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For a reversible Markov chain with invariant law $\mu$, its Markov kernel acts as a self-adjoint operator $P$ on $L^2(\mu )$. An absolute spectral gap means that $\Vert P\Vert$ on the mean-zero functions is at most $r<1$. Geometric ergodicity implies this gap for a reversible chain. Indeed, for bounded mean-zero $g$ supported where the geometric prefactor is bounded, the total-variation estimate gives $⟨g,P^{2n}g⟩\le C_g{r}^{2n}$. The spectral theorem for normal operators puts its spectral measure in $[-r,r]$. Such test functions are dense in the mean-zero subspace.

For a diagonalizable matrix, the active spectrum consists of eigenvalues whose eigenspaces have nonzero components of the starting vector in its spectral decomposition. These components are orthogonal projections for a self-adjoint operator, but need not be orthogonal projections for a general diagonalizable matrix. With a unique largest modulus in this active spectrum, the power method selects its eigendirection; an exactly absent dominant component is never created in exact arithmetic.

Bounded self-adjoint operators on separable infinite-dimensional complex Hilbert spaces are unitarily equivalent modulo compact operators exactly when they have the same essential spectrum in the Calkin algebra. The Weyl-von Neumann theorem diagonalizes each modulo a compact error. Their diagonal cluster sets are the essential spectra; cluster-set matching of bounded sequences pairs the two diagonal lists with differences tending to zero, giving a compact diagonal difference. The separability assumption is essential: projections with countable and uncountable range dimensions can have the same essential spectrum but cannot differ by a compact operator after unitary conjugation.

If A and B are commuting self-adjoint operators with $A^2=B^2=I$, then the displayed combination is a unitary operator. Indeed,and the reverse product is also I. The factors are involutions, and their commutation is essential to cancel the cross terms. Operators $A=U_A\otimes I$ and $B=I\otimes U_B$ on separate tensor factors provide a natural example. More generally, $\cos \theta A+i\sin \theta B$ is unitary for real $\theta$ by the same proof. Replacing i by a real coefficient usually loses this property.

A complete set of commuting observables consists of mutually commuting quantum observables whose joint eigenvalues distinguish all vectors of a simultaneous eigenbasis, up to phase. For a finite-dimensional Hilbert space, commuting self-adjoint operators admit a simultaneous orthonormal basis of eigenvectors; completeness means that every joint eigenspace is one-dimensional. An additional quantum observable that is already a function of the others supplies no independent quantum number.

For the addition of angular momentum, the uncoupled labels $(j_1,j_2,m_1,m_2)$ correspond to $\{J_1^2,J_2^2,J_{1z},J_{2z}\}$, whereas the coupled labels $(j_1,j_2,j,m)$ correspond to $\{J_1^2,J_2^2,J^2,J_z\}$. Within a fixed pair of irreducible angular momentum spaces each choice gives a natural eigenbasis. The Clebsch-Gordan coefficients implement the change between them.

For a self-adjoint operator $T$, a real number $\lambda$ lies in the continuous spectrum when $T-\lambda I$ is injective with dense range but is not surjective. Such spectral values need not be eigenvalues of normalizable states. The free-particle Hamiltonian on the real line has a continuous energy spectrum.

For an ordered basis $\mathcal{B}=(b_1,\dots ,b_n)$, the coordinate vector $[v{]}_{\mathcal{B}}=(a_1,\dots ,a_n{)}^T$ records the unique coefficients satisfying $v={\sum }_i{a}_i{b}_i$. Change of basis changes this representation while preserving the underlying vector.

For $b_1=(1,0{)}^T$, $b_2=(1,1{)}^T$, the vector $v=(3,1{)}^T$ has coordinate vector $(2,1{)}^T$, since $2b_1+b_2=v$. These are the same coefficients extracted by the dual basis in its worked example. The vector is independent of the chosen basis; its list of coordinates is not. In an orthonormal eigenbasis, the coordinates additionally separate the action of a self-adjoint operator into independent scalar multiplications.

A smooth uniformly elliptic differential expression with symmetric Dirichlet bilinear form has an unbounded self-adjoint operator realization $A_D$ on $L^2(U)$, whose domain is $H^2(U)\cap H_0^1(U)$ on a smooth bounded domain. If the form is strictly positive, the Lax-Milgram theorem and elliptic regularity make $A_D^{-1}$ a bounded map from $L^2$ to $H^2\cap H_0^1$. Its action on $L^2$ is compact by the Rellich-Kondrachov compactness theorem and self-adjoint by symmetry of the form. The spectral theorem for compact Hermitian operators therefore gives an orthonormal basis of smooth Dirichlet eigenfunctions with positive eigenvalues tending to infinity.

