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Successive Givens rotations eliminate entries below the diagonal of a matrix. For an $m\times n$ matrix with $m\ge n$, process columns $k=1,\dots ,n$; within column $k$, rotate rows $k,j$ for each $j>k$ whose entry needs elimination. Apply each Givens row-pair update to columns $k,\dots ,n$. Both rows are zero in earlier columns, so these operations preserve earlier eliminations. No assumption of full column rank is needed to perform the eliminations.

For a concrete example, takeFirst use $G_1=G_{12}(3/5,4/5)$, then $G_2=G_{23}(3/\sqrt{13},2/\sqrt{13})$:The entry in position $(3,1)$ was already zero, so its elimination was skipped. The second Givens rotation combines the remaining pair $(3,2{)}^T$ in column two. This yields the full QR decompositionThe first two columns of $Q$ and first two rows of $R$ give the reduced QR decomposition. Direct multiplication verifies both $QR=A$ and $Q^{T}Q=I$. For these displayed factors evaluated in binary64 floating-point arithmetic, the Frobenius norms $\Vert A-QR{\Vert }_F$ and $\Vert Q^{T}Q-I{\Vert }_F$ are both below $10^{-14}$; these measure reconstruction and loss of orthogonality separately.

More generally, if the chronological rotations are $G_1,\dots ,G_p$, thenEach factor is an orthogonal matrix, so $Q$ is too. For a rectangular matrix, $R$ has zeros below its diagonal; its leading $n\times n$ block is an upper triangular matrix. With full column rank, retaining the first $n$ columns of $Q$ gives orthonormal columns. To impose a positive diagonal when necessary, flip matching columns of $Q$ and rows of $R$, preserving their product.

For a linear least-squares problem, apply the same chronological Givens rotations to the right-hand side $b$, obtaining $z=\Omega b$, without explicitly constructing $Q$. Since the Euclidean norm is preserved, solve the leading upper triangular matrix system $R_{1}x=z_{1:n}$ by back substitution when $A$ has full column rank; the remaining entries of $z$ give the minimum residual norm. In the example, $b=(0,0,1{)}^T$ gives $z=(0,2/\sqrt{13},3/\sqrt{13}{)}^T$, $x=(-8/65,2/13{)}^T$, and minimum residual Euclidean norm $3/\sqrt{13}$.

The locality is useful for a sparse matrix or an existing QR decomposition needing an extra row: only two rows change at each step, and the new row can be eliminated against the stored upper triangular matrix. For an $n\times n$ factor, this extra-row procedure takes $n$ rotations and $O(n^2)$ arithmetic operations. Locality does not guarantee preservation of sparsity: mixing rows can create nonzeros wherever either row had one. For a fully populated $m\times n$ input, the straightforward elimination uses ${\sum }_{k=1}^n(m-k)$ rotations, each touching $n-k+1$ columns, giving $O(m{n}^2)$ work. Storage layout, zero patterns and implementation affect speed; this count alone establishes no universal advantage over other QR decomposition algorithms.

The Euclidean norm preservation explains the appeal for numerical stability: exact rotations do not amplify a perturbation's length. Rounded coefficients and Givens row-pair updates still introduce roundoff error. Small reconstruction and orthogonality residuals check the computed factors, while the condition number of the original problem still controls how accurately a linear least-squares problem determines its solution.

For $b>0$ and $-b\le y\le b$, the even quadratic polynomial $b^2-y^2$ has the displayed expansion in even Dirichlet eigenfunctions. Put $k_n=(2n+1)\pi /(2b)$. Their squared norm on $[-b,b]$ is $b$. Two integrations by parts, using $\cos (k_{n}b)=0$, giveDivision by $b$ gives the displayed Fourier coefficients. The cubic decay gives uniform convergence. The half-integer frequencies arise from requiring both evenness and zero values at the endpoints, rather than from a periodic boundary condition. This expansion is useful in rectangular duct flow.

