A self-adjoint operator is a linear operator with special properties. Operators and especially self-adjoint operators are examined in the mathematical sub-area of functional analysis. The self-adjoint operator is a generalization of the self-adjoint matrix .

## definition

In this section the definition of the self-adjoint operator is given. In the first section it is only given for bounded operators and in the second also for unbounded operators. Since bounded operators can always be defined on the whole vector space, the bounded self-adjoint operator is a special case of the unbounded self-adjoint operator.

### Limited operators

Let be a Hilbert space consisting of the vector space and the scalar product and be a bounded linear operator . If the equation ${\ displaystyle (H, \ langle.,. \ rangle)}$ ${\ displaystyle H}$ ${\ displaystyle \ langle \ cdot, \ cdot \ rangle}$${\ displaystyle T \ colon H \ to H}$ ${\ displaystyle T}$

${\ displaystyle \ langle Tx, y \ rangle = \ langle x, Ty \ rangle}$

### Unlimited operators

Let be a Hilbert space consisting of the vector space and the scalar product and be a densely defined operator . Be the space of everyone so that the linear functional${\ displaystyle (H, \ langle.,. \ rangle)}$ ${\ displaystyle H}$ ${\ displaystyle \ langle \ cdot, \ cdot \ rangle}$${\ displaystyle T \ colon D (T) \ to H}$${\ displaystyle D (T ^ {*})}$${\ displaystyle y \ in H}$

${\ displaystyle x \ mapsto \ langle Tx, y \ rangle}$

is steady. This functional has the domain , so it is densely defined in . Hence it has a definite steady continuation throughout . According to Fréchet-Riesz's theorem, there is a clearly defined element , so that ${\ displaystyle D (T)}$${\ displaystyle H}$${\ displaystyle H}$${\ displaystyle T ^ {*} y \ in H}$

${\ displaystyle \ langle Tx, y \ rangle = \ langle x, T ^ {*} y \ rangle}$

applies to all . The operator with the domain is the uniquely determined adjoint operator . ${\ displaystyle x \ in H}$${\ displaystyle T ^ {*}}$${\ displaystyle D (T ^ {*})}$${\ displaystyle T}$

The operator is now called self-adjoint, if and apply, i.e. if the operator agrees with its adjoint operator and the corresponding domains of definition. ${\ displaystyle T}$${\ displaystyle T = T ^ {*}}$${\ displaystyle D (T) = D (T ^ {*})}$${\ displaystyle T}$${\ displaystyle T ^ {*}}$

## history

John von Neumann , who founded the theory of unbounded operators in 1929 , was also the first to recognize the need to distinguish between symmetric and self-adjoint operators. A spectral decomposition, as described in the last section of this article, can only be shown for the latter. Von Neumann called symmetric operators Hermitian . He stated that it was important for spectral decomposition, among other things, that an operator did not allow a symmetrical extension and called this class of operators maximally Hermitian . However, this requirement is not yet sufficient for the spectral theorem, which presupposes self-adjoint operators. At the suggestion of Erhard Schmidt, Von Neumann called self-adjoint operators hypermaximal . The term self-adjoint operator was coined by Marshall Harvey Stone .

## Related objects

Be the real or complex number field and be a scalar product on then is a Hilbert space. A matrix is called self-adjoint if ${\ displaystyle \ mathbb {K} \ in \ {\ mathbb {R}, \ mathbb {C} \}}$${\ displaystyle \ langle \ cdot, \ cdot \ rangle}$${\ displaystyle \ mathbb {K} ^ {n},}$${\ displaystyle (\ mathbb {K} ^ {n}, \ langle \ cdot, \ cdot \ rangle)}$${\ displaystyle A}$

${\ displaystyle \ langle Ay, x \ rangle = \ langle y, Ax \ rangle}$

applies to all . The matrix is understood here as a linear mapping on the . Since it maps between finite-dimensional vector spaces, it is therefore restricted to continuous and thus also densely defined. So a self-adjoint matrix is ​​also a self-adjoint operator. If one looks at the with its standard scalar product , the symmetrical matrices correspond to the self-adjoint ones. In the case of the with the corresponding canonical scalar product, the Hermitian matrices are the self adjoint ones . ${\ displaystyle x, y \ in \ mathbb {K} ^ {n}}$${\ displaystyle A}$${\ displaystyle \ mathbb {K} ^ {n}}$${\ displaystyle A}$${\ displaystyle A}$${\ displaystyle \ mathbb {R} ^ {n}}$${\ displaystyle \ mathbb {C} ^ {n}}$

