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The Born rule connects a quantum state to measurement probabilities. Let $\{|j⟩\}$ be a complete orthonormal basis, finite or countably infinite, for a Hilbert space, and letAn ideal measurement distinguishing these basis vectors returns label $j$ with probabilityThis assignment is a physical measurement rule. Its compatibility with total probability one follows from orthonormality and normalization:The probabilities are the squared moduli, not the amplitudes themselves or the squares $a_j^2$. ^[ref].

If the basis is an eigenbasis of an observable, the reported numerical value is its eigenvalue. Several vectors can share one eigenvalue; the probability of that value is then the sum over the corresponding basis labels. More generally, a projective measurement with orthogonal projections $P_r$ assignsfor a pure state and a general density matrix $\rho$, respectively. This also covers mixed states and degenerate eigenspaces. The Born rule determines outcome statistics; the ideal conditional state change is specified separately by the Lüders rule. See projective measurement for that formulation and its assumptions. ^[ref].

For a worked example, take the normalized qubit quantum stateThe squared moduli add to $3/4+1/4=1$. A computational-basis measurement, equivalently a measurement of the $Z$ Pauli matrix, hasThese probabilities do not depend on the relative quantum phase $\varphi$. Now choose the Hadamard basiswhose vectors are eigenstates of the $X$ Pauli matrix, also used by the Pauli X gate. The new probability amplitudes and probabilities areBoth probabilities are nonnegative and their sum is one. The relative quantum phase becomes visible because the two probability amplitudes add before their squared modulus is taken:

A Hadamard gate followed by a computational-basis measurement realizes these $X$ statistics, since $H|\pm ⟩=|0⟩,|1⟩$, respectively. Compare the two bases using fresh, identically prepared copies: a preceding measurement may change the state. The quantum superposition section shows why replacing the preparation by the mixed state $\mathrm{diag}(3/4,1/4)$ keeps the $Z$ statistics but gives $p_{+}=p_{-}=1/2$ for every $\varphi$.

The usefulness of the Born rule is this conversion from linear algebra to experimental predictions: calculate overlaps in the basis actually measured, then take squared moduli. Its appealing geometry is that changing basis redistributes probability while preserving total squared norm. The same rule relates an energy eigenstate expansion to an energy measurement and a position wavefunction to a probability density.

Born's 1926 paper on quantum scattering interpreted the outgoing-wave coefficients in terms of the probabilities of collision outcomes; a corrective footnote specified proportionality to the square of the amplitude. This is a historical scattering formulation, rather than the general density matrix formula above. ^[ref]; ^[ref].

Writing a CNOT gate as $P_0\otimes I+P_1\otimes X$ gives $I\otimes Q_{+}+Z\otimes Q_{-}$, where $Q_{\pm }=(I\pm X)/2$. In the Hadamard basis, $Z$ flips the sign label and $X$ is diagonal. Thus the same gate matrix reverses its control and target roles under the two local basis changes.

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.

The Hadamard gate is the unitary operatorwhere $I$ is the identity operator. It maps the computational basis to the Hadamard basis:

These outputs have equal computational-basis measurement probabilities, but their relative quantum phase differs. Applying another Hadamard gate sends them to different definite outputs. Calling this gate a classical coin toss would discard precisely the information that makes this possible. ^[ref].

Starting from $|0⟩$, compare two quantum circuits, reading matrix products from right to left:Here $Z$ is the Pauli Z gate. After the first Hadamard gate, the probability amplitudes are $(1/\sqrt{2},1/\sqrt{2})$. Inserting $Z$ changes them to $(1/\sqrt{2},-1/\sqrt{2})$ without changing their squared magnitudes. At the final gate, the contributions to output zero cancel and those to output one add. Without $Z$, cancellation occurs at output one instead. This is controlled interference, with the quantum phase determining which answer emerges.

The same calculation works for any phase gate $P_{\varphi }=\mathrm{diag}(1,e^{i\varphi })$:For example, $\varphi =\pi /3$ gives $P(1)=1/4$, while $\varphi =\pi /2$ gives $P(1)=1/2$. The output-one amplitude at $\varphi =\pi /2$ is $(1-i)/2$: its two complex contributions add before the Born rule is applied. A phase gate alone leaves the populations unchanged; recombination converts its relative quantum phase into observable probabilities.

If quantum decoherence completely removes the computational basis coherence between the gates, the intermediate density matrix becomes $I/2$. Both the phase gate and final Hadamard gate leave $I/2$ unchanged, so $P(1)=1/2$ for every $\varphi$. The disappearing fringe explains why preserving quantum coherence matters: it enables the gate sequence to carry phase information into the final measurement.

