# Hermitian Operator Question

1. Aug 10, 2008

### Domnu

Let us define $$\hat{R} = |\psi_m\rangle \langle \psi_n|$$ where $$\psi_n$$ denotes the $$n$$th eigenstate of some Hermitian operator. When is $$\hat{R}$$ 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?

2. Aug 10, 2008

### Mute

Note that you don't automatically find that <n| = 0 or <m| = 0 when you operate with the other operator. You have to include the possibility that n = m first. So really what you get is

$$\left|n\right> = \delta_{n,m}\left|m\right>$$

and

$$\left|m\right> = \delta_{n,m}\left|n\right>$$

as $\left<m\right|\left.n\right> = \delta_{n,m}$ - so really the solution you found 'by inspection' was actually already considered. I just wanted to make sure you remembered this so that it doesn't slip by you in the future.

As for whether or not that's all the solutions, I don't see anything wrong with it. Even if you consider $\left|n\right> \propto \left|m\right> + \left|l\right>$, for some state $l \neq m$, you would end up with either |l> = 0 or |l> = |m>. So unless I too have missed something I'd say those are your only solutions. (I guess you could consider |n> = |m> = 0 explicity).

Last edited: Aug 10, 2008