, including in mixtures;

• use reduced density operators to describe a subsystem;

• predict correlations in Bell states and quantify partial entanglement with the Schmidt decomposition;

• explain why entanglement does not enable faster-than-light signaling;

• state the Tsirelson bound and what it does and does not imply about nature; and

• analyze simple two-qubit gates and measurement protocols.

Sections 5.1–5.6 provide the core route: build joint states, calculate local and joint predictions, interpret Bell correlations, and create entanglement. Section 5.7 is an optional quantitative extension using the Schmidt decomposition. Section 5.8 begins with the essential distinction between correlation and entanglement, then marks its Werner-state test as an advanced extension.

5.1 Building a joint state space¶

Chapters 2 through 4 developed a complete quantum description of a single spin: a two-dimensional complex state space, Hermitian operators for measurable quantities, and unitary time evolution. But that formalism, on its own, does not say how to describe two spins that were prepared together and are later measured separately, perhaps in two different laboratories. A first guess might be that each spin simply carries its own independent ket, and a joint experiment just reports the two kets’ separate predictions side by side. That guess is sometimes right and sometimes badly wrong, and sorting out the difference is the subject of this chapter.

The correct joint state space is built from the two one-spin spaces by a specific mathematical construction called the tensor product. It is not merely “the two spaces considered together”: it is a larger space, big enough to hold every possible correlation between the two spins, including correlations with no classical counterpart at all. Learning to build this space, and learning to recognize when a joint state does or does not reduce to “two independent spins,” is the first step toward composite systems of any size—two electrons, a spin coupled to a field mode, or the many-particle systems of Chapter 11.

One spin-$\tfrac12$ system has a two-dimensional state space. Two distinguishable spins require four independent basis states:

The joint space is the tensor product $\mathcal H_{AB}=\mathcal H_A\otimes\mathcal H_B$. We abbreviate $|+z\rangle_A|-z\rangle_B$ as $|+z,-z\rangle$ or, for qubits, $|01\rangle$. In the ordered $z$ basis,

If $|\psi\rangle_A=a|0\rangle+b|1\rangle$ and $|\phi\rangle_B=c|0\rangle+d|1\rangle$, then

Notice that the coefficient array factors into one list for $A$ and one for $B$. This factorization is the defining feature of a pure product state, and it is exactly what will fail, in Section 5.2, for a state that cannot be built this way.

Quick calculation: multiplying out a concrete product state¶

Take $|\psi\rangle_A=|+x\rangle_A=(|0\rangle_A+|1\rangle_A)/\sqrt2$ and $|\phi\rangle_B=|0\rangle_B$, so in the notation of equation (3), $a=b=1/\sqrt2$ while $c=1$ and $d=0$. Multiplying out term by term,

Only the terms ending in $B$-label 0 survive, because $|\phi\rangle_B$ has no $|1\rangle_B$ component at all: the amplitude for any joint outcome with $B=1$ is forced to zero regardless of what $A$ does. Contrast this with the entangled state introduced next: here both surviving terms share the same $B$ label while $A$’s label is free to be either 0 or 1, exactly the signature of $A$ and $B$ having been prepared independently, with no correlation between their outcomes.

Operators on one subsystem¶

An operation on $A$ alone is represented by $\hat A\otimes\hat I_B$; an operation on $B$ alone is $\hat I_A\otimes\hat B$. Operators acting on different subsystems always commute:

To construct these matrices rather than merely name them, replace every entry of the first matrix by that entry times the entire second matrix. For example, in the ordered basis $|00\rangle,|01\rangle,|10\rangle,|11\rangle$,

The first matrix flips $A$’s label while leaving $B$’s unchanged: $|00\rangle\mapsto|10\rangle$ and $|01\rangle\mapsto|11\rangle$. The second multiplies a basis ket by the eigenvalue associated with $B$’s label while leaving $A$ alone. Multiplying these two explicit matrices in either order gives the same result, verifying the commutator above concretely. This block-matrix recipe is the Kronecker product used whenever subsystem operators must be written as a matrix on the joint space.

Because these operators commute, measurements made on the two distinct systems can always be assigned one common joint probability distribution, even when the individual components being measured on either spin, taken alone, would be incompatible. This is a stronger and more precise statement than just “the two measurements don’t interfere.” It says that a single joint probability distribution can be written down for the two experiments at all, which is exactly the kind of object that incompatible single-spin observables (Chapter 3) cannot generally be assigned to begin with.

