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John Preskill's quantum computation notes develop the distinction between a quantum state ray, a normalized wave function, and a density operator. This makes a practical difference for a subsystem of an entangled state: its predictions are described by a reduced density operator obtained with the partial trace, even when the joint quantum state is pure. His discussion also shows why different statistical preparations can give the same density operator and therefore identical predictions for measurements confined to that subsystem.^[ref]

A theoretical physicist working on quantum information theory.^[ref]

Homepage: www.preskill.caltech.edu/

Write ${\rho }_0=|\varphi ⟩⟨\varphi |=(I+r_{x}X+r_{y}Y+r_{z}Z)/2$. Conjugating by a Pauli matrix keeps its own Bloch component and reverses the other two; for example,In the average of the four conjugations, the coefficient of every nonidentity Pauli matrix is zero. ConsequentlyThe Pauli operator basis gives the Bloch representation of a qubit density operator.^[ref]

This is the qubit Pauli twirl, the dimension-two case of the Heisenberg-Weyl twirling channel. The argument uses only trace one, so it also works for mixed qubit inputs. The preparation label must be forgotten: conditioning on the label would retain a known pure branch rather than the maximally mixed state obtained by the ensemble average.

No test on Alice and Bob's systems gives any information about which global state was prepared, even with arbitrarily many independent copies. In the pure GHZ density matrix, the off-diagonal terms contain $|0⟩⟨1{|}_C$ or its adjoint. Their partial traces vanish. Hence both preparations have the same reduced density matrixFor any POVM effect $E$ acting jointly on $AB$, the outcome probability is $\mathrm{Tr}(E{\sigma }_{AB})$ in either case. Equal reduced states give identical local statistics, where local here means local to the entire accessible laboratory $AB$, not separate operations on $A$ and $B$.^[ref]

For $n$ independent copies, taking the partial trace of each inaccessible $C$ system yields ${\sigma }_{AB}^{\otimes n}$ under either hypothesis. Even an arbitrary collective quantum measurement on all $2n$ accessible qubits therefore has identical outcome probabilities. Adaptive local processing cannot change equality of the accessible density matrices. With equal prior probabilities the optimal guessing probability remains $1/2$; with other priors, no observation updates their ratio.

Let $p=|\alpha {|}^2$, so $|\beta {|}^2=1-p$. The projective measurement has outcome 0 with probability $p$ and leaves the state $|0⟩$; outcome 1 has probability $1-p$ and leaves $|1⟩$. Alice knows her actual result, so her conditional description is a pure state. Her fidelity of quantum states with the initial state is $p$ after outcome 0 or $1-p$ after outcome 1. Its outcome average is $p^2+(1-p{)}^2$.

Bob knows the preparation but not the outcome, so his density matrix is the nonselective postmeasurement stateIt is mixed for $0<p<1$ and pure only at the endpoints. The relative phase has disappeared because the classical outcome record is unavailable to him.

The undisclosed qubit measurement outcome with an unspecified preparation prior requires care: Charlene has no specified prior over preparations. Consequently the information given does not determine a unique posterior probability for the two outcomes. Under a symmetric ignorance or maximum-entropy prior, her best assignment isMore generally, if her prior mean of $|\alpha {|}^2$ is $q$, her assignment is ${\rho }_C=\mathrm{diag}(q,1-q)$ and its fidelity of quantum states with the actual initial state is $pq+(1-p)(1-q)$. Knowing that a measurement occurred guarantees diagonality in its basis, but does not by itself prove $q=1/2$. Different descriptions here reflect different available records, rather than different physical outcomes. ^[ref]

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.

A quantum state ray is the equivalence class of nonzero vectors in a complex Hilbert space under multiplication by a nonzero complex number. Its span is a one-dimensional linear subspace, but the zero vector is not a state representative. A unit vector fixes the norm and leaves a global phase free. ^[ref]

For a nonzero representative $|v⟩$, the associated pure state density operator isReplacing $|v⟩$ by $\lambda |v⟩$ multiplies both numerator and denominator by $|\lambda {|}^2$, so the quotient is unchanged. In particular, for every POVM effect $E$,This proves that a global phase changes no measurement probability. Changing the relative quantum phase of two nonzero components can change the density operator, as the following comparison shows.

