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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 qutrit is a three-dimensional quantum system, with pure states $\alpha |0⟩+\beta |1⟩+\gamma |2⟩$ satisfying $|\alpha {|}^2+|\beta {|}^2+|\gamma {|}^2=1$. A qubit-qutrit state is a bipartite state on ${\mathbb{C}}^2\otimes {\mathbb{C}}^3$.

Ramsey interferometry uses two coherent pulses with a free interval between them: the first creates a quantum superposition, the interval accumulates relative quantum phase, and the second converts that quantum phase into measurement probabilities. Its attraction is that an ordinary population readout can reveal the quantum coherence hidden from an energy measurement during the free interval.

Here is an ideal qubit model with an explicit pulse convention. Let $R_y(\theta )=\exp (-i\theta {\sigma }_y/2)$, prepare $|0⟩$, apply $R_y(\pi /2)$, evolve freely with $U_{\varphi }=\exp (-i\varphi {\sigma }_z/2)$, and apply the inverse pulse $R_y(-\pi /2)$ before a computational-basis measurement. The first pulse gives $|+⟩$. Direct matrix multiplication givesThus the Born rule gives $P(0)={\cos }^2(\varphi /2)$ and $P(1)={\sin }^2(\varphi /2)$. Before the second pulse, the same computational-basis measurement would give $1/2$ for both outcomes for every $\varphi$. At $\varphi =2\pi /3$, the second pulse changes these probabilities to $1/4$ and $3/4$. If the two pulses have the same sign instead, the output labels interchange: the pulse convention is part of the prediction.

In a pulse experiment the phase is measured relative to the oscillator supplying the coherent pulses. In a rotating frame, write the free Hamiltonian operator as $H_{\mathrm{rot}}=\hslash \delta {\sigma }_z/2$ with signed angular frequency difference $\delta$ in the chosen labeling convention. A free interval $T$ then accumulates $\varphi =\delta T$; changing the second pulse's phase changes the analysis direction. This comparison makes Ramsey interferometry useful for measuring frequency differences and diagnosing loss of quantum coherence.

For example, suppose the first pulse has prepared a density matrix with equal populations and transverse Bloch vector length $0\le V\le 1$:The same inverse analysis pulse yields $P(0)=(1+V\cos \varphi )/2$. The pure state case $V=1$ gives the full ideal fringe; the mixed state $I/2$ gives the flat result $1/2$. This is a controlled mathematical comparison, not a fit to an experiment. Experimental contrast alone does not identify quantum decoherence: imperfect preparation, pulse errors and readout errors can also reduce it.

A random unitary channel has $T(\rho )={\sum }_j{p}_j{U}_j\rho U_j^{†}$. Its Choi matrix is a convex combination of maximally entangled pure states.

For normalized linearly independent pure states, their Gram matrix $G_{ij}=⟨{\psi }_i|{\psi }_j⟩$ is positive definite. The displayed reciprocal vectors satisfy $⟨{\chi }_i|{\psi }_j⟩={\delta }_{ij}$. The operator $R={\sum }_i|{\chi }_i⟩⟨{\chi }_i|$ has nonzero eigenvalues equal to those of $G^{-1}$: use the polar factorization of the matrix whose columns are ${\psi }_i$. Choosing $0<ϵ\le {\lambda }_{\min }(G)$ gives the POVM effects $E_i$ and $E_{?}=I-ϵR$. Each conclusive outcome has probability $ϵ$ on its own state and zero on every other input. The inconclusive effect preserves positivity, so the construction is a physical measurement in quantum mechanics.

The Robertson uncertainty principle bounds the product of the standard deviations of two observables in the same quantum state. For a normalized pure state $|\psi ⟩$ and self-adjoint operators $A,B$, write their expectation values and variances asand similarly for $B$, where $I$ is the identity operator. The quantum uncertainties $\Delta A,\Delta B$ then obeyHere $[A,B]$ is the commutator, and every expectation value is taken in the same $|\psi ⟩$. These standard deviations describe the outcome spreads when $A$ and $B$ are measured on separate ensembles prepared in that same quantum state. They are not instrument errors or the change caused by a preceding measurement in quantum mechanics. The definition and bound are developed in Barton Zwiebach, Uncertainty Principle and Compatible Observables, 2013, §§1–2.

