# Orthonormal basis

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 = $$\begin{pmatrix} 2 & i \\ -i & 2 \end{pmatrix}$$

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

ShayanJ
Gold Member
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:

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

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}.##

1 person