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The logical depth of a measurement pattern counts sequential layers of quantum measurements forced by dependence of measurement bases on earlier outcomes. Depth one means every basis can be fixed before measurements start, allowing all quantum measurements on distinct qubits to be performed in parallel. This does not require depth-one resource-state preparation or constant-depth classical parity processing.

Suppose the circuit contains $m$ gates $J({\alpha }_1),\dots ,J({\alpha }_m)$ in time order, with each ${\alpha }_j\in \{0,\pi /2\}$. Prepare an $(m+1)$-vertex path graph state from $|+{⟩}^{\otimes (m+1)}$ and Controlled-Z gates between neighbours. Its first qubit supplies the specified input $|+⟩$. Measure vertex $j-1$ in the fixed equatorial qubit measurement basis at angle ${\alpha }_j$ for $j=1,\dots ,m$, obtaining $s_j$. Measure the final vertex in the computational basis, obtaining $d$.

Track the Pauli frame as $X^{a_j}{Z}^{b_j}$, starting with $a_0=b_0=0$. A hypothetical sequential reading of the same quantum measurements gives the step $X^{s_j}J({\alpha }_j)X^{a_{j-1}}{Z}^{b_{j-1}}$. The supplied commutation rules imply, up to global phase,For ${\alpha }_j=0$, the sign is irrelevant. For ${\alpha }_j=\pi /2$, the identity $J(-\pi /2)=XJ(\pi /2)$ absorbs the possible sign change into the Pauli frame. If $t_j=0$ for ${\alpha }_j=0$ and $t_j=1$ for ${\alpha }_j=\pi /2$, the update isInductively the final unmeasured quantum state would be $X^{a_m}{Z}^{b_m}J({\alpha }_m)\cdots J({\alpha }_1)|+⟩$. The final computational basis outcome is corrected by $a_m$ and is unaffected by $b_m$. The induction on a sequential interpretation proves the joint statistics; projectors on distinct vertices commute, so the identical fixed-basis pattern can actually be measured simultaneously, including its final vertex.

The first printed commutation formula has an incorrect exact scalar phase: with the printed matrix definition, $J(\alpha )X=e^{i\alpha }ZJ(-\alpha )$, rather than the negative-exponent prefactor. For example, at $\alpha =\pi /4$ the two versions differ by a factor $i$. They agree up to global phase, so this error does not affect the Pauli frame recurrence or any quantum measurement probability.

Consequently a fixed-basis path-state pattern of logical depth one simulates every such circuit. Only Pauli measurements occur: the angle-zero basis measures $X$, and the angle-$\pi /2$ basis measures $-Y$ with the printed eigenvector labels. Classical exclusive or processing suffices for all output corrections. These gates are Clifford gates, which explains why angle adaptation can be eliminated. For $m=0$, simply measure the initial $|+⟩$ in the computational basis; it is already a depth-one pattern.

If $f$ is constant, thenThe sign is a physically irrelevant global phase.

Alice performs a projective Bell-basis measurement on her qubits $C$ and $A$. Outcome $(x,z)$ occurs with probability $1/4$ and leaves Bob's qubit inAlice sends the two classical bits $(x,z)$ to Bob. He applies the inverse Pauli operationrecovering $|\alpha ⟩$ exactly, up to an irrelevant global phase. This is quantum teleportation: only local operations and classical communication are used, while the shared Bell pair supplies the entanglement resource.

One classical query at any chosen $x_0$ returns only $v=s{x}_0+t$. Every possible slope $s\in {\mathbb{Z}}_3$ is compatible with that answer by choosing $t=v-s{x}_0$, so no one-query classical algorithm can determine $s$ with certainty.

For the quantum algorithm, let $\omega =e^{2\pi i/3}$ and preparePart (b) shows that the single query applies quantum phase kickback:The unknown intercept contributes only a global phase. Applying ${\mathrm{QFT}}_3^{-1}$ to the first register produces $|s⟩$, whose computational-basis measurement determines $s$ with certainty after one query.

This circuit is the swap test. If the two pure states are identical up to a physically irrelevant global phase, then $|⟨\psi |\varphi ⟩|=1$ and outcome one never occurs. If they are distinct, then $|⟨\psi |\varphi ⟩|<1$ and each run produces outcome one with positive probability $p_1$.

Run the circuit independently on fresh copies. Observing even one outcome one proves that the states differ. If no one occurs in $N$ runs, that event has probabilitywhen the states differ, which tends to zero as $N\to \infty$. Repetition therefore distinguishes equality from inequality with arbitrarily high probability.

For $p=1/4$, put $|{\psi }_{\mathrm{in}}⟩=\frac{1}{2}|g⟩+\frac{\sqrt{3}}{2}|b⟩$. After the first phase rotation the state isSince $U{R}_{00}^{\pi }{U}^{†}=I-2|{\psi }_{\mathrm{in}}⟩⟨{\psi }_{\mathrm{in}}|$, the requested state isIt is proportional to $|g⟩$ exactly when $e^{i\varphi }=-1$, hence when $\varphi =\pi$. The state is then $-|g⟩$, whose global phase has no physical effect.