Example 5.1: independent preparations¶

For $|+x\rangle_A|-z\rangle_B$, measurements of $S_z$ on both spins give $P(0,1)=P(1,1)=1/2$ and zero for the other outcomes. Since the preparation is a product, its joint probabilities factor:

Learning $B$’s result supplies no new information about $A$. This factorization is the operational signature of a product state: every joint question about $A$ and $B$ can be answered by combining two separate one-spin calculations, and nothing is lost by treating the two laboratories as running independent experiments on a shared clock.

5.2 States that do not factor¶

Now consider the normalized state

Try to write it in the product form of equation (3). If $ad=bc=0$ while both $ac$ and $bd$ are nonzero, the required single-system coefficients contradict one another: there is no consistent choice of $a$, $b$, $c$, $d$ that works. A state like this, which cannot be factored, is called entangled.

A joint $z$ measurement on $|\Phi^+\rangle$ produces 00 or 11, each with probability $1/2$. Each individual outcome is random, yet the two records always agree. Rewriting the state in the $x$ basis gives

so the $x$ records also always agree. This is the key point: entanglement is not merely “both spins were prepared the same way.” No assignment of one definite pure state to each spin separately can reproduce all of these correlations at once.

The four Bell states form an orthonormal basis:

Relative signs that were invisible in a $z$ measurement become visible once we switch to another basis, exactly as relative phase did for a single spin in Chapter 2.

Being precise about what “entangled” means matters here, since the word invites an all-or-nothing reading it does not deserve. Entanglement is not a label attached loosely to “the two particles are correlated somehow”; it is a precise property of the joint state, and it comes in degrees. A state can fail to factor into a product yet still be much “closer” to a product than $|\Phi^+\rangle$ is. Section 5.7 makes this quantitative with the Schmidt decomposition. For now, treat the Bell states as one convenient orthonormal family of maximally entangled states, not as the only entangled states that exist.

Foundation check: factor or not?¶

Can $(|00\rangle+|10\rangle)/\sqrt2$ be written as a product? What about $(|00\rangle+|11\rangle)/\sqrt2$?

5.3 Joint, marginal, and conditional probabilities¶

For outcome kets $|a\rangle_A$ and $|b\rangle_B$, the joint Born rule is

The marginal probability for $A$ ignores $B$’s result:

If $B=b$ is recorded, the conditional probability is $P(a|b)=P(a,b)/P_B(b)$ when $P_B(b)\ne0$.

Example 5.2: joint, marginal, and conditional records¶

Begin with $|\Phi^+\rangle$ and measure both spins in the $z$ basis. Reading the four amplitudes directly gives

To ignore $B$’s record, sum across the two possibilities for $B$:

Thus $A$ alone looks random. If $B=0$ is known, however,

so the same data are perfectly predictable after conditioning. Keep the three questions distinct: a joint probability concerns both records, a marginal deliberately ignores one record, and a conditional probability uses a known record to sort the trials.

The arbitrary-angle singlet formulas used in the Bell discussion are stated next. Their derivation is optional on a first pass.

The resulting joint distribution is

and the correlation is

Aligned analyzers always disagree, perpendicular analyzers show no correlation, and oppositely aligned analyzers always agree, under these numerical labels. One more feature of equation (20) deserves a direct check: summing it over $s$ gives $P_A(r)=1/2$ for every $r$, independent of $\mathbf a$ or $\mathbf b$. Each side’s own marginal statistics are, on their own, featureless coin flips. All the physical content lives in how the two records compare, not in either record alone—a point Section 5.4 develops further.

5.4 Reduced states¶

Suppose an experimenter has access only to spin $A$, and never touches spin $B$ at all—perhaps $B$ has been carried off to a distant laboratory. What can that experimenter predict, given only the full joint state $\hat\rho_{AB}$? The answer is captured by the reduced density operator,

where the partial trace sums over any orthonormal basis of $B$. Concretely, for any orthonormal basis $\{|k\rangle_B\}$,

Each term ${}_B\langle k|\hat\rho_{AB}|k\rangle_B$ is itself still an operator, but now acting on $A$ alone: $B$’s basis ket has been contracted away on both sides, and the sum over $k$ adds up the resulting $A$-only pieces.

