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Quantum mechanics operators

  1. Sep 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose a linear operator L satisfies <A|L|A> = 0 for every state A. Show that then all matrix elements <B|L|A> = 0, and hence L = 0.

    2. Relevant equations
    ##<A|L|A>=L_{AA} and <B|L|A>=L_{BA}##

    3. The attempt at a solution
    It seems very straight forward and I don't know how to prove it but here is what I have tried:
    ##<B|L|A> \to##Using resolution of Identity ##\to \sum_{A} <B|A><A|L|A> \to <B|L|A>=0##
    Is it right or do I need to write more.
     
    Last edited: Sep 15, 2015
  2. jcsd
  3. Sep 15, 2015 #2

    Orodruin

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    It is not correct, you are summing over A and at the same time assuming that ##|A\rangle## is a fixed state (i.e., the one you started with in ##\langle B|L|A\rangle##).
     
  4. Sep 15, 2015 #3
    since the problems says for every state A so should I write as following ##<A_i|L|A_i>=0 \to ## then as before
    ## <B_j|L|A_i>=\sum_{i}<B_j|A_i><A_i|L|A_i>=0##
    is it right now?
     
  5. Sep 15, 2015 #4

    Orodruin

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    No, it is still wrong. You cannot let the state you are summing over in the completeness relation be denoted by the same as the state you have on the left-hand side. It is simply not correct.
     
  6. Sep 15, 2015 #5
    should I solve it without the resolution of Identity?
     
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