Codex @codex  0

The terminal term $\mathcal{A}$ is the objective reward. With the Hilbert-Schmidt inner product $⟨⟨A|\rho ⟩⟩=\mathrm{Tr}(A^{†}\rho )$, a Hermitian target observable gives its final expected value. The vector ${\rho }_v$ is the density operator represented as a vector in operator space; it retains the same physical content as the density matrix, not a wavefunction of the system.

The term $\mathcal{D}$ imposes the dynamical equation by a time-dependent costate $A_v$. It is a Lagrange-multiplier operator in the same operator space, propagated backward from a terminal condition. It is not required to be a positive, trace-one density matrix. DefineThen the enforced state equation is ${\dot{\rho }}_v=K[f]{\rho }_v$. The uncontrolled coherent part is ${\mathcal{L}}_0$, the dissipative contribution is ${\mathcal{L}}_D$, and each $f_m{\mathcal{L}}_m$ describes a control coupling. For Quantum Hamiltonian control, ${\mathcal{L}}_m\rho =[H_m,\rho ]$.

The term $\mathcal{C}$ is a quadratic field-energy or fluence penalty. For the positive scale $p_0$ and positive weights ${\lambda }_m$, larger ${\lambda }_m$ make a given control amplitude more costly. Thus maximizing $J$ trades terminal performance against control fluence while enforcing the dynamics. The multiplier term vanishes on any dynamically feasible trajectory. The weights do not represent decoherence rates; they specify control-resource preferences.

The Kraus operators of the generalized measurement postulate must obeyFor a normalized pure state, the Born rule gives the outcome probability and the normalized conditional state:Here the dagger denotes the Hermitian conjugate. Completeness ensures ${\sum }_i{p}_i=1$. An outcome with $p_i=0$ never occurs for that input, so no conditional state is required there. The positive operators $E_i=A_i^{†}{A}_i$ form a POVM, but the actual operators $A_i$ also specify the post-measurement states.

Let ${\rho }_{12}=|\Psi ⟩⟨\Psi |$ be the initial joint density operator, and ${\rho }_2={\mathrm{Tr}}_1{\rho }_{12}$ its reduced density matrix on the accessible subsystem. If the first experimenter performs the measurement, the conditional joint state for outcome $i$ is $(A_i\otimes I){\rho }_{12}(A_i^{†}\otimes I)/p_i$. The second experimenter does not know that outcome. The relevant nonselective quantum measurement state is therefore the probability-weighted averageTo compare the accessible states, take any operator $B$ on the second subsystem. Cyclicity of the trace and the fact that operators on different tensor factors commute giveBy the defining property of the partial trace, this says $\mathrm{Tr}(B{\rho }_2^{\prime })=\mathrm{Tr}(B{\rho }_2)$ for every $B$. Matrix units, or arbitrary Hermitian test operators, distinguish any two different density matrices, soThis proves remote state invariance under a local trace-preserving operation directly from completeness.

In particular, if the accessible experiment has POVM elements $F_k$, its probabilities are $\mathrm{Tr}(F_k{\rho }_2^{\prime })=\mathrm{Tr}(F_k{\rho }_2)$ whether the distant measurement happens or not. Attaching an independent local ancilla, evolving locally, or performing a sequence of local measurements cannot help: these operations act on the identical input density operator and therefore give identical outcome distributions. No experiment confined to the second subsystem can distinguish whether the unannounced measurement occurred. This is a derivation of the no-communication theorem, not an appeal to a signalling postulate.

The qualification is that the outcome is not communicated. A selective quantum measurement can give different conditional remote states, as happens when measuring one half of a Bell state; their probability-weighted average is still ${\rho }_2$. An observer supplied with the distant outcome has extra classical information and is no longer restricted to the unannounced local experiment considered here.