Let’s carry this out explicitly for $|\Phi^+\rangle=(|00\rangle+|11\rangle) /\sqrt2$, using the $z$ basis for $B$. Writing out the full operator,

Tracing over $B$ means sandwiching this between ${}_B\langle0|\cdots|0 \rangle_B$ and separately between ${}_B\langle1|\cdots|1\rangle_B$, then adding the two results. Because ${}_B\langle0|0\rangle_B= {}_B\langle1|1\rangle_B=1$ while ${}_B\langle0|1\rangle_B= {}_B\langle1|0\rangle_B=0$, any term whose two $B$ labels disagree vanishes immediately, including both off-diagonal terms, $|00\rangle\langle11|$ and $|11\rangle\langle00|$, which are exactly the terms carrying $|\Phi^+\rangle$’s coherence. What survives is only

and the identical calculation, tracing over $A$ instead, gives $\hat\rho_B=\hat I/2$ as well.

Notice exactly which piece of information the partial trace discarded. It kept the diagonal terms, which correctly predict that a $z$ measurement of $A$ alone is 50–50. It discarded the off-diagonal $|00\rangle\langle11|$ coherence—the very feature that made $|\Phi^+\rangle$ a pure joint state rather than the classical mixture of Section 5.8. That coherence is a property of the joint operator alone; it has no counterpart living inside either local factor, which is exactly why no measurement confined to $A$ by itself can ever recover it.

The joint state is pure, but each subsystem, considered alone, is maximally mixed. This is not ordinary ignorance about which pre-existing Bell component actually occurred; the coherence that makes $|\Phi^+\rangle$ a pure state resides in joint off-diagonal terms such as $|00\rangle\langle11|$, which the reduced state at $A$ alone cannot see.

If $B$ is measured in the $z$ basis and 0 is selected, $A$ is conditionally prepared in $|0\rangle$. If $B$ instead measures in the $x$ basis and selects $+x$, $A$ is conditionally prepared in $|+x\rangle$. Yet before learning which setting was used, or what its outcome was, an observer at $A$ still assigns exactly $\hat I/2$, no matter what choice is made at $B$.

This is a general feature, not a special property of $|\Phi^+\rangle$: for any joint state, the reduced density operator $\hat\rho_A=\operatorname{Tr}_B(\hat\rho_{AB})$ used to predict $A$’s local statistics does not depend on which measurement, if any, is later performed at $B$. Choosing a different analyzer at $B$—or none at all—only changes how an already-fixed set of joint records can be sorted into subsets once the two sides compare notes; it cannot change the unconditional distribution of $A$’s own outcomes.

Here is the algebra behind that statement. If $B$ performs a complete projective measurement and the outcome is not supplied to $A$, the joint state becomes

Taking the partial trace and cycling operators that act only on $B$ within that trace,

Completeness, $\sum_b\hat P_b=\hat I$, is the decisive step. The same result holds for any local operation at $B$ whose outcomes are pooled and whose probabilities sum to one. If one outcome is selected and communicated, however, $A$ uses a conditional state; that sorting requires the ordinary communication channel and cannot produce a signal before the message arrives.

Midpoint mastery check¶

Before moving to Bell inequalities, make sure you can do all four of the following without consulting the preceding pages:

1. expand a product ket in the four-state basis;

2. decide whether a pure two-qubit ket factors;

3. compute a joint probability and sum it to obtain a marginal; and

4. take a partial trace of a Bell-state density operator.

Concept check 5.1¶

If $\hat\rho_A=\hat I/2$, does that mean the spin at $A$ is definitely unpolarized and uncorrelated with $B$?

5.5 Bell’s test of local hidden variables¶

A local hidden-variable model supposes that a measurement’s result is determined by information carried along from the shared source, and that a setting freely chosen at one station cannot affect the result recorded at a distant one. One experimentally useful constraint on any such model is the CHSH inequality. For two settings $\mathbf a,\mathbf a'$ at $A$ and $\mathbf b,\mathbf b'$ at $B$, define

Every local hidden-variable model of this type satisfies $|S|\le2$.

The short argument makes several assumptions that should remain visible. The setting choices are statistically independent of the source variables (often called measurement independence); the result at either station does not depend on the distant setting (locality); and the recorded sample fairly represents the emitted pairs. Allowing a hidden-variable model to be stochastic rather than deterministic does not evade the bound: any local response probability can be treated as an average over additional local randomness, and averaging numbers that each lie between -2 and 2 cannot leave that interval. Real Bell experiments are designed to close setting, locality, and detection loopholes closely enough that these assumptions can be tested rather than silently presumed.

