Hermitian operator

In math, the operator $A$ ${\ displaystyle A}$ $A$ in a complex or real Hilbert space ${\mathfrak {H}}$ ${\ displaystyle {\ mathfrak {H}}}$ $\ mathfrak H$ called hermitian , symmetric , if it satisfies the equality $(Ax,y)=(x,Ay)$ ${\ displaystyle (Ax, y) = (x, Ay)}$ $(Ax, y) = (x, Ay)$ for all $x, y$ ${\ displaystyle x, y}$ $x, y$ from the scope $A$ ${\ displaystyle A}$ $A$ . Hereinafter, it is assumed that $(x,y)$ ${\ displaystyle (x, y)}$ $(x, y)$ - scalar product in ${\mathfrak {H}}$ ${\ displaystyle {\ mathfrak {H}}}$ $\ mathfrak H$ . The name is given in honor of the French mathematician Charles Hermite .

Operator in ${\mathfrak {H}}$ ${\ displaystyle {\ mathfrak {H}}}$ $\ mathfrak H$ called self-adjoint , or hypermaximal Hermitian , if it coincides with its conjugate .

The self-adjoint operator is symmetric; the converse, generally speaking, is not true. For continuous operators defined throughout the space, the concepts of symmetric and self-adjoint coincide.

Content

Properties

1. The spectrum (the set of eigenvalues ) of a self-adjoint operator is real .

Evidence

For every eigenvalue $\lambda$ ${\ displaystyle \ lambda}$ true by definition $A\left(x\right)=\lambda x$ ${\ displaystyle A \ left (x \ right) = \ lambda x}$ . Therefore, by the definition of a self-adjoint transformation, the following expressions are equal:

\left({A\left(x\right),x}\right)=\left({\lambda x,x}\right)=\lambda \left({x,x}\right)

{\ displaystyle \ left ({A \ left (x \ right), x} \ right) = \ left ({\ lambda x, x} \ right) = \ lambda \ left ({x, x} \ right)}

and

\left({x,A\left(x\right)}\right)=\left({x,\lambda x}\right)={\overline {\lambda }}\left({x,x}\right)

{\ displaystyle \ left ({x, A \ left (x \ right)} \ right) = \ left ({x, \ lambda x} \ right) = {\ overline {\ lambda}} \ left ({x, x} \ right)}

,

where from $\lambda ={\overline {\lambda }}$ ${\ displaystyle \ lambda = {\ overline {\ lambda}}}$ - number $\lambda$ ${\ displaystyle \ lambda}$ real.

2. In unitary finite-dimensional spaces, the matrix of a self-adjoint operator is Hermitian . (In particular, in Euclidean space, the matrix of a self-adjoint operator is symmetric.)

Evidence

In unitary space, the scalar product is defined as $\left({x,y}\right)=\xi ^{T}{\overline {\eta }}$ ${\ displaystyle \ left ({x, y} \ right) = \ xi ^ {T} {\ overline {\ eta}}}$ where $\xi$ ${\ displaystyle \ xi}$ and $\eta$ ${\ displaystyle \ eta}$ - coordinate columns of vectors $x$ ${\ displaystyle x}$ and $y$ ${\ displaystyle y}$ respectively. Hence, by the definition of a self-adjoint operator, the expressions are equal

\left({A\left(x\right),y}\right)=\left({A\xi }\right)^{T}{\overline {\eta }}=\xi ^{T}A^{T}{\overline {\eta }}

{\ displaystyle \ left ({A \ left (x \ right), y} \ right) = \ left ({A \ xi} \ right) ^ {T} {\ overline {\ eta}} = \ xi ^ { T} A ^ {T} {\ overline {\ eta}}}

and

\left({x,A\left(y\right)}\right)=\xi ^{T}{\overline {A\eta }}=\xi ^{T}{\overline {A}}{\overline {\eta }}

{\ displaystyle \ left ({x, A \ left (y \ right)} \ right) = \ xi ^ {T} {\ overline {A \ eta}} = \ xi ^ {T} {\ overline {A} } {\ overline {\ eta}}}

Consequently, $A^{T}={\overline {A}}$ ${\ displaystyle A ^ {T} = {\ overline {A}}}$ , which is the definition of a Hermitian matrix.

3. A Hermitian matrix always has an orthonormal basis of eigenvectors — eigenvectors corresponding to different eigenvalues are orthogonal.

Evidence

Lemma 1. The eigensubspaces of a self-adjoint transformation are pairwise orthogonal.

Proof of Lemma 1: There are two different eigenvalues.

\lambda

{\ displaystyle \ lambda}

and

\mu

{\ displaystyle \ mu}

. Accordingly, for vectors

x

{\ displaystyle x}

and

y

{\ displaystyle y}

of the corresponding own subspaces

A\left(x\right)=\lambda x

{\ displaystyle A \ left (x \ right) = \ lambda x}

and

A\left(y\right)=\mu y

{\ displaystyle A \ left (y \ right) = \ mu y}

. From here

\left({A\left(x\right),y}\right)=\left({\lambda x,y}\right)=\lambda \left({x,y}\right)

{\ displaystyle \ left ({A \ left (x \ right), y} \ right) = \ left ({\ lambda x, y} \ right) = \ lambda \ left ({x, y} \ right)}

equally

\left({x,A\left(y\right)}\right)=\left({x,\mu y}\right)={\overline {\mu }}\left({x,y}\right)

{\ displaystyle \ left ({x, A \ left (y \ right)} \ right) = \ left ({x, \ mu y} \ right) = {\ overline {\ mu}} \ left ({x, y} \ right)}

