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Use the normalized eigenfunctions and their weighted inner product ${\int }_a^{b}w\overline{{\psi }_m}{\psi }_{n}dx={\delta }_{mn}$. Orthogonality givesFor complex trials, $F$ means the corresponding Hermitian quadratic form with $|{\psi }^{\prime }{|}^2,|\psi {|}^2$; it agrees with the displayed functional on real trials. If $a_0\ne 0$, division by $|a_0{|}^2$ yieldsSince ${\lambda }_n\ge {\lambda }_0$, this is at least ${\lambda }_0$. The Rayleigh-Ritz variational principle therefore gives an upper bound from any admissible trial, improved by minimizing over trial parameters or enlarging the trial space. Equality holds when the trial is entirely in the lowest eigenspace.

The displayed ratio using $b_n$ presupposes $a_0\ne 0$. If $a_0=0$, use the undivided quotient $\sum {\lambda }_n|a_n{|}^2/\sum |a_n{|}^2$ instead; the same upper-bound argument still applies. No ground-state coefficient can be divided out in that case.

Since both real matrices are antisymmetric, $A^{†}=-A$ and $B^{†}=-B$. Thus $U^{†}=-A+iB$. Expand the two unitary identities:Subtracting and adding give $\boxed{AB=BA,A^2+B^2=-I}$. Because $A$ has simple eigenvalues, each of its complex eigenspaces is one-dimensional. Commutation makes each one invariant under $B$: if $Av=av$, then $A(Bv)=a(Bv)$. Hence $Bv=bv$ for some scalar $b$. The complete eigenbasis of $A$ is therefore a common eigenbasis, which also diagonalizes $U$.

The skew-Hermitian decomposition of a unitary matrix is explicit:If $Uv=uv$, unitarity gives $U^{†}v=u^{-1}v=\overline{u}v$. ConsequentlyEquivalently, for $u=e^{i\theta }$ the paired eigenvalues are $\boxed{a=i\sin \theta ,b=-i\cos \theta }$. The hypotheses can only occur in even dimension: $U^T=-U$ and $U$ is invertible, whereas an odd-dimensional skew-symmetric matrix has zero determinant.

The plane spanned by $u_1,v_1$ is an invariant subspace for $M$, sinceIts orthogonal complement is also invariant: for $w$ orthogonal to both, $u_1^{T}Mw=-(M{u}_1{)}^{T}w=0$ and $v_1^{T}Mw=-(M{v}_1{)}^{T}w=0$. The restriction of the real symmetric matrix $M^2$ to this nonempty complement therefore has a unit eigenvector $u_2$, by the spectral theorem for real symmetric matrices.

If its eigenvalue is $-{\mu }_2^2<0$, the argument in part (b) shows that $v_2=M{u}_2/{\mu }_2$ is a unit eigenvector orthogonal to $u_2$. It also lies in the same complement; explicitly,Repeating the construction decomposes every negative eigenspace into orthogonal pairs. The zero eigenspace has dimension $n-2r$ for some number $r$ of pairs, and therefore has the same parity as $n$. This is the paired eigenspaces of the square of a real skew-symmetric matrix property.

For $n=2m$, there can be $m$ distinct negative eigenvalues and no zero, or at most $m-1$ negative ones plus zero, because a nonempty zero eigenspace has dimension at least two. For $n=2m+1$, there are at most $m$ distinct negative ones plus the mandatory zero. HenceThe bound is attained by skew two-by-two blocks with distinct positive parameters, adding one zero direction when $n$ is odd.

For a real symmetric matrix $A$, find its real eigenvalues from $\det (A-\lambda I)=0$, then solve $(A-\lambda I)v=0$ for the corresponding eigenspaces. Distinct eigenspaces are orthogonal. Within each repeated eigenspace, apply the Gram-Schmidt process and normalize to obtain an orthonormal basis of eigenvectors.

Put these vectors in the columns of $O$. The spectral theorem for real symmetric matrices guarantees a full basis, so $O^{T}O=I$ andThis is the orthogonal diagonalization of $A$ and the associated orthogonal change of coordinates.

Use the standard regular Sturm-Liouville problem hypotheses: positive weight $w$, real coefficients, and real self-adjoint boundary conditions. The positive weight and boundary-domain assumptions are implicit in the requested real orthonormal basis; real differential coefficients alone would not guarantee them for arbitrary complex boundary conditions.

