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A density operator is a positive self-adjoint trace-class operator with trace one. In finite dimensions this is a positive semidefinite Hermitian matrix with $\mathrm{tr}\rho =1$. Its expectation of an observable is $\boxed{⟨A⟩=\mathrm{tr}(A\rho )}$ whenever the expectation is defined.

In the source's square-root notation, take the unique positive square root $W=\sqrt{\rho }$. It has Hilbert-Schmidt norm one and satisfies $W{W}^{\ast }=\rho$, soThe standard vector purification is the associated vector $|W⟩⟩={\sum }_{r,s}{W}_{rs}|{\xi }_r⟩|{\eta }_s⟩$ on a system and a chosen copy. Tracing out the copy gives $\rho$; measuring $A\otimes I$ gives the same trace formula. Thus the operator square root and the purified vector are two related objects, distinguished by this vectorization.

For a general coefficient matrix $C$, expansion of the expectation givesThus $\boxed{{\rho }_1=C{C}^{\ast }}$. Positivity is immediate from $v^{\ast }C{C}^{\ast }v=\Vert C^{\ast }v{\Vert }^2$, and its trace is one by normalization of $\Psi$.

If $C$ is nonsingular, it is square and ${\rho }_1$ is positive definite. DefineThen $U{U}^{\ast }=C^{-1}{\rho }_1(C^{\ast }{)}^{-1}=I$, which for a square matrix also implies $U^{\ast }U=I$. Thus $U$ is unitary, $\sqrt{{\rho }_1}=CU$, and $C=\sqrt{{\rho }_1}{U}^{\ast }$.

In an eigenbasis of ${\rho }_1$, its square root is $\mathrm{diag}(\sqrt{p_r})$. Hence $c_{rs}=\sqrt{p_r}\overline{U_{sr}}$. DefineUnitarity gives $⟨{\zeta }_r|{\zeta }_t⟩={\delta }_{rt}$, and substitution gives the Schmidt decompositionFor singular or rectangular $C$, the same conclusion follows on its support by a singular value decomposition; the stated nonsingular case needs only the unitary calculation above.

For the supplied Bell state, the requested matrices areThe Schmidt vectors can be chosen as ${\zeta }_1={\eta }_2$, ${\zeta }_2=-{\eta }_1$, and the two coefficients are $1/\sqrt{2}$. The square root's canonical vector purification is $(|{\xi }_1{\eta }_1⟩+|{\xi }_2{\eta }_2⟩)/\sqrt{2}$; the original Bell vector is a different purification of the same reduced state, related by a unitary on the second factor. This illustrates the unitary freedom of purification.

For classical events, $A\setminus C\subset (A\setminus B)\cup (B\setminus C)$: an outcome in $A$ but not in $C$ either fails $B$ or belongs to $B\setminus C$. Monotonicity and the union bound prove the Fréchet-Bell inequalityMalus' law gives transmission ${\cos }^2\gamma$ for a photon polarized by the first filter, so its conditional rejection probability at the second is $\boxed{{\sin }^2\gamma }$. Conditional and joint probabilities must be distinguished: $P(A\setminus B)=P(A)P(B^c\mid A)$.

For equal unbiased marginals $P(A)=P(B)=1/2$, the quantum two-setting rejection probabilities are $\frac{1}{2}{\sin }^2({\theta }_A-{\theta }_B)$. A classical common assignment of all three outcomes would therefore require ${\sin }^2({\theta }_A-{\theta }_B)+{\sin }^2({\theta }_B-{\theta }_C)\ge {\sin }^2({\theta }_A-{\theta }_C)$. Choose directions $0,\gamma ,2\gamma$. ThenFor example $\gamma =\pi /6$ gives $1/2<3/4$. More generally ${\sin }^{2}a+{\sin }^{2}b-{\sin }^2(a+b)=-2\sin a\sin b\cos (a+b)$. In a triangle whose side directions are these filters, the internal angle supplementary to $a+b$ is obtuse exactly in this violation range. This specifies the angle convention behind the triangle description: with polarization axes regarded as unoriented lines, a claim that every sufficiently large physical axis separation violates the inequality would not be correct. The contradiction concerns a single classical joint model for all settings, not the validity of ordinary probability or a comparison of arbitrarily renormalized sequential experiments.