For a nonnegative self-adjoint operator with compact resolvent, this function counts eigenvalues up to a given level with their multiplicities. For the positive Laplace-Beltrami operator on a closed $n$-dimensional Riemannian manifold, $N(\Lambda )\le C(1+\Lambda {)}^{n/2}$. To prove this bound, multiply each vector in the low-eigenvalue subspace by a finite set of chart cutoffs and record its local Fourier coefficients of frequency at most $R$. The total number of retained coefficients is $O(R^n)$, while the sum of discarded squared norms is at most $C{R}^{-2}(1+\Lambda )\Vert u{\Vert }_2^2$ by Parseval identity and the energy bound. For $R$ a sufficiently large multiple of $\sqrt{1+\Lambda }$, vanishing of all recorded coefficients forces $u=0$. The recording map is injective and proves the dimension bound. Inverting this estimate gives ${\lambda }_j\ge c(j+1{)}^{2/n}-1$.

An energy eigenvalue is an eigenvalue of the Hamiltonian operator $H$: there is a nonzero vector $\psi$ in its operator domain $D(H)$ withFor a self-adjoint operator, $E$ is real. Normalizing $\psi$ produces an energy eigenstate; an ideal energy measurement then returns $E$ with probability one. The associated energy eigenspace includes all vectors with this same energy eigenvalue, including the zero vector, which represents no quantum state. Different energy eigenvalues have orthogonal energy eigenspaces. ^[ref].

The energy eigenvalues form the point spectrum, which can be only part of the spectrum. The term discrete spectrum more specifically means isolated eigenvalues of finite multiplicity. The usual fixed-nucleus hydrogen atom model has discrete negative bound state energy eigenvalues together with a continuum at nonnegative energy. ^[ref]. A quantum system need not have a ladder of discrete energy eigenvalues: a whole-line free particle has continuous spectrum. A plane wave used for a free particle is a generalized eigenfunction of the Hamiltonian operator, rather than a normalized vector in the physical Hilbert space. ^[ref].

For the infinite square well $0<x<L$, the Dirichlet boundary conditions giveSubstitution into $H{\varphi }_n=E_n{\varphi }_n$ and wavefunction normalization verify these statements directly. Here $E_1$ is the ground-state energy, even though its index is one; the labeling convention does not determine which quantum state is lowest. Equal populations of ${\varphi }_1$ and ${\varphi }_2$ have energy expectation value $(E_1+E_2)/2=5E_1/2$, but an energy measurement returns $E_1$ or $4E_1$, each with probability $1/2$. An average of allowed energy eigenvalues need not itself be an energy eigenvalue.

For a bounded self-adjoint operator $L$ on a complex Hilbert space, defineThis equals the spectrum of a bounded operator after removing its isolated eigenvalues of finite multiplicity. To see isolation when the kernel is finite and the shifted range is closed, restrict the shifted operator to its kernel complement: it is invertible there by image-kernel orthogonality for an adjoint. A Neumann-series perturbation leaves nearby nonzero shifts invertible on that complement and on the finite kernel. The closed range must concern the shifted operator, not the original one.

The expectation value of a quantum observable is the expected value of its measurement outcome random variable in a specified quantum state. It predicts the mean of outcomes from many identical quantum state preparations. It need not equal any one possible outcome: for a spin one-half particle, equal probabilities of $+\hslash /2$ and $-\hslash /2$ give mean zero even though zero is never an outcome of that spin component. Here $\hslash$ is the reduced Planck constant.