The normalized Hadamard transform maps a vector indexed by binary strings through entries $2^{-n/2}(-1{)}^{x\cdot y}$, where the dot product is evaluated modulo two. It is its own inverse and an orthogonal transformation, hence also a unitary operator on complex input vectors. The Hadamard gate is the $n=1$ quantum case, and the Hadamard basis is the image of the computational basis. The Walsh-Hadamard transform gives the same tensor product construction.

For two qubits, applying one Hadamard gate to each prepares the uniform quantum superpositionEach probability amplitude is $1/2$, so each computational-basis measurement outcome has probability $1/4$. Applying the Hadamard transform again reconstructs $|00⟩$: signed contributions cancel at the other three outputs. Its usefulness lies in combining simple local quantum gates into an operation on exponentially many amplitudes. Extracting useful information still requires an appropriate subsequent transformation and measurement.

Suppose one qubit is prepared in either of the known pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, with equal prior probabilities. Measuring in the computational basis seems natural, but a tilted projective measurement gives better quantum state discrimination. The two density operators and their difference areThe Pauli matrices give ${\Delta }^2=I/2$, and the eigenvalues of $\Delta$ are ${\lambda }_{\pm }=\pm 1/\sqrt{2}$. Their absolute sum gives trace distance $D=1/\sqrt{2}$, agreeing with trace distance between qubit states.

The positive spectral projector points along the unit Bloch vector $\mathbf{n}=(-1,0,1)/\sqrt{2}$:The outcome labels $+$ and $-$ here refer to the sign of the eigenvalue of $\Delta$; they do not denote a Hadamard basis measurement. Guess $|0⟩$ after $P_{+}$ and $|+⟩$ after $P_{-}$. This sign choice is the equal-prior Helstrom measurement for two pure states, a case of the Holevo–Helstrom theorem. ^[ref]

An explicit orthonormal basis for this projective measurement iswith $P_{\pm }=|h_{\pm }⟩⟨h_{\pm }|$. Thus the real coefficient rotation angle is $\beta =-\pi /8$ relative to the computational basis, while the Bloch vector of $P_{+}$ is tilted by $\pi /4$ from the positive $z$ axis toward negative $x$. State-vector and Bloch sphere angles differ by a factor of two.

Write $a=(1+1/\sqrt{2})/2$ and $b=(1-1/\sqrt{2})/2$, so $a+b=1$. The Born rule gives the conditional outcome probabilitiesThe event $P_{+}$ has probability gap $a-b=1/\sqrt{2}=D$, attaining the variational characterization of trace distance. Both preparations are identified correctly with conditional probability $a$, so

For comparison, a $Z$ projective measurement in the computational basis has conditional outcome probabilities $(1,0)$ on ${\rho }_0$ and $(1/2,1/2)$ on ${\rho }_{+}$. Guess zero on outcome zero and plus on outcome one. The average success isand the measured total variation distance is only $1/2$. The optimal tilted projective measurement gains about $0.103553$ in absolute success probability by balancing the two conditional error rates instead of identifying $|0⟩$ perfectly at the expense of $|+⟩$.

These are statistics for alternative measurements on independently prepared copies. Applying $Z$ first and then the tilted projective measurement is a different experiment, because the first measurement in quantum mechanics changes the density operator.

A CNOT gate can correlate either the original control's computational label with a computational pointer, or the original target's sign-basis label with a sign-basis pointer. For the second interpretation prepare the original control in $|+⟩$ and read it in the Hadamard basis. Its two pointer states become $|+⟩$ and $|-⟩$, measuring $X$ on the original target. An apparatus role therefore depends on the ready state and readout as well as the measurement interaction.

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]

Apply the Born rule to the two amplitudes in the introductory calculation. In the computational basis, the probabilities areThe other basis printed in this question is $(|0⟩\pm i|1⟩)/\sqrt{2}$, the Pauli Y gate eigenbasis. It is not the real Hadamard basis. Write the output amplitudes as $a=(1+e^{i\varphi })/2$ and $b=(1-e^{i\varphi })/2$. The plus-outcome amplitude is $(a-ib)/\sqrt{2}$ and the minus-outcome amplitude is $(a+ib)/\sqrt{2}$. Since $\mathrm{Im}(\overline{a}b)=-\sin \varphi /2$, the Born rule givesIn particular, at $\varphi =\pi /2$ the output is the minus eigenstate, which checks the sign of the imaginary quadrature.

