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Prove Eigenvalues of an (operator)^2 are real and positive

  1. Oct 13, 2013 #1
    Q: Using Dirac notation, show that if A is an observable associated with the operator A then the eigenvalues of A^2 are real and positive.

    Ans: I know how to prove hermitian operators eigenvalues are real:
    A ket(n) = an ket(n)
    bra(n) A ket(n) = an bra(n) ket(n) = an
    [bra(n) A ket(n)]* = an* bra(n) ket(n) = an*
    [bra(n) A(dager) ket(n)] =[bra(n) A ket(n)] = an*
    therefore, an*=an

    But, I don't know how to treat an operator^2 and start this question.
     
  2. jcsd
  3. Oct 14, 2013 #2

    DrClaude

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    Staff: Mentor

    What is the relation between an observable and Hermetian operators?

    Can you think of a way to rewrite the operator in the expression ##\hat{A}^2 | n \rangle##?
     
  4. Oct 14, 2013 #3
    Relationship: it's a postulate of QM that every dynamical observable is represented by a linear hermitian operator.

    A*A ket(n) = ? but not sure what will be on the right hand side
     
  5. Oct 14, 2013 #4

    DrClaude

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    Staff: Mentor

    Correct.

    Yes, which you can write as
    $$
    \hat{A}^2 | n \rangle = \hat{A} \left( \hat{A} | n \rangle \right)
    $$
    What can you say about the contents of the parenthesis?
     
  6. Oct 14, 2013 #5
    it equals =an ket(n)
    than A(an ket(n)) = an A ket(n) = an (an ket (n))= an^2 ket(n)

    is that correct way to do it?
     
  7. Oct 14, 2013 #6

    DrClaude

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    Staff: Mentor

    Yes. Now you can use the fact that A is Hermitian to complete the answer to the problem.
     
  8. Oct 14, 2013 #7
    Thanks
    So for hermitian A= A(dager), but would u write (A^2)* = A*A* = A*(an* ket(n)) ....
     
  9. Oct 14, 2013 #8

    DrClaude

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    Staff: Mentor

    No, that's not what I meant. You should use the proof you gave in the OP (I don't think you need the put the proof itself in the answer, but rather use what the proof says about Hermetian operators).
     
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