Multiply the eigenfunction equation by $\overline{y_n}$ and integrate. By self-adjointness, the number ${\int }_a^b\overline{y_n}\mathcal{L}{y}_{n}dx$ is real, whereasThe second integral is strictly positive, so $\boxed{{\lambda }_n\in \mathbb{R}}$. Because the coefficients and boundary conditions are real, the real and imaginary parts of any complex eigenfunction separately obey the same equation and conditions. At least one is nonzero, so a real eigenfunction can be chosen. More generally, each eigenspace has a real basis, which can be orthonormalized using the weighted inner product.

The orthogonality of Sturm-Liouville eigenfunctions, with normalization chosen to one, isFor the real choice, omit the complex conjugate.

Choose a real unit eigenvector $e_1$ of the real symmetric matrix $Q$, with eigenvalue ${\lambda }_1$. Its orthogonal complement is an invariant subspace: for any $v$ with $e_1^{T}v=0$,The restriction of $Q$ to this $(n-1)$-dimensional space is again real and symmetric. Applying the same argument inductively produces an orthonormal basis $e_1,\dots ,e_n$ of real eigenvectors. The base case $n=1$ is immediate. This proves the spectral theorem for real symmetric matrices without assuming that distinct eigenvalues are available.

One can also check that eigenvectors with different eigenvalues are automatically orthogonal: ${\lambda }_i{e}_i^T{e}_j=(Q{e}_i{)}^T{e}_j=e_i^{T}Q{e}_j={\lambda }_j{e}_i^T{e}_j$. Within a repeated eigenspace, the Gram-Schmidt process preserves the eigenvector property. Thus repeated eigenvalues cause no obstruction, and $\boxed{Q=OD{O}^T\text{ for an orthogonal }O\text{ and real diagonal }D}$.

The printed wording calls ${\Sigma }_3$ cyclic, but its meaning in question 9 is the symmetric group $S_3$. Under that interpretation, take its planar dihedral group representation, withHere $r^3=s^2=1$ and $srs=r^{-1}$. This representation is irreducible over the complex numbers: the two distinct rotation eigenspaces are exchanged by the reflection, so neither is invariant under the whole group. Its character is $2$ on the identity, $-1$ on the two 3-cycles and $0$ on the three transpositions. Squaring maps each transposition to the identity and permutes the 3-cycles. Its Frobenius-Schur indicator is thereforeIf “cyclic” were intended literally, no two-dimensional complex irreducible representation would exist, since an irreducible representation of a finite abelian group is one-dimensional.

The power iteration starts from $v_0\ne 0$, normally scaled to unit Euclidean norm, and repeatsNormalization prevents the rapidly growing or decaying magnitudes of $A^k{v}_0$ from obscuring its direction. The iteration requires $w_{k+1}\ne 0$.

Because $A$ is a diagonalizable matrix, choose an eigenbasis $u_1,\dots ,u_n$, with corresponding eigenvalues ${\lambda }_1=\lambda ,{\lambda }_2,\dots ,{\lambda }_n$, and expand $v_0={\sum }_j{c}_j{u}_j$. First suppose the dominant eigenvalue is simple and $c_1\ne 0$. ThenForthe sum in parentheses after $u_1$ is $O(q^k)$ in Big O notation. Consequently the power iteration direction error is $O(q^k)$: the iterates approach the line spanned by $u_1$. For a real matrix, a uniquely dominant eigenvalue is real, since otherwise its complex conjugate would have the same modulus. If $\lambda >0$, $v_k$ approaches one unit eigenvector; if $\lambda <0$, successive $v_k$ alternate signs. In either case,This is the sign and phase in power iteration qualification. Also, the necessary assumption is a nonzero coefficient $c_1$ in the eigenbasis, rather than merely a nonzero ordinary inner product with $u_1$: those tests can differ for a nonsymmetric matrix. If $c_1=0$, the method cannot generate the missing component.

If the dominant value $\lambda$ has a higher-dimensional eigenspace, the same argument replaces $c_1{u}_1$ by the nonzero sum of all starting components in that eigenspace. The limiting direction is then that of this sum, which is still an eigenvector for $\lambda$. The usual algorithm assumes $\lambda \ne 0$; a strict dominance gap guarantees this whenever another eigenvalue is present.

An estimate retaining the sign of $\lambda$ is the Rayleigh quotientContinuity of this quotient at a nonzero eigenvector proves the limit even when $A$ is a nonsymmetric matrix. Alternatively, $(A{v}_k{)}_i/(v_k{)}_i\to \lambda$ for a coordinate whose limiting eigenvector component is nonzero. The scaling factor $\Vert A{v}_k{\Vert }_2$ alone tends only to $|\lambda |$.