A qubit, or quantum bit, is a quantum system whose state space is a two-dimensional complex Hilbert space. Choose an orthonormal basis $|0⟩,|1⟩$, called the computational basis. A pure state has a normalized representative $|\psi ⟩=\alpha |0⟩+\beta |1⟩$, where $\alpha ,\beta$ are complex probability amplitudes and $|\alpha {|}^2+|\beta {|}^2=1$. General qubit states, including mixed states, require a density matrix. ^[ref].

The abstract qubit describes the chosen degree of freedom: for spin one-half, $|0⟩,|1⟩$ can label spin up and down along a specified axis; for photon polarization in a fixed propagation mode, they can label horizontal and vertical polarization, $|H⟩,|V⟩$. These implementations share the same two-dimensional linear algebra, but their preparation and measurement in quantum mechanics use different apparatus. A qubit therefore need not be an entire two-level particle. ^[ref].

The usefulness of a qubit lies in controlling quantum superpositions and their relative quantum phases, then converting them into outcome probabilities with quantum gates and an appropriate measurement. One ideal computational-basis measurement returns one label, zero or one. It does not reveal both probability amplitudes, their relative quantum phase, or all the bits of a larger register's expansion. Quantum state tomography instead estimates a qubit state using many repetitions of the preparation and suitable measurement settings. ^[ref].

An actual preparation and measurement example is Dederick and Beck's 2014 teaching-laboratory experiment. A 405-nm laser pumped two nonlinear crystals, producing signal and idler photons near 810 nm. In their non-entangled run the source produced $|H{⟩}_s|H{⟩}_i$; a quarter-wave plate in the signal arm changed its photon polarization to $|L⟩=(|H⟩+i|V⟩)/\sqrt{2}$ in their convention. The plate supplies a relative quantum phase between polarization components. An idler detection selected the signal trials; signal detectors measured horizontal/vertical, diagonal, and circular polarization bases on separate runs. Before the polarization analyzer, a half-wave plate at $22.5$ degrees selected the diagonal basis, and a quarter-wave plate at $45$ degrees selected the circular basis. Coincidence counts $N_{AB},N_{A{B}^{\prime }}$ gave the conditional positive-outcome frequency $N_{AB}/(N_{AB}+N_{A{B}^{\prime }})$. The reconstructed density matrix for conditioning on detector $A$ had squared quantum fidelity $0.96$ with $|L⟩$ and purity of a density operator $0.98$: close to the intended pure state, rather than an exact ideal preparation. ^[ref].

The model diagram explains why those settings matter. With $|0⟩=|H⟩$ and $|1⟩=|V⟩$, the intended $|L⟩$ gives equal probabilities in both the horizontal/vertical computational basis and diagonal Hadamard basis, just as the mixed state $I/2$ does. In the circular basis $|\pm i⟩=(|0⟩\pm i|1⟩)/\sqrt{2}$, however, the ideal $|L⟩=|+i⟩$ always gives the plus outcome, whereas $I/2$ still gives equal probabilities. These are deductions from the Born rule, not the paper's measured counts. Unequal detector efficiencies and finite samples must be accounted for when interpreting experimental frequencies; the ideal comparison assumes equally efficient detection.

A qubit state specifies measurement probabilities, rather than a pre-existing answer to every possible measurement. For a pure state, normalized vectors differing by a global phase represent the same physical state:Equivalently, the pure state is a ray: the one-dimensional complex linear subspace spanned by any nonzero representative. Dividing a nonzero vector by its norm fixes its length, but leaves the global phase free. The density operator is unchanged because $(e^{i\gamma }|\psi ⟩)(e^{i\gamma }|\psi ⟩{)}^{†}=|\psi ⟩⟨\psi |$. In contrast, multiplying only one nonzero component by $e^{i\varphi }$ changes a relative quantum phase and can change measurement probabilities. ^[ref].

After choosing a global phase, every pure state of a qubit can be writtenAt $\theta =0$ or $\pi$, the parameter $\varphi$ does not distinguish physical states: only one component survives. Elsewhere it is the relative quantum phase. The corresponding unit Bloch vector is $\mathbf{r}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ on the existing Bloch sphere. Its angles describe a space of states; for a photon polarization qubit they are not the photon's position or direction of travel. ^[ref].

