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Hermitian Operator Question

  1. Aug 10, 2008 #1
    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?

    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. jcsd
  3. Aug 10, 2008 #2


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    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

    [tex]\left|n\right> = \delta_{n,m}\left|m\right>[/tex]


    [tex]\left|m\right> = \delta_{n,m}\left|n\right>[/tex]

    as [itex]\left<m\right|\left.n\right> = \delta_{n,m}[/itex] - 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 [itex]\left|n\right> \propto \left|m\right> + \left|l\right>[/itex], for some state [itex]l \neq m[/itex], 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
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