Choose a shift $\mu$ which is not an eigenvalue. Inverse iteration applies power iteration to $B=(A-\mu I{)}^{-1}$, implemented by a linear system solve rather than by forming an explicit matrix inverse:A single LU decomposition of $A-\mu I$ can be reused for all the solves at this fixed shift. An eigenpair $A{u}_j={\lambda }_j{u}_j$ transforms intoThus an eigenvalue ${\lambda }_{\ast }$ uniquely closest to $\mu$ becomes the dominant eigenvalue of $B$. For a diagonalizable matrix, the argument in the preceding subpart gives convergence of diagonalizable fixed-shift inverse iteration, provided the starting coefficient in the target eigenspace is nonzero, with power iteration direction error $O(q_{\mu }^k)$, whereFor any prescribed simple eigenvalue, taking a sufficiently close but unequal shift makes the numerator small while the denominator stays close to its positive separation from the remaining eigenvalues. This makes $q_{\mu }$ small and gives very fast convergence per iteration. The Rayleigh quotient of $v_k$ estimates ${\lambda }_{\ast }$ directly. If ${\beta }_k$ estimates the associated eigenvalue of $B$, the corresponding estimate is $\mu +{\beta }_k^{-1}$.

The selection principle also applies when the other eigenvalues have nontrivial blocks in the Jordan normal form: their contributions acquire powers of $k$, multiplying strictly smaller geometric factors. A simple uniquely closest eigenvalue still attracts the direction when the Riesz projection of $v_0$ onto its generalized eigenspace is nonzero. For nonreal target eigenvalues, allow complex shifts and arithmetic; a real shift cannot distinguish a conjugate pair of equal distances. Setting $\mu ={\lambda }_{\ast }$ makes the solve singular, and an exactly zero starting target component prevents convergence to that target in exact arithmetic. When ${\lambda }_{\ast }<\mu$ is real, account for the sign behavior of fixed-shift inverse iteration.

Let ${\chi }_{\pm }=(1,\pm 1{)}^T/\sqrt{2}$. The eigenvalues of ${\Sigma }_1$ are $\pm 1/2$, each with multiplicity two, and orthonormal bases for their eigenspaces areA general linear combination of the two positive-energy solutions has upper spinor $\xi =a{\chi }^1+b{\chi }^2$ and lower spinor $(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$. An x-spin eigenvector needs $\xi \propto {\chi }_{\pm }$ and the lower spinor to have the same ${\sigma }_1$ eigenvalue. The Fixed-axis spin obstruction for positive-energy Dirac spinors isIts norm is $2\sqrt{p_y^2+p_z^2}$, so it vanishes exactly when $p_y=p_z=0$. For example with $\mathbf{p}=p\hat{z}$, the upper ${\chi }_{+}$ produces lower ${\chi }_{-}$, excluding a common eigenvector for every $p\ne 0$. In the nonrelativistic limit, the lower components are $O(p/m)$, and $({\psi }_1\pm {\psi }_2)/\sqrt{2}$ approach x-spin eigenvectors. At rest they are exact.

The prompt's unrestricted claim of impossibility at relativistic energies needs qualification: for momentum parallel to x these combinations remain exact eigenvectors at any energy. For a beam transverse to x, such as the conventional z-directed beam, the intended impossibility holds, and the lower components are no longer negligible. A transported relativistic spin observable differs from this fixed matrix; helicity is the conserved projection along momentum for a free particle.

For nonzero momentum, physical helicity is the spin projection along the direction of motion,The convention without $1/2$ labels the same two eigenspaces by $\pm 1$. Write $\hat{\mathbf{p}}=(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$. Normalized two-component eigenvectors areEach can carry an arbitrary phase. Before imposing an energy equation, the general four-component helicity eigenvector is $(a{\chi }_{\pm },b{\chi }_{\pm }{)}^{\mathsf{T}}$, with arbitrary complex $a,b$; unit Norm imposes $|a{|}^2+|b{|}^2=1$. For a positive-energy free electron, the Dirac equation fixes their ratio and givesMultiplying by an arbitrary overall amplitude and the common plane wave supplies the corresponding free helicity eigenstate. Helicity has no preferred direction at $\mathbf{p}=0$. For a massive particle a boost can overtake its momentum and change its helicity; it should not be identified with Lorentz-invariant chirality.

