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For a quantum state ensemble of pure vectors and an orthogonal projection $P$, measure $\{P,I-P\}$, preserve the successful subspace, and decode failure to a fixed normalized vector $\phi$. The resulting quantum channel is $\Lambda (X)=PXP+\mathrm{Tr}[(I-P)X]|\phi ⟩⟨\phi |$. For ${\alpha }_k=\Vert P{\psi }_k\Vert$, its squared quantum fidelity on the $k$th source vector is ${\alpha }_k^4+(1-{\alpha }_k^2)|⟨{\psi }_k|\phi ⟩{|}^2$. The inequality $a^2\ge 2a-1$ on $[0,1]$ proves the displayed average bound. It depends only on the source density operator's weight in $P$ and does not require the emitted vectors to be orthogonal.

Choose the qubit computational basis $|0⟩,|1⟩$, also called the $Z$ basis, and the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, also called the $X$ basis. Compare a coherent preparation of $|+⟩$ with choosing $|0⟩$ or $|1⟩$ at random with equal probabilities and discarding that label:These are original ideal preparation models. Their $Z$-basis probabilities are both $(1/2,1/2)$, because their quantum state populations agree. Their $X$-basis probabilities are respectively $(1,0)$ and $(1/2,1/2)$. Indeed $⟨+|{\rho }_{+}|+⟩=1$, whereas $⟨+|I/2|+⟩=1/2$. The missing off-diagonal quantum coherence therefore changes a prediction that the common populations cannot determine.

For a relative quantum phase $\varphi$, the coherent family instead hasThe last formula follows by adding or subtracting the two probability amplitudes before taking the squared modulus. At $\varphi =\pi$, the state is $|-⟩$ and the $X$ outcomes switch to $(0,1)$, while the $Z$ outcomes remain $(1/2,1/2)$. In contrast, $-|+⟩$ changes both components by the same global phase and retains the $X$ outcomes $(1,0)$. In photon polarization, the identification $|0⟩=|H⟩$, $|1⟩=|V⟩$ makes $|+⟩$ and $|-⟩$ the two orthogonal diagonal linear polarizations; these labels describe measurement bases, not spatial axes for photon spin.

Even perfect statistics in one basis cannot certify a preparation. The above $Z$ data fit both density matrices. Conversely, distinct preparation histories can yield the same quantum state:The off-diagonal terms cancel in the right-hand ensemble. Every measurement on the qubit alone then has identical probabilities for those two histories. A retained classical label would allow conditioning on different members of the quantum state ensemble; it is extra information beyond the unlabelled density operator.

For a Bell pair $|{\varphi }_{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, Alice's basic measurement in the entrywise complex conjugate of any desired Bob orthonormal basis prepares each Bob pure state with probability $1/2$. Indeed, if $|u⟩={\sum }_{j=0}^1{u}_j|j⟩$, then $⟨\overline{u}|j⟩=u_j$, which proves the displayed contraction identity. The undisclosed-outcome density matrix stays $I/2$; quantum steering changes the conditional quantum state ensemble without violating the no-communication theorem.

A finite-dimensional density operator $\rho$ represents a quantum state on a Hilbert space. It is Hermitian, positive semidefinite, and has trace one:The density matrix is its matrix in a chosen orthonormal basis. In dimension $N$ it has $N^2-1$ real parameters: a Hermitian matrix has $N^2$, and fixing its trace removes one; positivity restricts the allowed values.