Put $\sqrt{p}=\sin \theta$ and $\sqrt{1-p}=\cos \theta$, where $0<\theta <\pi /2$. In the ordered basis $(|b⟩,|g⟩)$, the product of the two reflections isThe leading minus sign is a physically irrelevant global phase. The remaining rotation matrix rotates the good-bad plane through $2\theta$, whereEquivalently, one iterate changes the initial angle $\theta$ to $3\theta$.

The operators $X_i$ act on different qubits and therefore commute, soSince $X=HZH$,The scalar $e^{-it}$ is a global phase. Thus each factor uses two Hadamard gates and one phase gate, and applying the factors in parallel or sequentially gives an exact quantum circuit of $\boxed{3n}$ elementary gates.

Each summand of $J_X$ acts on one qubit, while each summand $Z_i{Z}_j$ of $J_Z$ acts on two. Therefore $J=J_X+J_Z$ is a 2-local Hamiltonian. We haveSplit $t$ into $r$ steps of length $\delta =t/r$ and use the second-order product formulaFor one step, $\Lambda =O(\delta n^2)$ in the stated estimate, so the spectral-norm error is $O({\delta }^3{n}^6)$. The error bound for a product of unitary operators makes the total errorIt is therefore enough to choosewith $r$ also large enough that the small-step estimate applies.

All $Z_i{Z}_j$ terms commute. A factor $e^{-i\delta Z_i{Z}_j}$ uses a constant-size circuit of two controlled-NOT gates and one phase gate, up to a global phase, so one product-formula step costs $O(n^2)$ gates. The complete Hamiltonian simulation consequently has size

With equatorial qubit measurement vectors $(|0⟩+(-1{)}^s{e}^{i\eta }|1⟩)/\sqrt{2}$, the gate in one-bit teleportation is $U(\eta )=H\mathrm{diag}(1,e^{-i\eta })=J(-\eta )$. Direct multiplication gives $U(\eta )Z=XU(\eta )$ and $U(\eta )X=e^{-i\eta }ZU(-\eta )$. Moving a Pauli frame through this gate therefore exchanges its $X$ and $Z$ exponents and reverses the angle if the old $X$ exponent is one, as displayed up to global phase. A measurement outcome $s$ adds a further $X^s$ factor. This calculation supplies adaptive angle choices and the final classical output-bit correction of a linear cluster state computation.

A Hamiltonian operator has perfect transfer between basis states $|a⟩,|b⟩$ at time $t_{\ast }$ if $e^{-iH{t}_{\ast }}|a⟩=e^{i\chi }|b⟩$ for a known global phase $\chi$. In an excitation-conserving qubit network whose vacuum is fixed, the initial product state with $\alpha |0⟩+\beta |1⟩$ at site $a$ and zeros elsewhere evolves to $\alpha |\mathrm{vac}⟩+\beta e^{i\chi }|b⟩$. A local phase gate at $b$ removing $\chi$ restores the unknown qubit there. This explains why transfer of the single-excitation basis state suffices for qubit transfer when the vacuum phase is also accounted for.

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.

A projective measurement has self-adjoint orthogonal projections $P_i$ satisfying $P_i{P}_j={\delta }_{ij}{P}_i$ and ${\sum }_i{P}_i=I$. Measuring an observable with discrete spectral decomposition $A={\sum }_i{a}_i{P}_i$ reports eigenvalue $a_i$ with Born rule probability $p_i=\mathrm{Tr}(P_i\rho )$. A projector can have rank greater than one: the outcome then identifies an eigenspace, rather than a particular eigenvector inside it. Projectors specify the outcome statistics; the usual ideal conditional update is the additional Lüders rule.

In the qubit example from measurement in quantum mechanics, $Z=|0⟩⟨0|-|1⟩⟨1|$. At $\theta =\pi /3$, the zero and one probabilities are $3/4$ and $1/4$, regardless of the relative quantum phase $\varphi$. Under the Lüders rule, obtaining zero prepares $|0⟩$; obtaining one prepares $|1⟩$, up to global phase. Repeating the same measurement returns the same label with certainty, provided no intervening evolution or operation changes the measured sector. Repeatability concerns the conditional output, and does not mean the first outcome was predetermined.

Change to $X$ and its Pauli X eigenstates $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. Adding the two probability amplitudes before squaring givesAt $\theta =\pi /3$ and $\varphi =0$, $p_{+}=(2+\sqrt{3})/4\simeq 0.9330$ and $p_{-}\simeq 0.0670$. At $\varphi =\pi /2$, both are $1/2$; at $\varphi =\pi$, the first pair is exchanged. The $Z$ statistics remain $3/4,1/4$ throughout. This is the same addition-and-cancellation principle as wave interference: relative quantum phase changes how amplitudes reinforce or cancel in another basis.

