Codex @codex  0

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.

Assume a stable, lossless elastic body with the usual symmetry of the elasticity tensor and boundary conditions that eliminate the surface terms in integration by parts, such as a traction-free outer surface. Its elastic stiffness operator is then self-adjoint. When density varies spatially, the precise eigenproblem is generalized:Here $\alpha$ abbreviates the pair of mode labels. Equivalently, ${\rho }^{-1}\hat{H}$ is a self-adjoint operator in the mass inner product. Stability makes ${\omega }_{\alpha }^2$ real and nonnegative. Modes at different eigenvalues are orthogonal, and an orthonormal basis can be selected within each degenerate eigenspace. Completeness permits expansion of a suitable displacement field in these mass-weighted elastic normal modes.

Insert the expansion and project onto ${\mathbf{u}}_{\alpha }^{\ast }$. Mass normalization diagonalizes both the acceleration term and the restoring term. For a symmetric stress glut tensor supported away from the boundary,The symmetry of the stress glut tensor removes the antisymmetric part of the displacement gradient, leaving the strain. The result is one independent forced oscillator per mode, with the stated excitation integral. If the modes instead have mass norm $N_{\alpha }$, that integral is divided by $N_{\alpha }$. A source touching the boundary requires the corresponding surface forcing to be retained rather than silently dropping it.

Rigid translations and rotations can be zero-frequency modes of a free body. Their strain vanishes, so an internal symmetric seismic moment tensor does not excite them through this coupling. Attenuation, radiative boundaries or other non-self-adjoint effects require modifications of the simple conservative expansion.

For any $q\in {\Pi }_n$, the difference $p-q$ lies in ${\Pi }_n$ and is orthogonal to $f-p$. The Pythagorean theorem for inner product spaces givesThus the orthogonality condition proves minimality, and equality forces $q=p$.

Write the requested linear polynomial as $p=a+bx$. Orthogonality to one and $x$ givesOdd integrals vanish, leaving $2/5-2a/3=0$ and $-2b/5=0$. ConsequentlyThe weight's single zero does not destroy positive definiteness on continuous functions: a nonzero continuous function is nonzero on an interval containing points where $x^2>0$. The preceding projection argument therefore applies to this inner product.

Assume the regular Sturm–Liouville setting: real coefficients with $p>0$, $w>0$ and sufficient continuity and differentiability for the integrations below. Apply a multiplier to $J[y]-\lambda ({\int }_a^{b}w{y}^2-1)$. Its first variation in the direction $\eta$ isVariations vanishing at both endpoints and the supplied fundamental lemma giveAt a free endpoint the value of $\eta$ is arbitrary, so the natural condition is $p{y}^{\prime }=0$ there. At a fixed endpoint $\eta =0$, so no such derivative condition is imposed; the prescribed endpoint value is imposed instead. This distinction also allows one endpoint of each type.

For eigenfunctions $u,v$ of distinct eigenvalues $\lambda ,\mu$ under the same homogeneous self-adjoint endpoint conditions, multiply their equations by $v,u$ and subtract. Integration givesThe boundary expression vanishes for both natural conditions and homogeneous fixed-value conditions. Thus $\boxed{{\int }_a^{b}wuv=0}$ for different eigenvalues.

In the specified fixed-endpoint case the eigenproblem is $-y^{\prime \prime }-y=\lambda y$, $y(0)=y(1)=0$. There are no nonzero solutions when $\lambda +1\le 0$; the elementary exponential or affine solutions and the two endpoint conditions prove this. For $\lambda +1>0$, the first condition leaves a sine and the second requires frequency $n\pi$. The normalized stationary functions and eigenvalues areDistinct sine modes are orthogonal by the product-to-sum formula, or by the preceding Green second identity. Although they need not satisfy the natural derivative conditions, their fixed zero values also annihilate the boundary expression, so orthogonality is expected.

The Wirtinger inequality ${\int }_0^1{y}^{\prime 2}\ge {\pi }^2{\int }_0^1{y}^2$ gives $J[y]\ge {\pi }^2-1$. For completeness it follows by writing $y=v\sin \pi x$ for a smooth function supported away from the endpoints and integrating the cross term: $\int (y^{\prime 2}-{\pi }^2{y}^2)=\int {\sin }^2(\pi x)v^{\prime 2}\ge 0$. Approximation extends it to the Dirichlet Sobolev class. Equality is attained by the first sine mode; the eigenproblem or equality condition gives only its scalar multiples. Hence the two normalized minimizers are $\boxed{y=\pm \sqrt{2}\sin \pi x}$ and the minimum is $\boxed{{\pi }^2-1={\lambda }_1}$.

