Properties of bounded self-adjoint operators

Let H be a Hilbert space and let ${\displaystyle A:H\to H}$ be a bounded self-adjoint linear operator defined on ${\displaystyle \operatorname {D} \left(A\right)=H}$.

• ${\displaystyle \left\langle h,Ah\right\rangle }$ is real for all ${\displaystyle h\in H}$.^[4]
• ${\displaystyle \left\|A\right\|=\sup \left\{|\langle h,Ah\rangle |:\|h\|=1\right\}}$^[4] if ${\displaystyle \operatorname {dim} H\neq 0.}$
• If the image of A, denoted by ${\displaystyle \operatorname {Im} A}$, is dense in H then ${\displaystyle A:H\to \operatorname {Im} A}$ is invertible.
• The eigenvalues of A are real and eigenvectors belonging to different eigenvalues are orthogonal.^[4]
• If ${\displaystyle \lambda }$ is an eigenvalue of A then ${\displaystyle |\lambda |\leq \|A\|}$; in particular, ${\displaystyle |\lambda |\leq \sup \left\{|\langle h,Ah\rangle |:\|h\|\leq 1\right\}}$.^[4]
  • In general, there may not exist any eigenvalue ${\displaystyle \lambda }$ such that ${\displaystyle |\lambda |=\sup \left\{|\langle h,Ah\rangle |:\|h\|\leq 1\right\}}$, but if in addition A is compact then there necessarily exists an eigenvalue ${\displaystyle \lambda }$, equal to either ${\displaystyle \|A\|}$ or ${\displaystyle -\|A\|}$,^[5] such that ${\displaystyle |\lambda |=\sup \left\{|\langle h,Ah\rangle |:\|h\|\leq 1\right\}}$,^[4]
• If a sequence of bounded self-adjoint linear operators is convergent then the limit is self-adjoint.^[3]
• There exists a number ${\displaystyle \lambda }$, equal to either ${\displaystyle \|A\|}$ or ${\displaystyle -\|A\|}$, and a sequence ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }\subseteq H}$ such that ${\displaystyle \lim _{i\to \infty }Ax_{i}-\lambda x_{i}=0}$ and ${\displaystyle \|x_{i}\|=1}$ for all i.^[5]

Statement of the spectral theorem

Partially defined operators A, B on Hilbert spaces H, K are unitarily equivalent if and only if there is a unitary transformation U : H → K such that

• ${\displaystyle U}$ maps ${\displaystyle \operatorname {Dom} A}$ bijectively to ${\displaystyle \operatorname {Dom} B,}$
• ${\displaystyle BU\xi =UA\xi ,\qquad \forall \xi \in \operatorname {Dom} A.}$

Let ${\displaystyle (X,\Sigma ,\mu )}$ be a σ-finite measure space and ${\displaystyle h}$ a real-valued measurable function on ${\displaystyle X}$. An operator ${\displaystyle T_{h}}$ of the form

${\displaystyle [T_{h}\psi ](x)=h(x)\psi (x)}$

whose domain is the space of ${\displaystyle \psi }$ for which the right-hand side is in ${\displaystyle L^{2}}$ is called a multiplication operator.

More complete versions of the spectral theorem exist as well that involve direct integrals and carry with it the notion of "generalized eigenvectors".^[8]

The spectral theorem for unbounded self-adjoint operators can be proved by reduction to the spectral theorem for unitary (hence bounded) operators.^[9] This reduction uses the Cayley transform for self-adjoint operators which is defined in the next section. We might note that if ${\displaystyle T}$ is multiplication by ${\displaystyle h}$, then the spectrum of ${\displaystyle T}$ is just the essential range of ${\displaystyle h}$.

Functional calculus

One application of the spectral theorem is to define a functional calculus. That is, if ${\displaystyle f}$ is a function on the real line and ${\displaystyle T}$ is a self-adjoint operator, we wish to define the operator ${\displaystyle f(T)}$. The spectral theorem shows that if ${\displaystyle T}$ is represented as the operator of multiplication by ${\displaystyle h}$, then ${\displaystyle f(T)}$ is the operator of multiplication by the composition ${\displaystyle f\circ h}$.