A quantum state ensemble of normalized pure states $|{\psi }_i⟩$ with positive probabilities $p_i$ givesZero-weight terms can be omitted. Conversely, the spectral theorem gives such an ensemble for every density operator, using its orthonormal eigenvectors and nonnegative eigenvalues, with zero eigenvalues omitted. A pure state has $\rho =|\psi ⟩⟨\psi |$; a mixed state needs at least two nonzero eigenvalues. ^[ref]

For a projective measurement with orthogonal projections ${\Pi }_a$, the Born rule isThis is the average of the component pure state probabilities. Positivity makes each outcome probability nonnegative, and ${\sum }_a{\Pi }_a=I$ makes them sum to one. The same density operator therefore predicts every measurement on the system without requiring a unique preparation history. ^[ref]

For a closed system, Schrödinger picture evolution under a unitary operator iswhen ${\rho }_H={\rho }_S(0)$ denotes the time-independent Heisenberg picture state and $U(0)=I$. In the Schrödinger picture the density operator evolves; in the Heisenberg picture the observables evolve instead, with identical measurement predictions. This unitary operator preserves the eigenvalues and hence the purity of a density operator. ^[ref]

A historical source for the probabilistic formulation of quantum mechanics is von Neumann's 1927 paper Wahrscheinlichkeitstheoretischer Aufbau der Quantenmechanik. ^[ref]

If the two honest quantum state ensembles have the same density operator, their independently prepared blocks are perfectly hiding. A dishonest sender can instead send one half of a purification of a density operator and retain the other. The Hughston–Jozsa–Wootters theorem allows a measurement of the retained system to prepare either honest ensemble after commitment. If verification checks only the declared states, either opening is then accepted with the honest probabilities. This gives a concrete failure of binding while preserving perfect hiding.

The expectation value of a quantum observable is the expected value of its measurement outcome random variable in a specified quantum state. It predicts the mean of outcomes from many identical quantum state preparations. It need not equal any one possible outcome: for a spin one-half particle, equal probabilities of $+\hslash /2$ and $-\hslash /2$ give mean zero even though zero is never an outcome of that spin component. Here $\hslash$ is the reduced Planck constant.

Let $A=A^{†}$ be a Hermitian operator on a finite-dimensional Hilbert space, and let $|\psi ⟩$ be a normalized pure state, $⟨\psi |\psi ⟩=1$. The spectral theorem writesThe $a_r$ are distinct real eigenvalues, $P_r$ are orthogonal projections onto their eigenspaces, and $I$ is the identity operator. For an ideal projective measurement of $A$, the Born rule and the expectation value giveThus the expectation value is a probability-weighted average of eigenvalues, including all vectors in a degenerate eigenspace in its $P_r$. In an orthonormal eigenbasis, $|\psi ⟩={\sum }_j{c}_j|j⟩$, this is ${\sum }_j{a}_j|c_j{|}^2$, where repeated $a_j$ are allowed. The weighted average lies between the smallest and largest eigenvalues; it can lie in a gap between them. The statistical meaning and operator expression are connected in this way in the authored treatment of average energies.^[ref]

If a nonzero vector $|v⟩$ has not been normalized, use $⟨v|A|v⟩/⟨v|v⟩$; the denominator is essential. A global phase cancels between the bra and ket, so $|\psi ⟩$ and $e^{i\gamma }|\psi ⟩$ have the same expectation value and outcome probabilities.

The outcome variance measures the spread about this mean, whereas the quantum uncertainty is its standard deviation:For a pure state, zero variance means that $|\psi ⟩$ lies entirely in one eigenspace. That eigenspace need not be one-dimensional. Knowing only a mean generally does not determine the entire outcome probability distribution.

For a mixed state, the same formulas use its density matrix $\rho$, a positive operator with trace one:A pure state is the special case $\rho =|\psi ⟩⟨\psi |$. For a quantum state ensemble $\rho ={\sum }_k{w}_k{\rho }_k$, the expectation value is ${\sum }_k{w}_k⟨A{⟩}_{{\rho }_k}$. This linearity is useful when preparations vary between trials; different quantum state ensembles can have the same density matrix.

These formulas require no operator-domain qualifications in finite dimension. For an unbounded self-adjoint operator, the spectral theorem instead gives a spectral probability measure; a finite mean requires an integrable absolute first moment, and finite variance requires a finite second moment. Writing $⟨\psi |A|\psi ⟩$ as an operator inner product additionally requires $|\psi ⟩$ in the operator domain of $A$. The quadratic form of a positive quantum Hamiltonian explains the useful wider form-domain version.

