- #1
dyn
- 773
- 62
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 ?
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 ?