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A simple Lie algebra is a nonabelian Lie algebra whose only ideals of a Lie algebra are $0$ and the whole algebra.

Use the standard basis $e,f,h$ of the sl2 Lie algebra, withLet $I\ne 0$ be an ideal and choose $0\ne x=ae+bf+ch\in I$. Since $I$ is invariant under the Adjoint representation, it is invariant under the linear operator $\mathrm{ad}h$. The three basis vectors are eigenvectors of $\mathrm{ad}h$ with distinct eigenvalues $2,-2,0$. Applying the corresponding polynomial spectral projections to $x$ shows that $I$ contains at least one nonzero multiple of $e$, $f$, or $h$.

If $e\in I$, then $[f,e]=-h\in I$ and $[f,h]=2f\in I$; the cases $f\in I$ and $h\in I$ are identical after taking brackets with the other basis vectors. Hence $e,f,h\in I$, so $I={\mathfrak{sl}}_2(\mathbb{C})$. Therefore ${\mathfrak{sl}}_2(\mathbb{C})$ is simple.

Use the singular system of a compact operator convention $A{x}_j={\sigma }_j{y}_j$ and $A^{\ast }{y}_j={\sigma }_j{x}_j$. Starting from zero gives no component in $\ker A$. Write $x(t)={\sum }_j{c}_j(t)x_j$. The coefficient equations areSolving these scalar linear ordinary differential equations givesConsequently asymptotic regularization has the spectral regularization method representationThe component of $f$ orthogonal to the closure of the range of $A$ is annihilated by $A^{\ast }$ and contributes nothing. The scalar filter satisfies $(1-e^{-{\sigma }^2/\alpha })/\sigma \le \min (\sigma /\alpha ,1/\sigma )\le {\alpha }^{-1/2}$, so every fixed-$\alpha$ reconstruction is a bounded linear operator. Its long-time limit is the Moore–Penrose inverse of an operator on admissible data.

The notation $P_0=\mathcal{N}(0,I)$ denotes generalized Gaussian white noise over $H=L^2({\mathbb{R}}^2)$, not a Gaussian measure supported on $H$: the identity is not a trace-class operator there. For example, realize the observation on a suitable space of tempered distributions. Its Cameron-Martin space of a Gaussian measure is $H$. For deterministic $h\in H$, the stochastic pairing $W_m(h)$ has normal distribution $\mathcal{N}(0,\Vert h{\Vert }_H^2)$ under $P_0$ and is the isonormal Gaussian process indexed by $H$.

Interpret the stated maps between Sobolev spaces as bounded linear operators, as usual. In particular $K:H\to H^2({\mathbb{R}}^2)\subset H$ is bounded, so its adjoint operator $K^{\ast }$ is bounded on $H$. Hence $A=K^{\ast }K$ maps $H$ into $H$, which is the property needed here. The stronger smoothing follows by Sobolev duality: boundedness of $K:H^{-4}\to H^{-2}$ gives $K^{\ast }:H^2\to H^4$, and therefore $A:H\to H^4\subset H$. As $\Pi (H)=1$, the shift $Au$ belongs to the noise Cameron-Martin space of a Gaussian measure for $\Pi$-almost every $u$.

The white-noise likelihood for a square-integrable shift, obtained from the Cameron-Martin theorem for a Gaussian measure in its white-noise form, isHere $P_u$ is the law of $Au+\eta$. In an orthonormal basis $(e_j)$ of $H$, with white-noise coordinates $m_j$, the pairing under $P_0$ isFor fixed $u$ this has mean-square convergence and converges almost surely, since ${\sum }_j|⟨Au,e_j⟩{|}^2=\Vert Au{\Vert }_H^2$. The stipulated joint measurability allows the likelihood to be used under ${\nu }_0(du,dm)=\Pi (du)P_0(dm)$. One must not replace this expression by a finite $\Vert m-Au{\Vert }_H^2$, because white noise is not $H$-valued. On the unbounded domain ${\mathbb{R}}^2$, membership of white noise in a global unweighted negative Sobolev space must not be assumed either; the stochastic pairing avoids that issue.

Let $\nu (du,dm)=\Pi (du)P_u(dm)$ be the actual joint law. The shift formula gives $\nu \ll {\nu }_0$ with Radon-Nikodym derivative $L$. DefineFor each fixed admissible $u$, the moment-generating function of a normal distribution givesBy Tonelli theorem, $\int Z(m)P_0(dm)=1$, so $Z(m)<\infty$ for $P_0$-almost every $m$. The likelihood is finite and strictly positive for ${\nu }_0$-almost every $(u,m)$; Fubini's theorem then gives $Z(m)>0$ for $P_0$-almost every $m$. The marginal observation law is $Z(m)P_0(dm)$ and is an equivalent probability measure to $P_0$.

