Inmathematics, aself-adjoint operator on acomplex vector space${\displaystyle V}$ withinner product${\displaystyle ⟨\cdot ,\cdot ⟩}$ is alinear map${\displaystyle A}$ (from${\displaystyle V}$ to itself) that is its ownadjoint. That is,${\displaystyle ⟨Ax,y⟩=⟨x,Ay⟩}$ for all${\displaystyle x,y\in V}$. If${\displaystyle V}$ isfinite-dimensional with a givenorthonormal basis, this is equivalent to the condition that thematrix of${\displaystyle A}$ is aHermitian matrix, i.e., equal to itsconjugate transpose${\displaystyle A^{\ast }}$. By the finite-dimensionalspectral theorem,${\displaystyle V}$ has anorthonormal basis such that the matrix of${\displaystyle A}$ relative to this basis is adiagonal matrix with entries in thereal numbers. This article deals with applying generalizations of this concept to operators onHilbert spaces of arbitrary dimension.

Self-adjoint operators are used infunctional analysis andquantum mechanics. In quantum mechanics their importance lies in theDirac–von Neumann formulation of quantum mechanics, in which physicalobservables such asposition,momentum,angular momentum andspin are represented by self-adjoint operators on a Hilbert space. Of particular significance is theHamiltonian operator${\displaystyle \hat{H}}$ defined by

${\displaystyle \hat{H}\psi =-\frac{{\hslash }^2}{2m}{\mathrm{\nabla }}^2\psi +V\psi ,}$

which as an observable corresponds to the totalenergy of a particle of mass${\displaystyle m}$ in a realpotential field${\displaystyle V}$.Differential operators are an important class ofunbounded operators.

The structure of self-adjoint operators on infinite-dimensional Hilbert spaces essentially resembles the finite-dimensional case. That is to say, operators are self-adjoint if and only if they areunitarily equivalent to real-valuedmultiplication operators. With suitable modifications, this result can be extended to possibly unbounded operators on infinite-dimensional spaces. Since an everywhere-defined self-adjoint operator is necessarily bounded, one needs to be more attentive to the domain issue in the unbounded case. This is explained below in more detail.

Let${\displaystyle H}$ be aHilbert space and${\displaystyle A}$ anunbounded (i.e. not necessarily bounded)linear operator with adensedomain${\displaystyle \mathrm{Dom}A\subseteq H.}$ This condition holds automatically when${\displaystyle H}$ isfinite-dimensional since${\displaystyle \mathrm{Dom}A=H}$ for every linear operator on a finite-dimensional space.

Thegraph of an (arbitrary) operator${\displaystyle A}$ is the set${\displaystyle G(A)=\{(x,Ax)\mid x\in \mathrm{Dom}A\}.}$ An operator${\displaystyle B}$ is said toextend${\displaystyle A}$ if${\displaystyle G(A)\subseteq G(B).}$^[1] This is written as${\displaystyle A\subseteq B.}$

Let the inner product${\displaystyle ⟨\cdot ,\cdot ⟩}$ beconjugate linear on thesecond argument. Theadjoint operator${\displaystyle A^{\ast }}$ acts on the subspace${\displaystyle \mathrm{Dom}{A}^{\ast }\subseteq H}$ consisting of the elements${\displaystyle y}$ such that

${\displaystyle ⟨Ax,y⟩=⟨x,A^{\ast }y⟩,\mathrm{\forall }x\in \mathrm{Dom}A.}$

Thedensely defined operator${\displaystyle A}$ is calledsymmetric (orHermitian) if${\displaystyle A\subseteq A^{\ast }}$, i.e., if${\displaystyle \mathrm{Dom}A\subseteq \mathrm{Dom}{A}^{\ast }}$ and${\displaystyle Ax=A^{\ast }x}$ for all${\displaystyle x\in \mathrm{Dom}A}$. Equivalently,${\displaystyle A}$ is symmetric if and only if

${\displaystyle ⟨Ax,y⟩=⟨x,Ay⟩,\mathrm{\forall }x,y\in \mathrm{Dom}A.}$

Since${\displaystyle \mathrm{Dom}{A}^{\ast }\supseteq \mathrm{Dom}A}$ is dense in${\displaystyle H}$, symmetric operators are alwaysclosable (i.e. the closure of${\displaystyle G(A)}$ is the graph of an operator).^[2] If${\displaystyle A^{\ast }}$ is a closed extension of${\displaystyle A}$, the smallest closed extension${\displaystyle A^{\ast \ast }}$ of${\displaystyle A}$ must be contained in${\displaystyle A^{\ast }}$. Hence,