For observables $A,B$, nonzero $[A,B]$ obstructs a complete common eigenbasis. In a finite-dimensional system, commuting Hermitian operators can be simultaneously diagonalized; quantum degeneracy may require choosing an appropriate basis inside the degenerate eigenspace. Noncommuting observables can still share particular eigenvectors, so incompatibility is not a claim that every individual state must have nonzero spreads in both quantities.

The Robertson uncertainty relation follows by applying Cauchy-Schwarz inequality to $(A-⟨A⟩)|\psi ⟩$ and $(B-⟨B⟩)|\psi ⟩$:In particular $[x,p_x]=i\hslash$ gives $\Delta x\Delta p_x\ge \hslash /2$. These are preparation spreads, not necessarily measurement errors. A zero commutator expectation can make this bound zero even when the linear operators do not commute.

Measurement order can matter: preparing $|+z⟩$, measuring $S_x$, then measuring $S_z$ gives both z outcomes with probability $1/2$ when the intermediate result is ignored. A direct z measurement would give $+\hslash /2$ certainly. Finally the Heisenberg equation of motion $dA/dt=(i/\hslash )[H,A]+\partial A/\partial t$ relates commutation to conservation: an observable with no explicit time dependence and $[H,A]=0$ is conserved. For unbounded self-adjoint operators these statements require the appropriate common domains; a formal commutator calculation alone need not justify simultaneous spectral measurements.

The remaining cyclic commutators are $[S_y,S_z]=i\hslash S_x$ and $[S_z,S_x]=i\hslash S_y$. In a spin-$s$ irreducible representation,For spin one-half, $S_i=\hslash {\sigma }_i/2$, appropriate to an electron's spin. Each $S_i^2={\hslash }^{2}I/4$, so this Hamiltonian is $H=A{\hslash }^{2}I/4$. The orthonormal vectors $(1,0{)}^T$ and $(0,1{)}^T$ are eigenstates with the same energy $A{\hslash }^2/4$; every normalized superposition is also an eigenstate. The $B$ term vanishes identically because spin-one-half quadratic anisotropy is constant.

For the supplied three-dimensional basis one consistent choice isThese Hermitian matrices obey the cyclic commutators with the given $S_x$, and $S^2=2{\hslash }^{2}I$, representing spin one. Their squares giveThus the three normalized Cartesian basis vectors $e_1,e_2,e_3$ have energies ${\hslash }^2(A-B)$, ${\hslash }^2(A+B)$ and $0$, respectively; degeneracies permit arbitrary orthonormal choices in the corresponding eigenspaces.

The parallel and antiparallel $S_x$ states are $|+x⟩=(0,1,-i{)}^T/\sqrt{2}$ and $|-x⟩=(0,1,i{)}^T/\sqrt{2}$. Evolution from the first givesThis spin-one reversal under quadratic anisotropy is complete whenprovided $A+B\ne 0$. If $A+B=0$, its initial state is stationary up to phase and complete reversal never occurs. Assume real $A,B$ for a Hermitian Hamiltonian.

A pure quantum state is a unit ray in a complex Hilbert space: normalization fixes total probability, a common phase is unobservable, and superpositions retain relative phases. Observables are self-adjoint operators. For an ideal projective measurement with spectral projector $P_a$, the Born rule gives $p(a)=⟨\psi |P_a|\psi ⟩$ and the Lüders rule gives conditional state $P_a\psi /\sqrt{p(a)}$. This preserves an eigenspace, not necessarily a unique vector when the outcome is degenerate. Between measurements, a closed system evolves unitarily according to $i\hslash {\partial }_t\psi =H\psi$. Composite systems use tensor products; quantum statistics additionally restrict identical-particle states. These are a framework supported by many experiments, not separate logical deductions from any one experiment.

In the photoelectric effect, monochromatic light illuminates a clean metal and a retarding potential measures the largest emitted electron kinetic energy. frequency controls the stopping energy, $e{V}_s=h\nu -\Phi$, while in the one-photon regime intensity mainly controls the emission rate. A threshold frequency and a stopping-energy slope $h/e$ support quantized energy exchange rather than gradual absorption of arbitrary classical-wave energy by each electron. Contact potentials, surface contamination, dark current and spectral impurities require controls. Robert Andrews Millikan tested the frequency–stopping-potential relation with careful surface and electrical control.^[ref]

The Stern-Gerlach experiment sends a collimated neutral silver beam through an inhomogeneous magnetic field onto a detector. The force is approximately $F_z={\mu }_z{\partial }_{z}B$ in an appropriate strong-field electronic description. Two resolved components rather than a continuous classical orientation spread support discrete angular-momentum projections. Field-off beam profiles and magnetic-gradient/beam-width controls distinguish splitting from ordinary broadening. In the modern description the silver ground electronic state has $J=1/2$; the 1922 experiment preceded the electron-spin interpretation. Ideal sequential analyzers along different axes illustrate incompatible observables and state preparation, but were not the original apparatus.^[ref]