### Symmetric operator

An operator is called symmetric if ${\ displaystyle T \ colon D (T) \ to H}$

${\ displaystyle \ langle Ty, x \ rangle = \ langle y, Tx \ rangle}$

applies to all . In contrast to the self-adjoint operator, it is not required here that the operator must be defined densely (but this is not uniform in the literature). If densely defined (and thus the adjoint operator is well defined), then is symmetric if and only if holds. For bounded operators, the terms self-adjoint and symmetric coincide. Therefore symmetric, not self-adjoint operators are always unbounded . In addition, Hellinger-Toeplitz's theorem says that every symmetric operator that is entirely defined on is continuous and therefore self-adjoint. ${\ displaystyle x, \, y \ in D (T)}$${\ displaystyle T}$${\ displaystyle T}$${\ displaystyle T}$${\ displaystyle T \ subseteq T ^ {*}}$${\ displaystyle H}$

An operator is said to be essentially self-adjoint if it is symmetric, densely defined and its closure is self-adjoint. An essentially self-adjoint operator can therefore always be extended to a self-adjoint operator. ${\ displaystyle T \ colon D (T) \ to H}$${\ displaystyle T}$

## Examples

### Symmetrical matrix

A symmetric matrix can be understood as an operator . With regard to the standard scalar product, every symmetric matrix is ​​a self-adjoint matrix or a self-adjoint operator. ${\ displaystyle A \ in \ mathbb {R} ^ {n \ times n}}$${\ displaystyle A \ colon \ mathbb {R} ^ {n} \ to \ mathbb {R} ^ {n}}$

### The operator -i d / dx

If an operator is bounded, the terms symmetric operator, essentially self-adjoint operator and self-adjoint operator are equivalent, as mentioned. In the case of unbounded operators, self adjointness implies symmetry, but the converse is not true. The following couple provides a counterexample:

1. In the following, the Hilbert space and the differential operator with the Dirichlet boundary conditions are considered.${\ displaystyle C ^ {\ infty} (] 0.1 [) \ cap L ^ {2} (] 0.1 [)}$ ${\ displaystyle p_ {1}: = - {\ rm {i}} \, {\ tfrac {\ rm {d}} {{\ rm {d}} x}} = {\ tfrac {1} {\ rm {i}}} \, {\ tfrac {\ rm {d}} {{\ rm {d}} x}}}$ ${\ displaystyle \ psi (0) = \ psi (1) = 0}$
2. And its extension in one only "periodicity" calls .${\ displaystyle p_ {2},}$${\ displaystyle \ psi (1) = \ psi (0)}$

From the chain of equations

${\ displaystyle \ langle u, p_ {i} v \ rangle _ {L ^ {2}} - \ langle p_ {i} u, v \ rangle _ {L ^ {2}} = \ int _ {0} ^ {1} {\ overline {u (x)}} \ cdot p_ {i} v (x) - {\ overline {p_ {i} u (x)}} \ cdot v (x) \ mathrm {d} x = - {\ rm {i}} \ cdot \ left ({\ overline {u}} (1) \ cdot v (1) - {\ overline {u}} (0) \ cdot v (0) \ right) = 0}$

it follows that the operators for are symmetric. However, only the operator is self adjoint, because in the first case the domain of definition is unnecessarily restricted. It then has no eigenfunctions at all , because they are all of the form , i.e. would violate the required condition . ${\ displaystyle p_ {i}}$${\ displaystyle i \ in \ {1,2 \}}$${\ displaystyle p_ {2}}$${\ displaystyle \ exp (i \ lambda _ {n} \ cdot x)}$${\ displaystyle \ psi (0) = 0}$

### Laplace operator

The Laplace operator is an unbounded operator. It is self adjoint with respect to the -Scalar product. That means it is symmetrical with respect to this scalar product, what ${\ displaystyle \ Delta \ colon D (\ Delta) \ to L ^ {2} (\ mathbb {R} ^ {n})}$${\ displaystyle L ^ {2}}$

