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Hermiticity and expectation value

  1. Oct 7, 2006 #1
    is there a better way to check for hermicity than doing expecation values? for example, what if you had xp (operators) - px (operators), or pxp (operators again); how can you tell if these combos are hermetian or not, without going through the clumsy integration (that doesn't give a solid result, as far as i can tell)? thanks.
  2. jcsd
  3. Oct 7, 2006 #2
    [tex](AB+CD)^\dagger = B^\dagger A^\dagger + D^\dagger C^\dagger[/tex]
  4. Oct 7, 2006 #3
    heh, thanks. i suck.
  5. Oct 7, 2006 #4
    although, actually, i still have a question.. where does that come from, is it just an accepted property? i can't seem to derive it or find a place where it has been derived.
    Last edited: Oct 7, 2006
  6. Oct 8, 2006 #5
    [tex]\left< (AB)x,y \right> = \left< A(Bx),y \right> = \left< Bx, A^\dagger y \right> = \left< x, (B^\dagger A^\dagger) y \right>[/tex]

    By definition, this shows that [tex](AB)^\dagger=B^\dagger A^\dagger[/tex].
    Last edited: Oct 8, 2006
  7. Oct 9, 2006 #6


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    What Euclid wrote is valid only for bounded operators. The "x" and "p_x" operators are unbounded, so care is needed in order not to write some garbage.

  8. Oct 17, 2006 #7
    However, in the context of what holden wrote, what Euclid stated is valid enough, because holden wants to check for hermicity and hermitian operators are symmetric and bounded.

    On the other note, I want to see a somewhat rigorous proof that
    [tex]\left(A+B\right)^{\dagger} = A^{\dagger}+B^{\dagger} [/tex]

    All the sources say this property, but they don't bother proving it.
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