The Bayes formula for a dominated observation model therefore definesTo verify that it is the conditional distribution, for measurable sets $B$ of unknowns and $C$ of data,Thus the posterior distribution is well defined for almost every observation under the actual data law. This is the appropriate almost-everywhere assertion supplied by Bayes' theorem; the stated assumptions do not prescribe a canonical posterior at every exceptional generalized datum.

For a bounded linear operator $T:H\to H$, the spectrum iswhile the point spectrum is the set of eigenvalues, equivalently the $\lambda$ for which $T-\lambda I$ is not injective.

A compact operator maps the unit ball to a relatively compact set. A self-adjoint operator satisfies $T=T^{\ast }$, or $⟨Tx,y⟩=⟨x,Ty⟩$ for all $x,y\in H$. A finite-rank operator has finite-dimensional image.

For fixed $f\in L^2$, the functionalis bounded on $\Sigma$ by the continuous embedding from part 2:The Riesz representation theorem gives a unique $T(f)\in \Sigma$ such that $⟨v,T(f){⟩}_{\Sigma }={\ell }_f(v)$ for every $v\in \Sigma$. It also givesso the linear operator $T:L^2\to \Sigma$ is bounded.

The infinitesimal generator of a semigroup is the linear operatorwith generator domainFor example, the Bochner integralbelongs to $D(A)$ for every $u\in H$ and $t>0$, since $A{u}_t=U(t)u-u$.

To prove that $A$ is a closed linear operator, suppose $u_n\in D(A)$, $u_n\to u$, and $A{u}_n\to v$. For vectors in the generator domain,Passing to the limit in the Banach space givesAfter division by $t$, strong continuity makes the right side converge to $v$ as $t\downarrow 0$. Hence $u\in D(A)$ and $Au=v$, so $A$ is closed.

For $\mathrm{Re}z>\omega$, define the Bochner integralIt converges absolutely becauseIntegrating the semigroup difference quotient shows that $R(z)u\in D(A)$ and $(zI-A)R(z)u=u$. The same computation for $u\in D(A)$ gives $R(z)(zI-A)u=u$. Thus the Laplace-transform formula for a semigroup resolvent proves

The familyinherits the identity, semigroup property, and strong continuity from $U$, whileIts difference quotient satisfiesfor $u\in D(A)$. Conversely, existence of this limit implies existence of the generator limit for $U$, so $D(\hat{A})=D(A)$ andThis is the exponentially shifted semigroup construction.

The Hille-Yosida theorem in the uniformly bounded case says that a linear operator $B$ on a Banach space generates a C0-semigroup with $\Vert e^{tB}\Vert \le M$ if and only if:

A real linear operator $\mathcal{L}$ on an inner product space is a positive-definite operator when it is self-adjoint andfor every nonzero $u$ in its domain. In the variational setting one normally requires the stronger uniform estimate $⟨\mathcal{L}u,u⟩\ge c\Vert u{\Vert }_V^2$ for some $c>0$, which is coercivity.

A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.

Because $\mathbf{b}\cdot \mathbf{x}$ is a scalar, applying the linear operator again givesThis is the first nontrivial case of the powers of a rank-one operator; its matrix is the outer product $\mathbf{a}{\mathbf{b}}^T$.

The displayed power series is the matrix exponential $G=e^H$. From $H^3=-{\alpha }^{2}H$, its odd powers and nonconstant even powers sum separately:The scalar power series are $\sin \alpha /\alpha$ and $(1-\cos \alpha )/{\alpha }^2$, with continuous values $1$ and $1/2$ at zero. Substituting $H\mathbf{x}=\alpha \mathbf{n}\times \mathbf{x}$ and the preceding formula for $H^2\mathbf{x}$ yieldsThis is the Rodrigues rotation formula: the component parallel to $\mathbf{n}$ is fixed, while the perpendicular component undergoes a rotation through $\alpha$. For $\alpha =0$ the linear operator is the identity operator.

On an eigenvector of $A$ with eigenvalue $\lambda$, the linear operator $B=I-A$ has eigenvalue $1-\lambda$. ThereforeThe multiplicities are exchanged: if the matrix rank of $A$ is $r$, then $B$ has $n-r$ copies of $1$ and $r$ copies of $0$. Again, a value with zero multiplicity is absent.