For a positive self-adjoint Hamiltonian operator, finite energy means that a normalized state lies in $D(H^{1/2})$, its form domain, and has expectation value $q_H(\psi )$. This domain can be larger than $D(H)$. With a complete orthonormal basis of energy eigenstates, $\psi ={\sum }_n{c}_n{e}_n$ and $H{e}_n=E_n{e}_n$ give $q_H(\psi )={\sum }_n{E}_n|c_n{|}^2$ by applying $H^{1/2}$ and taking its squared norm. Membership in $D(H)$ instead requires ${\sum }_n{E}_n^2|c_n{|}^2<\infty$. For a Dirichlet free-particle Hamiltonian operator on an interval, the energy form is ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)$ on functions with square-integrable weak derivative and zero endpoint values.

For an admissible quantum state with zero position and momentum means, positivity of the squared norm $\Vert (p-isx)\psi {\Vert }^2$ for every real $s$ givesThe discriminant is nonpositive, hence $\Delta x\Delta p\ge \hslash /2$. Subtracting the expectation values from the position operator and momentum operator extends this outline to nonzero means.

A subset $A$ of a finite abelian group $G$ with subset density $\eta =|A|/|G|$ is quadratically $\alpha$-uniform, in the norm convention, if its balanced subset indicator $g=1_A-\eta$ has Gowers U3 norm at most $\alpha$. This controls correlations on additive three-dimensional cubes after the subset density contribution is removed. If a convention instead bounds the eighth power of the norm by $\alpha$, its parameter corresponds to the norm parameter ${\alpha }^{1/8}$. The convention must therefore be stated when giving quantitative thresholds.

The ensemble density of actual configurations equals the squared norm of the guiding wave function. For spinless particles with $v_k=j_k/|\Psi {|}^2$, the Schrödinger continuity equation is ${\partial }_t|\Psi {|}^2+{\sum }_k{\nabla }_k\cdot (|\Psi {|}^2{v}_k)=0$. A density transported by the same guiding field satisfies the same equation, so initial equality is preserved. This equivariance is a consistency property, not a proof that every possible initial ensemble is in equilibrium. The physical account of the equilibrium assumption and subsystem statistics requires an additional argument.

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.

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 quantum superposition is a linear combination of quantum state vectors in a complex Hilbert space. In a specified orthonormal basis, a normalized pure state has the displayed expansion, where $a_j$ are probability amplitudes. A nonzero linear combination can always be normalized; an arbitrary sum of normalized vectors need not already have norm one. Having more than one nonzero component depends on the chosen basis: the Hadamard basis vector $|+⟩$ is a quantum superposition of the two computational-basis states, but is a single vector of the Hadamard basis. ^[ref].

The distinction from classical ignorance is quantitative. Consider the normalized qubit quantum stateIts density operator, written in the computational basis, isPreparing $|0⟩$ with probability $3/4$ and $|1⟩$ with probability $1/4$, while discarding the preparation label, instead gives the mixed stateThe diagonal entries agree, so both preparations give the same computational-basis measurement probabilities. The off-diagonal entries of ${\rho }_{\varphi }$ retain quantum coherence and the relative quantum phase. In the Hadamard basis, the Born rule givesThe density matrix representation retains this distinction between a pure state and a mixed state. ^[ref].

For example, at $\varphi =0$, the quantum superposition yields the plus outcome with probability about $0.9330$, whereas the mixed state yields it with probability $0.5$. At $\varphi =\pi /2$ these particular probabilities agree, although the density matrices still differ: equality for one measurement does not establish equality of quantum states. The probability amplitude figure shows the comparison throughout a full cycle of relative quantum phase.

This is why a quantum superposition is useful: quantum gates act linearly on its probability amplitudes, and unitary operators preserve the full pure state information, including relative quantum phases. A Hadamard gate can turn otherwise hidden quantum coherence into different outcome probabilities. The beautiful feature is that the same basis weights can support different predictions when amplitudes recombine. The expansion specifies a preparation and its measurement predictions; it does not assign every unmeasured outcome a simultaneous classical value. See projective measurement for the separate question of the conditional state after a measurement.