The Born rule in the computational basis gives $P(0)={\cos }^2(\theta /2)$ and $P(1)={\sin }^2(\theta /2)$. These probabilities discard $\varphi$. In the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, adding or subtracting the probability amplitudes instead givesThis last equality follows by substituting the angular representative. In the basis $|\pm i⟩=(|0⟩\pm i|1⟩)/\sqrt{2}$, the same calculation givesThus both the real and imaginary parts of quantum coherence can be observed, using different measurement settings. For example, $(|0⟩+|1⟩)/\sqrt{2}$ and $(|0⟩-|1⟩)/\sqrt{2}$ both give $P(0)=P(1)=1/2$, but the first gives $P_X(+)=1$ and the second $P_X(+)=0$. A global phase could never cause this difference.

For a general qubit state, write its density matrix in the computational basis asThese are exactly the Hermitian, unit-trace, positive semidefinite constraints. Positivity means every eigenvalue is nonnegative, not that every matrix entry is nonnegative. The general definition is given in density matrix. ^[ref].

For this two-by-two matrix, the explicit constraint follows from its determinant, $\det \rho =p(1-p)-|c{|}^2$: with trace one, a nonnegative determinant and real eigenvalues rule out a negative eigenvalue. Its eigenvalues areThe pure states have $|c{|}^2=p(1-p)$ and eigenvalues $1,0$; the mixed states have strict inequality and two positive eigenvalues. For a normalized representative, $p=|\alpha {|}^2$ and $c=\alpha {\beta }^{\ast }$. A statistical preparation $\rho ={\sum }_j{w}_j|{\psi }_j⟩⟨{\psi }_j|$, with $w_j\ge 0$ and ${\sum }_j{w}_j=1$, averages probabilities, rather than adding probability amplitudes. Such a preparation is mixed if the positive-weight vectors do not all represent the same ray. A reduced density matrix of an entangled pair can also be mixed even when the pair has a pure state. ^[ref].

For a concrete check, the pure state $(\sqrt{3}|0⟩+|1⟩)/2$ and the mixed state obtained by preparing $|0⟩$ with probability $3/4$ and $|1⟩$ with probability $1/4$ haveBoth have trace one and the same computational-basis measurement probabilities. Their eigenvalues are respectively $(1,0)$ and $(3/4,1/4)$; the respective values of their purity of a density operator $\mathrm{Tr}({\rho }^2)$ are $1$ and $5/8$. Their Hadamard basis plus probabilities are respectively $1/2+\sqrt{3}/4\approx 0.9330$ and $1/2$. By contrast, setting $p=3/4,c=1/2$ would give determinant $-1/16$ and eigenvalue $(2-\sqrt{5})/4<0$: normalized diagonal entries alone do not make a valid qubit state.

The same constraints become especially beautiful in the Bloch ball representation. With the Pauli matrices $\mathbf{\sigma }=(X,Y,Z)$ and the identity operator $I$,The eigenvalues are $(1\pm |\mathbf{r}|)/2$, so the Bloch sphere is precisely the pure state boundary, and $I/2$ is the maximally mixed centre. For each axis $j$, the Born rule gives $P_j(\pm )=(1\pm r_j)/2$. This unifies a geometric picture, matrix positivity, and experimental frequencies in one description. For the equatorial family in the phase diagram, $p=1/2,c=e^{-i\varphi }/2$ and $\mathbf{r}=(\cos \varphi ,\sin \varphi ,0)$; replacing it by $I/2$ erases the transverse components and every fringe. See quantum state tomography for how independent ensembles measured along the three axes estimate these components. ^[ref].

A uniform quantum superposition assigns the same probability amplitude to each computational-basis state. For $n$ qubits, there are $2^n$ such states.

The uniform quantum superposition of all $n$-bit strings is

The factor $2^{-n/2}$ follows from normalization: $2^n$ squared amplitudes must sum to one. This state is the product state $|+{⟩}^{\otimes n}$, so a large number of nonzero basis amplitudes does not by itself imply entanglement. A computational-basis measurement returns one string with probability $2^{-n}$; it does not reveal all amplitudes. ^[ref]