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Consider the operators on all of $C_0(X)$ or $L^2(\mu )$, extending the displayed positive-function formula linearly. Here $C_0(X)$ is the space of continuous functions vanishing at infinity. The multiplier has modulus at most one, so $\Vert P_{t}f\Vert \le \Vert f\Vert$, and multiplication of exponentials gives $P_{s+t}=P_s{P}_t$ and $P_0=I$. For strong continuity on $C_0(X)$, choose a compact $K$ outside which $|f|$ is small. On $K$ the continuous $k$ is bounded, making $e^{-tk}\to 1$ uniform; outside $K$, $|(e^{-tk}-1)f|\le |f|$. For $L^2(\mu )$, use pointwise convergence and the dominated convergence theorem with bound $|(e^{-tk}-1)f{|}^2\le |f{|}^2$. Thus both are contraction semigroups, even when $k$ is unbounded.

The generator of a multiplication semigroup isIndeed pointwise difference quotients converge to $-kf$. A norm limit therefore forces the displayed domain condition, using an almost-everywhere convergent subsequence for the $L^2$ case. Conversely $|(e^{-tk}-1)/t|\le k$, so the dominated convergence theorem proves the $L^2$ limit when $kf\in L^2$. In $C_0(X)$ write the error as $kf$ times $1-(1-e^{-tk})/(tk)$, with the value at $k=0$ supplied by continuity. That second factor tends uniformly to zero where $k$ is bounded and has modulus at most one; the compact-set and small-tail argument applied to $kf$ proves uniform convergence. To prove directly that $L$ is a closed operator, take $f_n\to f$ and $L{f}_n\to g$. Pointwise limits in $C_0$, or a common almost-everywhere convergent subsequence in $L^2$, give $g=-kf$. Thus $f$ is in the appropriate generator domain and $Lf=g$. In $L^2$ the domain is dense because $f{1}_{\{k\le n\}}\to f$; in $C_0$ compactly supported continuous functions are a dense subspace contained in the domain.

A real multiplication operator on its maximal $L^2$ domain is an unbounded self-adjoint operator. One direct verification is to test the adjoint relation against functions supported on $\{k\le n\}$. If $h$ is in the adjoint operator domain with representing vector $g$, those tests force $g=-kh$ on each such set, hence $kh\in L^2$. The converse follows by integration. On the sigma-finite measure spaces used below, the spectrum of a real multiplication operator is its essential range; spectral values are detected by unit vectors supported where the multiplier is arbitrarily close to that value, while outside the essential range its reciprocal is a bounded resolvent multiplier.

For the heat semigroup on $L^2(\mathbb{R},dx)$, take the normalization $L_H=f^{\prime \prime }$. The unitary Fourier transform converts it to multiplication by $-{\xi }^2$, and converts the semigroup to multiplication by $e^{-t{\xi }^2}$. ConsequentlyThe Sobolev space domain is exactly $\{f:{\xi }^2\hat{f}\in L^2\}$. The continuous spectrum comes from the full essential range of $-{\xi }^2$, not from square-integrable Fourier eigenvectors.

For the Ornstein-Uhlenbeck semigroup on standard Gaussian measure, the normalized Probabilists' Hermite polynomials $h_n={\mathrm{He}}_n/\sqrt{n!}$ form an orthonormal basis. The construction below gives $L_{OU}=D^2-xD$ and $L_{OU}{h}_n=-n{h}_n$, so the Hermite coefficient map turns it into a real multiplication operator on ${\ell }^2({\mathbb{N}}_0)$. ThereforeFinally, the heat semigroup on the whole real line has no positive spectral gap, whereas the standard Ornstein-Uhlenbeck semigroup has gap one. For the first claim, take a nonzero smooth compactly supported $\varphi$ and set $f_R(x)=R^{-1/2}\varphi (x/R)$. Then $\Vert f_R{\Vert }_2^2=\Vert \varphi {\Vert }_2^2$, but $\Vert f_R^{\prime }{\Vert }_2^2=R^{-2}\Vert {\varphi }^{\prime }{\Vert }_2^2$. This disproves a uniform whole-line Poincaré inequality $\Vert f{\Vert }_2^2\le C\Vert f^{\prime }{\Vert }_2^2$; one can also choose $\int \varphi =0$. Lebesgue measure here is infinite, so it has no normalized probability mean to subtract. For Gaussian measure, the Hermite expansion instead gives the sharp Gaussian Poincaré inequalityEquality holds for affine functions. The identity on the right is interpreted on the energy-form domain, which is larger than the full operator domain.

For a densely defined operator $A$ on a complex Hilbert space, unbounded self-adjointness means equality $A=A^{\ast }$ including equality of the operator domains. Symmetry alone asserts only $A\subseteq A^{\ast }$ and is insufficient. Use an inner product linear in its first argument. If $\mathrm{Im}z\ne 0$, symmetry givesThus $A-z$ is bounded below. Since $A$ is closed, its range is closed; its orthogonal complement is $\ker (A^{\ast }-\overline{z})=\ker (A-\overline{z})=0$, so the range is also dense and therefore all of $H$. The inverse is bounded by $|\mathrm{Im}z{|}^{-1}$. The nonreal resolvent estimate for a self-adjoint operator proves $\sigma (A)\subseteq \mathbb{R}$.