For the nonzero spectral decomposition $\rho ={\sum }_{a=1}^r{\lambda }_a|e_a⟩⟨e_a|$, every pure-state quantum state ensemble of $\rho$ is specified by the displayed column linear isometry. To prove this, write $w_j=\sqrt{p_j}{\varphi }_j$. Positivity forces each $w_j$ into the support of $\rho$, and the coefficients $U_{ja}=⟨e_a|w_j⟩/\sqrt{{\lambda }_a}$ have orthonormal columns because ${\sum }_j|w_j⟩⟨w_j|=\rho$. Conversely this column identity reconstructs $\rho$; the squared vector norms are the probabilities and sum to one. Complete the columns to unitary bases after adding zero ensemble vectors to obtain the usual unitary-mixing formulation.

A pure-state source emits independent states from a fixed quantum state ensemble. Its average block density operator is the tensor power of the one-use average. Unassisted reliable quantum source compression preserves the emitted quantum states, rather than only their classical labels. Schumacher compression gives the optimum asymptotic qubit rate $S(\rho )$ for such a source, with every larger rate achievable. Nonorthogonal equally likely pure states need not have entropy one: their classical labels have entropy one but their average density operator can have lower Von Neumann entropy.

A mixed state is a density operator that is not a rank-one orthogonal projection. In finite dimension it is characterized by $\mathrm{Tr}({\rho }^2)<1$, so its purity of a density operator distinguishes it from a pure state.

Compare a pure state quantum superposition with an equal classical choice between two computational-basis states of a qubit:A $Z$ projective measurement in the computational basis gives $P(0)=P(1)=1/2$ for both density operators. An $X$ projective measurement in the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ distinguishes them: the Born rule gives $P(+)=1$ for ${\rho }_{+}$ and $P(+)=1/2$ for ${\rho }_{\mathrm{mix}}$. These comparisons use separate copies of each preparation; measuring $Z$ first changes the state tested by a subsequent $X$ measurement.

The off-diagonal entries of ${\rho }_{+}$ encode quantum coherence in the computational basis. Such quantum coherence depends on the basis: ${\rho }_{+}$ is diagonal in the Hadamard basis but remains a pure state. Mixedness is independent of the basis, because a unitary change of basis preserves the eigenvalues and the purity of a density operator.

Even the quantum state ensemble is not unique:The off-diagonal entries of the two Hadamard basis projectors cancel. Thus the density operator retains precisely what measurements on the system can reveal, while an accessible preparation label carries additional information. A mixed state can also be the reduced density matrix of an entangled state, even when the whole system is in a pure state; classical ignorance about a selected component is not its only physical origin. ^[ref]

The state is separable. The off-diagonal terms of the two Bell state projectors have opposite signs and cancel:This is a quantum state ensemble of two pure states that are product states, with probability $1/2$ each, establishing a separable quantum state. Its correlations are classical; an ensemble decomposition into entangled states does not itself establish quantum entanglement of the density matrix.

The state is separable. The identity matrix resolves into the four projectors of the computational basis, giving the explicit quantum state ensembleEach member is a pure state that is a product state, so this is a separable quantum state by its defining decomposition.

Write the proposed quantum state ensemble as $\{(|u⟩,p),(|v⟩,1-p)\}$ with normalized pure states, so thatTaking the expectation values in $|u⟩$ and $|v⟩$, and setting $c=|⟨u|v⟩{|}^2$, givesConsequently $p\le 1/2$ and $p\ge 1/2$. Substitution then gives $c=0$. The two pure states are therefore orthogonal, with equal probabilities:Conversely, any orthonormal basis of a qubit has projectors summing to $I$, so such a quantum state ensemble does give $I/2$. This proves the characterization of two-state ensembles of a maximally mixed qubit.

