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Write $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. Before any interaction the system is in the pure state $|{\psi }_S⟩=\alpha |0⟩+\beta |1⟩$, soOne interaction leaves the environmental state $|+⟩$ unchanged on the system branch $|0⟩$, while on the branch $|1⟩$ it producesAfter interactions with $n$ distinct environment qubits, the joint state isand the decoherence factor is the product $(\cos ϵ{)}^n$. Taking the partial trace givesFor small $ϵ$, $(\cos ϵ{)}^n\simeq e^{-n{ϵ}^2/2}$: the populations remain fixed while the phase coherence decays. This is environmental decoherence, caused by entanglement with unobserved records rather than by a nonunitary evolution of the complete state.

A measurement in the computational basis cannot reveal the decay because its probabilities remain $|\alpha {|}^2$ and $|\beta {|}^2$. An interference measurement can. For example, measuring in the Hadamard basis givesChanging the measurement phase similarly accesses the imaginary part, so the shrinking interference visibility directly displays decoherence.

If the same environment qubit is reused, the conditional rotation accumulates coherently. After $n$ interactions the overlap is $\cos (nϵ)$ rather than $(\cos ϵ{)}^n$, and the off-diagonal entries oscillate. Coherence vanishes at some times but returns periodically: this finite environment exhibits quantum recoherence rather than effectively irreversible decay.

If a positive operator-valued measure distinguishes normalized pure states $|u⟩,|v⟩$ with certainty, the effect $E$ for the first answer satisfies $⟨u|E|u⟩=1$ and $⟨v|E|v⟩=0$. Positivity of $E$ and $I-E$ implies $E|v⟩=0$ and $E|u⟩=|u⟩$, because a positive operator with zero expectation annihilates the vector. Consequently $⟨u|v⟩=⟨u|E|v⟩=0$. Conversely, orthogonal states can be measured with their rank-one linear projections. A common unitary operator preserves this inner product and cannot remove the obstruction.

After the quantum quench, hold $H_f$ fixed and evolve without an intervening energy measurement. The unitary time evolution isProjection onto the original $|0⟩$ gives probability amplitude $c^2{e}^{+i{\Delta }_{f}t/(2\hslash )}+s^2{e}^{-i{\Delta }_{f}t/(2\hslash )}$. Squaring its modulus yields the two-level quantum return probabilityTheir sum is one. For $\theta =\pi /3$, the first minimum of $P_0$ is $1/4$ at $t=\pi \hslash /{\Delta }_f$, and the return is complete at $T=2\pi \hslash /{\Delta }_f$. At $\theta =0$ or $\pi$ there is no oscillation, because the initial quantum state is already a final energy eigenstate.

Changing measurement basis reveals the evolving relative quantum phase. Measuring the final energy first instead destroys this quantum coherence between final energy eigenstates in the unconditioned ensemble: its later original-basis return probability is $c^4+s^4=1-\frac{1}{2}{\sin }^2\theta$, equal to $5/8$ for this example. That constant differs from the oscillation of the unmeasured pure state. The work protocol and the oscillation protocol therefore use separate preparations.

A probability amplitude is a complex number that describes the contribution of a pure state to a specified measurement outcome. For a normalized quantum state $|\psi ⟩$ and a rank-one outcome associated with a normalized vector $|j⟩$, it is the inner product $a_j=⟨j|\psi ⟩$. The Born rule gives the corresponding probability $p_j=|a_j{|}^2=a_j^{\ast }{a}_j$, using the complex conjugate. Thus an amplitude can be negative or complex, while a probability is real and nonnegative. A wavefunction is the analogous position amplitude: its squared modulus gives a probability density, which must be integrated over a region to obtain a probability. ^[ref].

The extra information is the quantum phase. If $a=r{e}^{i\theta }$, its modulus $r$ fixes $|a{|}^2=r^2$, but its angle $\theta$ controls how it adds to another contribution. In the complex plane, picture amplitudes as arrows: add the arrows, then square the resulting length. For coherent contributions $A$ and $B$ to the same outcome,The last term is interference. Adding probabilities first loses it. Contributions with different relative quantum phases can reinforce or cancel even when their individual squared moduli are fixed.

For the qubit quantum superposition $|{\psi }_{\varphi }⟩=(\sqrt{3}/2)|0⟩+(e^{i\varphi }/2)|1⟩$, project onto the plus vector of the Hadamard basis. By linearity, the two contributions combine asHere $|A{|}^2=3/8$ and $|B{|}^2=1/8$, while the interference term is $(\sqrt{3}/4)\cos \varphi$. At $\varphi =0$, the arrows align and the plus probability is $1/2+\sqrt{3}/4\simeq 0.9330127$. At $\varphi =\pi$, they oppose and that probability is $1/2-\sqrt{3}/4\simeq 0.0669873$; cancellation is incomplete because the lengths differ. At $\varphi =\pi /2$, the arrows are perpendicular and the cross term vanishes. The minus amplitude uses $A-B$, reversing the cross term. These are alternate projective measurements on identically prepared copies, rather than a sequence of measurements on one copy.

A global phase multiplies every contribution by the same factor $e^{i\gamma }$, givingIt therefore changes no outcome probability. Changing a relative quantum phase can change the angle between contributions and hence their interference. Keeping amplitudes until the last step makes quantum circuits and matter-wave interference calculable; discarding their angles prematurely discards physically testable information. The Born rule example compares two measurement bases explicitly.

An ontological model of a quantum system is psi-epistemic if some pair of distinct pure states has preparation measures with nonzero overlap. The same physical state can then be compatible with either preparation. This precise measure-overlap definition is stronger than merely interpreting a wave function as information or a computational tool.