For the singlet, suitable coplanar settings separated by $45^\circ$ instead give $|S|=2\sqrt2$. Quantum theory violates the classical bound while still preserving the no-signaling marginal probabilities from Section 5.4.

The conclusion this supports is precise, and it is worth stating carefully: the observed correlations cannot be reproduced by local models in which every result is fixed by shared pre-existing variables, independent of the later settings. Bell tests do not show that experimenters can send information backward in time or faster than light; Section 5.4 already ruled that out on its own.

The Tsirelson bound¶

Equation (28) shows that quantum mechanics can exceed the classical bound of 2, but it cannot exceed it arbitrarily. One can show that for the singlet, and more generally for any quantum state and any choice of $\pm1$-valued observables, $|S|\le2\sqrt2\approx2.828$. This number is the Tsirelson bound. The coplanar, $45^\circ$-separated setting above is not a lucky combination that happens to violate the classical bound; it is the specific configuration that saturates the strongest violation quantum mechanics allows.

Example 5.3: a setting that does not reveal a violation¶

A $2\sqrt2$ violation requires a specific relationship between the four analyzer directions; it does not happen automatically for any four directions we might pick. Suppose instead the four directions are chosen $22.5^\circ$ apart in sequence: $\theta_{\mathbf a}=0^\circ$, $\theta_{\mathbf b}=22.5^\circ$, $\theta_{\mathbf a'}=45^\circ$, $\theta_{\mathbf b'}=67.5^\circ$. Using $E(\mathbf a,\mathbf b)=-\cos(\theta_{\mathbf a}-\theta_{\mathbf b})$ from equation (21),

Here $|S|\approx1.31$, well below even the classical bound of 2. The very same singlet, analyzed with a different—equally legitimate—choice of four directions, shows no sign of any nonclassical correlation at all. Demonstrating a Bell violation is therefore an experiment-design problem as much as a fact about the state: the directions must be chosen, as in the $45^\circ$ configuration above, to make the classical and quantum predictions disagree as sharply as possible. This is the same lesson Chapter 1 drew from sequential Stern–Gerlach analyzers: a single measurement rarely distinguishes rival explanations, and a well-chosen combination is needed to force them apart.

Concept check 5.2¶

Could a future theory find correlations stronger than quantum mechanics’s $2\sqrt2$, while still forbidding faster-than-light signaling?

5.6 Creating entanglement with gates¶

In quantum-information notation, the Hadamard gate is

and the controlled-NOT gate acts as $|a,b\rangle\mapsto|a,b\oplus a\rangle$. Starting from $|00\rangle$,

The Hadamard creates a superposition in $A$; the controlled operation then correlates each alternative with a distinguishable state of $B$. A reversible unitary process has produced a pure entangled state, with no measurement involved anywhere.

This last point deserves emphasis: entanglement is created by ordinary unitary evolution acting on a product input, just as a beam splitter followed by a phase shift can entangle two photons, or a spin–spin interaction term in a Hamiltonian can entangle two initially independent spins. Chapter 4’s rotations and this section’s entangling gates are the same kind of mathematical object—unitary operators—differing only in how many subsystems they act on jointly.

5.7 Optional extension: Schmidt decomposition and amount of entanglement¶

Section 5.2 promised a precise, quantitative way to say “how entangled” a state is. Every pure state of two finite systems can be expressed in suitable local orthonormal bases as

This is the Schmidt decomposition. The number $r$ of nonzero coefficients is the Schmidt rank. A pure state is a product exactly when $r=1$; it is entangled precisely when $r>1$. The two reduced density operators share the same nonzero eigenvalues $\lambda_k$:

Worked method: finding a Schmidt decomposition¶

When a state is not already in Schmidt form, arrange its amplitudes as a coefficient matrix $C$, with $A$ labeling rows and $B$ labeling columns:

Then $\hat\rho_A=CC^\dagger$. Its eigenvalues are the Schmidt weights $\lambda_k$, and its eigenvectors are the $|u_k\rangle_A$. The corresponding $B$ vectors can be recovered from

For example, consider a normalized state that is not visibly in Schmidt form,

The reduced state is

Its eigenvectors are $|+x\rangle$ and $|-x\rangle$, with eigenvalues $9/10$ and $1/10$. Applying the recovery formula gives the same two vectors on $B$, so

This is the Schmidt decomposition. It immediately shows Schmidt rank two and therefore entanglement. In matrix language, the same procedure is the singular value decomposition of $C$; its singular values are $\sqrt{\lambda_k}$.