One example from quantum mechanics is the case where ${\displaystyle T}$ is the Hamiltonian operator ${\displaystyle {\hat {H}}}$. If ${\displaystyle {\hat {H}}}$ has a true orthonormal basis of eigenvectors ${\displaystyle e_{j}}$ with eigenvalues ${\displaystyle \lambda _{j}}$, then ${\displaystyle f({\hat {H}}):=e^{-it{\hat {H}}/\hbar }}$ can be defined as the unique bounded operator with eigenvalues ${\displaystyle f(\lambda _{j}):=e^{-it\lambda _{j}/\hbar }}$ such that:

${\displaystyle f({\hat {H}})e_{j}=f(\lambda _{j})e_{j}.}$

The goal of functional calculus is to extend this idea to the case where ${\displaystyle T}$ has continuous spectrum (i.e. where ${\displaystyle T}$ has no normalizable eigenvectors).

It has been customary to introduce the following notation

${\displaystyle \operatorname {E} (\lambda )=\mathbf {1} _{(-\infty ,\lambda ]}(T)}$

where ${\displaystyle \mathbf {1} _{(-\infty ,\lambda ]}}$ is the indicator function of the interval ${\displaystyle (-\infty ,\lambda ]}$. The family of projection operators E(λ) is called resolution of the identity for T. Moreover, the following Stieltjes integral representation for T can be proved:

${\displaystyle T=\int _{-\infty }^{+\infty }\lambda d\operatorname {E} (\lambda ).}$

Formulation in the physics literature

In quantum mechanics, Dirac notation is used as combined expression for both the spectral theorem and the Borel functional calculus. That is, if H is self-adjoint and f is a Borel function,

${\displaystyle f(H)=\int dE\left|\Psi _{E}\rangle f(E)\langle \Psi _{E}\right|}$

with

${\displaystyle H\left|\Psi _{E}\right\rangle =E\left|\Psi _{E}\right\rangle }$

where the integral runs over the whole spectrum of H. The notation suggests that H is diagonalized by the eigenvectors Ψ_E. Such a notation is purely formal. The resolution of the identity (sometimes called projection-valued measures) formally resembles the rank-1 projections ${\displaystyle \left|\Psi _{E}\right\rangle \left\langle \Psi _{E}\right|}$. In the Dirac notation, (projective) measurements are described via eigenvalues and eigenstates, both purely formal objects. As one would expect, this does not survive passage to the resolution of the identity. In the latter formulation, measurements are described using the spectral measure of ${\displaystyle |\Psi \rangle }$, if the system is prepared in ${\displaystyle |\Psi \rangle }$ prior to the measurement. Alternatively, if one would like to preserve the notion of eigenstates and make it rigorous, rather than merely formal, one can replace the state space by a suitable rigged Hilbert space.

If f = 1, the theorem is referred to as resolution of unity:

${\displaystyle I=\int dE\left|\Psi _{E}\right\rangle \left\langle \Psi _{E}\right|}$

In the case ${\displaystyle H_{\text{eff}}=H-i\Gamma }$ is the sum of an Hermitian H and a skew-Hermitian (see skew-Hermitian matrix) operator ${\displaystyle -i\Gamma }$, one defines the biorthogonal basis set

${\displaystyle H_{\text{eff}}^{*}\left|\Psi _{E}^{*}\right\rangle =E^{*}\left|\Psi _{E}^{*}\right\rangle }$

and write the spectral theorem as:

${\displaystyle f\left(H_{\text{eff}}\right)=\int dE\left|\Psi _{E}\right\rangle f(E)\left\langle \Psi _{E}^{*}\right|}$

(See Feshbach–Fano partitioning method for the context where such operators appear in scattering theory).

Formulation for symmetric operators

The spectral theorem applies only to self-adjoint operators, and not in general to symmetric operators. Nevertheless, we can at this point give a simple example of a symmetric (specifically, an essentially self-adjoint) operator that has an orthonormal basis of eigenvectors. Consider the complex Hilbert space L²[0,1] and the differential operator

${\displaystyle A=-{\frac {d^{2}}{dx^{2}}}}$

with ${\displaystyle \mathrm {Dom} (A)}$ consisting of all complex-valued infinitely differentiable functions f on [0, 1] satisfying the boundary conditions

${\displaystyle f(0)=f(1)=0.}$

Then integration by parts of the inner product shows that A is symmetric.^{[nb 1]} The eigenfunctions of A are the sinusoids

${\displaystyle f_{n}(x)=\sin(n\pi x)\qquad n=1,2,\ldots }$

with the real eigenvalues n²π²; the well-known orthogonality of the sine functions follows as a consequence of A being symmetric.