Use $|0⟩=|{\alpha }_0⟩$ and $|1⟩=|{\alpha }_1⟩$. Projecting the first photon with $|0⟩⟨0|$ gives the unnormalized product vector $c_0|00⟩$, with success probability $|c_0{|}^2$. Projection of the second onto $|\theta ⟩=\cos \theta |0⟩+\sin \theta |1⟩$ givesConditioned on success, normalize these vectors; a zero amplitude means that outcome never occurs. For the maximally entangled equal-amplitude pair, rotation of both real polarization bases gives the same correlations for arbitrary settings. Comparing the appropriately normalized coincidence/rejection frequencies at three setting pairs tests the inequality above. A general unequal-amplitude state need not reproduce the unbiased rotationally invariant probabilities, so the maximally entangled choice supplies the clean explicit example.

For swapping, label the four systems in order ${\xi }_1,{\eta }_1,{\xi }_2,{\eta }_2$ and let $|\Phi {⟩}_{\eta }=2^{-1/2}{\sum }_{j=0}^1|{\eta }_j{\eta }_j⟩$. Its overlap with $|{\eta }_r{\eta }_s⟩$ is ${\delta }_{rs}/\sqrt{2}$. Applying its rank-one projector to Alice's two systems givesAfter regrouping the remote and local systems, this factorizes into $[{\sum }_r{b}_r^2|{\xi }_r{\xi }_r⟩/\sqrt{2}]\otimes |\Phi {⟩}_{\eta }$. The selected outcome has probability $p=\frac{1}{2}{\sum }_r|b_r{|}^4$. Therefore the normalized conditional remote state isIt has Schmidt rank at least two whenever the original pair was entangled, since squaring a nonzero coefficient does not make it zero. Thus the two photons that never interacted become entangled conditional on Alice's joint measurement outcome. This is entanglement swapping. Alice is projecting onto a specified Bell state, not performing a universal test of whether an arbitrary unknown state is entangled. Averaging all her outcomes leaves the remote reduced density operator unchanged, so no faster-than-light signal is available without her classical outcome label.

A density operator is a positive self-adjoint trace-class operator with trace one. In finite dimensions this is a positive semidefinite Hermitian matrix with $\mathrm{tr}\rho =1$. Its expectation of an observable is $\boxed{⟨A⟩=\mathrm{tr}(A\rho )}$ whenever the expectation is defined.

In the source's square-root notation, take the unique positive square root $W=\sqrt{\rho }$. It has Hilbert-Schmidt norm one and satisfies $W{W}^{\ast }=\rho$, soThe standard vector purification is the associated vector $|W⟩⟩={\sum }_{r,s}{W}_{rs}|{\xi }_r⟩|{\eta }_s⟩$ on a system and a chosen copy. Tracing out the copy gives $\rho$; measuring $A\otimes I$ gives the same trace formula. Thus the operator square root and the purified vector are two related objects, distinguished by this vectorization.

For a general coefficient matrix $C$, expansion of the expectation givesThus $\boxed{{\rho }_1=C{C}^{\ast }}$. Positivity is immediate from $v^{\ast }C{C}^{\ast }v=\Vert C^{\ast }v{\Vert }^2$, and its trace is one by normalization of $\Psi$.

If $C$ is nonsingular, it is square and ${\rho }_1$ is positive definite. DefineThen $U{U}^{\ast }=C^{-1}{\rho }_1(C^{\ast }{)}^{-1}=I$, which for a square matrix also implies $U^{\ast }U=I$. Thus $U$ is unitary, $\sqrt{{\rho }_1}=CU$, and $C=\sqrt{{\rho }_1}{U}^{\ast }$.