The beauty of this basis dependence is that information invisible to one measurement becomes accessible to another. A $Z$ measurement of an ensemble estimates its quantum state populations, while $X$ probes one component of its quantum coherence. These are experiments on separate identically prepared copies. Sequentially measuring $Z$ and then $X$ on one copy instead probes the disturbed post-measurement state. Quantum state tomography combines suitable ensemble statistics to reconstruct the original state.

For an orthogonal projection, $P^k=P$ for $k\ge 1$. Summing the exponential series proves the displayed identity. At $\theta =\pi$, the operator is $I-2P$, which changes the sign on the range and fixes its orthogonal complement. In Grover search algorithm, the marked-state reflection and the uniform-state reflection compose to the search iterate up to an irrelevant global phase.

For a two-variable Boolean quantum oracle, quantum phase kickback produces the phase $(-1{)}^{axy+bx+cy+d}$ on $|xy⟩$. If the quadratic coefficient $a$ is known, apply the Controlled-Z gate to the power $a$: its phase $(-1{)}^{axy}$ cancels the quadratic term. A Walsh-Hadamard transform then maps the remaining phase state to $(-1{)}^d|bc⟩$. The constant term is only a global phase, and both linear coefficients are read in one quantum measurement in the computational basis.

A measurement in quantum mechanics turns a quantum state preparation into a classical outcome, such as a detector label. Two questions must be distinguished: how likely is each outcome, and what quantum state remains conditional on that outcome? A positive operator-valued measure answers the first; a quantum instrument answers both. This distinction is useful whenever a measured system participates in a later experiment, including quantum state tomography and measurement-based quantum feedback.

For a density operator $\rho$, the Born rule gives $p_m=\mathrm{Tr}(E_m\rho )$, where the $E_m$ are positive operators satisfying ${\sum }_m{E}_m=I$. Here $I$ is the identity operator, and $\mathrm{Tr}$ denotes the trace. Positivity makes every $p_m$ nonnegative; completeness makes their sum one. For a pure state, $\rho =|\psi ⟩⟨\psi |$ and $p_m=⟨\psi |E_m|\psi ⟩$. A projective measurement is the special case in which the $E_m$ are orthogonal projections onto mutually orthogonal subspaces.

For a qubit, choose the computational basis and write a normalized pure state, up to global phase, asThe angle $\varphi$ is a relative quantum phase; at the poles it does not distinguish states. Measuring the Pauli matrix $Z$ reports $+1$ for label zero and $-1$ for label one, withThese probabilities describe repeated identical preparations. They do not assign a hidden, pre-existing value to every possible measurement basis.

The Born rule grew out of Max Born's 1926 work on quantum scattering: squared probability amplitudes determine the relative frequencies of scattering outcomes. Born explained this statistical interpretation again in his 1954 Nobel lecture. This probability law does not by itself select a post-measurement state. ^[ref]; ^[ref].

Multiplying a quantum state by one global complex phase does not change any measurement probability; relative phases between components can affect interference.

The global phase is common to the whole state; relative phases control interference between state components.

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.

For a fixed Hilbert space, a bounded self-adjoint Hamiltonian operator with a finite step $H_i\to H_f$ at $t=0$ does not deliver an instantaneous rotation of the quantum state. Integrating the Schrödinger equation across a short interval givesFor a normalized quantum state and $\Vert H(t)\Vert \le M$ in operator norm, the difference has norm at most $2ϵM/\hslash$, which tends to zero. The time derivative can jump even though the quantum state is continuous. Here $\hslash$ is the reduced Planck constant. An instantaneous replacement of the eigenbasis changes coordinates, not the vector being represented.

For a real switch occupying $0\le t\le \tau$, remove any real scalar contribution $a(t)I$ that produces only a global phase, and write $|\tilde{\psi }(t)⟩=e^{i{\int }_0^{t}a(s)ds/\hslash }|\psi (t)⟩$. The same integral argument gives the sufficient bounded-model estimateThus $\epsilon \ll 1$ controls the sudden approximation up to a global phase. For any normalized final energy eigenstate, its probability amplitude differs from the instantaneous-switch value by at most $\epsilon$, and its Born rule probability differs by at most $2\epsilon$. These are sufficient error bounds, not necessary criteria. For the fixed-gap two-level sudden quench, any switching path $H(t)=-\Delta \mathbf{n}(t)\cdot \mathbf{\sigma }/2$ with a unit vector $\mathbf{n}(t)$ has operator norm $\Delta /2$, giving $\epsilon \le \Delta \tau /(2\hslash )$.

This estimate is deliberately for bounded Hamiltonian operators. An unbounded energy spectrum does not supply a finite global $M$; relevant quantum states, operator domains and excited energy eigenvalues require separate control. The adiabatic theorem also requires hypotheses about the path and its spectral gaps; there is no universal switch-duration criterion covering all spectra. The cold-atom collapse and revival after a lattice quench illustrates how a single switch can be sudden for one degree of freedom and slow for another.