The operator A can be seen to have a compact inverse, meaning that the corresponding differential equation Af = g is solved by some integral (and therefore compact) operator G. The compact symmetric operator G then has a countable family of eigenvectors which are complete in L². The same can then be said for A.

Pure point spectrum

A self-adjoint operator A on H has pure point spectrum if and only if H has an orthonormal basis {e_i}_{i ∈ I} consisting of eigenvectors for A.

Example. The Hamiltonian for the harmonic oscillator has a quadratic potential V, that is

${\displaystyle -\Delta +|x|^{2}.}$

This Hamiltonian has pure point spectrum; this is typical for bound state Hamiltonians in quantum mechanics. As was pointed out in a previous example, a sufficient condition that an unbounded symmetric operator has eigenvectors which form a Hilbert space basis is that it has a compact inverse.

Boundary conditions

In the case where the Hilbert space is a space of functions on a bounded domain, these distinctions have to do with a familiar issue in quantum physics: One cannot define an operator—such as the momentum or Hamiltonian operator—on a bounded domain without specifying boundary conditions. In mathematical terms, choosing the boundary conditions amounts to choosing an appropriate domain for the operator. Consider, for example, the Hilbert space ${\displaystyle L^{2}([0,1])}$ (the space of square-integrable functions on the interval [0,1]). Let us define a "momentum" operator A on this space by the usual formula, setting Planck's constant equal to 1:

${\displaystyle Af=-i{\frac {df}{dx}}.}$

We must now specify a domain for A, which amounts to choosing boundary conditions. If we choose

${\displaystyle \operatorname {Dom} (A)=\left\{{\text{smooth functions}}\right\},}$

then A is not symmetric (because the boundary terms in the integration by parts do not vanish).

If we choose

${\displaystyle \operatorname {Dom} (A)=\left\{{\text{smooth functions}}\,f\mid f(0)=f(1)=0\right\},}$

then using integration by parts, one can easily verify that A is symmetric. This operator is not essentially self-adjoint,^[10] however, basically because we have specified too many boundary conditions on the domain of A, which makes the domain of the adjoint too big (see also the example below).

Specifically, with the above choice of domain for A, the domain of the closure ${\displaystyle A^{\mathrm {cl} }}$ of A is

${\displaystyle \operatorname {Dom} \left(A^{\mathrm {cl} }\right)=\left\{{\text{functions }}f{\text{ with two derivatives in }}L^{2}\mid f(0)=f(1)=0\right\},}$

whereas the domain of the adjoint ${\displaystyle A^{*}}$ of A is

${\displaystyle \operatorname {Dom} \left(A^{*}\right)=\left\{{\text{functions }}f{\text{ with two derivatives in }}L^{2}\right\}.}$

That is to say, the domain of the closure has the same boundary conditions as the domain of A itself, just a less stringent smoothness assumption. Meanwhile, since there are "too many" boundary conditions on A, there are "too few" (actually, none at all in this case) for ${\displaystyle A^{*}}$. If we compute ${\displaystyle \langle g,Af\rangle }$ for ${\displaystyle f\in \operatorname {Dom} (A)}$ using integration by parts, then since ${\displaystyle f}$ vanishes at both ends of the interval, no boundary conditions on ${\displaystyle g}$ are needed to cancel out the boundary terms in the integration by parts. Thus, any sufficiently smooth function ${\displaystyle g}$ is in the domain of ${\displaystyle A^{*}}$, with ${\displaystyle A^{*}g=-i\,dg/dx}$.^[11]

Since the domain of the closure and the domain of the adjoint do not agree, A is not essentially self-adjoint. After all, a general result says that the domain of the adjoint of ${\displaystyle A^{\mathrm {cl} }}$ is the same as the domain of the adjoint of A. Thus, in this case, the domain of the adjoint of ${\displaystyle A^{\mathrm {cl} }}$ is bigger than the domain of ${\displaystyle A^{\mathrm {cl} }}$ itself, showing that ${\displaystyle A^{\mathrm {cl} }}$ is not self-adjoint, which by definition means that A is not essentially self-adjoint.