In an eigenbasis of ${\rho }_1$, its square root is $\mathrm{diag}(\sqrt{p_r})$. Hence $c_{rs}=\sqrt{p_r}\overline{U_{sr}}$. DefineUnitarity gives $⟨{\zeta }_r|{\zeta }_t⟩={\delta }_{rt}$, and substitution gives the Schmidt decompositionFor singular or rectangular $C$, the same conclusion follows on its support by a singular value decomposition; the stated nonsingular case needs only the unitary calculation above.

For the supplied Bell state, the requested matrices areThe Schmidt vectors can be chosen as ${\zeta }_1={\eta }_2$, ${\zeta }_2=-{\eta }_1$, and the two coefficients are $1/\sqrt{2}$. The square root's canonical vector purification is $(|{\xi }_1{\eta }_1⟩+|{\xi }_2{\eta }_2⟩)/\sqrt{2}$; the original Bell vector is a different purification of the same reduced state, related by a unitary on the second factor. This illustrates the unitary freedom of purification.

The normalized Bell statehas equal outcomes for any common real linear-polarization basis. In the stated rotated basis it is $(|{\theta }^{(+)}{\theta }^{(+)}⟩+|{\theta }^{(-)}{\theta }^{(-)}⟩)/\sqrt{2}$: the cross terms cancel when the inverse rotation is substituted.

Thus the coincidence probability for the transmitted ports is $p(A,B)=\frac{1}{2}{\cos }^2({\theta }_A-{\theta }_B)$. At equal angles it is $1/2$ per emitted pair, while Bob transmits with probability one conditional on Alice's transmission. Both photons are blocked for the other half of ideal pairs. The phrase always in coincidence must therefore mean matching outcomes or conditional transmission, rather than unit unconditional transmission through both polarizers.

The classical mixture has no coherence between $HH$ and $VV$; it gives equal-angle coincidence probability $1/4$ at $45^{\circ }$, rather than $1/2$. The coherent quantum superposition maintains correlation in this rotated basis. Bob's unconditional transmission remains $1/2$, so the conditional correlation does not permit signalling by Alice's polarizer choice.

Suppose two orthonormal qubit bases have disjoint physical rays. Alice chooses which conjugate basis to measure on her half of a Bell state. A device that perfectly labels Bob's individual member of either basis would reveal Alice's choice without her outcome being communicated. Spacelike operation would therefore violate relativistic causality, even though ordinary projective measurements see the same reduced density matrix $I/2$.

Let $H,V$ be orthogonal linear polarizations, and prepare the Bell state $|{\Phi }^{+}⟩=(|HH⟩+|VV⟩)/\sqrt{2}$. The $+1$ analyzer vector is $|a⟩=\cos a|H⟩+\sin a|V⟩$; its orthogonal companion has outcome $-1$. Direct inner products with the Bell state give $P_{++}=P_{--}={\cos }^2(a-b)/2$ and $P_{+-}=P_{-+}={\sin }^2(a-b)/2$. Thus the expectation value of the outcome product is $E(a,b)=\cos 2(a-b)$. These are physical polarizer angles; their corresponding Bloch sphere angles are doubled.

For a stipulated isotropic admixture, use the density matrixThe maximally mixed state contributes zero correlation, so $P_{st}(a,b)=[1+stv\cos 2(a-b)]/4$ for $s,t\in \{-1,1\}$. Each local marginal is $1/2$, independently of the other setting, illustrating quantum no-signalling. In this model $v$ is the interferometric visibility of the coincidence fringe; other noise models need not have one common visibility for every basis.

Choose $a_0=0^{\circ }$, $a_1=45^{\circ }$, $b_0=22.5^{\circ }$ and $b_1=-22.5^{\circ }$. The four correlations in the displayed CHSH inequality are $v/\sqrt{2},v/\sqrt{2},v/\sqrt{2},-v/\sqrt{2}$. Consequently $S=2\sqrt{2}v$, exceeding the local bound exactly when $v>1/\sqrt{2}$. Perfect correlation at a single common setting would not yield this comparison between incompatible quantum measurements.

