Quantum mechanics operators

  • Thread starter sayebms
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  • #1
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Homework Statement


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.

Homework Equations


##<A|L|A>=L_{AA} and <B|L|A>=L_{BA}##

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.
 
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Answers and Replies

  • #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##).
 
  • #3
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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?
 
  • #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.
 
  • #5
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should I solve it without the resolution of Identity?
 

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