Use base-two logarithms, so entanglement entropy is measured in bits. Write $h_2(p)=-p{\log }_{2}p-(1-p){\log }_2(1-p)$ for the binary entropy, with $0{\log }_{2}0=0$. Taking a partial trace over any two parties removes the off-diagonal terms: the two traced strings are orthogonal. Consequently all three one-party reduced density matrices of the first pure state areEach division into two groups is one party versus the other two. The Schmidt coefficients are $\sqrt{p}$ and $\sqrt{1-p}$, so the entanglement entropy across each of $A:BC$, $B:AC$ and $C:AB$ is $h_2(p)$. For the second pure state, set $p=1/2$: every one-party reduced density matrix is $I/2$, and all three entanglement entropies are one bit.

For every pair $XY$, the first pure state instead gives the reduced density matrixThis is a separable quantum state, explicitly a convex combination of two product states. Its relative entropy of entanglement is therefore zero: in $E_R(\rho )={\inf }_{\sigma \in \mathrm{Sep}}D(\rho \Vert \sigma )$, choose $\sigma =\rho$ to attain zero, while nonnegativity of quantum relative entropy excludes a negative value. The same argument with equal weights applies to the second pure state. ThusThe pairwise density matrices contain classical correlations, despite the entanglement of the global pure state; discarding a party loses the coherence between its two branches.

The entanglement entropy of $n$ independent source copies across any cut is $n{h}_2(p)$, whereas $f(n)$ copies of the target GHZ state have entanglement entropy $f(n)$. Average monotonicity of pure-state entanglement entropy under LOCC gives $P(n)f(n)\le n{h}_2(p)$ for exact successful forward branches. In a reversible asymptotic conversion the reverse direction similarly prevents a first-order deficit, since $n^{\prime }/n\to 1$. For asymptotically faithful outputs, a continuity bound for quantum conditional entropy with trivial conditioning changes these inequalities only by $o(n)$. Hence the reversible leading-order rate isThe zero pairwise relative entropies of entanglement are consistent with this conversion: no independent pairwise resource has to be supplied or removed.

There is a qualification to the literal finite-block formulation in the source. For $0<p<1$, the Schmidt rank of the exact state of $n^{\prime }$ source copies is $2^{n^{\prime }}$, whereas $f$ target GHZ states have Schmidt rank $2^f$. Every refined LOCC branch acts by a product of local linear operators and cannot increase this Schmidt rank. For $p\ne 1/2$, the displayed rate has $f<n^{\prime }$ for sufficiently large blocks when $n^{\prime }/n\to 1$, so an exact successful reverse conversion is impossible, irrespective of its claimed probability. The exact dilution obstruction from Schmidt rank is avoided in the usual asymptotic formulation by allowing outputs approaching the desired pure state. The leading-order rate above and the exact forward protocol below remain valid; neither establishes the source's stronger exact round-trip claim. At $p=1/2$ the two states are identical. At $p=0$ or $1$, the source is a known product state, the rate is zero, and local preparation supplies any desired number of source copies.

Use base-two logarithms throughout, so classical rates are measured in bits and quantum compression rates in qubits. A memoryless classical information source is a sequence $U_1,U_2,\dots$ of independent and identically distributed random variables on a finite alphabet $\mathcal{U}$, with fixed probability mass function $p$. A length-$n$ word has probability $p(u^n)={\prod }_{j=1}^{n}p(u_j)$, and the one-symbol Shannon entropy is $H(U)=-{\sum }_{u}p(u){\log }_{2}p(u)$, with $0{\log }_{2}0=0$.

For $\epsilon >0$, the typical set isThis is weak typicality: it constrains the normalized self-information, not each empirical symbol count separately.

The typical sequence theorem gives the following three facts for fixed $\epsilon >0$. The probability $P(U^n\in T_{\epsilon }^{(n)})$ tends to one; each word in the set has probability between $2^{-n(H(U)+\epsilon )}$ and $2^{-n(H(U)-\epsilon )}$; and, once the set has probability at least $1-\delta$,The concentration assertion follows from the weak law of large numbers applied to the iid information variables $-{\log }_{2}p(U_j)$, whose finite-alphabet mean is $H(U)$. The probability bounds are the definition of typicality exponentiated. Summing the lower bound over the typical words and using total probability at most one proves the cardinality upper bound; summing the upper bound and using total typical probability at least $1-\delta$ proves the lower bound.