For any normalized vector $|u⟩$, the Born rule assigns the probability $⟨u|\rho |u⟩$ to the corresponding rank-one orthogonal projection. This must be nonnegative for every $u$, which is exactly positive semidefiniteness. For any complete orthonormal basis $\{|j⟩\}$, the outcome probabilities must sum to one:Thus a Hermitian matrix represents a physical density matrix only ifThese conditions are also sufficient in finite dimension. The spectral theorem writes $\rho ={\sum }_j{p}_j|u_j⟩⟨u_j|$ with $p_j\ge 0$ and ${\sum }_j{p}_j=1$, describing a quantum state ensemble of pure states with valid mixing probabilities. A Hermitian matrix alone could have negative eigenvalues, hence negative measurement probabilities.

Compute the honest one-qubit density operators in the computational basis. Writing $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|u⟩=(5|0⟩+|1⟩)/\sqrt{26}$, the two quantum state ensembles giveThe equality includes the off-diagonal coherence; comparing only the basis outcome probabilities would not suffice.

The protocol is perfectly hiding against Bob when Alice is honest. Independence of the honest preparations makes his full received density operator ${\rho }_0^{\otimes N}={\rho }_1^{\otimes N}$. For any POVM $\{E_z\}$, including collective measurements and measurements using ancillas, the Born rule gives the same outcome probabilities $\mathrm{Tr}(E_z{\rho }_0^{\otimes N})=\mathrm{Tr}(E_z{\rho }_1^{\otimes N})$. Thus no measurement supplies information about the committed bit.

It is not binding against Alice. Here is an explicit ensemble-steering attack on quantum bit commitment. For each position, Alice prepares the normalized two-qubit pure stateretains $A$ and sends $B$. Its partial trace on $A$ is the honest common density operator. She sends the $B$ halves of $N$ independent copies while retaining all the $A$ halves, and chooses her bit only at unveiling.

To open one, she makes a projective measurement of each retained qubit in the computational basis. Outcome zero has probability $13/20$ and prepares Bob's state $|u⟩$; outcome one has probability $7/20$ and prepares $|1⟩$. She announces that list. It has exactly the honest bit-one distribution, and every declared-state test passes.

To open zero, she instead measures each retained qubit using a three-outcome POVM with effects $E_j=|v_j⟩⟨v_j|$, whereAll effects are positive. Their diagonal sums are $4/5+1/5=1$ and $4/70+16/70+5/7=1$, and the off-diagonal sum is zero, so $E_0+E_{+}+E_1=I$. This establishes that the proposed POVM is physically valid.

To verify the conditional states directly, the coefficient matrix of $|\Omega ⟩$, with Bob's index as the row and Alice's as the column, isA real rank-one effect $|v_j⟩⟨v_j|$ on Alice's half gives the unnormalized Bob state vector $C{v}_j$. For the three effects above,Their squared norms are therefore $1/2,1/4,1/4$, exactly the honest bit-zero probabilities, and Bob's conditional pure states are the corresponding declared states. Independent measurements on the retained halves produce the honest distribution of the entire list, so even checks on the declared frequencies do not defeat the attack. Every declared-state projector test again succeeds.

Alice can consequently unveil either bit with acceptance probability one, without having chosen it when sending the qubits. This is the Hughston–Jozsa–Wootters theorem mechanism implemented by an explicit measurement, rather than merely an appeal to the general impossibility of perfectly hiding and binding quantum bit commitment.

The Schrödinger–Jaynes–Hughston–Jozsa–Wootters theorem classifies all finite quantum state ensembles of a density matrix. Write its nonzero spectral decomposition as $\rho ={\sum }_{a=1}^r{\lambda }_a|e_a⟩⟨e_a|$, with ${\lambda }_a>0$ and orthonormal eigenvectors. An ensemble with subnormalized vectors $|w_j⟩=\sqrt{p_j}|{\varphi }_j⟩$, $j=1,\dots ,m$, represents $\rho$ exactly whenHere $U$ is an $m\times r$ column linear isometry. Zero-weight vectors may be included or discarded. Equivalently, any two ensembles represent the same density matrix if and only if, after padding the shorter list with zero vectors, their subnormalized vectors are related by a unitary matrix acting on the ensemble index. This is the isometry parametrization of a density-matrix ensemble.