Use separate fresh runs for each measurement basis. If ${\hat{P}}_j$ denotes the empirical outcome frequency, thenThe two differences estimate $\cos \varphi$ and $\sin \varphi$, resolving the ambiguity that would remain from the first basis alone. Each Bernoulli frequency converges by the strong law of large numbers, with standard error of order the inverse square root of the number of runs. One may project the noisy estimate onto the unit circle if a phase estimate is desired, but that projection is not necessary for consistency.

Assume the three physical phase-flip channels act independently, with the same probability $p$ on each qubit, and encoding, recovery and decoding are ideal. Zero or one physical error is corrected. HenceThe probability of an uncorrectable two- or three-site error is $p_L=3p^2(1-p)+p^3=3p^2-2p^3$.

One must also choose the logical convention consistently with the output phase error specified here. The usual phase-flip repetition code with codewords $|+++⟩$ and $|---⟩$ has residual operator $Z_1{Z}_2{Z}_3$ after two or three physical errors: their error syndrome equals that of the complementary zero- or one-error pattern. This operator exchanges those codewords, so plain inverse encoding would give logical $X$, not $Z$.

To obtain the stated logical phase-flip convention, use the logical Hadamard convention for a phase-flip repetition code. Put a logical Hadamard gate before that encoding and after its decoding. The encoded state iswhere the plus and minus states in these codewords are the real Hadamard basis states. The uncorrectable exchange is now conjugated to $HXH=Z$. With that convention the decoded density operator and its trace distance are exactlyFor $0<p<1/2$, $p-p_L=p(1-p)(1-2p)>0$, so coding reduces the trace distance. The displayed recovery probability means success of correction as a quantum operation on an arbitrary unknown input. If the particular input is a $Z$ eigenstate, even the logical phase-error branch represents the same physical state, and the probability of recovering that particular state is one.

Independence is an extra noise assumption needed for the numerical probability. If all three physical qubits flip together with probability $p$, each marginal still has the given phase-flip channel, but the logical failure probability is $p$, not $3p^2-2p^3$. More generally, for a classical joint phase-error distribution with a random number $K$ of flips, $P_{\mathrm{rec}}=\mathrm{Pr}(K\le 1)$ and $p_L=\mathrm{Pr}(K\ge 2)$; the same decoded mixture holds in the stated logical convention. Thus marginal error probabilities alone do not determine the requested recovery probability. A complex environment overlap likewise requires the coherent rotation to be retained or separately compensated before this stochastic phase-error analysis applies.

Secure in the intended ideal setting, with a sound entanglement-purification procedure. Bob confirms receipt before Alice reveals either the random basis string or the randomly chosen check positions. Eve must therefore release the travelling qubits before knowing which basis to use and which positions to protect from disturbance. Bob can then decode because $H^2=I$.

The random check sample tests the same ensemble of encoded transmissions as the retained data. Random computational basis and Hadamard basis encodings make the test sensitive to complementary disturbances; Eve cannot use the targeted zero-error attack available in Protocol 1. For example, measuring every intercepted qubit in one fixed one of these two bases uses the wrong basis half the time, and then disagrees with Alice half the time, producing a bit error rate of $1/4$. A sound entanglement purification protocol uses the tested error bounds, corrects both relevant types of error, and aborts if the noise is too high; secrecy is not claimed for every possible attack without an abort option.

When the protocol succeeds in producing nearly perfect Bell pairs, purity decouples a subsystem from its purification explains why they are nearly independent of Eve, and matching computational basis measurements produce an approximately uniform shared secret key. The disclosed check outcomes are discarded rather than included in that key. This is the standard entanglement-based quantum key distribution ordering: receipt first, basis and sample disclosure second, testing and purification next, and private measurements last.

The Pauli Z gate exchanges the two Hadamard basis states: $Z|+⟩=|-⟩$ and $Z|-⟩=|+⟩$. Multiplying the displayed gate by the Pauli X gate givesOn the other hand,The two operators agree, provingAlso $U(\theta )=H\mathrm{diag}(1,e^{-i\theta })=J(-\theta )$ in the J gate convention. Keeping this angle sign explicit avoids confusing the two conventions for equatorial qubit measurements.

