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Take any $q\in {\Pi }_n$. Write $f-q=(f-p_n)+(p_n-q)$. The second term belongs to ${\Pi }_n$, so the assumed orthogonality gives $⟨f-p_n,p_n-q⟩=0$. Expanding the squared norm yields the Pythagorean theorem in this inner product:Taking square roots provesEquality can hold only when $\Vert p_n-q\Vert =0$, hence $p_n=q$, so the best polynomial is unique. Completeness of the ambient space is unnecessary for this orthogonality argument.

Positive definiteness gives a positive smallest eigenvalue $c$ with $x^{t}Ax\ge c|x{|}^2$. Hence the set $x^{t}Ax=y^{t}Ay=1$ is compact, and $h$ attains a maximum there. Its two constraint gradients are $(2Ax,0)$ and $(0,2Ay)$; they are linearly independent because $x,y$ are nonzero and $A$ is invertible.

Put $s=x^{t}Ay$. At any constrained extremum the multiplier equations areMultiplication by $x^t$ and $y^t$ gives $\lambda =\mu =s^2$. If $s=0$ the objective is zero. Otherwise $Ay=sAx$, so $y=sx$ and normalization gives $s^2=1$. Thus every maximum is at most one, and the value one is attained at $x=y$. This proves $\boxed{(x^{t}Ay{)}^2\le 1}$ on the normalized constraints.

For nonzero arbitrary vectors apply that result to $x/\sqrt{x^{t}Ax}$ and $y/\sqrt{y^{t}Ay}$. If either vector is zero the conclusion is immediate. ThereforeEquality for nonzero vectors holds precisely when they are linearly dependent. This is the Cauchy-Schwarz inequality for the inner product induced by $A$.

A real inner product is a bilinear symmetric form satisfying $(v,v)>0$ for every nonzero $v$, and $(0,0)=0$. Its norm is $\Vert v\Vert =\sqrt{(v,v)}$. An orthogonal polynomial family has nonzero members of exact degrees $0,1,\dots ,k$, with $({\varphi }_i,{\varphi }_j)=0$ for different indices. Exact degree is part of this definition; merely belonging to the indicated polynomial vector spaces would not suffice.

Distinct degrees make ${\varphi }_0,\dots ,{\varphi }_{\ell -1}$ linearly independent. There are $\ell$ of them in the $\ell$-dimensional space ${\Pi }_{\ell -1}$, so they form a basis. Any lower-degree polynomial is their linear combination, and bilinearity therefore gives $({\varphi }_{\ell },p)=0$ for every $p\in {\Pi }_{\ell -1}$.

Use the orthogonal-projection characterization: $p_{\ast }$ is the best approximation to $v$ in a finite-dimensional subspace $W$ exactly when $v-p_{\ast }$ is orthogonal to $W$. Here $v={\varphi }_{\ell }$ is already orthogonal to $W={\Pi }_{\ell -1}$, so its unique best approximation is zero. For $\ell \ge 1$, ${\varphi }_{\ell -1}$ is nonzero and therefore cannot be that best approximation. Directly,For the specified weight, symmetry makes every odd moment vanish, whileChoose ${\varphi }_0=1$ and ${\varphi }_1=x$. A monic quadratic orthogonal to $x$ has no linear term by symmetry, and orthogonality to 1 requires $\int |x|(x^2-c)dx=1/2-c=0$. Thus a valid answer isThe weight vanishes only at one point, so a nonzero polynomial has positive squared norm; the form is indeed positive definite on the polynomial vector space. Nonzero rescalings of individual family members give equally valid unnormalized answers.

The norm induced by a real inner product isPositive definiteness and homogeneity follow from the inner product axioms; Cauchy-Schwarz gives the triangle inequality by expanding $\Vert u+v{\Vert }^2$.

The norm induced by a real inner product isPositive definiteness and homogeneity follow from the inner product axioms; Cauchy-Schwarz gives the triangle inequality by expanding $\Vert u+v{\Vert }^2$.

Expand both squared norms using the inner product:Adding cancels the cross terms, giving the parallelogram lawThe real part may be omitted in a real inner-product space.