For necessity, if $|v⟩$ lies in the kernel of $\rho$, thenEvery $w_j$ therefore lies in the support of $\rho$. Define $U_{ja}=⟨e_a|w_j⟩/\sqrt{{\lambda }_a}$. The ensemble identity givesso these columns are orthonormal and the displayed parametrization follows.

Conversely, start with any such isometry and define $w_j$ by the boxed equation. ThenThe weights are $p_j=\Vert w_j{\Vert }^2={\sum }_a{\lambda }_a|U_{ja}{|}^2\ge 0$ and sum to $\mathrm{Tr}\rho =1$. Normalize every nonzero $w_j$ to obtain its pure state. This proves both directions and includes degenerate eigenvalues without any preferred choice of basis in their eigenspaces.

To prove the unitary-mixing version, pad two ensembles to a common length $L$. Their matrices $U$ and $V$ have $r$ orthonormal columns in ${\mathbb{C}}^L$. Extend each column family to an orthonormal basis, obtaining square unitaries $\tilde{U}$ and $\tilde{V}$. Then $W=\tilde{V}{\tilde{U}}^{†}$ is unitary and $V=WU$, so the second ensemble vectors satisfy $v_k={\sum }_j{W}_{kj}{w}_j$. Conversely this unitary mixing preserves ${\sum }_j|w_j⟩⟨w_j|$, because ${\sum }_k{W}_{ki}{W}_{kj}^{\ast }={\delta }_{ij}$. Independent phase choices for ensemble kets are absorbed in the row phases of these matrices.

The classification also has a measurement interpretation. Purify $\rho$ as $|\Psi ⟩={\sum }_a\sqrt{{\lambda }_a}|e_a⟩|a⟩$. On the ancillary system define $|{\eta }_j⟩={\sum }_a{U}_{ja}^{\ast }|a⟩$. The effects $E_j=|{\eta }_j⟩⟨{\eta }_j|$ sum to $I_r$, hence form a POVM. The conditional unnormalized system ket is $(I\otimes ⟨{\eta }_j|)|\Psi ⟩=w_j$, with probability $p_j$. Thus every ensemble can be realized by measuring a purifying system; with a sufficiently enlarged ancilla the measurement can be projective. The density matrix specifies the common statistics without selecting one particular ensemble decomposition.

For the quantum state ensemble $\mathcal{E}=\{p_i,{\rho }_i\}$ with average $\overline{\rho }={\sum }_i{p}_i{\rho }_i$, the Holevo quantity isThe second equality follows by expanding the trace definition: ${\sum }_i{p}_i\mathrm{Tr}{\rho }_i\log \overline{\rho }=\mathrm{Tr}\overline{\rho }\log \overline{\rho }$. All positive-weight states are supported within $\overline{\rho }$, so the expression is finite.

For a CPTP map $\Lambda$, the output average is $\Lambda (\overline{\rho })$, with unchanged probabilities $p_i$. The data-processing inequality for quantum relative entropy givesThus the Holevo quantity cannot increase under a quantum channel.

The underlying monotonicity can also be seen directly from Strong subadditivity of Von Neumann entropy. Attach the classical label $X$, and realize $\Lambda$ by a Stinespring representation of a completely positive map with output $B$ and discarded environment $E$. The isometry preserves $I(X:BE)$, whose input value is $\chi (\mathcal{E})$. The difference from the output value iswhich is precisely strong subadditivity. This explains why discarding an environment loses accessible ensemble information.

