Re: Diagonality of an Observable Operator

On 2005-05-30, Ali <ph_question@xxxxxxxxx> wrote:
> Dear Members,
>
> why the square matrix representation of an observable operator A is
> diagonal?
> cheers,

An operator is a map from one vector space into another one, or from the
same vector space into itself. In the latter case, with a choice of
basis, the operator can be represented by a square matrix. For different
choices of basis, the matrix representing A may or may not be diagonal.

Observables in quantum mechanics are assumed to be self adjoint
(hermitian) operators. The spectral theorem for self adjoint operators
says that for any such operator there exists an orthonormal basis in
which the matrix representation of the operator is diagonal.

If this is news to you. I suggest supplementing Sakurai with a good book
on linear algebra.

Igor

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