I have a question about the matrix representation of quantum operators, in one of the books I'm reading I found this:

Let ${\{\psi_{n}\}}$ be a complete orthonormal system and $\bf A$ a operator. If ${\{\psi_{n}\}}$ form a base for $\bf A$, then: $A_{nm}=a_{n}\delta _{nm}$, $A_{nm}$ is the element of the matrix for $\bf A$, and $a_{n}$ is the eigenvalue corresponding to $\psi_n$ for $\bf A$.

Does this imply that if a set of orthonormal functions forms a base for $\bf A$, that does functions must automatically be eigenfunctions of $\bf A$? Because I know that a base is not necessarily made up of eigenfunctions.

  • $\begingroup$ $\uparrow$ Which book? Which page? $\endgroup$ – Qmechanic Oct 23 '16 at 10:47
  • $\begingroup$ Schwabl: Quantum mechanics , german version, chapter 8 $\endgroup$ – Luka8281 Oct 23 '16 at 10:51

This is not a complete answer but the condition: $A_{nm}=a_{n}\delta _{nm}$ implies that the matrix is diagonalized and from wikipedia:

An $n\times n$ matrix $A$ over the field $F$ is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to $n,$ which is the case if and only if there exists a basis of $F^n$ consisting of eigenvectors of $A$. If such a basis has been found, one can form the matrix $P$ having these basis vectors as columns, and $P^{−1} AP$ will be a diagonal matrix. The diagonal entries of this matrix are the eigenvalues of A.


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