Eigenstates of Hermitian Operators

Question

In quantum mechanics I learned about Hermitian operators and eigenstates. It seems that Hermitian operators are always mentioned together with eigenstates. Why is that so? Does it mean that a Hermitian operator acting on a wavefunction always give a corresponding eigenvalue and eigenfunction?

Answer 1

There is no specefic reason for Hermitian operator to be mentioned with eigenstates the real importance of Hermitian operators is that their eigenvalues are real and thus they correspond to physical observables. So a study of Q.M with no mention of Hermitian operators is very limited. It is not always the case that a Hermitian operator acting on an eigenfunction always give the eigenvalue. For instance the operator representing linear momentum is Hermitian operator but when applied to atomic wave functions will not give an eigenvalue.

Answer 2

Generally, any linear operator will provide corresponding eigenvalues and eigenvectors (eigenstates), but they may not have the mathematical properties desirable for a physical theory. Hermitian operators are specifically nice to work with and well behaved. For instance, one may define an orthogonality relation via Lagrange's formula (usually already done for most operators), see the Sturm-Liouville chapter of any math physics textbook for more info on this, and Hermitian operators provide eigenvalues that are real-valued, whose eigenvectors are guaranteed to form an orthogonal basis. John Rennie commented this post, which points out that orthogonality of the eigenbasis is the more important feature for defining physical observables.