Self-adjoint operator


In mathematics, a self-adjoint operator on an infinite-dimensional complex vector space V with inner product is a linear map A (from V to itself) that is its own adjoint. If V is finite-dimensional with a given orthonormal basis, this is equivalent to the condition that the matrix of A is a Hermitian matrix, i.e., equal to its conjugate transpose A. By the finite-dimensional spectral theorem, V has an orthonormal basis such that the matrix of A relative to this basis is a diagonal matrix with entries in the real numbers. This article deals with applying generalizations of this concept to operators on Hilbert spaces of arbitrary dimension.

Self-adjoint operators are used in functional analysis and quantum mechanics. In quantum mechanics their importance lies in the Dirac–von Neumann formulation of quantum mechanics, in which physical observables such as position, momentum, angular momentum and spin are represented by self-adjoint operators on a Hilbert space. Of particular significance is the Hamiltonian operator defined by

which as an observable corresponds to the total energy of a particle of mass m in a real potential field V. Differential operators are an important class of unbounded 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 are unitarily equivalent to real-valued multiplication 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.

Definitions edit

Let   be a Hilbert space and   an unbounded (i.e. not necessarily bounded) operator with a dense domain   This condition holds automatically when   is finite-dimensional since   for every linear operator on a finite-dimensional space.

The graph of an (arbitrary) operator   is the set   An operator   is said to extend   if   This is written as  

Let the inner product   be conjugate linear on the second argument. The adjoint operator   acts on the subspace   consisting of the elements   such that


The densely defined operator   is called symmetric (or Hermitian) if  , i.e., if   and   for all  . Equivalently,   is symmetric if and only if


Since   is dense in  , symmetric operators are always closable (i.e. the closure of   is the graph of an operator). If   is a closed extension of  , the smallest closed extension   of   must be contained in  . Hence,


for symmetric operators and


for closed symmetric operators.[1]

The densely defined operator   is called self-adjoint if  , that is, if and only if   is symmetric and  . Equivalently, a closed symmetric operator   is self-adjoint if and only if   is symmetric. If   is self-adjoint, then   is real for all  , i.e.,[2]


A symmetric operator   is said to be essentially self-adjoint if the closure of   is self-adjoint. Equivalently,   is essentially self-adjoint if it has a unique 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 self-adjoint operator.

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

Bounded self-adjoint operators edit

Let   be a Hilbert space and   a symmetric operator. According to Hellinger–Toeplitz theorem, if   then   is necessarily bounded.[3] A bounded operator   is self-adjoint if


Every bounded operator   can be written in the complex form   where   and   are bounded self-adjoint operators.[4]

Alternatively, every positive bounded linear operator   is self-adjoint if the Hilbert space   is complex.[5]

Properties edit

A bounded self-adjoint operator   defined on   has the following properties:[6][7]

  •   is invertible if the image of   is dense in  
  • The eigenvalues of   are real and the corresponding eigenvectors are orthogonal.
  • If   is an eigenvalue of   then  , where   if   and   is a compact self-adjoint operator.

Spectrum of self-adjoint operators edit

Let   be an unbounded operator.[8] The resolvent set (or regular set) of   is defined as


If   is bounded, the definition reduces to   being bijective on  . The spectrum of   is defined as the complement


In finite dimensions,   consists exclusively of (complex) eigenvalues.[9] The spectrum of a self-adjoint operator is always real (i.e.  ), though non-self-adjoint operators with real spectrum exist as well.[10][11] For bounded (normal) operators, however, the spectrum is real if and only if the operator is self-adjoint.[12] This implies, for example, that a non-self-adjoint operator with real spectrum is necessarily unbounded.

As a preliminary, define     and   with  . Then, for every   and every  



Indeed, let   By the Cauchy–Schwarz inequality,


If   then   and   is called bounded below.

Theorem — Self-adjoint operator has real spectrum


Let   be self-adjoint and denote   with   It suffices to prove that  

  1. Let   The goal is to prove the existence and boundedness of   and show that   We begin by showing that   and  
    1. As shown above,   is bounded below, i.e.   with   The triviality of   follows.
    2. It remains to show that   Indeed,
      1.   is closed. To prove this, pick a sequence   converging to some   Since
          is fundamental. Hence, it converges to some   Furthermore,   and   The arguments made thus far hold for any symmetric operator. It now follows from self-adjointness that   is closed, so     and consequently  
      2.   is dense in   The self-adjointness of   (i.e.  ) implies   and thus  . The subsequent inclusion   implies   and, consequently,  
  2. The operator   has now been proven to be bijective, so   exists and is everywhere defined. The graph of   is the set   Since   is closed (because   is), so is   By closed graph theorem,   is bounded, so  