${\ displaystyle \ int _ {\ mathbb {R} ^ {n}} \ Delta f (x) g (x) \ mathrm {d} x = \ int _ {\ mathbb {R} ^ {n}} f ( x) \ Delta g (x) \ mathrm {d} x}$

means for all and is densely defined. The derivation is to be understood here in the weak sense . Thus applies to the domain of definition ${\ displaystyle f, \, g \ in D (\ Delta)}$

${\ displaystyle D (\ Delta) = \ {u \ in L ^ {2} (\ mathbb {R} ^ {n}): \ Delta u \ in L ^ {2} (\ mathbb {R} ^ {n }) \}.}$

This corresponds to the Sobolev space of square integrable and twice weakly differentiable functions, this is close to . The symmetry of the Laplace operator follows from Green's formula . ${\ displaystyle H ^ {2} (\ mathbb {R} ^ {n})}$${\ displaystyle L ^ {2} (\ mathbb {R} ^ {n})}$

### Multiplication operator

Be a measure space and a measurable function . The multiplication operator with is defined by ${\ displaystyle (\ Omega, \ Sigma, \ mu)}$${\ displaystyle f \ colon \ Omega \ to \ mathbb {R}}$${\ displaystyle M_ {f} \ colon D (M_ {f}) \ to L ^ {2} (\ mu)}$${\ displaystyle D (M_ {f}) = \ {x \ in L ^ {2} (\ mu): f \ cdot x \ in L ^ {2} (\ mu) \} \ subset L ^ {2} (\ mu)}$

${\ displaystyle x \ mapsto M_ {f} x: = f \ cdot x.}$

This operator is unbounded and densely defined, because for contains all classes that vanish outside of and because is dense. It is also symmetrical with respect to the -Scalar product. The operator is also self adjoint. Since what applies to a symmetric operator is what and means, only needs to be shown for self-adjointness . Let be the characteristic function of , for and holds ${\ displaystyle \ Omega _ {n}: = \ {\ omega \ in \ Omega: | f (\ omega) | \ leq n \}}$${\ displaystyle D (M_ {f})}$${\ displaystyle L ^ {2}}$${\ displaystyle \ Omega _ {n}}$${\ displaystyle \ textstyle \ Omega = \ bigcup _ {n} \ Omega _ {n}}$${\ displaystyle D (M_ {f}) \ subset L ^ {2} (\ mu)}$${\ displaystyle M_ {f}}$${\ displaystyle L ^ {2}}$${\ displaystyle M_ {f} \ subset M_ {f} ^ {*}}$${\ displaystyle D (M_ {f}) \ subset D (M_ {f} ^ {*})}$${\ displaystyle M_ {f} ^ {*} | _ {D (M_ {f})} = M_ {f}}$${\ displaystyle D (M_ {f} ^ {*}) \ subset D (M_ {f})}$${\ displaystyle \ chi _ {n}}$${\ displaystyle \ Omega _ {n}}$${\ displaystyle z \ in D (M_ {f})}$${\ displaystyle x \ in D (M_ {f} ^ {*})}$

${\ displaystyle \ langle z, \ chi _ {n} M_ {f} ^ {*} x \ rangle _ {L ^ {2}} = \ langle \ chi _ {n} z, M_ {f} ^ {* } x \ rangle _ {L ^ {2}} = \ langle M_ {f} (\ chi _ {n} z), x \ rangle _ {L ^ {2}} = \ langle f \ chi _ {n} z, x \ rangle _ {L ^ {2}}.}$

That is, applies almost everywhere . Since it converges point by point, applies almost everywhere. Since there is now in there is what shows and thus proves self-adjointness. ${\ displaystyle \ chi _ {n} M_ {f} ^ {*} x = \ chi _ {n} fx}$${\ displaystyle \ chi _ {n} \ to 1}$${\ displaystyle M_ {f} ^ {*} x = fx}$${\ displaystyle M_ {f} ^ {*} x = fx}$${\ displaystyle L ^ {2}}$${\ displaystyle x \ in D (M_ {f})}$${\ displaystyle D (M_ {f}) = D (M_ {f} ^ {*})}$

## criteria

For an operator densely defined in a Hilbert space, the following criteria are repeatedly mentioned with regard to the question of self-adjointness . ${\ displaystyle (H, \ langle.,. \ rangle)}$ ${\ displaystyle T \ colon D (T) \ to H}$