To connect probabilities with expected counts, stipulate $M$ emitted pairs per setting pair and independent detection efficiencies ${\eta }_A,{\eta }_B$, independent also of settings, outcomes and hidden variables. The expected joint counts are $n_{st}=M{\eta }_A{\eta }_B{P}_{st}$. Their normalized correlation isFor original teaching inputs $M=1000$, ${\eta }_A=0.8$, ${\eta }_B=0.75$ and $v=0.8$, the $(a_0,b_0)$ expected counts are $n_{++}=n_{--}\simeq 234.85$ and $n_{+-}=n_{-+}\simeq 65.15$. They total $600$ and give $E_{\det }\simeq 0.56569$, hence $S\simeq 2.26274$. Fractional expected counts are not measured data. Efficiency cancels here because the loss model was specified to be independent; this algebra supplies no justification for discarding failures in an arbitrary Bell test with a detection loophole.

For $|{\Psi }^{+}⟩=(|01⟩+|10⟩)/\sqrt{2}$, the Pauli operator correlations are $⟨XX⟩=1$, $⟨ZZ⟩=-1$, and $⟨XZ⟩=⟨ZX⟩=0$. Substituting the measurement of linear photon polarization into $⟨M(\alpha )\otimes M(\beta )⟩$ gives $\sin 2\alpha \sin 2\beta -\cos 2\alpha \cos 2\beta =-\cos 2(\alpha +\beta )$. In particular, opposite physical angles give perfect anticorrelation. The sum of angles here depends on this choice of Bell state and basis conventions.

Quantum entanglement between electromagnetic polarization degrees of freedom. For example, $(|HV⟩+|VH⟩)/\sqrt{2}$ is a Bell state of two photons; neither photon separately has a definite horizontal polarization or vertical polarization in this description.

Quantum information theory asks what can be stored, learned, transmitted, and protected when information is carried by quantum states. Its basic objects are density operators, measurement in quantum measurements, and quantum channels; its resources include qubits, entanglement, and classical communication. Specifying the task matters: preserving an unknown quantum state, recovering a classical bit, and estimating a probability amplitude are different demands on the same physical system.

A particularly beautiful feature is that knowing the whole need not mean knowing its parts. A Bell state is a completely specified pure state, yet either qubit alone has a maximally uncertain reduced density matrix. Conversely, two systems can share perfectly matching measurement records without entanglement. The worked Bell correlations versus a classical mixture comparison separates these possibilities using only two choices of Pauli measurement.

This viewpoint makes limitations useful design rules. Schumacher compression asks how much quantum memory a source needs; Holevo's theorem limits the classical information extractable from an encoding; quantum error correction protects states against noise. Quantum teleportation uses shared entanglement and classical communication to transfer an unknown state. Its resource accounting respects both quantum no-signalling and the no-cloning theorem.

Matrix transposition preserves positivity and trace but fails complete positivity. For the Bell state $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, partial transposition givesThe singlet $(|01⟩-|10⟩)/\sqrt{2}$ has eigenvalue $-1/2$ for this matrix, so the output is not a density operator. Universal qubit transposition is therefore not a quantum channel, although restricted input families can admit exact physical approximations.

For the Bell state $|{\varphi }_{+}⟩$, let both zero-input measurements use the computational basis, and let the one-input bases be rotated by $\theta$ for Alice and $-\theta$ for Bob in the real qubit plane. In the CHSH game, the four conditional winning probabilities are $1,{\cos }^2\theta ,{\cos }^2\theta ,{\sin }^2(2\theta )$. They follow by contraction of the Bell state with vectors from the real orthonormal bases used for the basic measurements. Their average is the displayed expression. On $0\le \theta \le \pi /2$, this beats $3/4$ exactly when $0<\theta <\pi /4$ and reaches $13/16$ at $\theta =\pi /6$, since $t-t^2=1/4-(t-1/2{)}^2$.

A projective measurement on one half of a Bell state prepares the complex-conjugate basis on the other half, conditional on the outcome. Measuring Alice in the columns of $U^{\ast }$ prepares Bob in the columns of $U$. Ignoring the equally likely outcomes leaves Bob's reduced density matrix equal to $I/2$ in either basis.