${\displaystyle A\subseteq A^{\ast \ast }\subseteq A^{\ast }}$

for symmetric operators and

${\displaystyle A=A^{\ast \ast }\subseteq A^{\ast }}$

for closed symmetric operators.^[3]

The densely defined operator${\displaystyle A}$ is calledself-adjoint if${\displaystyle A=A^{\ast }}$, that is, if and only if${\displaystyle A}$ is symmetric and${\displaystyle \mathrm{Dom}A=\mathrm{Dom}{A}^{\ast }}$. Equivalently, a closed symmetric operator${\displaystyle A}$ is self-adjoint if and only if${\displaystyle A^{\ast }}$ is symmetric. If${\displaystyle A}$ is self-adjoint, then${\displaystyle ⟨x,Ax⟩}$ is real for all${\displaystyle x\in \mathrm{Dom}A}$, i.e.,^[4]

${\displaystyle ⟨x,Ax⟩=\overline{⟨Ax,x⟩}=\overline{⟨x,Ax⟩}\in \mathbb{R},\mathrm{\forall }x\in \mathrm{Dom}A.}$

A symmetric operator${\displaystyle A}$ is said to beessentially self-adjoint if the closure of${\displaystyle A}$ is self-adjoint. Equivalently,${\displaystyle A}$ is essentially self-adjoint if it has aunique self-adjoint extension. In practical terms, having an essentially self-adjoint operator is almost as good as having a self-adjoint operator, since we merely need to take the closure to obtain a self-adjoint operator.

In physics, the termHermitian refers to symmetric as well as self-adjoint operators alike. The subtle difference between the two is generally overlooked.

Let${\displaystyle H}$ be aHilbert space and${\displaystyle A:\mathrm{Dom}(A)\to H}$ a symmetric operator. According toHellinger–Toeplitz theorem, if${\displaystyle \mathrm{Dom}(A)=H}$ then${\displaystyle A}$ is necessarily bounded.^[5]Abounded operator${\displaystyle A:H\to H}$ is self-adjoint if

${\displaystyle ⟨Ax,y⟩=⟨x,Ay⟩,\mathrm{\forall }x,y\in H.}$

Every bounded operator${\displaystyle T:H\to H}$ can be written in thecomplex form${\displaystyle T=A+iB}$ where${\displaystyle A:H\to H}$ and${\displaystyle B:H\to H}$ are bounded self-adjoint operators.^[6]

Alternatively, everypositivebounded linear operator${\displaystyle A:H\to H}$ is self-adjoint if the Hilbert space${\displaystyle H}$ iscomplex.^[7]

A bounded self-adjoint operator${\displaystyle A:H\to H}$ defined on${\displaystyle \mathrm{Dom}\left(A\right)=H}$ has the following properties:^[8][9]

• ${\displaystyle A:H\to \mathrm{Im}A\subseteq H}$ is invertible if theimage of${\displaystyle A}$ is dense in${\displaystyle H.}$
• Theoperator norm is given by${\displaystyle \Vert A\Vert =sup\left\{|⟨x,Ax⟩|:\Vert x\Vert =1\right\}}$
• If${\displaystyle \lambda }$ is aneigenvalue of${\displaystyle A}$ then${\displaystyle |\lambda |\le sup\left\{|⟨x,Ax⟩|:\Vert x\Vert \le 1\right\}}$; the eigenvalues are real and the correspondingeigenvectors are orthogonal.

Bounded self-adjoint operators do not necessarily have an eigenvalue. If, however,${\displaystyle A}$ is acompact self-adjoint operator then it always has an eigenvalue${\displaystyle |\lambda |=\Vert A\Vert }$ and corresponding normalized eigenvector.^[10]

Let${\displaystyle A:\mathrm{Dom}(A)\to H}$ be an unbounded operator.^[11] Theresolvent set (orregular set) of${\displaystyle A}$ is defined as

${\displaystyle \rho (A)=\left\{\lambda \in \mathbb{C}:\mathrm{\exists }(A-\lambda I{)}^{-1}\text{bounded and densely defined}\right\}.}$

If${\displaystyle A}$ is bounded, the definition reduces to${\displaystyle A-\lambda I}$ beingbijective on${\displaystyle H}$. Thespectrum of${\displaystyle A}$ is defined as the complement