Let $v={\sum }_{k=1}^n⟨x,e_k⟩e_k$, using the convention that the inner product is linear in its first argument. Orthonormality givesfor every $j$. Hence $x-v$ is orthogonal to $U_n$. For any $u\in U_n$, the Pythagorean identity gives $\Vert x-u{\Vert }^2=\Vert x-v{\Vert }^2+\Vert v-u{\Vert }^2$, so $v$ is the unique best approximation. Therefore

Assume the usual nondegenerate interval $a<b$. The integral is bilinear and symmetric because the functions and weight are real. It is nonnegative on $(f,f)$. If $f$ is nonzero at a point, continuity makes $|f|$ bounded below by a positive number on a nontrivial relative interval nearby, and the positive continuous weight is also bounded below there. Hence ${\int }_a^b{f}^{2}w>0$. Thus the formula is positive definite and defines an inner product.

The functions are orthogonal precisely whenA singleton interval would make this integral zero for every function, so the condition $a<b$ is necessary.

The columns outside indices $i,j$ are unchanged coordinate vectors. The remaining two columns have nonzero entries only in rows $i,j$, with the $2\times 2$ block $M(\theta )$. They have squared norm ${\cos }^2\theta +{\sin }^2\theta =1$ and mutual inner product zero; they are also orthogonal to all the unchanged columns. Hence $J^{T}J=I$.

Let $r=\sqrt{x_i^2+x_j^2}$. If $r>0$, choose $\theta \in [-\pi ,\pi ]$ with $\cos \theta =x_i/r$ and $\sin \theta =x_j/r$. ThenEvery other coordinate is unchanged. If $r=0$, choose $\theta =0$; the same conclusions hold. Therefore the procedure can be fixed with the nonnegative pivot conventionThis is a Givens rotation.

The degree-zero polynomial is constant and therefore even. Assume inductively that $p_j(-x)=(-1{)}^j{p}_j(x)$ for $j<k$. Define $q_k(x)=(-1{)}^k{p}_k(-x)$, which has the same leading coefficient as $p_k$.

Changing variable $x\mapsto -x$ and using the even weight and inductive hypothesis gives $⟨q_k,p_j{⟩}_w=(-1{)}^{k+j}⟨p_k,p_j{⟩}_w=0$ for $j<k$. Hence $q_k-p_k$ has degree below $k$ and is orthogonal to every $p_0,\dots ,p_{k-1}$. Those polynomials form a basis for the polynomials of degree below $k$, so the difference is orthogonal to itself. Positivity of the weighted inner product forces it to be zero.

Thereforecompleting the induction and proving parity of orthogonal polynomials for an even weight. Thus even indices give even functions and odd indices give odd functions, regardless of the nonzero normalization chosen.

Two problems central to physics concern definite outcomes in quantum measurement and the temporal direction of entropy increase. They expose different gaps between a dynamical description and the phenomena explained; neither can be resolved merely by saying that physics predicts observations successfully.

For the measurement problem, combine a two-state system with an apparatus initially in a ready state. If an ideal unitary operator correlates basis states with distinct pointer states, its linearity givesThe same dynamics that gives reliable correlations therefore gives an entangled final state, rather than a rule choosing one pointer value. The problem is not that the state has failed to predict possible outcomes, but how definite outcomes and their probabilities are related to that state.

Quantum decoherence explains why interaction with an environment suppresses interference in the apparatus's reduced density matrix: its off-diagonal term is multiplied by the inner product of the corresponding environmental states. Near-orthogonal records make this term very small. Tracing out the environment, however, does not by itself select one term in the global superposition. An objective-collapse theory changes the dynamics, while a many-worlds interpretation changes what is required of an outcome. Distinguishable modifications can be tested; interpretations reproducing the same predictions raise additional explanatory and conceptual questions. Decoherence solves a substantial dynamical part of the problem without automatically settling every interpretation.

The second issue is the thermodynamic arrow of time. In a classical model, time reversal in classical mechanics can pair an entropy-increasing trajectory with a reversed entropy-decreasing one. Liouville theorem in Hamiltonian mechanics preserves phase-space volume, so reversible Hamiltonian dynamics does not imply a universal monotonic increase of fine-grained entropy. The Second law of thermodynamics therefore needs a statistical and boundary-condition account, not a deduction from reversibility alone.

Boltzmann entropy assigns larger entropy to a macrostate occupying more compatible phase-space volume. Starting from a suitable low-entropy condition, overwhelmingly more accessible microstates can evolve toward larger macrostates. But an unrestricted typicality argument applied at an arbitrary low-entropy instant generally points toward increasing entropy in both temporal directions. The past hypothesis supplies an asymmetric low-entropy boundary condition in the past; one must still justify the probability measure and the macroscopic description used.