Here “operation” means the unconditional channel, including the classical outcome when a measurement is retained. Conditioning on a selected outcome can increase the Holevo quantity. For instance, take equally likely vectors $\sqrt{a}|0⟩\pm \sqrt{1-a}|1⟩$, with $1/2<a<1$. Their input Holevo quantity is $h_2(a)<1$. The filter $K=\mathrm{diag}(\sqrt{(1-a)/a},1)$ succeeds on either state with probability $2(1-a)$ and produces the orthogonal states $|+⟩,|-⟩$ after normalization. The success-conditioned quantity is $1$. This does not violate the unconditional statement.

The construction is a convex roof extension. First establish its convexity, without assuming that the minimizing quantum state ensemble exists. If $\tau ={\sum }_j{w}_j{\tau }_j$, choose a pure state ensemble for each ${\tau }_j$ whose average value of $\mu$ is within $\epsilon$ of $F({\tau }_j)$. Combining these ensembles, with their weights multiplied by $w_j$, gives a pure state ensemble for $\tau$. Taking arbitrarily small $\epsilon >0$ provesAlso, $F(\pi )=\mu (\pi )$ for a pure state $\pi$. Indeed, if $\pi ={\sum }_j{p}_j{\pi }_j$ and $v$ is orthogonal to its one-dimensional support, then $0={\sum }_j{p}_j⟨v|{\pi }_j|v⟩$. All terms are nonnegative, so every positive-weight ${\pi }_j$ has that same support and equals $\pi$.

Represent outcome $k$ of the operation by a linear unnormalized map ${\mathcal{I}}_k$, as for a quantum instrument. Normalization is performed only after applying this map: ${\mathcal{I}}_k(\rho )=P_k{\rho }_k$. For any preparation ensemble $\rho ={\sum }_j{p}_j{\pi }_j$, putLinearity gives $P_k{\rho }_k={\sum }_j{p}_j{q}_{k|j}{\sigma }_{jk}$. For $P_k>0$, Bayes' theorem identifies the conditional preparation weightsOutcomes with zero probability contribute nothing; normalized states on such branches need not be defined. The conditional states ${\sigma }_{jk}$ may be mixed, so convexity of the convex roof is essential here.

Applying that convexity separately for each recorded outcome, and then the assumed average monotonicity for each pure preparation, yieldsThe left side depends only on the initial density operator and the operation, not on the chosen preparation ensemble. Taking the infimum of the last expression over all its pure state ensembles provesEquivalently, choose an ensemble within $\epsilon$ of the infimum and then let $\epsilon \downarrow 0$. This is the monotonicity of a convex roof under a quantum instrument. It uses linearity of the unnormalized branches, not a false assumption that normalized conditional states depend linearly on the input. As usual, the averages are assumed well defined; finite-valued roofs and finitely many outcomes suffice. If an input roof is $-\infty$, taking ensembles with averages tending to $-\infty$ gives the same conclusion in the extended-real interpretation whenever the output average is defined.

The printed normalization has its factors reversed. For the displayed map $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$, the condition ${\sum }_k{A}_k{A}_k^{†}=I$ says $\Phi (I)=I$, which is unitality. Trace preservation instead requiresThese are distinct trace-preserving and unital Kraus conditions. To disprove the printed equivalence, take $0<\gamma <1$ andThey obey ${\sum }_j{B}_j{B}_j^{†}=I$, butThus their completely positive map is not trace preserving. They are the adjoints of the operators of an amplitude damping channel. That channel is trace preserving but sends $I$ to $\mathrm{diag}(1+\gamma ,1-\gamma )$, so it cannot satisfy the printed normalization in any Kraus representation either. The corrected characterization is proved next.

Work in finite-dimensional input and output Hilbert spaces, with a linear map $\Phi$ on operators. Suppose first that $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$ and ${\sum }_k{A}_k^{†}{A}_k=I$. For every auxiliary dimension $m$ and every positive operator $W$ on the extended space,Each summand is positive because conjugation preserves positivity. This is exactly the definition of a completely positive map. Cyclicity of the trace givesso the map is also trace preserving.