Theorem — Symmetric operator with real spectrum is self-adjoint

  1.   is symmetric; therefore   and   for every  . Let   If   then   and the operators   are both bijective.
  2.   Indeed,  . That is, if   then   would not be injective (i.e.  ). But   and, hence,   This contradicts the bijectiveness.
  3. The equality   shows that   i.e.   is self-adjoint. Indeed, it suffices to prove that   For every   and  

Spectral theorem edit

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 sense or some continuous analog thereof. In the case of the momentum operator  , for example, physicists would say that the eigenvectors are the functions  , which are clearly not in the Hilbert space  . (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  , after replacing the usual Kronecker delta   by a Dirac delta function  .[13]

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

Multiplication operator form of the spectral theorem edit

Firstly, let   be a σ-finite measure space and   a measurable function on  . Then the operator  , defined by




is called a multiplication operator.[14] Any multiplication operator is a self-adjoint operator.[15]

Secondly, two operators   and   with dense domains   and   in Hilbert spaces   and  , respectively, are unitarily equivalent if and only if there is a unitary transformation   such that:[16]


If unitarily equivalent   and   are bounded, then  ; if   is self-adjoint, then so is  .

Theorem — Any self-adjoint operator   on a separable Hilbert space is unitarily equivalent to a multiplication operator, i.e.,[17]


The spectral theorem holds for both bounded and unbounded self-adjoint operators. Proof of the latter follows by reduction to the spectral theorem for unitary operators.[18] We might note that if   is multiplication by  , then the spectrum of   is just the essential range of  .

More complete versions of the spectral theorem exist as well that involve direct integrals and carry with it the notion of "generalized eigenvectors".[19]

Functional calculus edit

One application of the spectral theorem is to define a functional calculus. That is, if   is a function on the real line and   is a self-adjoint operator, we wish to define the operator  . The spectral theorem shows that if   is represented as the operator of multiplication by  , then   is the operator of multiplication by the composition  .

One example from quantum mechanics is the case where   is the Hamiltonian operator  . If   has a true orthonormal basis of eigenvectors   with eigenvalues  , then   can be defined as the unique bounded operator with eigenvalues   such that:


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

It has been customary to introduce the following notation


where   is the indicator function of the interval  . 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:


Formulation in the physics literature edit

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,




where the integral runs over the whole spectrum of H. The notation suggests that H is diagonalized by the eigenvectors ΨE. Such a notation is purely formal. The resolution of the identity (sometimes called projection-valued measures) formally resembles the rank-1 projections  . 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  , if the system is prepared in   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:


In the case   is the sum of an Hermitian H and a skew-Hermitian (see skew-Hermitian matrix) operator  , one defines the biorthogonal basis set


and write the spectral theorem as:


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

Formulation for symmetric operators edit

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


with   consisting of all complex-valued infinitely differentiable functions f on [0, 1] satisfying the boundary conditions


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


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 edit

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

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


This Hamiltonian has pure point spectrum; this is typical for bound state Hamiltonians in quantum mechanics.[clarification needed][20] 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.

Symmetric vs self-adjoint operators edit

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.

Boundary conditions edit

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   (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:


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


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

If we choose


then using integration by parts, one can easily verify that A is symmetric. This operator is not essentially self-adjoint,[21] 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   of A is


whereas the domain of the adjoint   of A is


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  . If we compute   for   using integration by parts, then since   vanishes at both ends of the interval, no boundary conditions on   are needed to cancel out the boundary terms in the integration by parts. Thus, any sufficiently smooth function   is in the domain of  , with  .[22]

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   is the same as the domain of the adjoint of A. Thus, in this case, the domain of the adjoint of   is bigger than the domain of   itself, showing that   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:


With this domain, A is essentially self-adjoint.[23]

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   for   are eigenvectors, with eigenvalues  , 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  . 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  .

Schrödinger operators with singular potentials edit

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


is not essentially self-adjoint on the space of smooth, rapidly decaying functions.[24] In this case, the failure of essential self-adjointness reflects a pathology in the underlying classical system: A classical particle with a   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   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   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


It is then possible to show that   is not a symmetric operator, which certainly implies that   is not essentially self-adjoint. Indeed,   has eigenvectors with pure imaginary eigenvalues,[25][26] which is impossible for a symmetric operator. This strange occurrence is possible because of a cancellation between the two terms in  : There are functions   in the domain of   for which neither   nor   is separately in  , but the combination of them occurring in   is in  . This allows for   to be nonsymmetric, even though both   and   are symmetric operators. This sort of cancellation does not occur if we replace the repelling potential   with the confining potential  .