These are genuine partial resolutions rather than interchangeable cures. Decoherence concerns the accessibility of interference and records; a past boundary condition concerns temporal asymmetry in a statistical explanation. A satisfactory account must say which extra assumptions it introduces and which observable consequences follow. Neither problem licenses dismissing successful physics, but both show why a predictive formalism and a complete interpretation of its success are different achievements.

The Pusey-Barrett-Rudolph theorem excludes overlapping preparation distributions for distinct pure states in an ontological model of a quantum system satisfying preparation independence and the specified quantum predictions. It excludes neither all hidden variables nor every informational use of a quantum state. The precise overlap criterion is essential: a psi-epistemic model lets at least two distinct pure-state preparations assign positive common probability mass to the same underlying states; a psi-ontic model has mutually singular measures for each distinct pair.^[ref]

Let preparation $x$ sample $\lambda \in \Lambda$ from ${\mu }_x$. A fixed quantum measurement $M$ has nonnegative response functions ${\xi }_M(j|\lambda )$ with ${\sum }_j{\xi }_M(j|\lambda )=1$. Averaging over the preparation must reproduce the Born rule. The response is fixed by $M$ and the complete underlying state, not by additional knowledge of the preparation label. No outcome determinism or Bell local causality is assumed; the joint measurement response need not factor into local responses. For independently operated preparations, preparation independence further identifies the joint state space with the appropriate product and assignsOperationally independent devices do not logically entail this ontic factorization; it is an additional physical premise.^[ref]

The two-qubit example can be checked directly. Set $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. In computational basis order $00,01,10,11$, take the following normalized columns as an orthonormal basis:Their pairwise inner products vanish. Direct multiplication shows that the respective preparations $00,0+,+0,++$ have zero probability amplitude for respective outcomes $1,2,3,4$. This is the PBR exclusion measurement for zero and plus: the measurement does not perfectly distinguish the preparations, but each outcome rules one out.

Suppose a common submeasure $\nu$ of ${\mu }_0$ and ${\mu }_{+}$ has mass $\omega >0$; with densities one can take $\nu (\lambda )=\min \{{\mu }_0(\lambda ),{\mu }_{+}(\lambda )\}$. Preparation independence makes $\nu \otimes \nu$ a common submeasure of all four product preparations, of mass ${\omega }^2$. The zero probability for outcome $j$ under its excluded preparation and nonnegativity forceSumming and using normalization yields ${\omega }^2=0$, a contradiction. This measure formulation avoids treating a set with merely nonempty topological support as a positive-probability overlap.

For arbitrary distinct nonorthogonal pure states, write $|{\psi }_{0,1}⟩=\cos (\theta /2)|0⟩\pm \sin (\theta /2)|1⟩$, with overlap $\cos \theta$ and $0<\theta <\pi /2$. The general proof constructs an exclusion measurement on $n$ independently prepared copies when $2^{1/n}-1\le \tan (\theta /2)$. A finite $n$ always exists. The same common-submeasure contradiction then excludes positive overlap. Orthogonal preparations are already disjoint if their perfectly distinguishing measurement is reproduced. The original paper also bounds residual overlap for imperfect exclusion data; exact disjointness concerns the ideal theorem, not a finite experiment's proof of exact zero.^[ref]

The scope is consequently sharp. Bohmian mechanics is not excluded: its underlying state includes the wave function, making distinct pure preparations disjoint even though extra configuration variables remain. A psi-ontic model need not be complete with the wave function alone, and the theorem does not identify the wave function with a three-dimensional material field. Models abandoning preparation independence, or accounts not representing preparations as uncertainty over observer-independent states of this specified kind, are outside this conclusion. Epistemic uncertainty between mixed preparations is not banned. Nor does the theorem establish a collapse mechanism, choose between Bohmian and Everettian theories, or solve the measurement problem. Its achievement is a conditional impossibility theorem about a precise form of pure-state epistemic overlap.

matrix multiplication gives ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$, hence $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$. For physical neutron spin, use $\mathbf{S}=(\hslash /2)\mathbf{\sigma }$, so $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$, $S^2=3{\hslash }^2/4$, and measurement along any axis gives $\pm \hslash /2$.