The problem with the preceding example is that we imposed too many boundary conditions on the domain of A. A better choice of domain would be to use periodic boundary conditions:

${\displaystyle \operatorname {Dom} (A)=\{{\text{smooth functions}}\,f\mid f(0)=f(1)\}.}$

With this domain, A is essentially self-adjoint.^[12]

In this case, we can understand the implications of the domain issues for the spectral theorem. If we use the first choice of domain (with no boundary conditions), all functions ${\displaystyle f_{\beta }(x)=e^{\beta x}}$ for ${\displaystyle \beta \in \mathbb {C} }$ are eigenvectors, with eigenvalues ${\displaystyle -i\beta }$, and so the spectrum is the whole complex plane. If we use the second choice of domain (with Dirichlet boundary conditions), A has no eigenvectors at all. If we use the third choice of domain (with periodic boundary conditions), we can find an orthonormal basis of eigenvectors for A, the functions ${\displaystyle f_{n}(x):=e^{2\pi inx}}$. Thus, in this case finding a domain such that A is self-adjoint is a compromise: the domain has to be small enough so that A is symmetric, but large enough so that ${\displaystyle D(A^{*})=D(A)}$.

Schrödinger operators with singular potentials

A more subtle example of the distinction between symmetric and (essentially) self-adjoint operators comes from Schrödinger operators in quantum mechanics. If the potential energy is singular—particularly if the potential is unbounded below—the associated Schrödinger operator may fail to be essentially self-adjoint. In one dimension, for example, the operator

${\displaystyle {\hat {H}}:={\frac {P^{2}}{2m}}-X^{4}}$

is not essentially self-adjoint on the space of smooth, rapidly decaying functions.^[13] In this case, the failure of essential self-adjointness reflects a pathology in the underlying classical system: A classical particle with a ${\displaystyle -x^{4}}$ potential escapes to infinity in finite time. This operator does not have a unique self-adjoint, but it does admit self-adjoint extensions obtained by specifying "boundary conditions at infinity". (Since ${\displaystyle {\hat {H}}}$ is a real operator, it commutes with complex conjugation. Thus, the deficiency indices are automatically equal, which is the condition for having a self-adjoint extension. See the discussion of extensions of symmetric operators below.)

In this case, if we initially define ${\displaystyle {\hat {H}}}$ on the space of smooth, rapidly decaying functions, the adjoint will be "the same" operator (i.e., given by the same formula) but on the largest possible domain, namely

${\displaystyle \operatorname {Dom} \left({\hat {H}}^{*}\right)=\left\{{\text{twice differentiable functions }}f\in L^{2}(\mathbb {R} )\left|\left(-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}f}{dx^{2}}}-x^{4}f(x)\right)\in L^{2}(\mathbb {R} )\right.\right\}.}$

It is then possible to show that ${\displaystyle {\hat {H}}^{*}}$ is not a symmetric operator, which certainly implies that ${\displaystyle {\hat {H}}}$ is not essentially self-adjoint. Indeed, ${\displaystyle {\hat {H}}^{*}}$ has eigenvectors with pure imaginary eigenvalues,^[14][15] which is impossible for a symmetric operator. This strange occurrence is possible because of a cancellation between the two terms in ${\displaystyle {\hat {H}}^{*}}$: There are functions ${\displaystyle f}$ in the domain of ${\displaystyle {\hat {H}}^{*}}$ for which neither ${\displaystyle d^{2}f/dx^{2}}$ nor ${\displaystyle x^{4}f(x)}$ is separately in ${\displaystyle L^{2}(\mathbb {R} )}$, but the combination of them occurring in ${\displaystyle {\hat {H}}^{*}}$ is in ${\displaystyle L^{2}(\mathbb {R} )}$. This allows for ${\displaystyle {\hat {H}}^{*}}$ to be nonsymmetric, even though both ${\displaystyle d^{2}/dx^{2}}$ and ${\displaystyle X^{4}}$ are symmetric operators. This sort of cancellation does not occur if we replace the repelling potential ${\displaystyle -x^{4}}$ with the confining potential ${\displaystyle x^{4}}$.

Self-adjoint extensions in quantum mechanics

In quantum mechanics, observables correspond to self-adjoint operators. By Stone's theorem on one-parameter unitary groups, self-adjoint operators are precisely the infinitesimal generators of unitary groups of time evolution operators. However, many physical problems are formulated as a time-evolution equation involving differential operators for which the Hamiltonian is only symmetric. In such cases, either the Hamiltonian is essentially self-adjoint, in which case the physical problem has unique solutions or one attempts to find self-adjoint extensions of the Hamiltonian corresponding to different types of boundary conditions or conditions at infinity.