${\displaystyle \sigma (A)=\mathbb{C}\setminus \rho (A).}$

In finite dimensions,${\displaystyle \sigma (A)\subseteq \mathbb{C}}$ consists exclusively of (complex)eigenvalues.^[12] The spectrum of a self-adjoint operator is always real (i.e.${\displaystyle \sigma (A)\subseteq \mathbb{R}}$), though non-self-adjoint operators with real spectrum exist as well.^[13][14] For bounded (normal) operators, however, the spectrum is realif and only if the operator is self-adjoint.^[15] This implies, for example, that a non-self-adjoint operator with real spectrum is necessarily unbounded.

As a preliminary, define${\displaystyle S=\{x\in \mathrm{Dom}A\mid \Vert x\Vert =1\},}$${\displaystyle m=\underset{x\in S}{inf}⟨Ax,x⟩}$ and${\displaystyle M=\underset{x\in S}{sup}⟨Ax,x⟩}$ with${\displaystyle m,M\in \mathbb{R}\cup \{\pm \mathrm{\infty }\}}$. Then, for every${\displaystyle \lambda \in \mathbb{C}}$ and every${\displaystyle x\in \mathrm{Dom}A,}$

${\displaystyle \Vert (A-\lambda )x\Vert \ge d(\lambda )\cdot \Vert x\Vert ,}$

where${\displaystyle d(\lambda )=\underset{r\in [m,M]}{inf}|r-\lambda |.}$

Indeed, let${\displaystyle x\in \mathrm{Dom}A\setminus \{0\}.}$ By theCauchy–Schwarz inequality,

${\displaystyle \Vert (A-\lambda )x\Vert \ge \frac{|⟨(A-\lambda )x,x⟩|}{\Vert x\Vert }=\left|⟨A\frac{x}{\Vert x\Vert },\frac{x}{\Vert x\Vert }⟩-\lambda \right|\cdot \Vert x\Vert \ge d(\lambda )\cdot \Vert x\Vert .}$

If${\displaystyle \lambda \notin [m,M],}$ then${\displaystyle d(\lambda )>0,}$ and${\displaystyle A-\lambda I}$ is calledbounded below.

In the physics literature, the spectral theorem is often stated by saying that a self-adjoint operator has an orthonormal basis of eigenvectors. Physicists are well aware, however, of the phenomenon of "continuous spectrum"; thus, when they speak of an "orthonormal basis" they mean either an orthonormal basis in the classic senseor some continuous analog thereof. In the case of themomentum operator$P=-i\frac{d}{dx}$, for example, physicists would say that the eigenvectors are the functions${\displaystyle f_p(x):=e^{ipx}}$, which are clearly not in the Hilbert space${\displaystyle L^2(\mathbb{R})}$. (Physicists would say that the eigenvectors are "non-normalizable.") Physicists would then go on to say that these "generalized eigenvectors" form an "orthonormal basis in the continuous sense" for${\displaystyle L^2(\mathbb{R})}$, after replacing the usualKronecker delta${\displaystyle {\delta }_{i,j}}$ by aDirac delta function${\displaystyle \delta \left(p-p^{\prime }\right)}$.^[16]

Although these statements may seem disconcerting to mathematicians, they can be made rigorous by use of the Fourier transform, which allows a general${\displaystyle L^2}$ function to be expressed as a "superposition" (i.e., integral) of the functions${\displaystyle e^{ipx}}$, even though these functions are not in${\displaystyle L^2}$. The Fourier transform "diagonalizes" the momentum operator; that is, it converts it into the operator of multiplication by${\displaystyle p}$, where${\displaystyle p}$ is the variable of the Fourier transform.

The spectral theorem in general can be expressed similarly as the possibility of "diagonalizing" an operator by showing it is unitarily equivalent to a multiplication operator. Other versions of the spectral theorem are similarly intended to capture the idea that a self-adjoint operator can have "eigenvectors" that are not actually in the Hilbert space in question.