For two qubits, the state

is already written in Schmidt form. At $\eta=0$ it is a product. At $\eta=\pi/4$ its coefficients are equal and each subsystem is maximally mixed: the fully entangled case.

A single number summarizing pure-state entanglement is the entropy of either reduced state:

It is zero for a product state and exactly one bit for a maximally entangled pair of qubits. For a pure joint state, this entropy measures entanglement by quantifying the mixedness of either reduced state. It is not, by itself, a general measure of all information or correlation in an arbitrary mixed state.

Example 5.4: a partially entangled state¶

Not every entangled state is maximally entangled. Take

already in Schmidt form with $\lambda_1=3/4$, $\lambda_2=1/4$ (compare equation (34); this is the $\eta=30^\circ$ member of the $|\Psi_\eta\rangle$ family above). It does not factor: no choice of single-system amplitudes reproduces both nonzero coefficients while leaving the $|01\rangle$ and $|10\rangle$ coefficients zero, so the state is entangled. Its reduced state is

which has $\operatorname{Tr}(\hat\rho_A^2)=9/16+1/16=5/8<1$: $A$ alone is mixed, confirming the entanglement directly. Its entropy is

less than the 1 bit of a Bell state. So this state is entangled, verifiably so, and yet quantifiably “less entangled” than $|\Phi^+\rangle$— exactly the kind of statement the all-or-nothing language of Section 5.2 could not make precise, but the Schmidt decomposition now can.

Example 5.5: conditional states in a rotated basis¶

Write $|\Phi^+\rangle$ as

The $y$ records always disagree, even though the $z$ and $x$ records always agree. Suppose $B$ obtains $+y$. Applying $\hat I\otimes|+y\rangle\langle+y|$ and normalizing leaves $A$ in $|-y\rangle$. The joint state determines this conditional preparation exactly, while the outcome itself is still random. Changing basis reveals a different correlation pattern without changing the entangled state one bit.

5.8 Entanglement, mixtures, and correlation¶

Correlation alone does not prove entanglement; here is a clean example where it does not. The mixed state

has perfect agreement in $z$, just like $|\Phi^+\rangle$. But it lacks the off-diagonal $|00\rangle\langle11|$ coherence that made $|\Phi^+\rangle$ special. Rewriting $\hat\rho_{\mathrm{class}}$ in the $x$ basis shows that its $x$ outcomes are completely uncorrelated, whereas the Bell state remains perfectly correlated there too.

A mixed state is called separable if it can be written

Such states can still contain shared classical randomness, but nothing more. Entangled mixed states cannot be written this way. Bell-inequality violation certifies entanglement when it occurs, but some entangled mixed states do not violate any particular Bell test: the choice of measurement matters.

Optional advanced extension: the Werner-state boundary¶

A cleaner illustration of the boundary between separable and entangled is the Werner state,

a mixture of the maximally entangled $|\Phi^+\rangle$ with the maximally mixed two-qubit background $\hat I/4$, which carries no correlation at all. At $p=0$ this is two independent, fully random spins; at $p=1$ it is the pure Bell state. To locate the entanglement boundary rather than merely quote it, use the partial-transpose test. Transposing only $B$ changes a matrix element by

Every separable state remains positive under this operation. For two qubits, the converse also holds: a positive partial transpose is sufficient for separability. The partial transpose of $|\Phi^+\rangle\langle\Phi^+|$ has three eigenvalues $+1/2$ and one eigenvalue $-1/2$. Adding the maximally mixed part therefore gives

The final eigenvalue becomes negative precisely when $p>1/3$, certifying entanglement. For $p\le1/3$, all four eigenvalues are nonnegative and the two-qubit sufficiency result guarantees separability. Mixing in only a little of the entangled state therefore leaves a classically explicable mixture; only past a specific threshold does the joint state require entanglement. “How entangled” a mixed state is does not scale simply with how much of an entangled state was mixed in.