Conversely, suppose $\Phi$ is completely positive and trace preserving, and let $|i⟩$ be an orthonormal input basis. Use the unnormalized entangled vector $|\Omega ⟩={\sum }_i|i⟩\otimes |i⟩$, with the reference factor first. Complete positivity makes the corresponding unnormalized Choi matrix positive:Its spectral decomposition can be written $J_{\Phi }={\sum }_k|v_k⟩⟨v_k|$, absorbing the nonnegative eigenvalues into the vectors. Define linear operators $A_k$ uniquely byExpanding the outer products and comparing their $(i,j)$ reference blocks yieldsThe matrix units span all operators, so linearity proves the Kraus representation $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$ for every $X$. Finally, trace preservation givesfor all $X$. Taking matrix units as test operators shows that the matrix in parentheses is zero. This proves both directions of the corrected CPT characterization, including the normalization rather than assuming it.

For an input quantum state ensemble $\{p_x,{\rho }_x\}$, write ${\sigma }_x=\Phi ({\rho }_x)$ and $\bar{\sigma }={\sum }_x{p}_x{\sigma }_x$. Its output Holevo quantity isThe Holevo capacity is the optimized one-use quantitywhere the supremum ranges over finite input ensembles and $S$ is the Von Neumann entropy in bits.

For $n$ uses of a memoryless quantum channel, the map is ${\Phi }^{\otimes n}$. An $n$-use code consists of a finite message set ${\mathcal{M}}_n$ of size $M_n$, an encoding state ${\rho }_m^{(n)}$ on the $n$ inputs for each message, and a decoding POVM $\{D_m^{(n)}:m\in {\mathcal{M}}_n\}$ on the output space, with $D_m^{(n)}\ge 0$ and ${\sum }_m{D}_m^{(n)}=I$. The encoding states may be entangled across uses. The measurement outcome is the receiver's decoded message, with conditional probabilityFor uniformly distributed messages the average error and rate areA rate $R$ is achievable if there is a sequence of such codes with $P_e^{(n)}\to 0$ and ${\mathrm{liminf}}_n{R}_n\ge R$. The unassisted classical capacity of a quantum channel is the supremum of achievable rates; the sender and receiver do not share prior entanglement.

The operational content of the Holevo-Schumacher-Westmoreland theorem is that ${\chi }^{\ast }(\Phi )$ is the capacity for product-state input encodings with collective output measurements. Allowing arbitrary entangled block inputs gives the regularized formulaThe limit equals the supremum because superadditivity of Holevo capacity follows by taking product ensembles for two blocks and using additivity of the Von Neumann entropy on product states. The coding theorem makes every rate below the one-block Holevo quantity achievable through many independently encoded blocks, with collective decoding.

The converse can be seen explicitly from the Holevo bound and Fano's inequality. For a uniform transmitted message $M$ and decoded message $\hat{M}$,The first inequality is obtained by subtracting the Fano bound on $H(M\mid \hat{M})$ from $H(M)={\log }_2{M}_n$. Dividing by $n$ and letting the average error vanish bounds every achievable rate by the regularized expression. The block-coding achievability just described supplies the reverse bound.

Under the assumed additivity of Holevo capacity across tensor powers, ${\chi }^{\ast }({\Phi }^{\otimes n})=n{\chi }^{\ast }(\Phi )$ for every $n$. ThusThe stated conditional assertion is true: under this additivity assumption, entangled input states cannot increase the asymptotic unassisted classical rate beyond the rate already achievable with product inputs. The decoder may still use collective measurements across channel outputs.

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.

If the preparation label $i$ is sampled with probability $p_i$ but is not supplied to the experimenter, the quantum state ensemble is represented byFor normalized states, this density operator is Hermitian, positive semidefinite and has trace one. The states in an ensemble need not be orthogonal or distinct. For any measurement in quantum mechanics effect $E$, averaging the Born rule over the preparation label givesThus the density operator summarizes all measurements performed on the prepared system alone. For an ensemble of mixed states ${\rho }_i$, the same reasoning gives $\rho ={\sum }_i{p}_i{\rho }_i$.