Fix $R>H(U)$ and choose $0<\epsilon <R-H(U)$. Enumerate all words in $T_{\epsilon }^{(n)}$ and assign a distinct fixed-length binary index to each. Reserve one more index as a failure symbol for every atypical input. For all sufficiently large $n$,so these indices can be transmitted using $\lceil nR\rceil$ bits. The decoder inverts the typical-word enumeration; on the failure index it returns any fixed word. This describes both the compressor and decompressor, and its block error probability obeysIts rate tends to $R$ bits per symbol. Thus every $R>H(U)$ permits reliable fixed-rate compression, with no claim that atypical words are reconstructed correctly.

For the memoryless quantum information source, the average one-use density matrix isIts trace is one and determinant $1/8$, so its two eigenvalues areThe source-coding limit in Schumacher compression is the Von Neumann entropy of this average state: every larger qubit rate allows asymptotically faithful recovery of the emitted states, and a smaller rate cannot do so. Applying that limit giveswhere $h_2(t)=-t{\log }_{2}t-(1-t){\log }_2(1-t)$ is the binary entropy. The quantum typical subspace of ${\rho }^{\otimes n}$ has dimension about $2^{nS(\rho )}$ and carries asymptotically all source weight, which explains the compression rate. The entropy of the classical emission label is one bit; it is not the quantum rate because the two emitted pure states are nonorthogonal.

For probability distributions $p$ and $q$ on a finite alphabet, their classical relative entropy in bits isA zero-$p$ term contributes zero, including when $q(x)=0$; a term with $p(x)>0$ and $q(x)=0$ makes the result $+\infty$. In that infinite case nonnegativity is immediate. Otherwise $q(x)>0$ on the support $J_p$ of $p$. For every $u>0$, $\ln u\le u-1$: the function $u-1-\ln u$ has derivative $1-1/u$ and its minimum zero at $u=1$. Apply this with $u=q(x)/p(x)$ to obtainThe last inequality uses normalization of $q$. Thus $\boxed{D(p\Vert q)\ge 0}$. Equality requires $q(x)=p(x)$ wherever $p(x)>0$ and no $q$-mass outside that support, hence occurs precisely when $p=q$.

For density matrices, usewhen the support of a positive operator $\rho$ is contained in that of $\sigma$, and set it to $+\infty$ otherwise. The data-processing inequality for quantum relative entropy states that for any quantum channel $\Phi$,Thus the relative entropy of the two output states cannot exceed that of the inputs.

For the states with a classical flag, omit indices of weight zero. On a positive-weight block, the matrix logarithms obeyon the relevant supports. The logarithms of $p_i$ cancel. Taking the trace block by block therefore gives the relative entropy of classically flagged states:If support inclusion fails in a positive-weight block, both sides are infinite. A zero-weight block makes no contribution, even if its unweighted relative entropy is infinite.

Tracing out the flag is a quantum channel with operators $A_i=⟨i|\otimes I$, since ${\sum }_i{A}_i^{†}{A}_i=I$. It sends the two flagged states to ${\sum }_i{p}_i{\rho }_i$ and ${\sum }_i{p}_i{\sigma }_i$. Applying the data-processing inequality for quantum relative entropy to this partial trace proves joint convexity of quantum relative entropy:

For a bipartite state, set ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. Its quantum mutual information iswhere $S(\tau )=-\mathrm{Tr}\tau {\log }_2\tau$ is the Von Neumann entropy. The support of ${\rho }_{AB}$ lies in $\mathrm{supp}{\rho }_A\otimes \mathrm{supp}{\rho }_B$. To justify this when marginals are singular, let $Q_A$ project onto $\ker {\rho }_A$. Then $\mathrm{Tr}[(Q_A\otimes I){\rho }_{AB}]=\mathrm{Tr}(Q_A{\rho }_A)=0$; positivity implies $(Q_A\otimes I){\rho }_{AB}=0$. Applying the same argument on $B$ proves the claimed support inclusion.