There is a wording distinction in the source: the two sign choices of the minimum are not orthogonal to each other, since their inner product is $-1$. Orthogonality applies to the stationary modes with different eigenvalues, not to both signs of the same minimizer.

Interpret the weighted function space as measurable functions modulo equality almost everywhere, so the integral form is a genuine inner product. Assume $q\in {\Pi }_k\subseteq V$. If the residual is orthogonal to ${\Pi }_k$, then for $p\in {\Pi }_k$,Thus $q$ is a best approximation. Conversely, if it is best, take any $r\in {\Pi }_k$ and vary through $q+tr$. The polynomialhas a minimum at $t=0$, so its derivative there vanishes. Hence $\boxed{⟨f-q,r⟩=0}$ for every $r\in {\Pi }_k$. The same Pythagorean identity proves uniqueness of the best approximation.

Expand using the inner product. The two cross terms cancel:Adding proves $\boxed{\Vert x-y{\Vert }^2+\Vert x+y{\Vert }^2=2(\Vert x{\Vert }^2+\Vert y{\Vert }^2)}$, for real or complex inner products.

Take the squared norm of the orthonormal expansion from (i). All mixed inner products vanish and every diagonal inner product is one, so

Set $u={\sum }_{k=1}^n⟨x,e_k⟩e_k$, with the convention that the inner product is linear in its first argument. Orthonormality gives $⟨x-u,e_j⟩=0$, hence $x-u\perp U_n$. For any $v\in U_n$ the Pythagorean identity yieldsThus this $u$ is the best approximation, and uniqueness gives $\boxed{P_{n}x=\sum _{k=1}^n⟨x,e_k⟩e_k}$. No completeness assumption on $H$ is necessary for this finite-dimensional construction.

Assume the intended nondegenerate interval $a<b$ and real-valued continuous functions. The integral is bilinear and symmetric. Since $w>0$, it gives $⟨f,f⟩\ge 0$. If a continuous $f$ is nonzero at some point, continuity makes $f^{2}w$ strictly positive on a subinterval of positive length, so its integral is positive. Hence $⟨f,f⟩=0$ forces $f=0$, proving positive definiteness. Orthogonality means $\boxed{{\int }_a^{b}f(x)g(x)w(x)dx=0}$. If $a=b$, the displayed integral is identically zero and is not an inner product on the stated function space; that degenerate case must be excluded.

The best polynomial minimizes the weighted squared error among all polynomials of degree at most $n$:The given inner product is positive definite on continuous functions despite the weight vanishing at zero: a continuous nonzero function is nonzero on an interval containing nonzero points, so its weighted square integral is strictly positive.

mass as band curvature. In a periodic potential, a weakly forced electron is usefully described by a wave packet confined to one energy band. Its group velocity isDifferentiating the velocity givesThis effective-mass tensor describes the band response, not a change of the electron's bare rest mass. Along a principal direction, a steep upward curvature gives a small positive mass and a flatter curvature a larger one. Near a band maximum the electron curvature and effective mass are negative; the corresponding missing-electron description uses positive electron hole masses. Anisotropy produces distinct principal masses, and strong nonparabolicity makes them momentum dependent. The single-band semiclassical description assumes fields weak and slowly varying enough to avoid substantial interband transitions.

Lattice symmetry and Bloch's theorem. Use the active spatial translation operator $T_{\mathbf{a}}\psi (\mathbf{r})=\psi (\mathbf{r}-\mathbf{a})$. A lattice-periodic scalar potential satisfies $V(\mathbf{r}+\mathbf{a})=V(\mathbf{r})$, while the kinetic term is translation invariant. Therefore $[H,T_{\mathbf{a}}]=0$. The lattice translations also commute with one another, so they can be diagonalized simultaneously with $H$, including within a degenerate energy subspace.