Firstly, let${\displaystyle (X,\mathrm{\Sigma },\mu )}$ be aσ-finite measure space and${\displaystyle h:X\to \mathbb{R}}$ ameasurable function on${\displaystyle X}$. Then the operator${\displaystyle T_h:\mathrm{Dom}{T}_h\to L^2(X,\mu )}$, defined by

${\displaystyle T_h\psi (x)=h(x)\psi (x),\mathrm{\forall }\psi \in \mathrm{Dom}{T}_h,}$

where

${\displaystyle \mathrm{Dom}{T}_h:=\left\{\psi \in L^2(X,\mu )|h\psi \in L^2(X,\mu )\right\},}$

is called amultiplication operator.^[17] Any multiplication operator is a self-adjoint operator.^[18]

Secondly, two operators${\displaystyle A}$ and${\displaystyle B}$ with dense domains${\displaystyle \mathrm{Dom}A\subseteq H_1}$ and${\displaystyle \mathrm{Dom}B\subseteq H_2}$ in Hilbert spaces${\displaystyle H_1}$ and${\displaystyle H_2}$, respectively, areunitarily equivalent if and only if there is aunitary transformation${\displaystyle U:H_1\to H_2}$ such that:^[19]

• ${\displaystyle U\mathrm{Dom}A=\mathrm{Dom}B,}$
• ${\displaystyle UA{U}^{-1}\xi =B\xi ,\mathrm{\forall }\xi \in \mathrm{Dom}B.}$

If unitarily equivalent${\displaystyle A}$ and${\displaystyle B}$ are bounded, then${\displaystyle \Vert A{\Vert }_{H_1}=\Vert B{\Vert }_{H_2}}$; if${\displaystyle A}$ is self-adjoint, then so is${\displaystyle B}$.

The spectral theorem holds for both bounded and unbounded self-adjoint operators. Proof of the latter follows by reduction to the spectral theorem forunitary operators.^[21] We might note that if${\displaystyle T}$ is multiplication by${\displaystyle h}$, then the spectrum of${\displaystyle T}$ is just theessential range of${\displaystyle h}$.

More complete versions of the spectral theorem exist as well that involve direct integrals and carry with it the notion of "generalized eigenvectors".^[22]

One application of the spectral theorem is to define afunctional 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 theHamiltonian operator${\displaystyle \hat{H}}$. If${\displaystyle \hat{H}}$ has a true orthonormal basis of eigenvectors${\displaystyle e_j}$ with eigenvalues${\displaystyle {\lambda }_j}$, then${\displaystyle f(\hat{H}):=e^{-it\hat{H}/\hslash }}$ can be defined as the unique bounded operator with eigenvalues${\displaystyle f({\lambda }_j):=e^{-it{\lambda }_j/\hslash }}$ 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

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

where${\displaystyle {\mathbf{1}}_{(-\mathrm{\infty },\lambda ]}}$ is theindicator function of the interval${\displaystyle (-\mathrm{\infty },\lambda ]}$. The family of projection operators E(λ) is calledresolution of the identity forT. Moreover, the followingStieltjes integral representation forT can be proved:

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

In quantum mechanics,Dirac notation is used as combined expression for both the spectral theorem and theBorel functional calculus. That is, ifH is self-adjoint andf is aBorel function,

${\displaystyle f(H)=\int dE\left|{\mathrm{\Psi }}_E⟩f(E)⟨{\mathrm{\Psi }}_E\right|}$

with

${\displaystyle H|{\mathrm{\Psi }}_E⟩=E|{\mathrm{\Psi }}_E⟩}$

where the integral runs over the whole spectrum ofH. The notation suggests thatH is diagonalized by the eigenvectors Ψ_E. Such a notation is purelyformal. The resolution of the identity (sometimes calledprojection-valued measures) formally resembles the rank-1 projections${\displaystyle |{\mathrm{\Psi }}_E⟩⟨{\mathrm{\Psi }}_E|}$. In the Dirac notation, (projective) measurements are described viaeigenvalues andeigenstates, 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 thespectral measure of${\displaystyle |\mathrm{\Psi }⟩}$, if the system is prepared in${\displaystyle |\mathrm{\Psi }⟩}$ 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 suitablerigged Hilbert space.

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

${\displaystyle I=\int dE|{\mathrm{\Psi }}_E⟩⟨{\mathrm{\Psi }}_E|}$

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

${\displaystyle H_{\text{eff}}^{\ast }|{\mathrm{\Psi }}_E^{\ast }⟩=E^{\ast }|{\mathrm{\Psi }}_E^{\ast }⟩}$

and write the spectral theorem as:

${\displaystyle f\left(H_{\text{eff}}\right)=\int dE|{\mathrm{\Psi }}_E⟩f(E)⟨{\mathrm{\Psi }}_E^{\ast }|}$

(SeeFeshbach–Fano partitioning for the context where such operators appear inscattering theory).