In the Davisson–Germer experiment, accelerated electrons scatter from a nickel single crystal and an angularly scanned collector measures the scattered intensity. Changing voltage and crystal orientation changes narrow diffraction maxima. The inferred wavelength agrees with $\lambda =h/p$, with nonrelativistic $p=\sqrt{2meV}$. vacuum, crystal cleaning, energy selection and background measurements matter: adsorbed surface gas can produce additional diffraction features. This supports matter-wave amplitudes and interference, with measured intensities related to their squared magnitude; it does not mean an electron must follow a classical orbit through the apparatus.^[ref] Together these experiments motivate quantized outcomes, wave amplitudes and probabilistic measurement while leaving a distinction between observed evidence and the general postulates used to organize it.

A continuous symmetry is represented by a unitary family $U(ϵ)=e^{-iϵG/\hslash }$ with a self-adjoint conserved generator of a continuous quantum symmetry $G$. If $UH{U}^{†}=H$ for all $ϵ$, differentiating gives $[H,G]=0$. Thus, for no explicit time dependence,translation invariance conserves the corresponding momentum component; rotational invariance about one axis conserves that angular-momentum component. A central potential conserves all components and $L^2$, but the components need not be simultaneously measurable because $[L_x,L_y]=i\hslash L_z$. Time-translation invariance of a closed system gives a time-independent Hamiltonian and energy conservation.

A discrete symmetry need not have a continuously varying generator. For parity, $[H,P]=0$ lets stationary states be chosen even or odd, and preserves parity sectors. symmetry does not force every state to be an eigenstate of its conserved observable. Explicit time dependence adds $⟨\partial G/\partial t⟩$ to the conservation equation, and an external symmetry-breaking potential generally permits exchange with the environment. Degenerate eigenspaces require choosing a compatible common basis rather than assuming every eigenvector already diagonalizes every commuting operator.

A density operator describes both a quantum state prepared in one pure state and a statistical preparation whose members occupy different pure states. If normalized vectors $|{\psi }_i⟩$ occur with probabilities $p_i$, defineThe resulting operator is Hermitian, positive semidefinite and has trace one. Positivity follows because $⟨v|\rho |v⟩={\sum }_i{p}_i|⟨{\psi }_i|v⟩{|}^2\ge 0$. Conversely, the spectral theorem gives such a representation of any finite-dimensional operator with these properties, using its eigenvectors and eigenvalues. The matrix representing $\rho$ depends on the chosen basis; the operator and its predictions do not.

One operator gives every measurement probability. For a projective measurement with orthogonal projections ${\Pi }_a$ and outcomes $a$,For example, the first formula follows by averaging $⟨{\psi }_i|{\Pi }_a|{\psi }_i⟩$ over the preparation. The trace expression is basis independent. A general measurement replaces ${\Pi }_a$ by positive measurement effects that sum to the identity, without changing the probability formula. Thus $\rho$ specifies the statistics accessible to measurements on that system.

The distinction between pure and mixed states is especially transparent in the eigenvalues. A pure state has $\rho =|\psi ⟩⟨\psi |$, ${\rho }^2=\rho$, and $\mathrm{Tr}{\rho }^2=1$. A mixed state has at least two positive eigenvalues and $\mathrm{Tr}{\rho }^2<1$. For a qubit,pure states lie on the Bloch sphere; mixed states lie inside the Bloch ball. The maximally mixed qubit $\rho =I/2$ can be prepared by choosing equally between the two $z$ spin eigenstates, or equally between the two $x$ spin eigenstates. These ensembles are different preparation procedures but give the same density operator. No measurement on the resulting qubit alone can identify which ensemble was used.

quantum coherence must be distinguished from merely having two populated levels. In the $z$ basis, the state $|+x⟩$ hasBoth give equal probabilities for the two $z$ outcomes. A measurement along $x$ gives the positive result with probability one for the first state and one half for the second. The off-diagonal entries encode the relative phase information responsible for this difference. Whether a matrix entry is off diagonal is basis dependent; purity is not.