Curiously, the reduced state $\hat\rho_A=\operatorname{Tr}_B(\hat\rho_W) =\hat I/2$ for every value of $p$: local measurements at $A$ alone cannot distinguish any Werner state from any other, entangled or not. Whether $\hat\rho_W$ is entangled is a fact about the joint state that no amount of data collected at $A$ alone, or at $B$ alone, can ever reveal.

Concept check 5.3¶

The reduced states of $\hat\rho_{\mathrm{class}}$, $|\Phi^+\rangle$, and every Werner state $\hat\rho_W$ are all $\hat I/2$. Does this mean “entangled” is not a meaningful physical distinction, since it cannot be measured locally?

Local unitary operations can rotate the Schmidt bases, but they cannot change the Schmidt coefficients, the reduced-state purity, or the entanglement entropy. Creating or destroying entanglement genuinely requires an interaction, a joint measurement, or discarding information into an external system; no amount of purely local fiddling will do it.

Concept check 5.4¶

Is $(|00\rangle+|01\rangle)/\sqrt2$ entangled?

Concept check 5.5¶

For $|\Phi^+\rangle$, does measuring $B$ create a random result at $A$?

Concept check 5.6¶

Can two different pure joint states have the same reduced state for $A$?

Summary¶

• Composite state spaces are tensor products, and their dimensions multiply.

• Product states factor into subsystem states; entangled states do not.

• Joint probabilities can contain perfect correlations while each marginal distribution remains random.

• A partial trace gives the state relevant to local measurements.

• Bell correlations violate a bound obeyed by local hidden-variable models but do not permit faster-than-light signaling, and quantum mechanics itself never exceeds the Tsirelson bound $2\sqrt2$.

• Superposition followed by a controlled gate can create entanglement.

• Entanglement is quantitative, not binary: Schmidt coefficients and entanglement entropy measure how far a state is from a product, and mixed states such as the Werner state are entangled only above a threshold mixing parameter.

Exercises¶

Exercises 1–12 are the foundational set. Exercises 13–18 consolidate the core ideas and begin using the optional Schmidt machinery. Exercises 19–32 are extensions and synthesis problems; assigning a representative subset is more useful than treating all fourteen as routine end-of-chapter practice.

Core practice¶

1. Expand $|+x\rangle_A|+y\rangle_B$ in the joint $z$ basis.

2. Determine which states factor: $(|00\rangle-|10\rangle)/\sqrt2$, $(|00\rangle+|11\rangle)/\sqrt2$, and $(|00\rangle+|01\rangle+|10\rangle+|11\rangle)/2$.

3. Calculate all joint $z$ probabilities for each Bell state.

4. Rewrite $|\Phi^-\rangle$ in the $x$ basis and describe its correlations.

5. For the singlet, use equation (20) with an angle of $60^\circ$ to find all four joint probabilities.

6. Compute $\hat\rho_A$ and the entanglement entropy for $\sqrt{2/5}|00\rangle+\sqrt{3/5}|11\rangle$. Is $A$ pure?

7. Show that a product pure state has a pure reduced density operator.

8. Prove from equation (20) that both local marginal distributions are $1/2$ for every pair of settings.

9. Use singlet correlations to evaluate the CHSH expression for directions with planar angles $0^\circ,90^\circ,45^\circ,-45^\circ$.

10. Write the $4\times4$ matrix of CNOT in the ordered computational basis and verify its action on all four basis kets.

11. Start with each computational basis state, apply Hadamard to $A$ and then CNOT, and identify the resulting Bell state.

12. Explain why the phrase “each entangled spin has a definite but unknown state” fails for $|\Phi^+\rangle$ when both $z$ and $x$ correlations are considered.

13. Find the reduced density operators and entanglement entropy of $\cos\eta|00\rangle+\sin\eta|11\rangle$.

14. Verify the $y$-basis expansion in Example 5.5 and its anticorrelations.

15. Compare all joint $x$ probabilities of $\hat\rho_{\mathrm{class}}$ and $|\Phi^+\rangle$.

16. Prove that applying $\hat U_A\otimes\hat U_B$ does not change the eigenvalues of either reduced density operator.

17. For the singlet, evaluate $E(\mathbf a,\mathbf b)$ at relative angles $30^\circ$ and $150^\circ$, and explain the sign of each result in terms of typical agreement or disagreement between the two records.