Non-self-adjoint operators in quantum mechanics edit

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  , defined initially on smooth compactly supported functions, is essentially self-adjoint for 0 < α ≤ 2 but not for α > 2.[27][28]

The failure of essential self-adjointness for   has a counterpart in the classical dynamics of a particle with potential  : The classical particle escapes to infinity in finite time.[29]

Example. There is no self-adjoint momentum operator   for a particle moving on a half-line. Nevertheless, the Hamiltonian   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.[30]

Examples edit

A symmetric operator that is not essentially self-adjoint edit

We first consider the Hilbert space   and the differential operator


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


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


which are in L2[0, 1]. One can show that each one of these solution spaces is 1-dimensional, generated by the functions xe−x and xex respectively. This shows that D is not essentially self-adjoint,[31] 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  . Since   is a first-order operator, only one boundary condition is needed to ensure that   is symmetric. If we replaced the boundary conditions given above by the single boundary condition


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  .

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


where Pdist is the distributional extension of P.

Constant-coefficient operators edit

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


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




We also use the notation


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


is essentially self-adjoint on L2(Rn).

Theorem — Let P a polynomial function on Rn with real coefficients, F the Fourier transform considered as a unitary map L2(Rn) → L2(Rn). Then F*P(D)F is essentially self-adjoint and its unique self-adjoint extension is the operator of multiplication by the function P.

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


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


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


Theorem — The adjoint P* of P is a restriction of the distributional extension of the formal adjoint to an appropriate subspace of  . Specifically:


Spectral multiplicity theory edit

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 operators A and B are unitarily equivalent. The finest grained representation which we now discuss involves spectral multiplicity. This circle of results is called the HahnHellinger theory of spectral multiplicity.

Uniform multiplicity edit

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 Mf of multiplication by the function f(λ) = λ on


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


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)

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

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 edit

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

Theorem — [32] Any self-adjoint operator on a separable Hilbert space is unitarily equivalent to multiplication by the function λ ↦ λ on


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   is determined almost everywhere with respect to μ.[33] The function   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.[34]

Example: structure of the Laplacian edit

The Laplacian on Rn is the operator


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).

Theorem — If n = 1, then −Δ has uniform multiplicity  , otherwise −Δ has uniform multiplicity  . Moreover, the measure μmult may be taken to be Lebesgue measure on [0, ∞).

See also edit

Remarks edit

  1. ^ The reader is invited to perform integration by parts twice and verify that the given boundary conditions for   ensure that the boundary terms in the integration by parts vanish.

Notes edit

  1. ^ Reed & Simon 1980, p. 255-256.
  2. ^ Griffel 2002, pp. 224.
  3. ^ Hall 2013 Corollary 9.9
  4. ^ Griffel 2002, p. 238.
  5. ^ Reed & Simon 1980, p. 195.
  6. ^ Rudin 1991, pp. 326–327.
  7. ^ Griffel 2002, pp. 224–230, 241.
  8. ^ Hall 2013, pp. 133, 177.
  9. ^ de la Madrid Modino 2001, pp. 95–97.
  10. ^ Hall 2013 Section 9.4
  11. ^ Bebiano & da Providência 2019.
  12. ^ Rudin 1991, pp. 327.
  13. ^ Hall 2013, pp. 123–130.
  14. ^ Hall 2013, p. 207.
  15. ^ Akhiezer 1981, p. 152.
  16. ^ Akhiezer 1981, pp. 115–116.
  17. ^ Hall 2013, p. 127,207.
  18. ^ Hall 2013 Section 10.4
  19. ^ Hall 2013, pp. 144–147, 206–207.
  20. ^ Ruelle 1969.
  21. ^ Hall 2013 Proposition 9.27
  22. ^ Hall 2013 Proposition 9.28
  23. ^ Hall 2013 Example 9.25
  24. ^ Hall 2013 Theorem 9.41
  25. ^ Berezin & Shubin 1991 p. 85
  26. ^ Hall 2013 Section 9.10
  27. ^ Berezin & Shubin 1991, pp. 55, 86.
  28. ^ Hall 2013, pp. 193–196.
  29. ^ Hall 2013 Chapter 2, Exercise 4
  30. ^ Bonneau, Faraut & Valent 2001.
  31. ^ Hall 2013 Section 9.6
  32. ^ Hall 2013 Theorems 7.19 and 10.9
  33. ^ Hall 2013 Proposition 7.22
  34. ^ Hall 2013 Proposition 7.24

References edit

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