For an isolated system, each component evolves with the same unitary time-evolution operator. Therefore $\rho (t)=U(t)\rho (0)U^{†}(t)$ andUnitary evolution preserves the eigenvalues and purity of $\rho$. A density operator commuting with a time-independent Hamiltonian operator is stationary. A thermal equilibrium example is the canonical ensemble ${\rho }_{\beta }=e^{-\beta H}/Z$, where $Z=\mathrm{Tr}(e^{-\beta H})$ and $\beta =1/(k_{\mathrm{B}}T)$. expectation values then reproduce quantum statistical mechanics without assigning the system a definite energy eigenstate on every run.

A mixed state can also arise without classical randomness in the joint preparation. For a bipartite quantum system, the reduced density operator ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ gives every measurement probability accessible to subsystem $A$. For the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, the joint state is pure, whereas ${\rho }_A=I/2$. Its local mixedness reflects entanglement with the inaccessible system; the partial trace discards correlations that matter for joint measurements. More generally, interaction with an environment can suppress off-diagonal entries in a selected basis, producing decoherence while the combined system still evolves unitarily.

Finally, state evolution conditional on a measurement outcome differs from isolated evolution. An ideal projective measurement yielding $a$ updates the state by the Lüders rule $\rho \mapsto {\Pi }_a\rho {\Pi }_a/P(a)$. Discarding the outcome gives ${\sum }_a{\Pi }_a\rho {\Pi }_a$, which generally removes coherence between different eigenspaces. The Von Neumann entropy $-\mathrm{Tr}(\rho \ln \rho )$ quantifies mixedness: it vanishes for pure states and reaches $\ln d$ for $I/d$. Multiplying by $k_{\mathrm{B}}$ gives thermodynamic entropy units. These uses make the density operator a common language for preparation uncertainty, measurement, thermal states and entangled subsystems, while retaining their distinct physical origins.

An observable is represented by a self-adjoint operator. For an ideal projective measurement, let $P_a$ be its spectral projector for outcome $a$. The Born rule gives $p(a)=⟨\psi |P_a|\psi ⟩$, or $p(a)=\mathrm{tr}(\rho P_a)$ for a density matrix $\rho$. Conditional on obtaining $a$, the Lüders rule givesFor a degenerate outcome this projects onto the whole associated eigenspace, rather than arbitrarily selecting a vector within it. An immediate repeated ideal measurement returns the same outcome, provided there is no intervening evolution or disturbance.

Commuting finite-dimensional observables admit a common orthonormal eigenbasis; their projectors commute and ideal joint measurements are compatible. Noncommuting observables generally have no complete common eigenbasis, and measurement order can affect the subsequent statistics. Particular common eigenvectors can nevertheless exist, so a nonzero commutator does not mean that every state has a nonzero uncertainty in both quantities.

For a quantitative constraint, apply the Cauchy-Schwarz inequality to $(A-⟨A⟩)|\psi ⟩$ and $(B-⟨B⟩)|\psi ⟩$. The imaginary part of their inner product is $⟨[A,B]⟩/(2i)$, giving the Robertson uncertainty principleThe canonical commutation relation $[x,p]=i\hslash$ gives $\Delta x\Delta p\ge \hslash /2$. Retaining the real part of the same inner product gives the stronger Robertson-Schrödinger uncertainty relation, including a covariance term. These are constraints on the preparation of a quantum state, not simply statements about a poorly calibrated instrument.

A sequential spin one-half example makes the change in statistics explicit. Prepare $|+z⟩$. An immediate $S_z$ measurement is certainly positive. If an $S_x$ measurement is inserted first, its outcomes each have probability $1/2$ and leave states $|\pm x⟩$. Each then yields either $S_z$ result with probability $1/2$, so the final positive fraction is $1/2$. The change reflects projection onto incompatible spin eigenstates, even with ideal measuring apparatus. The theory of measurement in quantum mechanics allows more general state-update rules, but the operator description and the Born rule still constrain their probabilities.

Orthogonality means zero inner product, not merely that two objects look different. For complex vectors, $⟨u,v⟩={\sum }_i{u}_i^{\ast }{v}_i$; for functions a common choice is $⟨u,v⟩=\int w(x)u(x{)}^{\ast }v(x)dx$ with positive weight $w$. An orthonormal set has $⟨u_m,u_n⟩={\delta }_{mn}$.

For an orthogonal set, the component of $f$ along $u_n$ has coefficientA complete orthonormal set gives $f={\sum }_n{c}_n{u}_n$ in the appropriate norm and Parseval's identity $\Vert f{\Vert }^2={\sum }_n|c_n{|}^2$. Without completeness there can be a nonzero residual orthogonal to the entire chosen set. Orthogonal nonzero vectors are linearly independent, but independence alone does not imply orthogonality.