### First criterion

${\ displaystyle T}$is a self-adjoint operator in if and only if the following condition is met: ${\ displaystyle H}$

1. It applies .${\ displaystyle T = T ^ {*} = T ^ {**}}$

### Second criterion

${\ displaystyle T}$is a self-adjoint operator in if and only if the following conditions are met: ${\ displaystyle H}$

1. ${\ displaystyle T}$ is symmetrical.
2. ${\ displaystyle T}$is complete .
3. The null spaces of the two operators and are equal .${\ displaystyle T ^ {*} - \ mathrm {i} \ cdot Id_ {H}}$${\ displaystyle T ^ {*} + \ mathrm {i} \ cdot Id_ {H}}$${\ displaystyle \ {0 \}}$

For the null spaces occurring in the last-mentioned condition, one often considers their Hilbert space dimensions . In the case of a symmetric operator , this is also called its defect index . The last-mentioned condition can therefore also be expressed in such a way that the defect indices are equal to 0 . ${\ displaystyle T}$${\ displaystyle T}$

### Third criterion

Conditions 2 and 3 of the second criterion can be reinterpreted as a single one and in this way one obtains another equivalent criterion with regard to the question of self adjointness of : ${\ displaystyle T}$

${\ displaystyle T}$is a self-adjoint operator in if and only if the following conditions are met: ${\ displaystyle H}$

1. ${\ displaystyle T}$ is symmetrical.
2. The image spaces of the two operators and are the same .${\ displaystyle T- \ mathrm {i} \ cdot Id_ {H}}$${\ displaystyle T + \ mathrm {i} \ cdot Id_ {H}}$${\ displaystyle H}$

### Fourth criterion

The fourth criterion shows that the self adjointness of a tightly defined operator is essentially determined by the position of its spectrum within the real numbers :

${\ displaystyle T}$is a self-adjoint operator in if and only if the following conditions are met: ${\ displaystyle H}$

1. ${\ displaystyle T}$ is symmetrical.
2. The spectrum of consists solely of real numbers, so .${\ displaystyle T}$${\ displaystyle \ sigma (T) \ subset \ mathbb {R}}$

## properties

Let be a tightly defined operator on the Hilbert space${\ displaystyle T}$${\ displaystyle (H, \ langle.,. \ rangle),}$

• then is a self-adjoint operator with${\ displaystyle T ^ {*} T}$${\ displaystyle \ langle Tx, x \ rangle \ geq 0.}$

Let be a self-adjoint operator on the Hilbert space${\ displaystyle T}$${\ displaystyle (H, \ langle.,. \ rangle).}$

• The following applies to the spectrum of: There are no spectral values ​​that are truly complex numbers . In particular, a self-adjoint matrix has only real spectral or eigenvalues .${\ displaystyle \ sigma (T)}$${\ displaystyle T}$${\ displaystyle \ sigma (T) \ subset \ mathbb {R}.}$
• An operator is positive , that is, it applies to all if and only if the inclusion applies to the spectrum .${\ displaystyle T}$${\ displaystyle \ langle Tx, x \ rangle \ geq 0}$${\ displaystyle x \ in D (T)}$${\ displaystyle \ sigma (T)}$${\ displaystyle \ sigma (T) \ subset [0, \ infty]}$
• If holds for all , then there exists a self-adjoint operator with for all , so that holds.${\ displaystyle \ langle Tx, x \ rangle \ geq 0}$${\ displaystyle x \ in H}$${\ displaystyle B}$${\ displaystyle \ langle Bx, x \ rangle \ geq 0}$${\ displaystyle x \ in H}$${\ displaystyle B \ circ B = T}$

## Friedrichs expansion

Let be a Hilbert space and a densely defined semi-bounded operator . To be semi-bounded for an operator means that the operator satisfies either the inequality or the inequality for one and for all . Then there exists a self-adjoint extension of which satisfies the same estimate. ${\ displaystyle (H, \ langle, \ rangle _ {H})}$${\ displaystyle T \ colon D (T) \ to H}$${\ displaystyle T}$${\ displaystyle \ langle Tx, x \ rangle _ {H} \ geq C \ | x \ | _ {H} ^ {2}}$${\ displaystyle \ langle Tx, x \ rangle _ {H} \ leq C \ | x \ | _ {H} ^ {2}}$${\ displaystyle C \ in \ mathbb {R}}$${\ displaystyle x \ in D (T)}$${\ displaystyle T}$${\ displaystyle T}$