On that support, product eigenvectors showThe definition of the partial trace therefore givesConsequently $\boxed{S(A:B)=S({\rho }_{AB}\Vert {\rho }_A\otimes {\rho }_B)}$.

For completeness, nonnegativity of quantum relative entropy also follows from the classical proof above. Diagonalize $\rho$ in a basis $|j⟩$, with eigenvalues $r_j$, and put $q_j=⟨j|\sigma |j⟩$. These are two classical probability distributions. Scalar concavity of the logarithm applied to the spectral weights of $\sigma$ gives $⟨j|{\log }_2\sigma |j⟩\le {\log }_2{q}_j$ whenever $r_j>0$ and support inclusion holds. HenceThe support-failure case is again infinite. Applying this to the product of the marginals proves $\boxed{S(A:B)\ge 0}$ without requiring an additional entropy inequality.

Take the vectors in the quantum state ensemble to be normalized, with $p_i\ge 0$ and ${\sum }_i{p}_i=1$. The corresponding density matrix isEach rank-one projector is a self-adjoint operator, so ${\rho }^{†}=\rho$. For every $|v⟩$ in the Hilbert space,proving that $\rho$ is positive semidefinite. The trace is $\mathrm{Tr}\rho ={\sum }_i{p}_i⟨{\psi }_i|{\psi }_i⟩=1$.

Conversely, diagonalize a self-adjoint operator $\rho$ in an orthonormal eigenbasis, writing $\rho ={\sum }_j{\lambda }_j|e_j⟩⟨e_j|$. Its positive semidefiniteness implies ${\lambda }_j=⟨e_j|\rho |e_j⟩\ge 0$, and its unit trace gives ${\sum }_j{\lambda }_j=1$. Thus $\{{\lambda }_j,|e_j⟩\}$ is a quantum state ensemble with precisely this density matrix. Terms with zero eigenvalue may be omitted. This argument includes rank-one pure states; an ensemble representation need not describe a genuinely mixed state.

The stated condition is the definition of an extreme point in the convex set of density matrices. Suppose first that $\rho =|\psi ⟩⟨\psi |$ with $\Vert \psi \Vert =1$, and consider any convex combination $\rho ={\sum }_i{a}_i{\rho }_i$ with all $a_i>0$. For every $|w⟩$ orthogonal to $|\psi ⟩$,Every term is nonnegative, so $⟨w|{\rho }_i|w⟩=0$ for each $i$. To see that this implies ${\rho }_{i}w=0$, expand ${\rho }_i={\sum }_j{\mu }_j|u_j⟩⟨u_j|$, ${\mu }_j\ge 0$. The equality ${\sum }_j{\mu }_j|⟨u_j|w⟩{|}^2=0$ implies ${\mu }_j⟨u_j|w⟩=0$ for every $j$, which kills ${\rho }_{i}w$. Thus each density matrix ${\rho }_i$ vanishes on ${\psi }^{\perp }$. By self-adjointness its range lies in the line $\mathbb{C}\psi$, so ${\rho }_i=c_i|\psi ⟩⟨\psi |$. Its unit trace forces $c_i=1$, establishing the required purity.