Fourier coefficients are projections becauseLikewise, for a self-adjoint operator, $A{u}_n={\lambda }_n{u}_n$ gives $({\lambda }_n-{\lambda }_m)⟨u_m,u_n⟩=0$, so distinct eigenvalues have orthogonal eigenvectors. Degeneracy does not itself force a chosen pair to be orthogonal, but an orthonormal basis can be selected within each eigenspace. The differential version uses the correct Sturm-Liouville weight and boundary conditions.

For mechanical normal modes with unequal masses the relevant inner product is $⟨u,v{⟩}_{\mathsf{M}}=u^{†}\mathsf{M}v$. This makes the modes in D14 orthogonal even when an ordinary unweighted dot product does not. Expanding in such coordinates diagonalizes the kinetic and potential quadratic forms, explaining why modal motions evolve independently and their energies add.

Normal-mode idea. Near a stable equilibrium, expand the potential energy to quadratic order and the kinetic energy as a quadratic form in generalized velocities. The Euler-Lagrange equations then take the linear form $\mathsf{M}\ddot{\mathbf{x}}+\mathsf{K}\mathbf{x}=0$. A normal mode has fixed displacement ratios, $\mathbf{x}=\mathbf{q}\cos (\omega t+\phi )$, with $\mathsf{K}\mathbf{q}={\omega }^2\mathsf{M}\mathbf{q}$. Real systems admit this description for sufficiently small motion, before higher-order nonlinearities or strong damping matter. Superposition permits a general motion to be assembled from the modes. A free symmetry direction may instead be a zero mode, with constant or linearly changing coordinate.

Lagrangian and bead equations. Take each $x_i$ as an arc-length displacement positive in the same sense around the fixed circular wire. Spring length changes are $x_2-x_1$, $x_3-x_2$ and $x_1-x_3$. Equal equilibrium spring tensions give no net tangential force; their linear energy terms sum to zero. Dropping an irrelevant constant, the Lagrangian isFor example, differentiating the potential with respect to $x_1$ gives $k(2x_1-x_2-x_3)$. The three Euler-Lagrange equations therefore areThe wire supplies normal constraint forces but does no tangential work. These are the equal-mass case of the normal modes of three beads connected by springs on a circular hoop.

Frequencies. The stiffness matrix divided by $k$ isIt annihilates $1$. On the two-dimensional subspace $q_1+q_2+q_3=0$, its action is multiplication by three. Thus the angular frequencies of the normal modes areThe repeated frequency is a degeneracy of normal modes, while the zero frequency expresses free rigid rotation of all beads along the wire.

General solution. Two independent vectors spanning the vibrating subspace are ${\mathbf{q}}_A=(1,-1,0{)}^{\mathsf{T}}$ and ${\mathbf{q}}_B=(1,0,-1{)}^{\mathsf{T}}$. Each amplitude satisfies $\ddot{Q}+{\omega }^{2}Q=0$, with $\omega =\sqrt{3k/m}$. The uniform coordinate satisfies ${\ddot{Q}}_0=0$. HenceThe six freely specifiable real constants $A,\varphi ,B,\psi ,C,D$ give the six initial data; $\omega$ is fixed by $k,m$ and is not an additional independent integration constant. The linear zero-mode term must be included to allow nonzero total tangential momentum.

Shapes and uniqueness. In the first term beads 1 and 2 have opposite equal displacements and bead 3 remains at its equilibrium position. In the second term beads 1 and 3 do this while bead 2 stays fixed. The last term moves all three identically, leaving spring extensions unchanged: $D$ is a rigid phase shift and $C$ a uniform drift around the wire. The sketches show positive modal amplitudes; all arrows reverse half a vibration period later.

The two nonzero-frequency eigenvectors are not unique, because any independent linear combinations in their degenerate two-dimensional eigenspace form an equally valid basis. In particular the displayed pair is independent but not orthogonal; one could use $(1,-1,0)$ and $(1,1,-2)$ instead. The zero eigenspace is one-dimensional, so its vector is fixed up to scale and sign.

Specified release. Zero initial velocities permit $\varphi =\psi =0$ and require $C=0$. At $t=0$, the three displacement conditions become $A+B+D=a$, $-A+D=0$, $-B+D=0$. Therefore one convenient set of constants isEquivalently,Their sum remains $a$, as required by zero initial total tangential momentum. Alternative amplitude signs with compensating phase shifts represent the same motion.