Thespectral 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 spaceL²[0,1] and thedifferential operator

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

with${\displaystyle \mathrm{D}\mathrm{o}\mathrm{m}(A)}$ consisting of all complex-valued infinitelydifferentiable functionsf on [0, 1] satisfying the boundary conditions

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

Thenintegration by parts of the inner product shows thatA is symmetric.^{[nb 1]} The eigenfunctions ofA are the sinusoids

${\displaystyle f_n(x)=\sin (n\pi x)n=1,2,\dots }$

with the real eigenvaluesn²π²; the well-known orthogonality of the sine functions follows as a consequence ofA being symmetric.

The operatorA can be seen to have acompact inverse, meaning that the corresponding differential equationAf =g is solved by some integral (and therefore compact) operatorG. The compact symmetric operatorG then has a countable family of eigenvectors which are complete inL². The same can then be said forA.

A self-adjoint operatorA onH has purepoint spectrum if and only ifH has an orthonormal basis {e_i}_{i ∈ I} consisting of eigenvectors forA.

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

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

This Hamiltonian has pure point spectrum; this is typical for bound stateHamiltonians in quantum mechanics.^{[clarification needed][23]} 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.

Although the distinction between a symmetric operator and a (essentially) self-adjoint operator is subtle, it is important since self-adjointness is the hypothesis in the spectral theorem. Here we discuss some concrete examples of the distinction.

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 specifyingboundary 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 operatorA on this space by the usual formula, setting the Planck constant to 1:

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

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

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

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

If we choose

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

then using integration by parts, one can easily verify thatA is symmetric. This operator is not essentially self-adjoint,^[24] however, basically because we have specified too many boundary conditions on the domain ofA, which makes the domain of the adjoint too big (see also theexample below).

Specifically, with the above choice of domain forA, the domain of the closure${\displaystyle A^{\mathrm{c}\mathrm{l}}}$ ofA is

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

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

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

That is to say, the domain of the closure has the same boundary conditions as the domain ofA itself, just a less stringent smoothness assumption. Meanwhile, since there are "too many" boundary conditions onA, there are "too few" (actually, none at all in this case) for${\displaystyle A^{\ast }}$. If we compute${\displaystyle ⟨g,Af⟩}$ for${\displaystyle f\in \mathrm{Dom}(A)}$ 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${\displaystyle A^{\ast }}$, with${\displaystyle A^{\ast }g=-idg/dx}$.^[25]

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${\displaystyle A^{\mathrm{c}\mathrm{l}}}$ is the same as the domain of the adjoint ofA. Thus, in this case, the domain of the adjoint of${\displaystyle A^{\mathrm{c}\mathrm{l}}}$ is bigger than the domain of${\displaystyle A^{\mathrm{c}\mathrm{l}}}$ itself, showing that${\displaystyle A^{\mathrm{c}\mathrm{l}}}$ is not self-adjoint, which by definition means thatA is not essentially self-adjoint.

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

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

With this domain,A is essentially self-adjoint.^[26]

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 forA, the functions${\displaystyle f_n(x):=e^{2\pi inx}}$. Thus, in this case finding a domain such thatA is self-adjoint is a compromise: the domain has to be small enough so thatA is symmetric, but large enough so that${\displaystyle D(A^{\ast })=D(A)}$.

A more subtle example of the distinction between symmetric and (essentially) self-adjoint operators comes fromSchrödinger 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.^[27] 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 aunique 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.)

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

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

It is then possible to show that${\displaystyle {\hat{H}}^{\ast }}$ is not a symmetric operator, which certainly implies that${\displaystyle \hat{H}}$ is not essentially self-adjoint. Indeed,${\displaystyle {\hat{H}}^{\ast }}$ has eigenvectors with pure imaginary eigenvalues,^[28][29] which is impossible for a symmetric operator. This strange occurrence is possible because of a cancellation between the two terms in${\displaystyle {\hat{H}}^{\ast }}$: There are functions${\displaystyle f}$ in the domain of${\displaystyle {\hat{H}}^{\ast }}$ for which neither${\displaystyle d^{2}f/d{x}^2}$ nor${\displaystyle x^{4}f(x)}$ is separately in${\displaystyle L^2(\mathbb{R})}$, but the combination of them occurring in${\displaystyle {\hat{H}}^{\ast }}$ is in${\displaystyle L^2(\mathbb{R})}$. This allows for${\displaystyle {\hat{H}}^{\ast }}$ to be nonsymmetric, even though both${\displaystyle d^2/d{x}^2}$ 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${\displaystyle x^4}$.

