- #1
Domnu
- 178
- 0
Let us define [tex]\hat{R} = |\psi_m\rangle \langle \psi_n|[/tex] where [tex]\psi_n[/tex] denotes the [tex]n[/tex]th eigenstate of some Hermitian operator. When is [tex]\hat{R}[/tex] Hermitian?
Solution?
Well, let us just call |psi_m> = |m> and |psi_n> = |n>. Now, we need
|m><n| = |n><m|
If we left multiply by <m| then we find that
<n| = 0
By symmetry, if we left multiply by <n| we find that
<m| = 0
But, clearly, by inspection, we find that R is Hermitian if |m> = |n>. Are these all the solutions?
Solution?
Well, let us just call |psi_m> = |m> and |psi_n> = |n>. Now, we need
|m><n| = |n><m|
If we left multiply by <m| then we find that
<n| = 0
By symmetry, if we left multiply by <n| we find that
<m| = 0
But, clearly, by inspection, we find that R is Hermitian if |m> = |n>. Are these all the solutions?