Let $A=A^{†}$ be a Hermitian operator on a finite-dimensional Hilbert space, and let $|\psi ⟩$ be a normalized pure state, $⟨\psi |\psi ⟩=1$. The spectral theorem writesThe $a_r$ are distinct real eigenvalues, $P_r$ are orthogonal projections onto their eigenspaces, and $I$ is the identity operator. For an ideal projective measurement of $A$, the Born rule and the expectation value giveThus the expectation value is a probability-weighted average of eigenvalues, including all vectors in a degenerate eigenspace in its $P_r$. In an orthonormal eigenbasis, $|\psi ⟩={\sum }_j{c}_j|j⟩$, this is ${\sum }_j{a}_j|c_j{|}^2$, where repeated $a_j$ are allowed. The weighted average lies between the smallest and largest eigenvalues; it can lie in a gap between them. The statistical meaning and operator expression are connected in this way in the authored treatment of average energies.^[ref]

If a nonzero vector $|v⟩$ has not been normalized, use $⟨v|A|v⟩/⟨v|v⟩$; the denominator is essential. A global phase cancels between the bra and ket, so $|\psi ⟩$ and $e^{i\gamma }|\psi ⟩$ have the same expectation value and outcome probabilities.

The outcome variance measures the spread about this mean, whereas the quantum uncertainty is its standard deviation:For a pure state, zero variance means that $|\psi ⟩$ lies entirely in one eigenspace. That eigenspace need not be one-dimensional. Knowing only a mean generally does not determine the entire outcome probability distribution.

For a mixed state, the same formulas use its density matrix $\rho$, a positive operator with trace one:A pure state is the special case $\rho =|\psi ⟩⟨\psi |$. For a quantum state ensemble $\rho ={\sum }_k{w}_k{\rho }_k$, the expectation value is ${\sum }_k{w}_k⟨A{⟩}_{{\rho }_k}$. This linearity is useful when preparations vary between trials; different quantum state ensembles can have the same density matrix.

These formulas require no operator-domain qualifications in finite dimension. For an unbounded self-adjoint operator, the spectral theorem instead gives a spectral probability measure; a finite mean requires an integrable absolute first moment, and finite variance requires a finite second moment. Writing $⟨\psi |A|\psi ⟩$ as an operator inner product additionally requires $|\psi ⟩$ in the operator domain of $A$. The quadratic form of a positive quantum Hamiltonian explains the useful wider form-domain version.

Every self-adjoint operator on a finite-dimensional real or complex inner product space has an orthonormal eigenbasis, and all its eigenvalues are real. In orthonormal coordinates, a real symmetric matrix has the form $A=Q\Lambda Q^T$, with $Q$ an orthogonal matrix and $\Lambda$ a real diagonal matrix. A complex Hermitian matrix instead has $A=U\Lambda U^{\ast }$, with $U$ a unitary matrix and $U^{\ast }$ its conjugate transpose. These are the self-adjoint cases of the spectral theorem; Axler gives the real and complex finite-dimensional statements in Section 7B.^[ref]

An orthonormal eigenbasis turns a seemingly coupled linear map into independent scalar multiplications while preserving lengths and angles. If $x={\sum }_j{c}_j{e}_j$ and $A{e}_j={\lambda }_j{e}_j$, thenThe last identity removes every cross term from the quadratic form. The eigenvalues tell us whether it is positive, negative, or mixed in sign, and the eigenvectors identify the corresponding directions. This connects positive-definite matrices, quadratic forms, and the Hermitian Rayleigh quotient range.

For a fully computed example, takeDirect multiplication gives $A{e}_{+}=3e_{+}$ and $A{e}_{-}=e_{-}$. Their dot product is zero and each has norm one. ThusWriting $x=c{e}_{+}+d{e}_{-}$ changes $x^{T}Ax=2x_1^2+2x_1{x}_2+2x_2^2$ into $3c^2+d^2$. Its level set $x^{T}Ax=1$ is therefore an ellipse with semiaxis lengths $1/\sqrt{3}$ and $1$, along $e_{+}$ and $e_{-}$. The coordinates change, but the ellipse does not: orthogonal diagonalization reveals its existing geometry.

For distinct eigenvalues, orthogonality follows at once from self-adjointness: $\lambda ⟨u,v⟩=⟨Au,v⟩=⟨u,Av⟩=\mu ⟨u,v⟩$. Within a repeated eigenspace, choose an orthonormal basis; arbitrarily chosen eigenvectors of the same eigenvalue need not be orthogonal. One outline of the full argument finds an eigenvector, observes that its orthogonal complement is invariant, and repeats in that smaller space. Merely having real eigenvalues is weaker: $\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has two real eigenvalues, but its eigenvectors $(1,0{)}^T$ and $(-1,1{)}^T$ are not orthogonal.