Example. The one-dimensional Schrödinger operator with the potential ${\displaystyle V(x)=-(1+|x|)^{\alpha }}$, defined initially on smooth compactly supported functions, is essentially self-adjoint (that is, has a self-adjoint closure) for 0 < α ≤ 2 but not for α > 2. See Berezin and Schubin, pages 55 and 86, or Section 9.10 in Hall.

The failure of essential self-adjointness for ${\displaystyle \alpha >2}$ has a counterpart in the classical dynamics of a particle with potential ${\displaystyle V(x)}$: The classical particle escapes to infinity in finite time.^[18]

Example. There is no self-adjoint momentum operator p for a particle moving on a half-line. Nevertheless, the Hamiltonian ${\displaystyle p^{2}}$ of a "free" particle on a half-line has several self-adjoint extensions corresponding to different types of boundary conditions. Physically, these boundary conditions are related to reflections of the particle at the origin (see Reed and Simon, vol.2).

A symmetric operator that is not essentially self-adjoint

We first consider the Hilbert space ${\displaystyle L^{2}[0,1]}$ and the differential operator

${\displaystyle D:\phi \mapsto {\frac {1}{i}}\phi '}$

defined on the space of continuously differentiable complex-valued functions on [0,1], satisfying the boundary conditions

${\displaystyle \phi (0)=\phi (1)=0.}$

Then D is a symmetric operator as can be shown by integration by parts. The spaces N₊, N₋ (defined below) are given respectively by the distributional solutions to the equation

${\displaystyle {\begin{aligned}-iu'&=iu\\-iu'&=-iu\end{aligned}}}$

which are in L²[0, 1]. One can show that each one of these solution spaces is 1-dimensional, generated by the functions x → e^−x and x → e^x respectively. This shows that D is not essentially self-adjoint,^[19] but does have self-adjoint extensions. These self-adjoint extensions are parametrized by the space of unitary mappings N₊ → N₋, which in this case happens to be the unit circle T.

In this case, the failure of essential self-adjointenss is due to an "incorrect" choice of boundary conditions in the definition of the domain of ${\displaystyle D}$. Since ${\displaystyle D}$ is a first-order operator, only one boundary condition is needed to ensure that ${\displaystyle D}$ is symmetric. If we replaced the boundary conditions given above by the single boundary condition

${\displaystyle \phi (0)=\phi (1)}$,

then D would still be symmetric and would now, in fact, be essentially self-adjoint. This change of boundary conditions gives one particular essentially self-adjoint extension of D. Other essentially self-adjoint extensions come from imposing boundary conditions of the form ${\displaystyle \phi (1)=e^{i\theta }\phi (0)}$.

This simple example illustrates a general fact about self-adjoint extensions of symmetric differential operators P on an open set M. They are determined by the unitary maps between the eigenvalue spaces

${\displaystyle N_{\pm }=\left\{u\in L^{2}(M):P_{\operatorname {dist} }u=\pm iu\right\}}$

where P_dist is the distributional extension of P.

Constant-coefficient operators

We next give the example of differential operators with constant coefficients. Let

${\displaystyle P\left({\vec {x}}\right)=\sum _{\alpha }c_{\alpha }x^{\alpha }}$

be a polynomial on Rⁿ with real coefficients, where α ranges over a (finite) set of multi-indices. Thus

${\displaystyle \alpha =(\alpha _{1},\alpha _{2},\ldots ,\alpha _{n})}$

and

${\displaystyle x^{\alpha }=x_{1}^{\alpha _{1}}x_{2}^{\alpha _{2}}\cdots x_{n}^{\alpha _{n}}.}$

We also use the notation

${\displaystyle D^{\alpha }={\frac {1}{i^{|\alpha |}}}\partial _{x_{1}}^{\alpha _{1}}\partial _{x_{2}}^{\alpha _{2}}\cdots \partial _{x_{n}}^{\alpha _{n}}.}$

Then the operator P(D) defined on the space of infinitely differentiable functions of compact support on Rⁿ by

${\displaystyle P(\operatorname {D} )\phi =\sum _{\alpha }c_{\alpha }\operatorname {D} ^{\alpha }\phi }$

is essentially self-adjoint on L²(Rⁿ).