In quantum mechanics, observables correspond to self-adjoint operators. ByStone's theorem on one-parameter unitary groups, self-adjoint operators are precisely the infinitesimal generators of unitary groups oftime 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 for0 <α ≤ 2 but not forα > 2.^[30][31]

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.^[32]

Example. There is no self-adjoint momentum operator${\displaystyle p}$ 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.^[33]

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

${\displaystyle D:\varphi \mapsto \frac{1}{i}{\varphi }^{\prime }}$

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

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

ThenD is a symmetric operator as can be shown byintegration by parts. The spacesN₊,N₋ (defined below) are given respectively by thedistributional solutions to the equation

which are inL²[0, 1]. One can show that each one of these solution spaces is 1-dimensional, generated by the functionsx →e^−x andx →e^x respectively. This shows thatD is not essentially self-adjoint,^[34] but does have self-adjoint extensions. These self-adjoint extensions are parametrized by the space of unitary mappingsN₊ →N₋, which in this case happens to be the unit circleT.

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

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

thenD 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 ofD. Other essentially self-adjoint extensions come from imposing boundary conditions of the form${\displaystyle \varphi (1)=e^{i\theta }\varphi (0)}$.

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

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

whereP_dist is the distributional extension ofP.

We next give the example of differential operators withconstant coefficients. Let

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

be a polynomial onRⁿ withreal coefficients, where α ranges over a (finite) set ofmulti-indices. Thus

${\displaystyle \alpha =({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)}$

and

${\displaystyle x^{\alpha }=x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\cdots x_n^{{\alpha }_n}.}$

We also use the notation

${\displaystyle D^{\alpha }=\frac{1}{i^{|\alpha |}}{\mathrm{\partial }}_{x_1}^{{\alpha }_1}{\mathrm{\partial }}_{x_2}^{{\alpha }_2}\cdots {\mathrm{\partial }}_{x_n}^{{\alpha }_n}.}$

Then the operatorP(D) defined on the space of infinitely differentiable functions of compact support onRⁿ by

${\displaystyle P(\mathrm{D})\varphi =\sum _{\alpha }{c}_{\alpha }{\mathrm{D}}^{\alpha }\varphi }$

is essentially self-adjoint onL²(Rⁿ).

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

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

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

${\displaystyle C_0^{\mathrm{\infty }}(M)\to C_0^{\mathrm{\infty }}(M).}$

Corresponding toP there is another differential operator, theformal adjoint ofP

${\displaystyle P^{\ast \mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}}\varphi =\sum _{\alpha }{D}^{\alpha }\left(\overline{a_{\alpha }}\varphi \right)}$

The multiplication representation of a self-adjoint operator, though extremely useful, is not a canonical representation. This suggests that it is not easy to extract from this representation a criterion to determine when self-adjoint operatorsA andB are unitarily equivalent. The finest grained representation which we now discuss involves spectral multiplicity. This circle of results is called theHahn–Hellinger theory of spectral multiplicity.

We first defineuniform multiplicity:

Definition. A self-adjoint operatorA has uniform multiplicityn wheren is such that 1 ≤n ≤ω if and only ifA is unitarily equivalent to the operator M_f of multiplication by the functionf(λ) =λ on

whereH_n is a Hilbert space of dimensionn. The domain of M_f consists of vector-valued functionsψ onR such that

Non-negative countably additive measuresμ,ν aremutually singular if and only if they are supported on disjoint Borel sets.

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

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

Unlike the multiplication-operator version of the spectral theorem, the direct-integral version is unique in the sense that the measure equivalence class ofμ (or equivalently its sets of measure 0) is uniquely determined and the measurable function${\displaystyle \lambda \mapsto \mathrm{d}\mathrm{i}\mathrm{m}(H_{\lambda })}$ is determined almost everywhere with respect toμ.^[36] The function${\displaystyle \lambda \mapsto \dim \left(H_{\lambda }\right)}$ is thespectral 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.^[37]

The Laplacian onRⁿ is the operator

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

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

• Compact operator on Hilbert space
• Unbounded operator
• Hermitian adjoint
• Normal operator
• Positive operator
• Helffer–Sjöstrand formula

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• Hilbert spaces
• Operator theory
• Linear operators

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