. But the eigenvalues of the self-adjoint transformation are real; we can derive from the last expression

\left({x,A\left(y\right)}\right)=\mu \left({x,y}\right)

{\ displaystyle \ left ({x, A \ left (y \ right)} \ right) = \ mu \ left ({x, y} \ right)}

. Thus, by the definition of a self-adjoint transformation, we can obtain

\left({\lambda -\mu }\right)\left({x,y}\right)=0

{\ displaystyle \ left ({\ lambda - \ mu} \ right) \ left ({x, y} \ right) = 0}

, whence with the difference in eigenvalues

\lambda \neq \mu

{\ displaystyle \ lambda \ neq \ mu}

it's clear that

\left({x,y}\right)=0

{\ displaystyle \ left ({x, y} \ right) = 0}

, as required.

Lemma 2. If the subspace

E^{'}

{\ displaystyle E '}

invariant under the self-adjoint transformation

A

{\ displaystyle A}

, then the orthogonal complement of this subspace is also invariant under

A

{\ displaystyle A}

.

Proof of Lemma 2: It is known that the image of any vector

x

{\ displaystyle x}

owned by subspace

E^{'}

{\ displaystyle E '}

lies in it. Therefore, for any vector

y\in \left({E'}\right)^{\bot }

{\ displaystyle y \ in \ left ({E '} \ right) ^ {\ bot}}

performed

\left({A\left(x\right),y}\right)=0

{\ displaystyle \ left ({A \ left (x \ right), y} \ right) = 0}

. Since the conversion

A

{\ displaystyle A}

self-adjoint, it follows that

\left({x,A\left(y\right)}\right)=0

{\ displaystyle \ left ({x, A \ left (y \ right)} \ right) = 0}

, that is, the image of any vector

y

{\ displaystyle y}

of

\left({E'}\right)^{\bot }

{\ displaystyle \ left ({E '} \ right) ^ {\ bot}}

belongs

\left({E'}\right)^{\bot }

{\ displaystyle \ left ({E '} \ right) ^ {\ bot}}

, which means that the subspace

\left({E'}\right)^{\bot }

{\ displaystyle \ left ({E '} \ right) ^ {\ bot}}

is invariant under the transformation A, as required.

Proof of Property 3:

For an operator R in n-dimensional space, there is at least one eigenvalue

\lambda _{1}

{\ displaystyle \ lambda _ {1}}

. By property 1, this eigenvalue is real. One can find the eigenvector e ₁ corresponding to it. Without loss of generality, we can assume that

|e_{1}|=1

{\ displaystyle | e_ {1} | = 1}

. If n = 1, then the proof is complete.

Consider E ₁ , the linear span of the element e ₁ , which is a one-dimensional invariant eigenspace of R. Let E ^n-1 be the orthogonal complement of E ₁ . Then, by Lemma 2, E ^{n-1 is} invariant under the operator in question. We now consider it as R ', as acting only in E ^n-1 . Then it is obvious that it will be a self-adjoint operator defined in E ^n-1 , since E ^{n-1 is} invariant with respect to R by Lemma 2 and, moreover, for

\forall

{\ displaystyle \ forall}

x, y

\in

{\ displaystyle \ in}

E ⁿ : (Rx, y) = (x, Ry), including for

\forall

{\ displaystyle \ forall}

x, y

\in

{\ displaystyle \ in}

E ^n-1 .

Applying the above reasoning, we find a new eigenvalue

\lambda _{2}

{\ displaystyle \ lambda _ {2}}

and its corresponding eigenvector

e_{2}

{\ displaystyle e_ {2}}

. Without loss of generality, we can assume that

|e_{2}|=1

{\ displaystyle | e_ {2} | = 1}

. Wherein

\lambda _{2}

{\ displaystyle \ lambda _ {2}}

may coincide with

\lambda _{1}

{\ displaystyle \ lambda _ {1}}

However, it is clear from the construction that

(e_{1},e_{2})=0

{\ displaystyle (e_ {1}, e_ {2}) = 0}

. If n = 2, then the proof is complete. Otherwise, consider E, the linear shell

{(e_{1},e_{2})}

{\ displaystyle {(e_ {1}, e_ {2})}}

and its orthogonal complement E ^n-2 . Find a new eigenvalue

\lambda _{3}

{\ displaystyle \ lambda _ {3}}

and its corresponding eigenvector

e_{3}

{\ displaystyle e_ {3}}

etc.

We carry out similar considerations until exhaustion of E ⁿ .

The proof is complete.

Matrices

The matrix Hermitian conjugate to a given is called the matrix $A^{\dagger },$ ${\ displaystyle A ^ {\ dagger},}$ obtained from the original matrix $A$ ${\ displaystyle A}$ by transposing it and moving to a complex conjugate, i.e. $(A^{\dagger })_{ij}=A_{ji}^{*}$ ${\ displaystyle (A ^ {\ dagger}) _ {ij} = A_ {ji} ^ {*}}$ . This is a natural definition: if we write a linear mapping and a Hermitian conjugate operator in any basis in the form of matrices, then their matrices will be Hermitian conjugate. A matrix equal to its Hermitian conjugation is called Hermitian, or self-adjoint: for it $A^{\dagger }=A$ ${\ displaystyle A ^ {\ dagger} = A}$ .

Application

Hermitian operators play an important role in quantum mechanics , where they represent the observable physical quantities, see Heisenberg Uncertainty Principle .

Hermitian operator

Content

Properties

Matrices

Application

See also

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