For each mode in Cartesian coordinates, $[a_i,a_j^{†}]={\delta }_{ij}$ and $a_i^{†}{a}_i=H_i/(\hslash \omega )-1/2$. Consequently the two-dimensional isotropic harmonic oscillator hasThe normalized actions of the creation and annihilation operators areInvert their definitions as $x=\sqrt{\hslash /(2m\omega )}(a_x+a_x^{†})$ and $p_x=i\sqrt{m\hslash \omega /2}(a_x^{†}-a_x)$, and similarly for $y$. linear operators belonging to different coordinates commute, so substitution into $L_z=x{p}_y-y{p}_x$ givesEach term transfers one excitation between the two modes: $[N_x+N_y,a_x^{†}{a}_y]=(1-1)a_x^{†}{a}_y=0$, and likewise for its adjoint. Thus $[H,L_z]=0$ and, with no explicit time dependence, $d{L}_z/dt=(i/\hslash )[H,L_z]=0$. This is conservation of angular momentum under planar rotations.

Finally $L_z|1,0⟩=i\hslash |0,1⟩$ and $L_z|0,1⟩=-i\hslash |1,0⟩$. HenceThe angular momentum is $+\hslash$; division by $\sqrt{2}$ performs wavefunction normalization on this state without changing its eigenvalue. It is one excitation of the positive circular mode.

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.

Use the canonical commutation relation $[\hat{x},\hat{p}]=i\hslash I$ on a common invariant domain of the operator products. Write $A=\hat{p}\hat{x}$, so $\hat{x}\hat{p}=A+i\hslash I$. Associativity then giveswhereasThereforeAs a direct position-representation check, the momentum operator is $\hat{p}=-i\hslash d/dx$. Acting on a smooth function $f$ givesThis is the equality of two quadratic position-momentum orderings; it does not make arbitrary rearrangements of noncommuting linear operators interchangeable.

For arbitrary linear operators, expansion of the anticommutators givesNow use the commutator product rule $[VX,W]=V[X,W]+[V,W]X$, with $W=YZ$, twice. The result isNo grading assumption is needed for these algebraic identities.

For the Dirac field, supplement the supplied equal-time canonical anticommutation relations with $\{{\psi }_a(x),{\psi }_b(y)\}=\{{\psi }_a^{†}(x),{\psi }_b^{†}(y)\}=0$. Define the fermion number density $n(\mathbf{x})={\sum }_a{\psi }_a^{†}(\mathbf{x}){\psi }_a(\mathbf{x})$. In the preceding identity take $V={\psi }_a^{†}(\mathbf{x})$, $X={\psi }_a(\mathbf{x})$, $Y={\psi }_b^{†}(\mathbf{y})$, $Z={\psi }_c(\mathbf{z})$ and sum over $a$. Only its first and last anticommutators survive, givingThis is an identity of operator-valued distributions. Point splitting, or smearing with test functions, gives a precise meaning to coincident bilinears. Normal ordering only subtracts a constant from the density and does not alter this commutator.

The commutator of two linear operators is $\boxed{[A,B]=AB-BA}$. Write $T^{†}$ for the adjoint operator, using the inner product convention linear in its second argument. Since $A^{†}=A$ and $B^{†}=B$,Thus the commutator is anti-self-adjoint. With $C=-i[A,B]$, conjugation of the scalar givesAll products are understood on the stated common operator domain; for physical unbounded linear operators, the state must lie in the relevant product domains and have the finite moments used below.

Centre the position operator and momentum operator by their real expectation values:Scalar multiples of the identity operator commute with every linear operator, so the canonical commutation relation givesThe second moments of the centred linear operators are the variances, ${\mathbb{E}}_{\psi }(A^2)={\Delta }_{\psi }(X{)}^2$ and ${\mathbb{E}}_{\psi }(B^2)={\Delta }_{\psi }(P{)}^2$. Applying the preceding inequality and taking nonnegative square roots proves the Heisenberg uncertainty relation:Subtracting the means is essential: the uncentred second moments alone do not equal the variances. The equality case of the Heisenberg uncertainty relation requires the centred vectors to be imaginary multiples, as in the preceding equality conditions.

For a differentiable family of invertible linear operators, differentiate the identity without changing the factor order:Multiplication on the right by $A^{-1}$ gives the derivative of matrix inversion,Commuting $A^{-1}$ past the derivative would generally be invalid.

In a finite-dimensional basis, interchange the two finite sums:This cyclic property of the trace does not assert $AB=BA$ or permit arbitrary reordering of three factors. For infinite-dimensional linear operators, the traces and rearrangements need hypotheses, for example one trace-class operator and one bounded operator; an arbitrary pair of unbounded linear operators is not covered.