Inmathematics, specifically inoperator theory, eachlinear operator${\displaystyle A}$ on aninner product space defines aHermitian adjoint (oradjoint) operator${\displaystyle A^{\ast }}$ on that space according to the rule

${\displaystyle ⟨Ax,y⟩=⟨x,A^{\ast }y⟩,}$

where${\displaystyle ⟨\cdot ,\cdot ⟩}$ is theinner product on thevector space.

The adjoint may also be called theHermitian conjugate or simply theHermitian^[1] afterCharles Hermite. It is often denoted byA^† in fields likephysics, especially when used in conjunction withbra–ket notation inquantum mechanics. Infinite dimensions where operators can be represented bymatrices, the Hermitian adjoint is given by theconjugate transpose (also known as the Hermitian transpose).

The above definition of an adjoint operator extends verbatim tobounded linear operators onHilbert spaces${\displaystyle H}$. The definition has been further extended to include unboundeddensely defined operators, whose domain is topologicallydense in, but not necessarily equal to,${\displaystyle H.}$

Consider alinear map${\displaystyle A:H_1\to H_2}$ betweenHilbert spaces. Without taking care of any details, the adjoint operator is the (in most cases uniquely defined) linear operator${\displaystyle A^{\ast }:H_2\to H_1}$ fulfilling

${\displaystyle {⟨A{h}_1,h_2⟩}_{H_2}={⟨h_1,A^{\ast }{h}_2⟩}_{H_1},}$

where${\displaystyle ⟨\cdot ,\cdot {⟩}_{H_i}}$ is theinner product in the Hilbert space${\displaystyle H_i}$, which is linear in the first coordinate andconjugate linear in the second coordinate. Note the special case where both Hilbert spaces are identical and${\displaystyle A}$ is an operator on that Hilbert space.

When one trades the inner product for thedual pairing, one can define the adjoint, also called thetranspose, of an operator${\displaystyle A:E\to F}$, where${\displaystyle E,F}$ areBanach spaces with correspondingnorms${\displaystyle \Vert \cdot {\Vert }_E,\Vert \cdot {\Vert }_F}$. Here (again not considering any technicalities), its adjoint operator is defined as${\displaystyle A^{\ast }:F^{\ast }\to E^{\ast }}$ with

${\displaystyle A^{\ast }f=f\circ A:u\mapsto f(Au),}$

i.e.,${\displaystyle \left(A^{\ast }f\right)(u)=f(Au)}$ for${\displaystyle f\in F^{\ast },u\in E}$.

The above definition in the Hilbert space setting is really just an application of the Banach space case when one identifies a Hilbert space with its dual (via theRiesz representation theorem). Then it is only natural that we can also obtain the adjoint of an operator${\displaystyle A:H\to E}$, where${\displaystyle H}$ is a Hilbert space and${\displaystyle E}$ is a Banach space. The dual is then defined as${\displaystyle A^{\ast }:E^{\ast }\to H}$ with${\displaystyle A^{\ast }f=h_f}$ such that

${\displaystyle ⟨h_f,h{⟩}_H=f(Ah).}$

Let${\displaystyle \left(E,\Vert \cdot {\Vert }_E\right),\left(F,\Vert \cdot {\Vert }_F\right)}$ beBanach spaces. Suppose${\displaystyle A:D(A)\to F}$ and${\displaystyle D(A)\subset E}$, and suppose that${\displaystyle A}$ is a (possibly unbounded) linear operator which isdensely defined (i.e.,${\displaystyle D(A)}$ is dense in${\displaystyle E}$). Then its adjoint operator${\displaystyle A^{\ast }}$ is defined as follows. The domain is

${\displaystyle D\left(A^{\ast }\right):=\left\{g\in F^{\ast }:\mathrm{\exists }c\ge 0:\text{ for all }u\in D(A):|g(Au)|\le c\cdot \Vert u{\Vert }_E\right\}.}$

Now for arbitrary but fixed${\displaystyle g\in D(A^{\ast })}$ we set${\displaystyle f:D(A)\to \mathbb{R}}$ with${\displaystyle f(u)=g(Au)}$. By choice of${\displaystyle g}$ and definition of${\displaystyle D(A^{\ast })}$, f is (uniformly) continuous on${\displaystyle D(A)}$ as${\displaystyle |f(u)|=|g(Au)|\le c\cdot \Vert u{\Vert }_E}$. Then by theHahn–Banach theorem, or alternatively through extension by continuity, this yields an extension of${\displaystyle f}$, called${\displaystyle \hat{f}}$, defined on all of${\displaystyle E}$. This technicality is necessary to later obtain${\displaystyle A^{\ast }}$ as an operator${\displaystyle D\left(A^{\ast }\right)\to E^{\ast }}$ instead of${\displaystyle D\left(A^{\ast }\right)\to (D(A){)}^{\ast }.}$ Remark also that this does not mean that${\displaystyle A}$ can be extended on all of${\displaystyle E}$ but the extension only worked for specific elements${\displaystyle g\in D\left(A^{\ast }\right)}$.