An ontological model of a quantum system is psi-ontic if distinct pure states have mutually singular preparation distributions. The physical state then determines which pure quantum state was prepared, almost surely in the pairwise sense. Extra underlying variables may still exist. The PBR theorem establishes this conclusion under preparation independence and exact quantum predictions.

A POVM whose nonzero effects have rank one, $E_i=w_i|{\psi }_i⟩⟨{\psi }_i|$, with normalized pure states and $w_i>0$. Completeness is ${\sum }_i{w}_i|{\psi }_i⟩⟨{\psi }_i|=I$. This use of pure concerns effect rank, and does not mean an extremal point in the convex set of all POVMs.

For a bipartite pure state with finite Von Neumann entropy, the joint Von Neumann entropy is zero, so its quantum conditional entropy is $-S({\rho }_A)$. The Schmidt decomposition makes the nonzero eigenvalues of ${\rho }_A$ the squared Schmidt coefficients. Their Shannon entropy vanishes exactly when there is one coefficient, equivalently the joint pure state is a product state. Thus negative quantum conditional entropy detects exactly the entangled bipartite pure states; it is only a sufficient criterion for entanglement of general mixed density operators.

For normalized pure states $\rho =|\psi ⟩⟨\psi |$ and $\sigma =|\phi ⟩⟨\phi |$, the trace distance and quantum fidelity satisfyThe last two expressions distinguish the unsquared quantum fidelity from the squared quantum fidelity. Identical pure states give $D=0$, and orthogonal pure states give $D=1$. The equality need not hold for mixed states; the Fuchs-van de Graaf inequalities give bounds in that case.

Brief proof: for two distinct pure states, their two-dimensional span contains the nonzero part of $\rho -\sigma$. This restriction has trace zero and determinant $-(1-F^2)$, hence eigenvalues $\pm \sqrt{1-F^2}$. Its trace norm is twice the positive eigenvalue, giving trace distance $D=\sqrt{1-F^2}$. For identical pure states, $\rho -\sigma =0$ directly.

For a density operator $\rho$ and pure state $P=|\psi ⟩⟨\psi |$, the diagonal absolute-sum bound for the trace norm in a basis containing $|\psi ⟩$ gives $D(\rho ,P)\ge 1-⟨\psi |\rho |\psi ⟩=1-F(\rho ,P{)}^2$. The quantum fidelity is unsquared. Block-diagonal states with respect to the target and its complement attain equality.

For two normalized pure states, the difference of their projectors has two eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$, giving that trace distance. Write a mixed density operator as $\rho ={\sum }_j{q}_j|u_j⟩⟨u_j|$. Convexity of the trace norm and weighted Cauchy-Schwarz inequality give $D(\rho ,|v⟩⟨v|)\le {\sum }_j{q}_j\sqrt{1-|⟨u_j|v⟩{|}^2}\le \sqrt{1-⟨v|\rho |v⟩}$. Thus a small probability of failing the pure-state projector test forces trace-distance proximity, including for correlated multipartite outputs.

If subsystem $S$ has pure state $|u⟩⟨u|$ in a normalized joint pure vector $|\Omega {⟩}_{SE}$, then $|\Omega {⟩}_{SE}=|u{⟩}_S\otimes |e{⟩}_E$. Indeed, for an orthonormal basis $\{v_j\}$ of $u^{\perp }$, each squared norm $\Vert (⟨v_j|\otimes I)|\Omega ⟩{\Vert }^2=⟨v_j|{\rho }_S|v_j⟩$ is zero. Therefore the joint vector has no component outside $\mathbb{C}u\otimes {\mathcal{H}}_E$. In particular, if Alice and Bob share a known pure entangled state, an adversary holding a purifying system has no correlations with it. This does not make every pure bipartite state useful for key distribution: a pure product state has no shared random correlations.

The purity of a density operator is $\gamma =\mathrm{Tr}({\rho }^2)$. It equals one exactly for a pure state; in finite dimension $d$ it lies between $1/d$ and $1$. The minimum occurs exactly at the maximally mixed state $I/d$.

A worked qubit family interpolates between the mixed state $I/2$ and the pure state ${\rho }_{+}=|+⟩⟨+|$, where $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$:The eigenvectors are $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, with eigenvalues ${\lambda }_{\pm }=(1\pm p)/2$. Squaring and summing these eigenvalues givesHere $P_X(+)$ is the Born rule probability of the positive outcome in an $X$ projective measurement. The Bloch vector is $(p,0,0)$, since ${\rho }_p=(I+p{\sigma }_x)/2$ with ${\sigma }_x$ one of the Pauli matrices. At $p=0$ the purity is $1/2$; at $p=1/2$ it is $5/8$, with $P_X(+)=3/4$; at $p=1$ it is one. Within this family, $p=1$ is a pure state and every $p<1$ is a mixed state. Extending the formula to $p>1$ produces a negative eigenvalue $(1-p)/2$, so it ceases to be a density operator despite remaining Hermitian with trace one.

The Bloch vector length measures polarization: for any qubit density operator, $\gamma =(1+|\mathbf{r}{|}^2)/2$. This connects the interior of the Bloch ball with mixed states and its boundary, the Bloch sphere, with pure states. ^[ref]

For a short proof of the general bound, use the spectral theorem: the nonnegative eigenvalues ${\lambda }_j$ sum to one, so $\gamma ={\sum }_j{\lambda }_j^2$. ThenThe lower bound is attained only when all eigenvalues equal $1/d$; the upper bound is attained only when exactly one eigenvalue is nonzero, and hence equals one. Purity is therefore a simple quantitative test of mixedness, invariant under a unitary operator. It does not specify the entire density operator: even distinct pure states all have purity one. ^[ref]