Orthonormal Basis: Show A is Self-Adjoint

[dyn]

Hi. Been looking at a question and its solution and I'm confused. Question is -

Let $ψ_n$ ,n=1,2,... be an orthonormal basis consisting of eigenstates of a Hamiltonian operator H with non-degenerate eigenvalues $E_n$. Let A be a linear operator which acts on the energy eigenstates $ψ_n$ as
A$ψ_1$=2$ψ_1$ - i $ψ_2$
A$ψ_2$= i $ψ_1$ + 2$ψ_2$
A$ψ_n$ = 0 , n=3,4,5,... Show that A is self-adjoint

The answer is - In the basis $ψ_n$ , A can be written as a matrix having zeroes everywhere except in the left upper 2 x 2 block where A = $$\begin{pmatrix} 2 & i \\ -i & 2 \end{pmatrix}$$
This makes self-adjointness obvious.

What I'm confused about is where this matrix comes from and what does it operate on ?

 

[ShayanJ]

The matrix elements are calculated using:

$A_{mn}=\langle \psi_m|A|\psi_n\rangle=\int_{-\infty}^\infty \psi_m^* A \psi_n dx$

And the matrix acts on column vectors coming from the procedure below:

$|\Phi\rangle=\sum_{s=0}^\infty a_s |\psi_s\rangle \Rightarrow |\Phi\rangle \rightarrow \left( \begin{array} \\ a_1 \\ a_2 \\ a_3 \\ . \\ . \\ . \end{array} \right)$

 

[dyn]

Thanks. In the example I have given ,what is the column vector $ψ_1$ ? Because when matrix A acts on it it also produces a $ψ_2$ term ?

 

[Avodyne]

$\psi_1=\pmatrix{1 \cr 0 },\quad\psi_2=\pmatrix{0 \cr 1}.$