Hermitian operator

An Hermitian operator is the physicist's version of an object that mathematicians call a self-adjoint operator. It is a linear operator on a vector space V hat is equipped with positive definite inner product. In physics an inner product is usually notated as a bra and ket, following Dirac. Thus, the inner product of &Phi; and &Psi; is written as,

\langle \Phi | \Psi \rangle \in \mathbb{C},\quad \Phi, \Psi \in V. $$ Physicists use the convention

\langle c\, \Phi | \Psi \rangle = c^* \langle \Phi | \Psi \rangle, \quad \langle \Phi | c \Psi \rangle = c \langle  \Phi | \Psi \rangle, \quad\hbox{and} \quad \langle \Phi | \Psi \rangle = \langle \Psi | \Phi \rangle^*, $$ where $$c^*\,$$ is the complex conjugate of $$c\,$$.

Physicists name the pair of linear operators $$Q,\;Q^\dagger$$  each other's Hermitian adjoint when they are connected through the turn-over rule,

\langle \Phi | Q | \Psi \rangle = \langle Q^\dagger \Phi | \Psi \rangle. $$ This is expressed by the statement: "the operator Q dagger is the Hermitian adjoint of Q". Note that the second upright bar surrounding an operator in a bra-ket does not serve any function, some physicists omit it.

The linear operator Q is called Hermitian if it is equal to its Hermitian adjoint, i.e., if

Q^\dagger = Q ,\, $$ Q is Hermitian.

Except for problems associated with scattering of (nearly) free particles, it is common in physics not to consider domain and range of the adjoint pair. Range and domain are usually taken to be the whole vector space V. Doing this, physicists assume implicitly that V is of finite dimension.

Properties
Hermitian operators have real eigenvalues. Indeed, let

Q |\Phi\rangle = q |\Phi\rangle \quad\Longrightarrow\quad \langle \Phi | Q | \Phi \rangle = q\langle \Phi | \Phi \rangle = \langle Q \Phi | \Phi \rangle = \langle q\Phi | \Phi \rangle = q^* \langle \Phi | \Phi \rangle, $$ from which follows $$ q = q^*\, $$, that is,  the eigenvalue q is real.

Eigenvectors of a Hermitian operator associated with different eigenvalues are orthogonal. Indeed, let

Q |\Phi\rangle = q |\Phi\rangle, \quad Q |\Phi'\rangle = q' |\Phi'\rangle, \quad q\ne q'. $$ Then

\langle \Phi | Q | \Phi' \rangle = q' \langle \Phi | \Phi' \rangle = \langle Q \Phi | \Phi' \rangle = q \langle \Phi | \Phi' \rangle , $$ or, since q - q&prime; &ne; 0,

(q-q') \langle \Phi | \Phi' \rangle = 0 \;\Longrightarrow \; \langle \Phi | \Phi' \rangle = 0, $$ which is expressed by stating that &Phi; and &Phi;&prime; are orthogonal (have zero inner product). In case of finite degeneracy (finite number of linearly independent eigenvectors associated with the same eigenvalue) orthogonality is not assured. However, it is always possible to orthogonalize a finite set of linearly independent set of vectors, i.e., to form linear superpositions such that the new set is orthogonal.

It can be shown that a Hermitian operator on a finite dimensional vector space has as many linearly independent eigenvectors as the dimension of the space. This means that its eigenvectors can serve as a basis of the space. Physicists often assume this to be true for operators on infinite dimensional spaces, but here one should be careful. For instance, the set of normalizable eigenstates of a hydrogen-like Hamilton operator does not span the whole vector space that they belong to.

Commuting Hermitian operators
Two operators are said to commute, if their commutator is the zero operator, i.e., if

\big[ Q, R \big] \equiv QR-RQ = 0. $$ The zero operator maps any vector of the space onto the zero vector.

Commuting operators can have common eigenvectors. A set of vectors can be found that are simultaneously eigenvectors of both Hermitian operators Q and R, if and only if Q and R commute.

To prove this, we let R have n orthogonal eigenvectors $$|\Phi_i\rangle\,$$ with eigenvalue r (the eigenvalue is n-fold degenerate, i = 1, ..., n). These eigenvectors span a linear space called the eigenspace of R belonging to r. (Note that the special case of no degeneracy follows by taking n = 1). Now,

R\left(Q | \Phi_i \rangle\right) = QR | \Phi_i \rangle = Q r  | \Phi_i \rangle = r \left(Q | \Phi_i \rangle\right). $$ If $$|\Phi_i \rangle$$ is an eigenvector of R with eigenvalue r, and R commutes with Q, then $$Q | \Phi_i \rangle$$ is also an eigenvector of R with eigenvalue r. In other words, $$Q | \Phi_i \rangle$$ belongs to the eigenspace associated with r, which is expressed by stating that the eigenspace is stable (invariant) under Q. That is, with any $$|\Psi \rangle$$ in the eigenspace also $$Q | \Psi \rangle$$ belongs to the eigenspace.

(To be continued)