Conversely, suppose $\rho$ is not rank one. In its spectral decomposition, at least two eigenvalues are positive. Select a normalized eigenvector $|e_1⟩$ with positive eigenvalue ${\lambda }_1$; then $0<{\lambda }_1<1$. DefineThe remaining eigenvalues show that $\sigma$ is a density matrix. The decomposition $\rho ={\lambda }_1|e_1⟩⟨e_1|+(1-{\lambda }_1)\sigma$ is a convex combination of two different density matrices, neither equal to $\rho$. This violates the given purity condition. Hence the extreme points of the density-operator state space obey

Let $R=|\psi ⟩⟨\psi |$ and choose an orthonormal basis $\{|e_{\mu }⟩\}$ of ${\mathcal{H}}_2$. The reduced density matrix on $S_1$ is the partial traceHere $I_1$ is the identity on ${\mathcal{H}}_1$, and the bras and kets in the sum contract only the second tensor factor. Equivalently, if $|\psi ⟩={\sum }_{a,\mu }{c}_{a\mu }|f_a⟩\otimes |e_{\mu }⟩$, then $({\rho }_1{)}_{ab}={\sum }_{\mu }{c}_{a\mu }{c}_{b\mu }^{\ast }$. The definition is independent of the chosen basis: it is characterized by $\mathrm{Tr}(M{\rho }_1)=\mathrm{Tr}[(M\otimes I_2)R]$ for every operator $M$ on ${\mathcal{H}}_1$.

The generalized measurement postulate takes $A_i$ to be operators on ${\mathcal{H}}_2$ satisfying ${\sum }_i{A}_i^{†}{A}_i=I_2$. The Born rule and the conditional state update giveIf the outcome is not supplied to $S_1$, the appropriate post-measurement density matrix is $R^{\prime }={\sum }_i(I_1\otimes A_i)R(I_1\otimes A_i^{†})$. For any local operator $M$, cyclicity of the full trace givesSince this holds for every $M$, $\boxed{{\mathrm{Tr}}_2{R}^{\prime }={\rho }_1}$. This is the no-communication theorem.

The invariance concerns the nonselective measurement in quantum mechanics. A conditional reduced density matrix ${\rho }_{1|i}={\mathrm{Tr}}_2|{\psi }_i⟩⟨{\psi }_i|$ can change, although ${\sum }_i{p}_i{\rho }_{1|i}={\rho }_1$. For example, measuring one half of a Bell pair in the computational basis changes the other half conditionally to $|0⟩$ or $|1⟩$; before learning the outcome, its density matrix remains $I/2$. Thus distant measurement in quantum measurements cannot transmit a controllable signal without classical communication. This quantum no-signalling is compatible with special relativity, even though some entangled states have correlations incompatible with a local hidden-variable theory.

A secure key can be produced, at one ideal secret bit per copy. Denote Alice's additional flag qubit by $F$. Alice can measure $F$ in the computational basis and, when its outcome is $1$, apply the Pauli Z gate to her half of the shared pair. The two flag outcomes each have probability $1/2$, and $Z_A|{\Phi }^{-}⟩=|{\Phi }^{+}⟩$; either branch therefore supplies the same pure Bell pair.

Equivalently, Alice applies the local controlled unitaryThis local flag correction of a Bell-state phase mixture changes the full density matrix intoThe corrected pair is pure, so every purification factors between $AB$ and $FE$. Eve may know the flag without knowing the key bit obtained from $AB$. Alice need not disclose the flag, but even disclosing it does not compromise the corrected pure Bell pair. Tracing out the flag before correction would give the nonprivate mixture in the preceding case; retaining Alice's local side information is essential.

Impossible to implement as written: Bob is never given the basis string. Step 7 requires him to know which incoming qubits were acted on by a Hadamard gate, but the announcement contains only the check positions. Since $H^{-1}=H$, knowing that a transformation is its own inverse does not identify which qubits need it.

Bob cannot recover the missing information from the qubits alone. For each encoded Bell pair, his reduced density matrix isfor either $b=0$ or $b=1$. His entire received register has the same density matrix $I/2^{2n}$ for every string $b$, so every local measurement in quantum mechanics has statistics independent of $b$. Nor can one fixed unitary undo both possibilities: preserving $|{\Phi }^{+}⟩$ requires that unitary to be proportional to $I$, whereas decoding $(I\otimes H)|{\Phi }^{+}⟩$ requires it to be proportional to $H$. Supplying the missing basis announcement after receipt would repair the protocol and make it the third one, but that communication is absent from the printed sequence.