In the Hadamard basis, $X_D$ is diagonal with projectors $Q_{\pm }=(I\pm X_D)/2$. Rewrite the same CNOT gate in the original tensor-factor order $S\otimes D$ asExpanding $P_0\otimes I+P_1\otimes X_D$ verifies this identity directly. Now $Z_S|+⟩=|-⟩$ and $Z_S|-⟩=|+⟩$, so $Z_S$ is ${\tilde{X}}_S$ in the complementary basis. If $D$ is plus, the identity acts on $S$; if $D$ is minus, its sign-basis label flips. Hence $D$ is the control and $S$ the target in this basis.

Equivalently,This CNOT control reversal in the Hadamard basis concerns the matrix in changed local bases, rather than a physical exchange of the two carriers.

The Hadamard basis changes the Pauli matrices by $Z_S={\tilde{X}}_S$ and $X_D={\tilde{Z}}_D$. Thus, keeping tensor factors ordered as $S,D$, the Hamiltonian operator isIf $D$ has new-basis label plus, the second factor vanishes. If $D$ has label minus, the integrated generator on $S$ is $\pi (I-{\tilde{X}}_S)/2$, whose exponential is ${\tilde{X}}_S$ by part (a). The generator therefore has the same controlled form with $D$ as control in the complementary bases.

To use $S$ as the apparatus, prepare it in the ready state $|+{⟩}_S$ and later read it in the Hadamard basis. For an arbitrary input $\alpha |+{⟩}_D+\beta |-{⟩}_D$, part (b) givesThe two apparatus states are orthogonal, so its readout measures the corresponding projectors on $D$ and preserves each input eigenstate. The observable iswith outcomes $+1$ and $-1$. This is the measurement direction of a CNOT interaction in the complementary basis.

The ready state and readout basis matter. Keeping $S$ in the original $|0{⟩}_S$ would make the old-basis controlled gate act as the identity on $D$, producing no record. Thus the same interaction can serve either measurement direction, with different apparatus preparation and pointer readout; the Hamiltonian operator alone does not assign an intrinsic system/apparatus role.

For each position $i$, Alice uses $y_i$ to choose a basis and $x_i$ to choose a vector in that basis. Explicitly, she sends the qubitThus $y_i=0$ uses the computational basis $\{|0⟩,|1⟩\}$, while $y_i=1$ uses the Hadamard basis $\{|+⟩,|-⟩\}$.

The pure state $|\Psi {⟩}_{AB}$ is entangled when it cannot be written as a product state $|a{⟩}_A|b{⟩}_B$, equivalently when its Schmidt rank exceeds one. Its subsystem states are the reduced density matriceswhere the partial trace is characterized by $\mathrm{Tr}(M_A{\rho }_A)=⟨\Psi |M_A\otimes I_B|\Psi ⟩$ for every observable $M_A$.

Let Bob's measurement have Kraus operators $M_b$ satisfying ${\sum }_b{M}_b^{†}{M}_b=I_B$. If Alice does not learn Bob's random outcome, her state after the measurement isThis no-communication theorem means that Bob can change Alice's conditional state after she learns his outcome, but cannot change any local outcome distribution available to her alone. Standard nonrelativistic quantum mechanics is therefore operationally compatible with the prohibition of superluminal signalling in special relativity, despite its nonlocal conditional-state updates.

An exact nondisturbing state readout would destroy this protection if it reported the globally collapsed state on an absolute-time slice. For example, Alice and Bob may share Bell states. At a prearranged time Bob encodes a bit by measuring his qubit in either the computational or Hadamard basis. The usual projection postulate assigns Alice respectively one of $|0⟩,|1⟩$ or one of $|+⟩,|-⟩$. A device returning the exact pure state lets Alice identify which basis Bob chose without waiting for his outcome, producing a superluminal signal. Equivalently, Bob may choose whether to measure, and the device distinguishes the resulting proper pure state from the original improper maximally mixed local state.

A causal alternative is a quantum state readout device located at a spacetime point $x$ that reports the local quantum state under objective collapse: take reduced states on spacelike hypersurfaces through $x$ and let those hypersurfaces approach the past light cone of $x$. The result includes localized collapses in $J^{-}(x)$ and excludes spacelike-separated collapses. This assumes a fixed Minkowski spacetime, localized collapse events, ordinary local unitary dynamics between them, and outputs that may control only operations in their causal future. The postulate is logically consistent because it adds a classical record of this local state without changing the state or any standard measurement probability. It also cannot support an indirect signalling algorithm. Inductively through the algorithm's events, every readout at $x$ depends only on operations, collapses, and earlier readouts in $J^{-}(x)$; any operation selected from that output lies in the readout's future light cone. Composing readouts, unitary evolutions, and measurements therefore never carries a controllable dependence outside a future light cone, so relativistic causality is preserved.

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.