For bounded self-adjoint $T$, its numerical range of an operator and numerical radius areEvery number in $W(T)$ is real. If $z\notin \overline{W(T)}$, its distance $\delta$ from that closed set is positive, and $\Vert (T-z)u\Vert \ge \delta \Vert u\Vert$. The same bound applies to $T-\overline{z}$, so the range is dense as well as closed and $T-z$ has a bounded inverse. Hence the numerical-range spectral enclosure isThe closure is necessary in infinite dimension: the endpoints of the numerical range need not themselves be attained.

Put $\beta =\sup W(T)$ and $B=\beta I-T$, a bounded positive semidefinite operator. Choose unit vectors $u_n$ with $⟨B{u}_n,u_n⟩\to 0$. The Cauchy-Schwarz inequality for the positive form $⟨B\cdot ,\cdot ⟩$ impliesTherefore $\beta$ is an approximate eigenvalue. Apply the same argument to $T-\alpha I$, where $\alpha =\inf W(T)$, to get the other endpoint. This proves the numerical-range endpoint approximate-eigenvalue theorem. Each endpoint is in the spectrum: a bounded inverse would forbid unit vectors with vanishing residual.

If an unbounded self-adjoint operator $A$ is positive semidefinite, then for every real $z<0$, $\Vert (A-z)u\Vert \ge (-z)\Vert u\Vert$. The range argument above again proves invertibility, excluding negative values from the spectrum. For the converse, suppose $\sigma (A)\subseteq [0,\infty )$. Its resolvent operator $R=(I+A{)}^{-1}$ is bounded and self-adjoint. The resolvent spectral mapping identity puts $\sigma (R)\subseteq [0,1]$: for $w\ne 0$, the relevant spectral parameter of $A$ is $1/w-1$, and nonreal values are excluded by self-adjointness. By the endpoint result just proved, $W(R)\subseteq [0,1]$. Positive-form Cauchy-Schwarz inequality then givesFor $u=Rf\in D(A)$ this yields $⟨Au,u⟩=⟨f-Rf,Rf⟩\ge 0$. Thus the spectral positivity criterion for a self-adjoint operator isThis resolvent proof uses the bounded numerical-range result to handle the full unbounded domain.

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

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

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

The Robertson uncertainty principle bounds the product of the standard deviations of two observables in the same quantum state. For a normalized pure state $|\psi ⟩$ and self-adjoint operators $A,B$, write their expectation values and variances asand similarly for $B$, where $I$ is the identity operator. The quantum uncertainties $\Delta A,\Delta B$ then obeyHere $[A,B]$ is the commutator, and every expectation value is taken in the same $|\psi ⟩$. These standard deviations describe the outcome spreads when $A$ and $B$ are measured on separate ensembles prepared in that same quantum state. They are not instrument errors or the change caused by a preceding measurement in quantum mechanics. The definition and bound are developed in Barton Zwiebach, Uncertainty Principle and Compatible Observables, 2013, §§1–2.

For unbounded self-adjoint operators, the operator domains matter. Assume $\psi \in D(A)\cap D(B)$, so that both second moments are finite; $⟨A^2⟩$ here means the second moment of the measured probability distribution, equal to $\Vert A\psi {\Vert }^2$, even if $A^2\psi$ is undefined. To use the displayed operator commutator, also assume $\psi \in D(AB)\cap D(BA)$. Without these product-domain assumptions, the always meaningful version on $D(A)\cap D(B)$ uses $|\mathrm{Im}⟨A\psi |B\psi ⟩|$ as its right-hand side. These qualifications prevent a formal commutator identity from being applied to an inadmissible quantum state.

The brief proof is Cauchy-Schwarz inequality applied to $u=(A-⟨A⟩I)\psi$ and $v=(B-⟨B⟩I)\psi$:The final equality uses self-adjointness and the product-domain assumptions. Thus a geometric statement about inner products becomes a constraint on experimental spreads. For nonzero $\Delta A,\Delta B$, equality requires $u,v$ to be proportional with a purely imaginary proportionality factor; the equality case of the Heisenberg uncertainty relation turns this condition into a Gaussian wave packet.

The bound is useful because computing a single commutator and its expectation value can constrain two entire outcome distributions. Its right-hand side depends on the quantum state: the spin uncertainty state dependence example gives both a positive saturated bound and a zero bound for the same noncommuting observables. Robertson's paper, titled "The Uncertainty Principle", appeared on 1 July 1929 in Physical Review, volume 34, starting at page 163. ^[ref].

A densely defined operator $T$ on a Hilbert space is self-adjoint when it equals its adjoint operator, including equality of their operator domains:For a bounded linear operator defined on the whole Hilbert space, this is equivalent tofor all vectors $v,w$. For an unbounded self-adjoint operator, the domain condition is essential: the inner-product identity on $D(T)$ alone gives only a symmetric operator.

For a real measurable function $m$ on a sigma-finite measure space, multiplication on $D(M_m)=\{f\in L^2:mf\in L^2\}$ is an unbounded self-adjoint operator. Testing its adjoint relation on $\{|m|\le n\}$ identifies the same maximal domain. Outside the essential range, multiplication by $(m-z{)}^{-1}$ is a bounded resolvent. Inside it, normalized indicators of finite-positive-measure sets where $m$ is close to $z$ are approximate eigenvectors, so the inverse cannot be bounded.