For a photon in a fixed spatial mode, the two transverse photon polarizations span a qubit space. A pure photon polarization state is $\alpha |H⟩+\beta |V⟩$ with $|\alpha {|}^2+|\beta {|}^2=1$, and a general state is a two-dimensional density matrix. Its Bloch vector gives the normalized Stokes parameters. Polarization-dependent path separation can correlate this internal degree of freedom with a path qubit.

// Target: quantum-error-correction.bigb

A 2018 photonic experiment makes the distinction between populations and quantum coherence controllable. A 400-nm laser pumped a crystal to generate paired photons, filtered near 800 nm. One photon triggered collection; the other entered the apparatus. A half-wave plate and polarizing beam splitter selected horizontal photon polarization; another half-wave plate prepared balanced $H,V$ quantum superpositions. Separated polarization paths and diffraction gratings let a spatial light modulator control their frequency-dependent quantum phases and spectral weights. Recombined paths passed through quartz plates whose adjustable thickness controlled polarization–frequency coupling. Quarter- and half-wave plates plus a polarizing beam splitter selected tomography bases; jointly registered photodetector events supplied the readout. Quantum state tomography inferred polarization density matrices at each thickness. The dephasing model preserves diagonal populations while changing off-diagonal quantum coherence. Here the unobserved photon frequency acts as the environment for the polarization qubit. Finite modulator resolution and imperfect spectral shaping limited agreement with the intended dynamics; this experiment does not establish exact $I/2$ preparation or the ideal probabilities above. ^[ref]

Write each pure density matrix as $(I+r\cdot \sigma )/2$. The trace identities of the Pauli matrices, $\mathrm{Tr}{\sigma }_i=0$ and $\mathrm{Tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$, make the trace of their product equal to the displayed overlap. Antipodal unit vectors represent orthogonal states; vectors at right angles correspond to squared overlap one half. For mixed states the same trace formula gives Hilbert–Schmidt overlap, not in general quantum fidelity.

A quantum state describes a preparation through its predicted measurement in quantum mechanics probabilities. In the Hilbert space formulation, a pure state is a quantum state ray: a normalized vector representative $|\psi ⟩$ satisfies $⟨\psi |\psi ⟩=1$, and representatives differing by a global phase describe the same quantum state. ^[ref]

Both pure states and mixed states are represented by a density operator on the chosen complex Hilbert space:Thus $\rho$ is Hermitian, positive semidefinite and of unit trace; in infinite dimension it must also be a trace-class operator. For a normalized pure state, $\rho =|\psi ⟩⟨\psi |$ is a rank-one orthogonal projection. A mixed state has no single such vector representative. Its density matrix is the matrix of $\rho$ in a chosen orthonormal basis. ^[ref] For a discrete POVM with effects $E_a\ge 0$ and ${\sum }_a{E}_a=I$, the Born rule readsThese constraints and Born rule probabilities describe arbitrary finite-dimensional preparations. ^[ref]

Randomly choosing normalized pure states with probabilities $w_j$ and forgetting the label givesThis quantum state ensemble averages probabilities; a quantum superposition adds probability amplitudes. Several ensembles can give the same density operator. Quantum entanglement can make a reduced density matrix mixed while the joint quantum state is pure. ^[ref]

The preparation, the choice of basis, and the measurement are separate ingredients. A wavefunction $\psi (x)=⟨x|\psi ⟩$ is the position representation of a pure state; a general mixed state requires a density operator. For a closed quantum system, a unitary operator evolves it as $\rho \mapsto U\rho U^{†}$, preserving its eigenvalues and purity of a density operator. ^[ref] The Bloch sphere and Bloch ball give the qubit geometry; quantum state tomography connects repeated preparations to an estimated density matrix; projective measurement treats the conditional state after an outcome.