Take four congruent octagonal regular polygons with four alternate sides labelled $a,b,c,d$. The intervening sides have Neumann boundary conditions. Pair equal face parameters in the positive reference boundary direction. The complete gluing data areAll unpaired labelled faces not in a Dirichlet column also have Neumann boundary conditions. Encode these rules by signed permutation matrices, using a positive swap for a paired interface and diagonal $-1$ or $+1$ for an unpaired Dirichlet boundary condition or Neumann boundary condition. The four-by-four Hadamard matrix $C$ above satisfies $C{M}_s=N_{s}C$ for every label. Explicitly, if its columns are $h_0,h_1,h_2,h_3$, applying $N_a,N_b,N_c,N_d$ gives respectively $(h_0,h_1,h_3,h_2)$, $(h_0,h_2,h_1,h_3)$, $(h_1,h_0,-h_2,-h_3)$, and $(h_0,-h_1,h_2,-h_3)$, which are exactly the columns of the corresponding $C{M}_s$. On every intervening face the face matrix is the identity.

Since $C^{T}C=4I$, the orthogonal matrix $C/2$ preserves the $L^2$ inner product and the sum of Dirichlet energies. Its intertwining relations map matching Sobolev traces and zero Dirichlet traces bijectively to those of the other assembly. It therefore preserves the quadratic form domains and the forms themselves. The associated self-adjoint operators are unitarily equivalent, proving equality of all eigenvalues with their multiplicities for the specified mixed boundary conditions.

The topology of disk tiles glued along disjoint boundary arcs shows that $D$, whose gluing graph is $0,1,2,3$ in a path, is a closed disc and is simply connected and orientable. The gluing graph of $B$ has edges $13,12,02,01,23$; it is connected with two independent cycles, so $\chi (B)=-1$ and ${\pi }_1(B)\cong F_2$. Its odd cycle $1,2,3,1$ makes it a nonorientable surface. Each assembled surface has nonempty polygonal boundary. This explicitly realizes isospectrality between a simply-connected orientable bordered surface and a nonorientable, non-simply-connected one. The boundary corners and mixed boundary conditions are essential qualifications; it does not contradict the Euler characteristic from the heat trace of a bordered surface for smooth boundaries and pure boundary conditions.

For a self-adjoint operator $L$, solving $Lu=f$ requires $f$ to be orthogonal to every $v\in \ker L$, because $⟨v,Lu⟩=⟨Lv,u⟩=0$. When the range is closed, this condition is also sufficient: $(\mathrm{ran}L{)}^{\perp }=\ker L$ and closedness gives $\mathrm{ran}L=(\ker L{)}^{\perp }$. In finite dimensions the range is automatically closed. This Fredholm solvability condition determines amplitude equations by projecting perturbative forcing onto critical eigenfunctions.

A densely defined semibounded symmetric operator on a Hilbert space has a canonical semibounded extension that is a self-adjoint operator obtained by closing its quadratic form. After shifting the lower bound to zero, complete the initial form domain in the norm $(\Vert u{\Vert }^2+q[u]{)}^{1/2}$; the associated closed form determines the Friedrichs extension. For the positive Laplace-Beltrami operator on a polygonal surface, this specifies the finite-energy realization at corners. Different choices of the initial form domain impose Dirichlet boundary conditions, Neumann boundary conditions, or mixed boundary conditions; the extension does not choose the boundary condition independently of that domain.

For rigid walls $a<r<b$, a nonzero vertical wavenumber $k$ and radial Lagrangian fluid displacement $\xi$, set $y=r\xi$ and $\mathcal{D}={\kappa }^2-r(d/dr)(v_A^2/r^2)$. Axisymmetric modes satisfyThe operator on the left is a positive self-adjoint operator. Its Rayleigh quotient is ${\omega }^2=k^2\int \mathcal{D}|y{|}^2/rdr/\int (|y^{\prime }{|}^2+k^2|y{|}^2)/rdr$. All mode squares are real; a growing mode exists exactly when the continuous coefficient $\mathcal{D}$ is negative somewhere. Sufficiency follows by a test function supported in that negative interval and the compact operator and self-adjoint operator eigenproblem obtained by conjugating the multiplication operator with the inverse square root of the positive differential operator.