Quantum theory supplies a common language for quantum mechanics, measurement in quantum mechanics, and quantum information: prepare a quantum state, transform it, and ask which experimental outcomes occur. In a finite-dimensional idealization, a pure state is a normalized vector of complex numbers. Its entries are probability amplitudes; the Born rule turns their squared magnitudes into probabilities. Between measurements, an isolated system evolves by a unitary operator, preserving its norm and inner products. A quantum circuit makes this evolution explicit as a sequence of quantum gates.

The extra information carried by quantum phases is essential. Relative quantum phases affect how probability amplitudes add when different contributions reach the same output. Their interference can enhance one outcome and cancel another. A global phase, multiplying the entire quantum state by the same complex factor of magnitude one, changes neither these relative phases nor observable predictions. The Hadamard gate below provides a small experiment in which this distinction becomes a deterministic change of output.

Quantum coherence describes this capacity to interfere relative to a specified orthonormal basis. In a density matrix, it appears in off-diagonal entries; a diagonal mixed state retains the populations but lacks that basis coherence. Thus matching the probabilities of one measurement need not match the predictions of a later experiment. Baumgratz, Cramer, and Plenio made this basis-dependent viewpoint precise in their resource framework for quantum coherence. ^[ref].

This makes quantum information an operational question: which distinctions can a preparation, an available unitary operator, and measurement in quantum mechanics reveal? Amplitude amplification illustrates how carefully arranged interference converts a weakly represented answer into a likely observable result. Preserving the relevant quantum coherence throughout the computation is part of the physical task. ^[ref].

A quasi-norm on a vector space is nonnegative, vanishes only at zero, is absolutely homogeneous, and satisfies a triangle inequality with one constant factor $C\ge 1$. For $0<p<1$, $(\int |f{|}^p{)}^{1/p}$ on an Lp space is a quasi-norm: $\Vert f+g{\Vert }_p^p\le \Vert f{\Vert }_p^p+\Vert g{\Vert }_p^p$ gives a quasi-triangle inequality. Measure-preserving composition preserves this quantity even though the Minkowski inequality is unavailable below exponent one.

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].

For a unital irreducible action and $\xi \ne 0$, the annihilator $L_{\xi }=\{a:\pi (a)\xi =0\}$ is a maximal left ideal and is closed. Thus $A/L_{\xi }$ supplies the displayed complete norm on the module, with contractive action. Every module endomorphism is bounded for this quotient norm: its value at the cyclic vector is $a\xi$, so it is induced by bounded right multiplication by $a$ on the quotient. The Gelfand-Mazur theorem then makes the module's commutant consist of scalar operators. This constructs one compatible norm; it does not by itself establish continuity in a different preassigned norm.

For the radial Laplace operator $L=-r^{-2}(r^2{u}^{\prime }{)}^{\prime }$ on $0<r<R$, require a finite value at the origin and $u(R)=0$. The map $u\mapsto v=ru$ preserves the norm from $L^2((0,R),r^{2}dr)$ to $L^2(0,R)$ and sends $L$ to $-d^2/d{r}^2$. The regular endpoint condition becomes $v(0)=0$. The self-adjoint realization therefore corresponds to the ordinary Dirichlet boundary conditions for the one-dimensional second derivative. Its eigenvalues and eigenfunctions are displayed above, with $u_n(0)=n\pi /R$ understood by a limit. Their squared weighted norm is $R/2$. Completeness and orthogonality follow from the Fourier sine series on $(0,R)$.

A radial profile is a one-variable function $\bar{u}$ representing a radial function $u$ by $u(x)=\bar{u}(|x|)$. On an annulus $a<|x|<b$, with $a>0$, the spherical coordinate system identifies its Lp norm with the weighted profile norm through $\int |u{|}^p=|S^{n-1}|{\int }_a^b|\bar{u}(r){|}^p{r}^{n-1}dr$. Thus equality of two radial functions almost everywhere identifies their profiles almost everywhere in the ordinary one-dimensional sense.

