When are the solutions for \hat{R} being Hermitian?

[Domnu]

Let us define \hat{R} = |\psi_m\rangle \langle \psi_n| where \psi_n denotes the nth 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?

 

[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&gt; = \delta_{n,m}\left|m\right&gt;

and

\left|m\right&gt; = \delta_{n,m}\left|n\right&gt;

as \left&lt;m\right|\left.n\right&gt; = \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&gt; \propto \left|m\right&gt; + \left|l\right&gt;, 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).