Definition and equations. A normal mode is a motion in which every component of a linear system has the same time dependence with fixed amplitude ratios and phases. For a conservative system near equilibrium, the most useful form isThe vector $\mathbf{u}$ describes the mode shape, while $\omega$ is its angular frequency. Linear combinations of modes describe arbitrary sufficiently small motions when the modes form a complete basis.

Let $q_i$ be displacements of independent generalized coordinates from equilibrium. Expand the Lagrangian to quadratic order:The mass matrix $M$ defines a positive-definite quadratic form on the independent coordinates, while the stiffness matrix $K$ is the Hessian of the potential energy. Substituting a mode gives the generalized eigenvalue problem for small oscillationsA practical calculation identifies the equilibrium, chooses independent coordinates, computes the quadratic kinetic and potential energies, solves for the eigenvalues and then obtains their eigenvectors. One can equivalently diagonalize the real symmetric matrix $M^{-1/2}K{M}^{-1/2}$. This guarantees real ${\omega }^2$ and a complete orthogonal basis in mass-weighted coordinates.

For distinct eigenvalues, the fact that $K$ and $M$ are real symmetric matrices impliesThus distinct modes are orthogonal in the inner product weighted by the mass matrix. Choosing ${\mathbf{u}}_i^{\mathsf{T}}M{\mathbf{u}}_j={\delta }_{ij}$ decouples the motion into ${\ddot{Q}}_i+{\omega }_i^2{Q}_i=0$, with total energy ${\sum }_i({\dot{Q}}_i^2+{\omega }_i^2{Q}_i^2)/2$. A positive-definite matrix $K$ gives positive eigenvalues and oscillatory stability. Zero modes can describe rigid translations or rotations without restoring force; a negative eigenvalue gives exponential instability rather than an oscillatory mode.

Symmetry. A symmetry preserving both quadratic energies maps modes into modes of the same frequency. Coordinates adapted to symmetry divide the eigenvalue problem into smaller independent blocks. For two identical subsystems, exchanging them gives symmetric and antisymmetric modes. The degeneracy of normal modes means that independent modes share one frequency. Higher-dimensional group representations of a symmetry can enforce such degeneracies; symmetry does not require every pair of modes to be degenerate, and accidental degeneracies can also occur. Within a degenerate eigenspace, orthogonal linear combinations are equally valid mode shapes.

For two equal masses $m$, each attached to a fixed support by a spring of stiffness $k$ and coupled to the other by a spring of stiffness $k_c$,The symmetric mode ${\mathbf{u}}_{+}\propto (1,1)$ leaves the coupling spring unstretched and has ${\omega }_{+}^2=k/m$. The antisymmetric mode ${\mathbf{u}}_{-}\propto (1,-1)$ stretches it and has ${\omega }_{-}^2=(k+2k_c)/m$. Superposing the two gives beats when their frequencies are close.

Continuous systems. The same idea replaces matrices by differential operators and mode vectors by functions satisfying boundary conditions. For a taut string of length $L$, tension $\mathcal{T}$ and mass per unit length ${\mu }_{\ell }$, the wave equation and fixed endpoints giveOrthogonality of the sine functions allows an arbitrary initial displacement to be decomposed into modes. Changing the endpoints changes the spectrum: boundary conditions are part of the physical mode problem.

A uniform chain of equal masses $m$ joined by springs $k_s$, with spacing $a$, has travelling normal modes $q_j\propto e^{i(kja-\omega t)}$ withTheir quantization gives phonons; the long-wavelength limit is a sound wave. Rectangular electromagnetic cavity modes satisfy Maxwell's equations and conductor boundary conditions, with $\omega =c|\mathbf{k}|$ in vacuum. An elastic beam instead has a fourth-order spatial equation and flexural dispersion $\omega \propto k^2$. These examples show how restoring forces and geometry determine the frequency spectrum.

Real systems. Small-amplitude motion is often approximately linear even when the full equations are nonlinear. Large amplitudes produce frequency shifts, mode coupling and transfer of energy between modes. Friction broadens resonances and makes free oscillations decay. A damping matrix proportional to $M$ or $K$ can retain the undamped mode shapes; general damping need not, and the appropriate eigenfrequencies and modes can be complex. External periodic forcing predominantly excites modes near resonance, with amplitude set by coupling to the forcing and dissipation. Imperfections split ideal degeneracies, while radiation or coupling to a surrounding fluid can remove energy from the chosen subsystem. Normal modes remain useful when linearization, the boundary conditions and the treatment of damping accurately describe the motion being measured.