More generally, consider linear differential operators acting on infinitely differentiable complex-valued functions of compact support. If M is an open subset of Rⁿ

${\displaystyle P\phi (x)=\sum _{\alpha }a_{\alpha }(x)\left[D^{\alpha }\phi \right](x)}$

where a_α are (not necessarily constant) infinitely differentiable functions. P is a linear operator

${\displaystyle C_{0}^{\infty }(M)\to C_{0}^{\infty }(M).}$

Corresponding to P there is another differential operator, the formal adjoint of P

${\displaystyle P^{\mathrm {*form} }\phi =\sum _{\alpha }D^{\alpha }\left({\overline {a_{\alpha }}}\phi \right)}$

Theorem — The adjoint P* of P is a restriction of the distributional extension of the formal adjoint to an appropriate subspace of ${\displaystyle L^{2}}$. Specifically:

$${\displaystyle \operatorname {dom} P^{*}=\left\{u\in L^{2}(M):P^{\mathrm {*form} }u\in L^{2}(M)\right\}.}$$

Uniform multiplicity

We first define uniform multiplicity:

Definition. A self-adjoint operator A has uniform multiplicity n where n is such that 1 ≤ n ≤ ω if and only if A is unitarily equivalent to the operator M_f of multiplication by the function f(λ) = λ on

${\displaystyle L_{\mu }^{2}\left(\mathbf {R} ,\mathbf {H} _{n}\right)=\left\{\psi$ :\mathbf {R} \to \mathbf {H} _{n}:\psi {\mbox{ measurable and }}\int _{\mathbf {R} }\|\psi (t)\|^{2}d\mu (t)<\infty \right\}}

where H_n is a Hilbert space of dimension n. The domain of M_f consists of vector-valued functions ψ on R such that

${\displaystyle \int _{\mathbf {R} }|\lambda |^{2}\ \|\psi (\lambda )\|^{2}\,d\mu (\lambda )<\infty .}$

Non-negative countably additive measures μ, ν are mutually singular if and only if they are supported on disjoint Borel sets.

Theorem — Let A be a self-adjoint operator on a separable Hilbert space H. Then there is an ω sequence of countably additive finite measures on R (some of which may be identically 0)

$${\displaystyle \left\{\mu _{\ell }\right\}_{1\leq \ell \leq \omega }}$$

such that the measures are pairwise singular and A is unitarily equivalent to the operator of multiplication by the function f(λ) = λ on

$${\displaystyle \bigoplus _{1\leq \ell \leq \omega }L_{\mu _{\ell }}^{2}\left(\mathbf {R} ,\mathbf {H} _{\ell }\right).}$$

This representation is unique in the following sense: For any two such representations of the same A, the corresponding measures are equivalent in the sense that they have the same sets of measure 0.

Direct integrals

The spectral multiplicity theorem can be reformulated using the language of direct integrals of Hilbert spaces:

Theorem — ^[20] Any self-adjoint operator on a separable Hilbert space is unitarily equivalent to multiplication by the function λ ↦ λ on

$${\displaystyle \int _{\mathbf {R} }^{\oplus }H_{\lambda }\,d\mu (\lambda ).}$$

Unlike the multiplication-operator version of the spectral theorem, the direct-integral version is unique in the sense that the measure equivalence class of μ (or equivalently its sets of measure 0) is uniquely determined and the measurable function ${\displaystyle \lambda \mapsto \mathrm {dim} (H_{\lambda })}$ is determined almost everywhere with respect to μ.^[21] The function ${\displaystyle \lambda \mapsto \operatorname {dim} \left(H_{\lambda }\right)}$ is the spectral multiplicity function of the operator.

We may now state the classification result for self-adjoint operators: Two self-adjoint operators are unitarily equivalent if and only if (1) their spectra agree as sets, (2) the measures appearing in their direct-integral representations have the same sets of measure zero, and (3) their spectral multiplicity functions agree almost everywhere with respect to the measure in the direct integral.^[22]

Example: structure of the Laplacian

The Laplacian on Rⁿ is the operator

${\displaystyle \Delta =\sum _{i=1}^{n}\partial _{x_{i}}^{2}.}$

As remarked above, the Laplacian is diagonalized by the Fourier transform. Actually it is more natural to consider the negative of the Laplacian −Δ since as an operator it is non-negative; (see elliptic operator).