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I What happens to the eigenvalue if an operator acts on a bra?

  1. Mar 23, 2016 #1
    I'm going through a derivation and it shows: (dirac notation)

    <una|VA-AV|unb>=(anb-ana)<una|V|unb>

    V and A are operators that are hermition and commute with each other and ana and anb are the eigenvalues of the operator A. I imagine it is trivial and possibly doesn't even matter but why does the sign flip when the operator acts on the eigenfunctions.

    i.e why is it not <una|VA-AV|unb>=(ana-anb)<una|V|unb>
     
    Last edited: Mar 23, 2016
  2. jcsd
  3. Mar 23, 2016 #2

    blue_leaf77

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    If V and A commute, then ##\langle u_{na}|VA-AV|u_{nb}\rangle=0##.
     
  4. Mar 23, 2016 #3
    Yeh that's part of the proof but that's not what I don't understand. Why do the operations flip the sign of the eigenvalues. I'd have thought <una|VA-AV|unb>=(ana-anb)<una|V|unb>
     
  5. Mar 23, 2016 #4

    blue_leaf77

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    No, the sign is not relevant. If you are referring to the minus sign in front of ##a_{nb}##, that's because of the minus sign in front of ##AV## in ##
    \langle u_{na}|VA-AV|u_{nb}\rangle=0##.
     
  6. Mar 23, 2016 #5
    I don't understand why the signs change at all.
     
  7. Mar 23, 2016 #6

    blue_leaf77

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    Ah I think I see what you mean, do you mean like this ##
    \langle u_{na}|VA-AV|u_{nb}\rangle= \langle u_{na}|VA|u_{nb}\rangle -\langle u_{na}|AV|u_{nb}\rangle = (a_{nb}-a_{na})\langle u_{na}|V|u_{nb}\rangle##?
    Well, I don't know where you get this equation but since V and A commute, you can write ##AV-VA## in place of ##VA-AV##, and you will get ##(a_{na}-a_{nb})##.
     
  8. Mar 23, 2016 #7
    That's exactly it, thank you :). I didn't realise that you could separate the operators out. It's part of the derivation for finding the energy changes in perturbation theory.
     
  9. Mar 23, 2016 #8

    blue_leaf77

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    I could because an inner product, ##\langle u_1|O|u_2\rangle##, is a linear operation.
     
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