For the specified field, $H=(\alpha B\hslash /2){\sigma }_y$. The Time-dependent Schrödinger equation is $i\hslash \dot{\psi }=H\psi$ with $\psi (0)=(1,0{)}^T$. The uniform-field spin precession follows from ${\sigma }_y^2=I$:

The sign of the precession is contained in the given $\alpha$. The normalized $x$-axis eigenstates are $(1,1{)}^T/\sqrt{2}$ and $(1,-1{)}^T/\sqrt{2}$.

The measurement direction is $\mathbf{n}=(0,\sin \theta ,\cos \theta )$, as confirmed by the original diagram. Write $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. The spinors ${\chi }_{+}=(c,is{)}^T$ and ${\chi }_{-}=(is,c{)}^T$ have unit norm and inner product $ics-isc=0$. Applying $\mathbf{n}\cdot \mathbf{\sigma }$ gives eigenvalues $+1$ and $-1$. At $\theta =0$ they are the usual $z$ basis; at $\pi /2$ they are the $y$ basis up to a phase; at $\pi$ the $z$ up/down roles are exchanged, with a factor $i$; and at $2\pi$ both return with a minus sign. A spinor changes sign under a full rotation, while its isolated-state measurement probabilities are unchanged.

For spin measurement in the yz plane, projection givesFor $\theta =0$ these are ${\cos }^2\varphi ,{\sin }^2\varphi$; for $\pi /2$ both are $1/2$ at all times; for $\pi$ they are exchanged; and for $2\pi$ they are the same as at zero angle. The constant $y$ probabilities also follow from conservation of $S_y$.

Factorization. Put $\alpha =\sqrt{m\omega /(2\hslash )}$ and use the canonical commutation relation $[x,p]=i\hslash$:It follows thatThe $-1/2$ arises from quantum operator ordering; treating $x$ and $p$ as commuting would lose the zero-point energy.

commutators. Direct use of $[x,p]=i\hslash$ gives $[a,a^{†}]=1$. The product rule for commutators then yields $[a^{†}a,a]=-a$ and $[a^{†}a,a^{†}]=a^{†}$, hence

Raising the energy. If $H|\varphi ⟩=E|\varphi ⟩$, thenThus the creation operator produces an energy eigenstate with energy $E+\hslash \omega$. It is nonzero: for a normalized physical oscillator eigenstate, $\Vert a^{†}\varphi {\Vert }^2=⟨a{a}^{†}⟩=E/(\hslash \omega )+1/2>0$. Normalize by this norm if a normalized state is wanted.

Lowest wavefunctions and orthogonality. In the position representation in quantum mechanics,The annihilation condition therefore gives ${\psi }_0^{\prime }=-(m\omega /\hslash )x{\psi }_0$. Applying $a^{†}$ to its solution givesTheir inner product is proportional to ${\int }_{-\infty }^{\infty }x{e}^{-m\omega x^2/\hslash }dx=0$, since the integrand is odd. These are the even and odd lowest wavefunctions shown in A2, with energies $\hslash \omega /2$ and $3\hslash \omega /2$.

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.

Time dependence in quantum mechanics describes how a prepared quantum state changes and thereby changes the probabilities of later measurements. For an isolated system, the Schrödinger picture assigns this evolution to a state vector $|\psi (t)⟩$, governed by the Time-dependent Schrödinger equationThe Hamiltonian is a self-adjoint operator; it has units of energy, while $\hslash$ has units of action, so $\hat{H}/\hslash$ has units of inverse time. For a particle with potential energy $V(\mathbf{r},t)$, this becomesGiven a suitable initial wavefunction and the physical boundary conditions, this first-order equation determines its subsequent propagation. Its linearity means that the evolution of a quantum superposition is the corresponding quantum superposition of evolved states.

For a real potential energy, this equation and its complex conjugate imply the local continuity equationHere $\rho$ is the probability density and $\mathbf{j}$ the probability current. Vanishing net boundary flux preserves wavefunction normalization, as required by the Born rule.

The time-evolution operator packages the solution for every initial state:Because ${\hat{H}}^{†}=\hat{H}$, differentiation givesTherefore $U^{†}U=I$: this is unitary time evolution, preserving inner products and all state norms. It is reversible, with $U(t,t_0{)}^{-1}=U(t_0,t)=U(t,t_0{)}^{†}$, and obeys the composition law $U(t_2,t_0)=U(t_2,t_1)U(t_1,t_0)$.