It should be noted that the expression must be real-valued for a semi-restricted operator , otherwise the order relations and are not defined; and operators that apply to all are symmetric. ${\ displaystyle T}$${\ displaystyle \ langle Tx, x \ rangle _ {H}}$ ${\ displaystyle \ geq}$${\ displaystyle \ leq}$${\ displaystyle \ langle Tx, x \ rangle _ {H} \ in \ mathbb {R}}$${\ displaystyle x \ in H}$

Be a closed and tightly defined operator. Then one can deduce from Friedrich's expansion that it is densely defined and self-adjoint. ${\ displaystyle T \ colon D (A) \ to H}$${\ displaystyle T ^ {*} T \ colon \ {x \ in D (T): Tx \ in D (T ^ {*}) \} \ to H}$

## Spectral theorem for unlimited operators

### Spectral decomposition

Let be a Hilbert space and the Borel σ-algebra . For every self-adjoint operator there is a unique spectral measure , so that ${\ displaystyle (H, \ langle.,. \ rangle _ {H})}$${\ displaystyle \ Sigma}$${\ displaystyle T \ colon D (T) \ to H}$ ${\ displaystyle E \ colon \ Sigma \ to L (H, H)}$

${\ displaystyle \ langle Tx, y \ rangle _ {H} = \ int _ {\ mathbb {R}} t \, \ mathrm {d} \ langle E_ {t} \, x, y \ rangle _ {H} }$

with and applies. This statement is the spectral theorem for unbounded self-adjoint operators. If one demands that the operators are bounded and self-adjoint or even compact and self-adjoint, the result is simplified. This is explained in more detail in the article Spectral Theorem. ${\ displaystyle x \ in D (T)}$${\ displaystyle y \ in H}$

### Multiplication operator

Be also a Hilbert space and let a self-adjoint operator. Then there exists a (in the separable case a -finite) measure space , a measurable function and a unitary operator with ${\ displaystyle H}$${\ displaystyle T \ colon H \ supset D (T) \ to H}$${\ displaystyle \ sigma}$${\ displaystyle (\ Omega, \ Sigma, \ mu)}$ ${\ displaystyle f \ colon \ Omega \ to \ mathbb {R}}$ ${\ displaystyle U \ colon H \ to L ^ {2} (\ mu)}$

1. ${\ displaystyle x \ in D (T) \ Leftrightarrow f \ cdot Ux \ in L ^ {2} (\ mu)}$ and
2. ${\ displaystyle UTU ^ {*} \ phi = f \ cdot \ phi}$for .${\ displaystyle \ phi \ in \ {\ phi \ in L ^ {2} (\ mu): f \ cdot \ phi \ in L ^ {2} (\ mu) \}}$

In essence, the multiplication operator is the only example of a self-adjoint operator. ${\ displaystyle \ phi \ mapsto f \ cdot \ phi}$

## Individual evidence

1. Dirk Werner : Functional Analysis . 6th, corrected edition. Springer-Verlag, Berlin 2007, ISBN 978-3-540-72533-6 , pp. 236-237 .
2. ^ Walter Rudin : Functional Analysis. McGraw-Hill, New York 1991, ISBN 0-07-054236-8 , pp. 347-348.
3. Dirk Werner : Functional Analysis . 6th, corrected edition. Springer-Verlag, Berlin 2007, ISBN 978-3-540-72533-6 , Chapter VII.6.
4. Dirk Werner : Functional Analysis . 6th, corrected edition. Springer-Verlag, Berlin 2007, ISBN 978-3-540-72533-6 , pp. 342-347 .
5. Friedrich Hirzebruch , Winfried Scharlau : Introduction to Functional Analysis. (=  BI university paperbacks . Volume 296 ). Bibliographisches Institut, Mannheim [u. a.] 1971, ISBN 3-411-00296-4 , pp. 158-159 .
6. Reinhold Meise, Dietmar Vogt: Introduction to Functional Analysis (=  Vieweg Studium - advanced course in mathematics . Volume 62 ). Vieweg Verlag, Braunschweig [u. a.] 1992, ISBN 3-528-07262-8 , pp. 204 ff .