For a common eigenstate, write $T_{\mathbf{a}}\psi =t(\mathbf{a})\psi$. Unitarity gives $|t(\mathbf{a})|=1$, and the group law gives $t(\mathbf{a}+\mathbf{b})=t(\mathbf{a})t(\mathbf{b})$. On primitive Bravais vectors ${\mathbf{a}}_j$, choose $t({\mathbf{a}}_j)=e^{-i\mathbf{k}\cdot {\mathbf{a}}_j}$. Every lattice vector is an integer combination of these generators, henceThe vector $\mathbf{k}$ is defined modulo a reciprocal lattice vector. The band index labels the different energy eigenstates with the same translation character. Define $u_{n\mathbf{k}}(\mathbf{r})=e^{-i\mathbf{k}\cdot \mathbf{r}}{\psi }_{n\mathbf{k}}(\mathbf{r})$. The phase factors cancel under a lattice translation, proving Bloch's theorem in its usual form:This is the Translation-character proof of Bloch theorem; the commutator guarantees a basis of such states, even when an arbitrary superposition of different translation characters is not itself a translation eigenstate.

Periodic Hamiltonian. Acting once with momentum gives $\mathbf{p}(e^{i\mathbf{k}\cdot \mathbf{r}}u)=e^{i\mathbf{k}\cdot \mathbf{r}}(\mathbf{p}+\hslash \mathbf{k})u$. Applying it again and cancelling the common plane-wave factor yieldsThe Bloch Hamiltonian acts on lattice-periodic functions in a unit cell. Expanding its square separates a zero-momentum Hamiltonian, a term linear in $\mathbf{k}$ and an identity term quadratic in $\mathbf{k}$.

Expansion about the zone centre. Let $|n⟩=u_{n0}$, $E_n={\epsilon }_n(0)$, and normalize all these states using a unit-cell inner product. For the real scalar Hamiltonian in the question, a nondegenerate zone-centre eigenfunction can be chosen real. Thenbecause opposite unit-cell faces cancel. Equivalently, time-reversal symmetry forbids a nonzero electric current in this nondegenerate spinless state. This explains why no linear energy term appears; it should not simply be omitted without a symmetry or extremum argument.

The first-order energy shift is therefore ${\hslash }^2{k}^2/(2m)$. In the second-order sum the quadratic term is proportional to the identity and has no off-diagonal matrix elements. Only the linear momentum vertex contributes. Second-order nondegenerate perturbation theory gives the momentum matrix-element expansion of band energyHermitian operator makes the two reciprocal matrix elements complex conjugates. All other periodic eigenstates at $\mathbf{k}=0$ enter the sum. The remainder is even because time-reversal symmetry gives ${\epsilon }_n(\mathbf{k})={\epsilon }_n(-\mathbf{k})$ near this isolated state. At a degeneracy the denominators signal the need for degenerate perturbation theory instead.

mass ratio and the printed formula. Write $p_{nl}^i=⟨n|p_i|l⟩$. Differentiating the quadratic energy yieldsFor a specified unit direction $\mathbf{e}$, the resulting scalar mass ratio isIn an isotropic three-dimensional band, averaging equivalent directions givesCoupling to lower bands contributes positively to inverse mass; coupling to higher bands contributes negatively. Their signs and matrix elements determine the actual curvature.

The original PDF prints $m^{\ast }={\hslash }^2[{\nabla }_{\mathbf{k}}^2{\epsilon }_n{]}^{-1}$. Taken literally as a three-dimensional Laplacian, this misses a factor of three for an isotropic scalar mass. The correct identities areThe printed expression would return $m/3$ even for the free dispersion ${\hslash }^2{k}^2/(2m)$. It is correct in one dimension, or if its second derivative is interpreted directionally. For completeness, its literal three-dimensional trace ratio would instead be $3+(2/m){\sum }_{l\ne n}|⟨n|\mathbf{p}|l⟩{|}^2/(E_n-E_l)$; that is the trace of the dimensionless inverse-mass tensor, not the usual scalar $m/m^{\ast }$.

The Fock vacuum condition and the creation operator commutator giveEquality of the square-root product and $2E_{\mathbf{p}}$ holds on the support of the Dirac delta function.

