# Quantum mechanics operators

1. Sep 15, 2015

### sayebms

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. Sep 15, 2015

### Orodruin

Staff Emeritus
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. Sep 15, 2015

### sayebms

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. Sep 15, 2015

### Orodruin

Staff Emeritus
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. Sep 15, 2015

### sayebms

should I solve it without the resolution of Identity?