18. Derive the elementary classical CHSH bound directly: for any four numbers $A,A',B,B'\in\{+1,-1\}$, show that $AB+AB'+A'B-A'B'=A(B+B')+A'(B-B')$, and that this expression cannot exceed 2 in magnitude because $B+B'$ and $B-B'$ cannot both be nonzero.

Extensions and synthesis¶

19. Verify the arithmetic of Example 5.3 by direct substitution into equation (21), and find one nearby setting—a small deviation from the $45^\circ$-spaced configuration of Section 5.5—that still violates the classical bound of 2.

20. Show that the reduced state $\hat\rho_A$ of the Werner state (49) equals $\hat I/2$ for every value of $p$, by computing the partial trace of both terms separately.

21. For $p=1/2$, decide whether the Werner state is separable or entangled and justify your answer using the threshold stated in Section 5.8.

22. Verify that the two-qubit reduced state of the GHZ state $(|000\rangle+|111\rangle)/\sqrt2$, obtained by tracing out any one qubit, equals $\hat\rho_{\mathrm{class}}$ from Section 5.8. Explain why this shows the pairwise correlations contain no entanglement even though the three-qubit state does not factor.

23. Two spins are prepared independently, each in $|+z\rangle$, so the joint state is a product. If $A$ is measured along $\mathbf a$ and $B$ along $\mathbf b$, find $E(\mathbf a,\mathbf b)$ and show that the resulting CHSH combination cannot exceed the classical bound of 2 for any four settings.

24. A student claims that because $\hat\rho_A=\hat I/2$ for $|\Phi^+\rangle$, the two spins “must actually be independently random, like a coin flip on each side.” Using the CHSH violation of Section 5.5, explain precisely what is wrong with this claim.

25. Show that applying a local phase gate $\hat S\otimes\hat I$, with $\hat S=\operatorname{diag}(1,i)$, to $|\Phi^+\rangle$ changes the state but not its Schmidt coefficients or entanglement entropy.

26. Construct the two-qubit state produced by applying CNOT (control $A$, target $B$) to $|+x\rangle_A|0\rangle_B$, and show it is a Bell state.

27. Compute the entanglement entropy for Schmidt coefficients $\lambda_1=0.9,\lambda_2=0.1$ and compare it with the 0.811-bit result of Example 5.4. Which state is closer to a product state, and does the entropy order match your intuition?

28. For a general two-qubit pure state written in Schmidt form with $\lambda_1=\cos^2\eta$, $\lambda_2=\sin^2\eta$, argue from the shape of $-\lambda\log_2\lambda-(1-\lambda)\log_2(1-\lambda)$ as a function of $\lambda\in[0,1]$ that the entanglement entropy is maximized at $\eta=\pi/4$ and vanishes at $\eta=0$ or $\pi/2$.

29. Two experimentalists share $10{,}000$ singlet pairs and measure CHSH correlations using the optimal $45^\circ$-spaced settings of Section 5.5. Each correlation estimate is subject to the same $1/\sqrt N$ statistical fluctuation discussed in Chapter 1. Estimate the order of magnitude of the statistical uncertainty in $S$, and explain why a modest violation just above 2, rather than any positive excess at all, is needed before ruling out statistical fluctuation as an explanation.

30. Explain, in your own words, why “the two spins are entangled” and “the two spins are correlated” are not synonyms, using $\hat\rho_{\mathrm{class}}$ and $|\Phi^+\rangle$ as your two contrasting examples.

31. A colleague proposes building a faster-than-light telegraph by having $A$ choose to measure in the $z$ basis to send a “0” and the $x$ basis to send a “1,” using a shared $|\Phi^+\rangle$ pair with $B$ far away. Using the results of Sections 5.3 and 5.4, explain specifically what $B$ would need to do to detect which basis $A$ used, and why the attempt fails.

32. Using the definition of a separable mixed state in Section 5.8, verify explicitly that $\hat\rho_{\mathrm{class}}=\tfrac12|00\rangle\langle00| +\tfrac12|11\rangle\langle11|$ has the separable form $\sum_jw_j\hat\rho_A^{(j)}\otimes\hat\rho_B^{(j)}$, and identify the weights $w_j$ and single-qubit states explicitly.

Selected exercise guidance¶

Use these answers to check method and conventions after making a complete attempt. They are intentionally selective rather than a substitute for working the full set.