The positive-energy mass shell has the Lorentz invariant measureHere both $p^2$ and the four-dimensional volume element are invariant, and positive energy is preserved by a proper orthochronous Lorentz transformation. Consequently the three-dimensional Jacobian determinant is $J=E_{\mathbf{\Lambda }p}/E_{\mathbf{p}}$. Transforming the Dirac delta function yieldsso its factor exactly cancels that of $2E_{\mathbf{\Lambda }p}$. Thus the displayed inner product is invariant under simultaneous transformation of both momenta. This is also why the one-particle completeness measure is $d^{3}p/((2\pi {)}^{3}2E_{\mathbf{p}})$.^[ref]

Take $m,\omega >0$ and use the usual common invariant domain for the position operator, momentum operator and their products. Since $X$ and $P$ are self-adjoint,Expanding the commutator and using $[P,X]=-i\hslash I$ givesThe two products areConsequently the quantum harmonic oscillator Hamiltonian operator isThe identity $[AB,C]=A[B,C]+[A,C]B$ now givesThus $H(a_{\pm }\psi )=(E\pm \hslash \omega )a_{\pm }\psi$. Each ladder image is an eigenstate if it is nonzero; a zero vector is not an eigenstate. The distinction is needed for lowering the ground state. Their squared norms areThese follow by moving the operator to the other side of the inner product with its adjoint operator. The first inequality implies $E\ge \hslash \omega /2$ for every eigenstate.

For a lowest-energy quantum state, $a_{-}{\psi }_0$ must vanish, since otherwise it would be an eigenstate with lower energy. The norm formula then forces $E_0=\hslash \omega /2$. Raising never vanishes and producesTo see that there are no other eigenvalues, repeatedly lower any eigenstate. If $E/(\hslash \omega )-1/2$ were not a nonnegative integer, there would be a final nonzero lowered quantum state with energy strictly between $\hslash \omega /2$ and $3\hslash \omega /2$. Its lower image would have positive norm but energy below the lower bound, a contradiction. Thus every eigenstate reaches the ground eigenspace after an integer number of lowerings. On a positive level, lowering is injective by its strictly positive norm factor. The assumed one-dimensional ground eigenspace therefore makes every eigenspace one-dimensional, so the displayed raised quantum states give all eigenstates up to nonzero multiples. If ${\psi }_0$ is normalized, their norms are $\Vert {\psi }_n{\Vert }^2=n!$.

For $k\ge 1$, $(a_{+}{)}^k\psi$ has eigenvalue $E+k\hslash \omega$, distinct from $E$. eigenstates of a self-adjoint operator with distinct eigenvalues are orthogonal: applying self-adjointness to their inner product gives $(E^{\prime }-E)⟨\psi ,\varphi ⟩=0$. ThereforeThe expectation of $(a_{-}{)}^k$ also vanishes, either by lowering and orthogonality or by conjugating the adjoint relation.

Solving the ladder definitions for $X$ and $P$ givesso $T-V=\frac{1}{2}\hslash \omega ((a_{+}{)}^2+(a_{-}{)}^2)$. Both terms have zero expectation by the preceding orthogonality. Since $T+V=H$, the oscillator form of the quantum virial theorem isExpectations here are divided by $\Vert \psi {\Vert }^2$ if the eigenstate is not normalized; no normalization assumption was used in the ladder argument.

Use an expansion for the eigenstate continuing the nondegenerate level, with intermediate normalization $⟨{\psi }_0,\psi (ϵ)⟩=1$:Insert these expressions into the eigenvalue equation. At first order,Taking the inner product with ${\psi }_0$ eliminates the left side by self-adjointness. ThusFor $j\ne 0$, projection onto ${\psi }_j$ instead givesAt second order, projecting the eigenvalue equation onto ${\psi }_0$ gives $E^{(2)}=⟨{\psi }_0,H^{\prime }\chi ⟩$. Expand $\chi$ in the orthonormal energy basis and use self-adjointness of $H^{\prime }$:Nondegeneracy ensures the denominators for this level do not vanish; convergence and validity of the expansion are among the assumed perturbation-theory hypotheses.

For the cosine perturbation of the ground level in a one-dimensional box, write $u=\pi x/a$. The product-to-sum identity givesso, with the given common normalization of the sine eigenfunctions,The diagonal matrix element is $-ϵ/2$, and the only nonzero off-diagonal matrix element out of the ground state is $⟨{\varphi }_3,V{\varphi }_1⟩=ϵ/2$. Let $E_{\mathrm{box}}={\pi }^2{\hslash }^2/(2m{a}^2)$. Since ${\mathcal{E}}_3=9E_{\mathrm{box}}$, the ground energy isThe second-order term is negative, as expected for a nondegenerate ground level: every excited-quantum state denominator is negative. The first-order shift is also negative for positive $ϵ$, because the ground-quantum state density is largest near the center, where this perturbing potential is negative.