For unbounded self-adjoint operators, the operator domains matter. Assume $\psi \in D(A)\cap D(B)$, so that both second moments are finite; $⟨A^2⟩$ here means the second moment of the measured probability distribution, equal to $\Vert A\psi {\Vert }^2$, even if $A^2\psi$ is undefined. To use the displayed operator commutator, also assume $\psi \in D(AB)\cap D(BA)$. Without these product-domain assumptions, the always meaningful version on $D(A)\cap D(B)$ uses $|\mathrm{Im}⟨A\psi |B\psi ⟩|$ as its right-hand side. These qualifications prevent a formal commutator identity from being applied to an inadmissible quantum state.

The brief proof is Cauchy-Schwarz inequality applied to $u=(A-⟨A⟩I)\psi$ and $v=(B-⟨B⟩I)\psi$:The final equality uses self-adjointness and the product-domain assumptions. Thus a geometric statement about inner products becomes a constraint on experimental spreads. For nonzero $\Delta A,\Delta B$, equality requires $u,v$ to be proportional with a purely imaginary proportionality factor; the equality case of the Heisenberg uncertainty relation turns this condition into a Gaussian wave packet.

The bound is useful because computing a single commutator and its expectation value can constrain two entire outcome distributions. Its right-hand side depends on the quantum state: the spin uncertainty state dependence example gives both a positive saturated bound and a zero bound for the same noncommuting observables. Robertson's paper, titled "The Uncertainty Principle", appeared on 1 July 1929 in Physical Review, volume 34, starting at page 163. ^[ref].

For the pure state ${\lambda }_0|00⟩+{\lambda }_1|11⟩$ with real normalized Schmidt coefficients, the two-body Pauli correlators form $T=\mathrm{diag}(s,-s,1)$, $s=2{\lambda }_0{\lambda }_1$. Thus arbitrary local axes give $⟨(\mathbf{a}\cdot \sigma )\otimes (\mathbf{b}\cdot \sigma )⟩=s(a_x{b}_x-a_y{b}_y)+a_z{b}_z$. The negative $yy$ component follows from the phases in the Pauli operators.

The Schmidt coefficients of a normalized bipartite pure state are the nonnegative numbers $s_j$ in its Schmidt decomposition; they satisfy ${\sum }_j{s}_j^2=1$ and their squares are the nonzero eigenvalues of either reduced density matrix.

Every finite-dimensional bipartite pure state has a decomposition $|\psi ⟩={\sum }_{j=1}^r{s}_j|a_j⟩|b_j⟩$ with positive Schmidt coefficients $s_j$ and orthonormal families on each subsystem.

For $n$ copies of a normalized pure state $\alpha |00⟩+\beta |11⟩$, regroup Alice's and Bob's tensor factors. Alice performs a projective measurement of the total number $k$ of zeroes, preserving all coherence among strings with that count. The outcome has binomial distributionEach retained string has the same complex amplitude ${\alpha }^k{\beta }^{n-k}$, so, up to a common phase, the conditional quantum state isIts equal Schmidt coefficients make it a maximally entangled state of Schmidt rank $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. Alice sends the count to Bob. They can relabel the corresponding local computational basis strings with local unitary operations, and convert the resulting resource to Bell pairs. Measuring the individual bits instead would reveal the entire string and leave a product state.

The Schmidt rank is the number of nonzero terms in a Schmidt decomposition, equivalently the rank of either reduced density matrix of a bipartite pure state.

The coefficient matrix of a bipartite pure state has rank equal to its Schmidt rank, by singular value decomposition. A local product operation $A\otimes B$ sends it to $AM{B}^{\mathsf{T}}$, whose rank cannot increase. A complete classical transcript of an LOCC protocol still selects one product Kraus operator, so every nonzero branch obeys the bound. Discarding the transcript gives a mixed output with Schmidt number at most the initial rank.

The nonzero eigenvalues, including multiplicities, of either reduced density operator of a bipartite pure state. A Schmidt decomposition $|\psi ⟩={\sum }_k\sqrt{p_k}|a_k{b}_k⟩$ gives ${\rho }_A={\sum }_k{p}_k|a_k⟩⟨a_k|$ and the corresponding expression for ${\rho }_B$ by partial trace. Thus unequal marginal spectra rule out a pure joint state. The amplitudes $\sqrt{p_k}$ are often called Schmidt coefficients; the spectrum here uses the probabilities $p_k$.