Now, we can define the adjoint of${\displaystyle A}$ as

The fundamental defining identity is thus

${\displaystyle g(Au)=\left(A^{\ast }g\right)(u)}$ for${\displaystyle u\in D(A).}$

SupposeH is a complexHilbert space, withinner product${\displaystyle ⟨\cdot ,\cdot ⟩}$. Consider acontinuouslinear operatorA :H →H (for linear operators, continuity is equivalent to being abounded operator). Then the adjoint ofA is the continuous linear operatorA^∗ :H →H satisfying

${\displaystyle ⟨Ax,y⟩=⟨x,A^{\ast }y⟩\text{for all }x,y\in H.}$

Existence and uniqueness of this operator follows from theRiesz representation theorem.^[2]

This can be seen as a generalization of theadjoint matrix of a square matrix which has a similar property involving the standard complex inner product.

The following properties of the Hermitian adjoint ofbounded operators are immediate:^[2]

1. Involutivity:A^∗∗ =A
2. IfA is invertible, then so isA^∗, with${\left(A^{\ast }\right)}^{-1}={\left(A^{-1}\right)}^{\ast }$
3. Conjugate linearity:
   • (A +B)^∗ =A^∗ +B^∗
   • (λA)^∗ =λA^∗, whereλ denotes thecomplex conjugate of thecomplex numberλ
4. "Anti-distributivity":(AB)^∗ =B^∗A^∗

If we define theoperator norm ofA by

${\displaystyle \Vert A{\Vert }_{\text{op}}:=sup\left\{\Vert Ax\Vert :\Vert x\Vert \le 1\right\}}$

then

${\displaystyle {\Vert A^{\ast }\Vert }_{\text{op}}=\Vert A{\Vert }_{\text{op}}.}$^[2]

Moreover,

${\displaystyle {\Vert A^{\ast }A\Vert }_{\text{op}}=\Vert A{\Vert }_{\text{op}}^2.}$^[2]

One says that a norm that satisfies this condition behaves like a "largest value", extrapolating from the case of self-adjoint operators.

The set of bounded linear operators on a complex Hilbert spaceH together with the adjoint operation and the operator norm form the prototype of aC*-algebra.

Let the inner product${\displaystyle ⟨\cdot ,\cdot ⟩}$ be linear in thefirst argument. Adensely defined operatorA from a complex Hilbert spaceH to itself is a linear operator whose domainD(A) is a denselinear subspace ofH and whose values lie inH.^[3] By definition, the domainD(A^∗) of its adjointA^∗ is the set of ally ∈H for which there is az ∈H satisfying

${\displaystyle ⟨Ax,y⟩=⟨x,z⟩\text{for all }x\in D(A).}$

Owing to the density of${\displaystyle D(A)}$ andRiesz representation theorem,${\displaystyle z}$ is uniquely defined, and, by definition,${\displaystyle A^{\ast }y=z.}$^[4]

Properties 1.–5. hold with appropriate clauses aboutdomains andcodomains.^{[clarification needed]} For instance, the last property now states that(AB)^∗ is an extension ofB^∗A^∗ ifA,B andAB are densely defined operators.^[5]

For every${\displaystyle y\in \ker A^{\ast },}$ the linear functional${\displaystyle x\mapsto ⟨Ax,y⟩=⟨x,A^{\ast }y⟩}$ is identically zero, and hence${\displaystyle y\in (\mathrm{im}A{)}^{\perp }.}$

Conversely, the assumption that${\displaystyle y\in (\mathrm{im}A{)}^{\perp }}$ causes the functional${\displaystyle x\mapsto ⟨Ax,y⟩}$ to be identically zero. Since the functional is obviously bounded, the definition of${\displaystyle A^{\ast }}$ assures that${\displaystyle y\in D(A^{\ast }).}$ The fact that, for every${\displaystyle x\in D(A),}$${\displaystyle ⟨Ax,y⟩=⟨x,A^{\ast }y⟩=0}$ shows that${\displaystyle A^{\ast }y\in D(A{)}^{\perp }={\overline{D(A)}}^{\perp }=\{0\},}$ given that${\displaystyle D(A)}$ is dense.