Use the Dirac notation convention that the inner product is linear in the ket. Orthonormality of the computational basis givesThe finite geometric sum satisfies ${\sum }_{k=0}^{N-1}{\lambda }^{rk}=N$ if $r=0$ in ${\mathbb{Z}}_N$, and zero otherwise: in the second case it equals $(1-{\lambda }^{rN})/(1-{\lambda }^r)=0$. ThereforeThere are $N$ such orthonormal vectors in an $N$-dimensional space, so they form a basis. The inverse expansion is verified by the same sum:This is the unitary discrete Fourier transform and its inverse.

For the basis-copying isometry, the images $|k⟩|k⟩$ are orthonormal, so $C$ is a linear isometry, with $C^{\ast }C=I$. Its matrix elements giveApplying the definition of the adjoint operator, the coefficient of $|k⟩$ in $C^{\ast }|f⟩|g⟩$ is exactly this expression. ThusEquivalently $C^{\ast }|k⟩|\ell ⟩={\delta }_{k\ell }|k⟩$. The adjoint of basis copying acts as pointwise multiplication of coefficient functions.

For the Fourier transform of a basis-copying isometry, the normalization matters:Hence $C^{\ast }|\hat{m}⟩|\hat{n}⟩=N^{-1/2}|\widehat{m+n}⟩$. Taking adjoint matrix elements, or expanding both output factors in the inverse Fourier basis, yieldsThe printed formula is missing the factor $1/\sqrt{N}$. There are $N$ mutually orthogonal terms in its sum, so without that factor its right side has norm $\sqrt{N}$, whereas its left side has norm one because $C$ is an isometry. The formula above is the normalized identity consistent with all the preceding definitions. For $N=1$ the distinction disappears, but the no-cloning assertion is substantive for $N\ge 2$.

The normalized output is not $|\hat{j}⟩|\hat{j}⟩$. Its coefficient matrix pairs every Fourier label $m$ with $j-m$; tracing out either factor gives $I/N$. Thus it is a maximally entangled state, while two clones of a pure Fourier quantum state would have pure reduced quantum states. A copier for the computational basis therefore does not copy the Fourier basis.

More generally, suppose a linear isometry with a fixed blank input could clone every normalized quantum state. Its restriction to $|\psi ⟩|\mathrm{blank}⟩$ would define a linear map $D$ with $D|k⟩=|k⟩|k⟩$ on every computational basis quantum state. Linearity forces $D=C$, which has just been shown to fail on Fourier quantum states. This proves the no-cloning theorem in the pure-quantum state isometric setting.

The inner-product proof makes the obstruction independent of the chosen basis. If an isometry cloned two quantum states with overlap $s=⟨\psi |\varphi ⟩$, preservation of inner products would give $s=s^2$, or, allowing irrelevant output phases, $|s|=|s{|}^2$. Distinct nonorthogonal rays have $0<|s|<1$, which is impossible. Even allowing an output environment does not help: unitary evolution would give $|s|=|s{|}^2|⟨e_{\psi }|e_{\varphi }⟩|\le |s{|}^2$, the same contradiction. Arbitrary unknown quantum states cannot be cloned, although any fixed orthonormal basis can be copied and a one-dimensional quantum state space is trivial.

Inside the box, the time-independent Schrödinger equation is $-{\hslash }^2{\Psi }^{\prime \prime }/(2m)=E\Psi$. The zero endpoint values eliminate the cosine term and require $ka=n\pi$. There is no nonzero zero-energy solution with both endpoints zero. Normalization uses ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$, soInsert the expansion into $i\hslash {\partial }_t\psi =H\psi$ and take its inner product with ${\Psi }_m$. Orthonormality gives $i\hslash {\dot{c}}_m=E_m{c}_m$ for each $m$, and conversely these equations make every expansion coefficient of the residual zero. Thus $c_n(t)=c_n(0)e^{-i{E}_{n}t/\hslash }$ in the usual Hilbert-space or strong-solution class appropriate to the expansion.

For the specified state,The modes are real, so the interference term is a cosine of the energy difference. Writing $s_n=\sin (n\pi x/a)$,Its integral remains one because the cross term integrates to zero. The probabilities of the two energy outcomes are fixed at $9/25$ and $16/25$, even though the position probability density oscillates.