For four uniformly distributed Bell states, measure one qubit along Bloch vector $\mathbf{n}$ and the other along $z$. The sign within each ${\Phi }^{\pm }$ or ${\Psi }^{\pm }$ pair is invisible and the displayed mutual information is at most one bit. A local Hadamard gate before a computational detector changes $n_z$ from one to zero and destroys the remaining information. An entangling Bell decoder falls outside this product-measurement restriction and recovers two bits.

Superdense coding uses a shared Bell state to communicate two classical bits by transmitting one qubit. The sender selects one of four orthogonal Bell states with a local Pauli operator, and the receiver performs a Bell-basis measurement. For an ideal shared Bell pair, the transmitted qubit alone has a maximally mixed state for every message and reveals no information about the chosen encoding.

For two equally coupled atoms initially in $|ee;0⟩$, excitation conservation confines the interaction picture to $|ee;0⟩$, $(|eg⟩+|ge⟩)|1⟩/\sqrt{2}$ and $|gg;2⟩$. The Hamiltonian operator divided by $\hslash$ has off-diagonal entries $\sqrt{2}g$ and $2g$. Its eigenvalues are $0,\pm \sqrt{6}g$, so the amplitudes are $(2+\cos \sqrt{6}gt)/3$, $-i\sin (\sqrt{6}gt)/\sqrt{3}$ and $\sqrt{2}(\cos \sqrt{6}gt-1)/3$. An ideal one-photon number projection therefore heralds a symmetric atomic Bell state with probability ${\sin }^2(\sqrt{6}gt)/3$, at most $1/3$.

Let $c,s\ge 0$ and $c^2+s^2=1$. Starting from the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, Alice applies the measurement in quantum mechanics with Kraus operators $M_0,M_1$ displayed above. Completeness follows from $M_0^{†}{M}_0+M_1^{†}{M}_1=I$. Each outcome has probability $1/2$. The normalized outputs are $c|00⟩+s|11⟩$ and $s|00⟩+c|11⟩$. Communicating the outcome allows both parties to apply a Pauli X gate in the second branch, making both outputs equal to the first. Thus this LOCC protocol deterministically converts one Bell pair into any two-qubit pure state with these Schmidt coefficients, after appropriate local unitary operations. It includes the product state endpoint without postselection.

For two collected spatial modes $a,b$, each with polarizations $H,V$, put $K^{†}=a_H^{†}{b}_V^{†}+a_V^{†}{b}_H^{†}$. The one-pair state is a Bell state. The normalized two-pair component isThis follows by expanding $(K^{†}{)}^2$ and using the bosonic creation operator identity $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$. The three orthogonal normalized Fock states have equal probability $1/3$. A tensor product of independently labelled pairs is a different multimode description; identifying its labels without renormalization changes its norm.

This collective-unitary-invariant two-qubit state has singlet weight $p$ and equal weights on the three triplet Bell states. With $P_s$ the spin singlet state projector, its partial transpose has eigenvalues $(1-2p)/2$ once and $(1+2p)/6$ three times. It is entangled exactly when $p>1/2$. At $p=1/2$, it is the equally weighted mixture of six antiparallel product states along the three Pauli axes. In the interval $1/4\le p\le 1/2$, convex mixing of this boundary state with $I/4$ explicitly proves separability. The weight $p$ here is the singlet probability, not the alternative coefficient multiplying a pure singlet in a white-noise mixture.

The ideal universal NOT map sends every qubit Bloch vector $\mathbf{s}$ to $-\mathbf{s}$, taking a pure state to its orthogonal pure state. It is a trace-preserving positive map but not a quantum channel: applied to one half of a Bell state, it produces an operator with a negative eigenvalue. A physical approximation can reverse all directions with factor $-1/3$ using the equal mixture of conjugations by the three Pauli matrices. The isotropic qubit map complete-positivity bound proves that a larger uniform reversal is impossible among such isotropic maps.