// Target: astrophysical-fluid-dynamics.bigb

For a real smooth function $\chi$ on Euclidean space, put $A_a={\partial }_a-{\partial }_a\chi$ and $q={\sum }_a{\psi }_a^{†}{A}_a$. The adjoint operator for the usual $L^2$ inner product is $A_a^{†}=-{\partial }_a-{\partial }_a\chi$. Since $[A_a,A_b]=0$, the canonical anticommutation relations give $q^2=0$. Expanding $H=\{q,q^{†}\}$ and using $[A_a,A_b^{†}]=-2{\partial }_a{\partial }_b\chi$ givesIn the empty fermionic Fock space sector the last term vanishes; in the filled sector ${\psi }_a^{†}{\psi }_b={\delta }_{ab}$. Consequently the two scalar Hamiltonian operators are $H_0=-\Delta +|\nabla \chi {|}^2+\Delta \chi$ and $H_d=-\Delta +|\nabla \chi {|}^2-\Delta \chi$. On the compatible self-adjoint operator domain the Hamiltonian operator is nonnegative because $⟨v,Hv⟩=\Vert qv{\Vert }^2+\Vert q^{†}v{\Vert }^2$.

A ground state is a normalized energy eigenstate that attains the lowest possible energy of a quantum system. Precisely, let the Hamiltonian operator $H$ be a self-adjoint operator bounded below on a Hilbert space, and write $E_{\ast }=\inf \sigma (H)$ for the infimum of its spectrum. A ground state is a unit vector ${\psi }_0$ in the operator domain $D(H)$ satisfyingThus a ground state exists exactly when the spectral bottom $E_{\ast }$ is an energy eigenvalue. The lowest energy eigenvalue among some listed bound states is insufficient if other parts of the spectrum lie below it. The ground-state subspace collects all minimizing energy eigenstates; it can have ground-state degeneracy. The quantum variational principle characterizes $E_{\ast }$ as the infimum of normalized energy expectation values. ^[ref].

For the one-dimensional quantum harmonic oscillator, the ground state has energy $\hslash \omega /2$ above the minimum of the potential energy. This zero-point energy is compatible with being the lowest allowed energy: a localized wavefunction cannot simultaneously have zero kinetic energy. For a free particle on the whole line, the spectral bottom is zero but is unattained by any normalizable wavefunction; see ground-state energy. A lower bound therefore need not describe a physically preparable energy eigenstate.

The ground state supplies a reference for excitation costs and for the quantum variational principle. A ground state is also a stationary state under a time-independent Hamiltonian operator: its quantum phase changes, while its position probability density stays fixed. A nonzero ground-state energy does not imply motion along a classical trajectory, and ground-state degeneracy means there need not be a unique lowest quantum state.

The ground-state energy $E_0$ is the energy eigenvalue attained by a ground state. For a self-adjoint operator $H$ bounded below, distinguish this attained value from the always-defined spectral bottom $E_{\ast }=\inf \sigma (H)$. When a ground state exists, $E_0=E_{\ast }$. Some usage calls $E_{\ast }$ the ground-state energy even when it is unattained; here an unattained bottom is explicitly identified as an infimum. The quantum variational principle statesApproaching this infimum with trial wavefunctions does not ensure that a minimizing wavefunction exists. ^[ref].

For the whole-line free particle with mass $m>0$,Its spectrum is continuous and contains no energy eigenvalues represented by normalizable wavefunctions. The Fourier transform makes $H$ multiplication by ${\hslash }^2{k}^2/(2m)$; a fixed energy restricts $k$ to a set of zero length, on which a nonzero square-integrable function cannot be supported. ^[ref]. An explicit family of normalized Gaussian wave packets demonstrates how the bottom is approached:The formula follows from ${\psi }_s^{\prime }=-x{\psi }_s/(2s^2)$, integration by parts, and $⟨x^2{⟩}_s=s^2$. Every finite width gives positive kinetic energy. At zero energy, $H\psi =0$ would require an affine function on the whole line, which is never a nonzero normalizable wavefunction. Spreading a Gaussian wave packet indefinitely produces an unattained infimum, rather than a zero-energy ground state.

An additive choice of energy origin changes the numerical ground-state energy without changing the ground-state subspace. For $H^{\prime }=H+CI$, every energy eigenvalue becomes $E^{\prime }=E+C$, while all differences $E_j-E_i$ stay fixed. Time evolution gains only the global phase $e^{-iCt/\hslash }$. This makes excitation costs and spectroscopic differences more intrinsic than the isolated number $E_0$.