This property shows that${\displaystyle \ker A^{\ast }}$ is a topologically closed subspace even when${\displaystyle D(A^{\ast })}$ is not.

If${\displaystyle H_1}$ and${\displaystyle H_2}$ are Hilbert spaces, then${\displaystyle H_1\oplus H_2}$ is a Hilbert space with the inner product

${\displaystyle ⟨(a,b),(c,d){⟩}_{H_1\oplus H_2}\stackrel{\text{def}}{=}⟨a,c{⟩}_{H_1}+⟨b,d{⟩}_{H_2},}$

where${\displaystyle a,c\in H_1}$ and${\displaystyle b,d\in H_2.}$

Let${\displaystyle J:H\oplus H\to H\oplus H}$ be thesymplectic mapping, i.e.${\displaystyle J(\xi ,\eta )=(-\eta ,\xi ).}$ Then the graph

${\displaystyle G(A^{\ast })=\{(x,y)\mid x\in D(A^{\ast }),y=A^{\ast }x\}\subseteq H\oplus H}$

of${\displaystyle A^{\ast }}$ is theorthogonal complement of${\displaystyle JG(A):}$

${\displaystyle G(A^{\ast })=(JG(A){)}^{\perp }=\{(x,y)\in H\oplus H:⟨(x,y),(-A\xi ,\xi ){⟩}_{H\oplus H}=0\mathrm{\forall }\xi \in D(A)\}.}$

The assertion follows from the equivalences

${\displaystyle ⟨(x,y),(-A\xi ,\xi )⟩=0\iff ⟨A\xi ,x⟩=⟨\xi ,y⟩,}$

and

An operator${\displaystyle A}$ isclosed if the graph${\displaystyle G(A)}$ is topologically closed in${\displaystyle H\oplus H.}$ The graph${\displaystyle G(A^{\ast })}$ of the adjoint operator${\displaystyle A^{\ast }}$ is the orthogonal complement of a subspace, and therefore is closed.

An operator${\displaystyle A}$ isclosable if the topological closure${\displaystyle G^{\text{cl}}(A)\subseteq H\oplus H}$ of the graph${\displaystyle G(A)}$ is the graph of a function. Since${\displaystyle G^{\text{cl}}(A)}$ is a (closed) linear subspace, the word "function" may be replaced with "linear operator". For the same reason,${\displaystyle A}$ is closable if and only if${\displaystyle (0,v)\notin G^{\text{cl}}(A)}$ unless${\displaystyle v=0.}$

The adjoint${\displaystyle A^{\ast }}$ is densely defined if and only if${\displaystyle A}$ is closable. This follows from the fact that, for every${\displaystyle v\in H,}$

${\displaystyle v\in D(A^{\ast }{)}^{\perp }\iff (0,v)\in G^{\text{cl}}(A),}$

which, in turn, is proven through the following chain of equivalencies:

Theclosure${\displaystyle A^{\text{cl}}}$ of an operator${\displaystyle A}$ is the operator whose graph is${\displaystyle G^{\text{cl}}(A)}$ if this graph represents a function. As above, the word "function" may be replaced with "operator". Furthermore,${\displaystyle A^{\ast \ast }=A^{\text{cl}},}$ meaning that${\displaystyle G(A^{\ast \ast })=G^{\text{cl}}(A).}$

To prove this, observe that${\displaystyle J^{\ast }=-J,}$ i.e.${\displaystyle ⟨Jx,y{⟩}_{H\oplus H}=-⟨x,Jy{⟩}_{H\oplus H},}$ for every${\displaystyle x,y\in H\oplus H.}$ Indeed,

In particular, for every${\displaystyle y\in H\oplus H}$ and every subspace${\displaystyle V\subseteq H\oplus H,}$${\displaystyle y\in (JV{)}^{\perp }}$ if and only if${\displaystyle Jy\in V^{\perp }.}$ Thus,${\displaystyle J[(JV{)}^{\perp }]=V^{\perp }}$ and${\displaystyle [J[(JV{)}^{\perp }]{]}^{\perp }=V^{\text{cl}}.}$ Substituting${\displaystyle V=G(A),}$ obtain${\displaystyle G^{\text{cl}}(A)=G(A^{\ast \ast }).}$