For a bipartite pure state, the squared Schmidt coefficients are the nonzero eigenvalues of either reduced density matrix. Their sum except the largest defines $E_2=1-{\lambda }_{\max }({\rho }_A)$. The largest eigenvalue is a convex function, since ${\lambda }_{\max }(\rho )={\max }_{\Vert v\Vert =1}⟨v|\rho |v⟩$; consequently $E_2$ is concave. If Alice measures locally and records outcome $r$, Bob's conditional reductions satisfy ${\sum }_r{p}_r{\rho }_{B,r}={\rho }_B$. Concavity gives ${\sum }_r{p}_r{E}_2({\chi }_r)\le E_2(\chi )$. For Bob's measurements use Alice's unchanged average reduction instead. Iteration proves average monotonicity under LOCC. For Schmidt rank at most two, $E_2$ is the smaller squared Schmidt coefficient. A conversion succeeding with probability $p$ therefore obeys $p{E}_2(\varphi )\le E_2(\psi )$, since failure branches contribute nonnegative amounts.

In Simon's algorithm, conditioning on a function value leaves the first quantum register in an equal coherent sum of its two preimages. This is a pure state, even though discarding the second quantum register without retaining its outcome generally leaves a mixed state. The label $x$ depends on the measured value; the period $s$ does not.

A unital quantum channel mixes its input spectrum with a doubly stochastic matrix, so its output spectrum satisfies the majorization relation with its input spectrum. The reverse inequality is generally false: a pure state can be sent to a maximally mixed state. The direction expresses the fact that unital noise can make eigenvalues more uniform. Equality of the two spectra is possible, for example under a unitary channel.

For spin one-half, choose the normalized spin eigenstates $|+z⟩,|-z⟩$ with $S_z|\pm z⟩=\pm (\hslash /2)|\pm z⟩$. The spin one-half along an axis pure state in the $xz$ plane isIts norm is one because the two eigenstates are orthogonal and ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$. Applying the Born rule to a $z$ projective measurement givesConsequently the expectation value, second moment, and variance areAt $\theta =\pi /3$, the outcome probabilities are $3/4$ and $1/4$, the mean is $\hslash /4$, and the variance is $3{\hslash }^2/16$. Every individual outcome remains $\pm \hslash /2$. At $\theta =\pi /2$, the mean is zero and the variance is maximal, ${\hslash }^2/4$. At either pole, one outcome has probability one and the variance vanishes.

The diagonal mixed state ${\rho }_{\mathrm{diag}}=p_{+}|+z⟩⟨+z|+p_{-}|-z⟩⟨-z|$ has the same $S_z$ probabilities, expectation value, and variance as $|{\psi }_{\theta }⟩$. It is nevertheless a different quantum state for $0<\theta <\pi$: the displayed pure state has $⟨S_x⟩=\hslash \sin \theta /2$, while ${\rho }_{\mathrm{diag}}$ has $⟨S_x⟩=0$. A single component's mean does not reveal all relative quantum phases or establish purity of a density operator.

This is the square of the unsquared quantum fidelity, another common convention for fidelity of quantum states. If one state is the normalized pure state $|\psi ⟩⟨\psi |$, thenBy the Born rule, the squared quantum fidelity here is the probability that a measurement in quantum mechanics with acceptance orthogonal projector $|\psi ⟩⟨\psi |$ accepts $\sigma$. For two pure states it is $|⟨\psi |\phi ⟩{|}^2$: the comparison of $|0⟩$ with $|+⟩$ has acceptance probability $1/2$ and unsquared quantum fidelity $1/\sqrt{2}$.

For two general mixed states, neither quantum fidelity nor squared quantum fidelity is a probability that the states are equal. Equality of two specified density operators is a definite mathematical condition. The measurement characterization of quantum fidelity instead describes overlap between outcome probability distributions. The ensemble average used in reliable quantum source compression is normally the average squared quantum fidelity with the emitted pure state; the convention must be stated when comparing formulas.

The swap test estimates the overlap of two pure states without requiring their classical descriptions. Prepare an auxiliary qubit in $|0⟩$, apply a Hadamard gate, a Fredkin gate controlled by that auxiliary qubit, and another Hadamard gate. Then measure the auxiliary qubit in the computational basis.

For target states $|a⟩,|b⟩$, the auxiliary-zero branch has unnormalized target vector $(|a⟩|b⟩+|b⟩|a⟩)/2$. Its squared norm is thereforeIdentical pure states give $p_0=1$, while pure states with zero inner product give $p_0=1/2$. A single zero outcome does not certify equality; repeated trials estimate the probability. The calculation is a direct projection onto the symmetric target subspace using the SWAP gate.

Ideal corrected quantum teleportation transfers an input system to the receiver while preserving its entire joint density operator with an arbitrary reference. Expand a joint pure state as ${\sum }_j|j⟩|r_j⟩$. Every teleportation branch acts with the same known correction on each system basis vector and does nothing to the reference vectors; undoing that correction restores the complete sum. Linearity extends the result to mixed states. The sender's measured carrier no longer retains the original reference correlations.