For a time-independent Hamiltonian, the solution is the operator exponentialIf the Hamiltonian varies with time, the analogous simple exponential of its integral is valid when $[\hat{H}(t),\hat{H}(t^{\prime })]=0$ for all relevant times. Otherwise time ordering is essential:The meaning is explicit in the first terms of the ordered series,the later-time operator stands on the left. In the position representation, the same propagation is expressed by the quantum-mechanical propagatorIts initial value is the Dirac delta function ${\delta }^3(\mathbf{r}-{\mathbf{r}}^{\prime })$.

A time-independent Hamiltonian has energy eigenstates satisfying $\hat{H}|n⟩=E_n|n⟩$. A single such initial state evolves asIt is a stationary state: its global phase changes, but its probability density and the distributions of time-independent observables do not. Its energy is sharp. Any quantum superposition entirely within one degenerate energy eigenspace also acquires one common global phase and is stationary. Stationarity does not require zero probability current; for instance, $u_n(\varphi )=e^{in\varphi }/\sqrt{2\pi }$ on the ring has a constant density and persistent circulation when $n\ne 0$.

An arbitrary normalized state can instead contain different energy eigenvalues:A continuous spectrum requires corresponding integrals. For the discrete case, an energy measurement gives $E_n$ with fixed probability $|c_n{|}^2$ (summed over degeneracies), but relative quantum phases change with time. ConsequentlyThe interference terms can oscillate at angular frequencies $(E_m-E_n)/\hslash$. For two populated energy levels, this supplies a characteristic period $2\pi \hslash /|E_m-E_n|$ when the interference term is nonzero. Thus fixed energy populations do not imply a stationary spatial distribution.

For a normalized state and an observable $\hat{A}(t)$, its expectation value isThis is the statistical mean over repeated preparations and measurements. In an eigenbasis of energy eigenstates for a time-independent Hamiltonian and a time-independent $\hat{A}$,Only matrix elements connecting different energies can generate these oscillations. If $[\hat{A},\hat{H}]=0$, its expectation value is constant for every initial state, provided it has no explicit time dependence.

Differentiating the expectation value and substituting the Schrödinger equation for both the bra and ket proves the general Ehrenfest theorem:Equivalently,These manipulations require the state to lie in the relevant operator domains. Taking $\hat{A}=\hat{H}$ gives $d⟨H⟩/dt=⟨{\partial }_t\hat{H}⟩$, so energy is conserved under a time-independent Hamiltonian.

For $\hat{H}={\hat{p}}^2/(2m)+V(\hat{x})$, the canonical commutation relation givesHence the particularly useful position-and-momentum form of Ehrenfest theorem isIt resembles Newton's second law, but the mean force generally differs from the force at the mean position. If a narrow wave packet samples a region where the force varies little, $⟨V^{\prime }(x)⟩\simeq V^{\prime }(⟨x⟩)$ gives approximate classical motion. For a quadratic potential energy this replacement is exact. In particular the quantum harmonic oscillator obeysThe centroid follows the classical harmonic oscillator, although this does not specify the full evolving wavefunction or its width.

The Heisenberg picture expresses the same physics by assigning evolution to observables instead:The expectation value is unchanged, and differentiation yields $d{\hat{A}}_H/dt=(i/\hslash )[{\hat{H}}_H,{\hat{A}}_H]+U^{†}({\partial }_t{\hat{A}}_S)U$. Thus state propagation, evolving expectation values, and the classical correspondence in Ehrenfest theorem are mutually consistent descriptions of the same unitary time evolution.

A normal mode has all displacement coordinates proportional to a common time dependence, with fixed relative amplitudes. For a nonzero frequency the time dependence is sinusoidal. A zero-frequency mode is a free displacement or uniform drift rather than an oscillation.