Removing any interior point $a$ from the Euclidean closed unit ball allows a retraction onto its unit sphere. For $v=x-a\ne 0$, the positive root of $|a+tv{|}^2=1$ isThen $r(x)=a+\tau (x)v$ is a continuous map on the punctured closed unit ball, has norm one, and fixes its boundary because $\tau (x)=1$ there. The quadratic equation has one positive and one negative root since $|a|<1$, so the chosen root is unambiguous.

For $H_R=\lambda (k_x{\sigma }_y-k_y{\sigma }_x)$ with $\lambda >0$, the lower- and upper-band spinors of unit norm can be chosen asTheir spin expectation values are $⟨\mathbf{S}{⟩}_{\pm }=\pm (\hslash /2)(-k_y,k_x,0)/k$. They are tangential to circles of constant momentum magnitude, opposite between branches, and reversed at opposite momenta as required by time-reversal symmetry. When $E_F>0$, the outer contour belongs to the lower branch and the inner one to the upper branch. For $E_F<0$ the two Fermi contours both belong to the lower branch and have the same tangential spin orientation at a given angle in polar coordinates. At $k=0$ the Hamiltonian has quantum degeneracy in spin, and no unique spin texture direction is selected.

The Rayleigh quotient measures the quadratic form of a self-adjoint operator per unit squared norm. For a nonzero vector $x$ and a finite-dimensional Hermitian matrix $A$,For a real symmetric matrix, replace the conjugate transpose $x^{\ast }$ by $x^T$. More generally, for a self-adjoint operator $H$ on a Hilbert space, use $R_H(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$ for nonzero $\psi$ in its operator domain. On an eigenvector it returns the exact eigenvalue. Multiplying the vector by any nonzero scalar leaves the quotient unchanged, so it measures a direction rather than a chosen length.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ from the finite-dimensional spectral theorem example,The first value is not an eigenvalue: it is an average of the two directional responses. To see every direction, write the unit vector $q(\theta )=\cos \theta e_{+}+\sin \theta e_{-}$ in the orthonormal eigenbasis. ThenThe largest value is $3$ along $e_{+}$; the smallest is $1$ along $e_{-}$. A direction halfway between them has quotient $2$. In the left panel, the color on the unit circle represents the quotient; in the right panel, the same values are plotted against the angle measured from $e_{+}$.

The Hermitian Rayleigh quotient range explains why this is useful: evaluating one trial vector gives an upper bound on the smallest eigenvalue and a lower bound on the largest one, without solving a characteristic polynomial. At an extreme eigenvector, a small angular error changes the quotient only quadratically. Near $e_{+}$ in this example, $3-R_A(q(\theta ))=2{\sin }^2\theta =2{\theta }^2+O({\theta }^4)$. This is the geometric reason for quadratic Rayleigh-quotient improvement for the power method and motivates the Rayleigh-Ritz method.

There is a concrete historical link with normal modes. In The Theory of Sound (1877), Volume I, Sections 88–89, Lord Rayleigh studies the period of a constrained trial vibration. He shows that the period has a stationary point at a normal mode and that a trial type bounds the longest natural period. These energy ratios are a mechanical setting for the modern Rayleigh quotient and generalized Rayleigh quotient; the book describes the result through vibration periods, rather than today's matrix notation.^[ref]

For $u_t-u_{xx}+u=0$ with homogeneous mixed Dirichlet and Neumann boundary conditions, testing the Backward Euler method finite element equation with the new solution gives the displayed identity. Both the squared derivative norm and the squared function norm are required in the dissipation, because diffusion and reaction contribute separately. Replacing the full $H^1$ norm by its seminorm omits the reaction term. Positivity gives unconditional decay of the $L^2$ norm.

A real-valued function takes each function input to a real number. Its codomain can be all of $\mathbb{R}$ or a specified subset. Thus $f(x)=x^2$ on the real line is a real-valued function and has image $[0,\infty )$, whether the declared target is that interval or all of $\mathbb{R}$.

The inputs need not be real numbers: the cardinality of a finite set and the squared norm of a vector are also real-valued functions. The adjective describes outputs, while scalar-valued function allows outputs in other fields, notably the complex numbers.