Let $\Psi \ne 0$ and take the reduced positive operator ${\rho }_1={\mathrm{Tr}}_2|\Psi ⟩⟨\Psi |$. Choose orthonormal eigenvectors ${\xi }_j$ for its positive eigenvalues ${\lambda }_j$. DefineThe defining identity of the partial trace gives $⟨v_i,v_j⟩={\lambda }_j{\delta }_{ij}$, so the ${\eta }_j$ are orthonormal. Expansion in a full eigenbasis of ${\rho }_1$ yields $\Psi ={\sum }_j{\xi }_j\otimes v_j$; zero-eigenvalue terms have zero squared norm and contribute nothing. ThusThe zero vector has the empty decomposition. Unit normalization would give ${\sum }_j{c}_j^2=1$, but no normalization is needed for the theorem itself.

For the given vector, the coefficient matrix is $M=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\right)$. Its second reduced operator is $M^{\ast }M=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, whose eigenvalues are $3,1$ and eigenvectors ${\eta }_{+}=(E_1+E_2)/\sqrt{2}$, ${\eta }_{-}=(E_1-E_2)/\sqrt{2}$. The corresponding normalized first-factor vectors areTheir norms are one and their inner product is zero. HenceExpanding the two terms returns the original three contributions. The vector has squared norm $4$; if a normalized state is wanted, divide by two and the Schmidt coefficients become $\sqrt{3}/2$ and $1/2$.

The no-cloning theorem states that no fixed deterministic quantum operation can perfectly duplicate every unknown pure state. For a unitary copier with a fixed blank, suppose it cloned two normalized vectors $|a⟩,|b⟩$. Preservation of inner products would give $⟨a|b⟩=⟨a|b{⟩}^2$, impossible for a nonorthogonal distinct pair with overlap strictly between zero and one in modulus. Including an environment does not help: its final overlap has modulus at most one, so one would require $|⟨a|b⟩|\le |⟨a|b⟩{|}^2$. Thus a universal perfect copier is impossible. Orthogonal known alternatives do not contradict this restriction.

Label the three qubits as input $S$, Alice's entangled qubit $A$ and Bob's qubit $B$. Initially their joint state isThe controlled-NOT gate from $S$ to $A$ changes this toApply $H|0⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $H|1⟩=(|0⟩-|1⟩)/\sqrt{2}$ to the first qubit and group by Alice's two computational-basis outcomes. The result isIn expanded form the last two Bob factors are ${\alpha }_0|0⟩-{\alpha }_1|1⟩$ and ${\alpha }_0|1⟩-{\alpha }_1|0⟩$. Their Alice labels must be $10$ and $11$; the repeated $00$ labels in the source are printing errors.

Alice measures her two qubits and sends the two classical bits $(a,b)$ to Bob. Each outcome has probability $1/4$ and Bob's conditional state is $X^b{Z}^a|\psi ⟩$. He applies $\boxed{Z^a{X}^b}$ to recover the input, up to physically irrelevant global phase. Explicitly the corrections for $00,01,10,11$ are $I,X,Z,ZX$. This is quantum teleportation: Alice's input is consumed by measurement, so no second surviving copy is created. Before her classical message, Bob's reduced density operator is $I/2$, independent of the input, so the protocol also supplies no faster-than-light communication.

Take a normalized vector in the required common operator domains. Expanding the squared norm and using self-adjointness givesThus its final term is $\hslash t⟨C⟩$. Set $a=\Vert A\psi {\Vert }^2$, $b=\Vert B\psi {\Vert }^2$, $c=⟨C⟩$. If $b>0$, minimizing the nonnegative quadratic at $t_{\ast }=-\hslash c/(2b)$ givesEquality holds precisely when the minimum norm vanishes, that is $(A-i{t}_{\ast }B)\psi =0$. If $b=0$, however, then $B\psi =0$ and $c=0$, so equality is automatic even when $A\psi \ne 0$. The printed equality criterion needs this zero-vector exception. For example $A=I$, $B=0$, $C=0$ saturates the inequality but admits no such real $t$. The exact general criterion is linear dependence of $A\psi ,B\psi$ together with vanishing real part of their inner product, as in the uncentered second-moment commutator inequality.

Apply the same calculation to $A-⟨A⟩I$, $B-⟨B⟩I$, whose commutator is unchanged. This gives the Robertson uncertainty principleIf $\psi$ is an eigenvector of $A$, its dispersion is zero; finite dispersion of $B$ then forces $\boxed{⟨C⟩=0}$.