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Homework Help: Proof about hermitian operators

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data

    This is something ive been trying to prove for a bit today. My quantum mechanics book claims that the following two definitions about hermitian operators are completely equivalent

    my operator here is Q (with a hat) and we have functions f,g

    [tex] \langle f \mid \hat Q f \rangle = \langle Q f \mid \hat f \rangle [/tex] for any function f in hilbert space


    [tex] \langle f \mid \hat Q g \rangle = \langle Q f \mid \hat g \rangle [/tex] for any functions f,g in hilbert space

    2. Related formulas

    [tex] \langle f \mid g \rangle = \int f^{\ast} g dx [/tex]

    3. The attempt at a solution

    Clearly the second definition implies the first, but I'm having trouble showing that the first implies the second.
    My quantum mechanics book has this as an exercise and as a hint it suggests to let f=g+h
    and then let f=g+ih with i being the square root of -1. I have done both of these things, expanding the inner products in terms of integrals. If i assume the first definition and let f=g+h, i can get

    [tex] \langle f \mid \hat Q g \rangle + \langle g \mid \hat Q f \rangle = \langle \hat Q f \mid g \rangle + \langle \hat Q g \mid f \rangle [/tex]

    doing a similar thing with f=g+ih i get the same result, and not sure where to go from here. Anybody have a better way to prove it or any ideas? thanks
    Last edited: Jan 21, 2012
  2. jcsd
  3. Jan 21, 2012 #2


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    I don't see any definitions in your post.
  4. Jan 21, 2012 #3
    Ah Sorry about that. I've just edited my post and corrected that
  5. Jan 21, 2012 #4


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    Start with (Qf,f)=(f,Qf). If you take f=g+h and simplify, you get (g,Qh) + (h,Qh) = (Qg,h) + (Qh,g). And if you take f=g-ih and simplify, you get (g,Qh) - (h,Qg) = (Qg,h) - (Qh,g).
  6. Jan 21, 2012 #5
    isn't it just that

    Q=Q^dagger where dagger is the transpose and complex conjugate

    for a matrix to be hermitian

    such that <Q| = |Q>
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