Use the printed mass ordering $M,m,M$. Displacements are measured from the unstretched equilibrium positions, givingEquivalently,This is the generalized eigenvalue problem for normal modes of three masses connected by two equal springs. Substitute $\mathbf{x}=\mathbf{u}{e}^{i\omega t}$ to obtain $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. Reflection symmetry gives an antisymmetric vector $(1,0,-1)$ and symmetric vectors $(1,b,1)$. The former has ${\omega }^2=k/M$. For a symmetric vector, the end and centre equations give $M{\omega }^2=k(1-b)$ and $m{\omega }^{2}b=2k(b-1)$. Their solutions are $b=1$, $\omega =0$, or $b=-2M/m$, ${\omega }^2=k/M+2k/m$. HenceThese are all three eigenmodes. The last two keep the centre of mass fixed and are orthogonal in the mass-weighted inner product. The general zero-mode contribution is $(A+Bt)(1,1,1)$, accounting for both a centre-of-mass shift and a conserved total momentum. There are no fixed end springs in this system.

The original amplitude profiles illustrate $M=2m$; the symmetric internal shape is rescaled for display. The formula above gives every positive mass ratio.

Two further physical examples are standing-wave modes of a string fixed at both ends, with spatial shapes $\sin (n\pi x/L)$ and frequencies $n\pi c/L$, and electrical normal modes of two coupled LC circuits, in which the capacitor charges oscillate at fixed relative amplitudes. Independent normal mode coordinates simplify both systems in the same way as the spring-chain eigenvectors.

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.

A normal mode is a small-oscillation motion with every coordinate proportional to one common sinusoid, at a fixed amplitude ratio and relative phase. Expanding an arbitrary small motion in normal mode coordinates decouples its linear equations into independent oscillators. With mechanical mass matrix $\mathsf{M}$ and mechanical stiffness matrix $\mathsf{K}$, the generalized eigenvalue equation is $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$.

Horizontal gravitational restoring force on each bob is $-m_i(g/l)x_i$. The two springs add $k(x_2-x_1)$ on the first bob, $k(x_1-2x_2+x_3)$ on the middle bob, and $k(x_2-x_3)$ on the last. Dividing by the bob masses gives the stated equations withReflection symmetry separates an antisymmetric end motion from two symmetric motions. The three coupled pendulums with equal end masses have independent modesCall their positive frequencies ${\omega }_0,{\omega }_a,{\omega }_s$ respectively. The middle amplitude in the last mode ensures $m{x}_1+M{x}_2+m{x}_3=0$. The modes are orthogonal in the mass-weighted inner product, rather than generally in the unweighted coordinate dot product.

The initially stationary displacement decomposes asThere are no sine terms because every initial velocity is zero. ThereforeThese give $(0,0,d)$ at $t=0$ and solve all three coupled equations. The spring constant is assumed positive; for $k=0$ the frequencies become degenerate and the same superposition reduces to uncoupled pendulum motion.

The original example uses $M=2m$ and $k/m=g/l$ to display the mode vectors and resulting superposition. Those illustrative parameter ratios are not additional supplied data.

A normal mode is a pattern in which all coordinates share one frequency and fixed relative amplitudes. For a conservative linear system, it solves a generalized eigenvalue equation with stiffness matrix $K$ and mass matrix $M$. For this free three-mass spring chain,Solving $K\mathbf{u}={\omega }^{2}M\mathbf{u}$, for $\alpha >0$, givesTheir overall normalization is arbitrary; distinct modes are orthogonal in the mass inner product. The zero mode is rigid translation because neither end is fixed. In the antisymmetric mode the middle mass stays still, so changing its mass cannot alter the frequency. The symmetric internal mode keeps the centre of mass fixed through its middle displacement $-2/\alpha$. Its frequency rises without bound as $\alpha \to 0$ and approaches $\sqrt{k/m}$ as $\alpha \to \infty$, becoming degenerate with the antisymmetric frequency in that limit.

For equal masses put ${\omega }_1=\sqrt{k/m}$ and ${\omega }_2=\sqrt{3k/m}$. Positions are equilibrium positions plus displacementsThe translational mode needs the linear-in-time term; a cosine at zero frequency alone would miss arbitrary centre-of-mass velocity.

Taking the springs initially at equilibrium, the impulse gives $\dot{\mathbf{q}}(0)=(v,0,0)=(v/3){\mathbf{u}}_0+(v/2){\mathbf{u}}_1+(v/6){\mathbf{u}}_2$. Their squared Euclidean norms are $3,2,6$, so their initial kinetic energies are $m{v}^2/6,m{v}^2/4,m{v}^2/12$. Relative to total $m{v}^2/2$,Subsequent motion conserves each mode's total energy, exchanging its kinetic and spring contributions. The zero-mode share remains centre-of-mass kinetic energy.