For a closable operator${\displaystyle A,}$${\displaystyle A^{\ast }={\left(A^{\text{cl}}\right)}^{\ast },}$ meaning that${\displaystyle G(A^{\ast })=G\left({\left(A^{\text{cl}}\right)}^{\ast }\right).}$ Indeed,

${\displaystyle G\left({\left(A^{\text{cl}}\right)}^{\ast }\right)={\left(J{G}^{\text{cl}}(A)\right)}^{\perp }={\left({\left(JG(A)\right)}^{\text{cl}}\right)}^{\perp }=(JG(A){)}^{\perp }=G(A^{\ast }).}$

Let${\displaystyle H=L^2(\mathbb{R},l),}$ where${\displaystyle l}$ is the linear measure. Select a measurable, bounded, non-identically zero function${\displaystyle f\notin L^2,}$ and pick${\displaystyle {\phi }_0\in L^2\setminus \{0\}.}$ Define

${\displaystyle A\phi =⟨f,\phi ⟩{\phi }_0.}$

It follows that${\displaystyle D(A)=\{\phi \in L^2\mid ⟨f,\phi ⟩\ne \mathrm{\infty }\}.}$ The subspace${\displaystyle D(A)}$ contains all the${\displaystyle L^2}$ functions with compact support. Since${\displaystyle {\mathbf{1}}_{[-n,n]}\cdot \phi \stackrel{L^2}{\to }\phi ,}$${\displaystyle A}$ is densely defined. For every${\displaystyle \phi \in D(A)}$ and${\displaystyle \psi \in D(A^{\ast }),}$

${\displaystyle ⟨\phi ,A^{\ast }\psi ⟩=⟨A\phi ,\psi ⟩=⟨⟨f,\phi ⟩{\phi }_0,\psi ⟩=⟨f,\phi ⟩\cdot ⟨{\phi }_0,\psi ⟩=⟨\phi ,⟨{\phi }_0,\psi ⟩f⟩.}$

Thus,${\displaystyle A^{\ast }\psi =⟨{\phi }_0,\psi ⟩f.}$ The definition of adjoint operator requires that${\displaystyle \text{Im}{A}^{\ast }\subseteq H=L^2.}$ Since${\displaystyle f\notin L^2,}$ this is only possible if${\displaystyle ⟨{\phi }_0,\psi ⟩=0.}$ For this reason,${\displaystyle D(A^{\ast })=\{{\phi }_0{\}}^{\perp }.}$ Hence,${\displaystyle A^{\ast }}$ is not densely defined and is identically zero on${\displaystyle D(A^{\ast }).}$ As a result,${\displaystyle A}$ is not closable and has no second adjoint${\displaystyle A^{\ast \ast }.}$

Abounded operatorA :H →H is called Hermitian orself-adjoint if

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

which is equivalent to

${\displaystyle ⟨Ax,y⟩=⟨x,Ay⟩\text{ for all }x,y\in H.}$^[6]

In some sense, these operators play the role of thereal numbers (being equal to their own "complex conjugate") and form a realvector space. They serve as the model of real-valuedobservables inquantum mechanics. See the article onself-adjoint operators for a full treatment.

For aconjugate-linear operator the definition of adjoint needs to be adjusted in order to compensate for the complex conjugation. An adjoint operator of the conjugate-linear operatorA on a complex Hilbert spaceH is an conjugate-linear operatorA^∗ :H →H with the property:

${\displaystyle ⟨Ax,y⟩=\overline{⟨x,A^{\ast }y⟩}\text{for all }x,y\in H.}$

The equation

${\displaystyle ⟨Ax,y⟩=⟨x,A^{\ast }y⟩}$

is formally similar to the defining properties of pairs ofadjoint functors incategory theory, and this is where adjoint functors got their name.

• Mathematical concepts
  • Conjugate transpose
  • Hermitian operator
  • Pullback § Functional analysis
  • Transpose of linear maps
• Physical applications
  • Operator (physics)
  • †-algebra

• Brezis, Haim (2011),Functional Analysis, Sobolev Spaces and Partial Differential Equations (first ed.), Springer,ISBN 978-0-387-70913-0.
• Reed, Michael; Simon, Barry (2003),Functional Analysis, Elsevier,ISBN 981-4141-65-8.
• Rudin, Walter (1991).Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY:McGraw-Hill Science/Engineering/Math.ISBN 978-0-07-054236-5.OCLC 21163277.

• Operator theory

• Articles with short description
• Short description is different from Wikidata
• Wikipedia articles needing clarification from May 2015