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Need help! Inverse of an operator

  1. Oct 26, 2003 #1
    Another quantum question.....I feel dumb :frown:.

    A Hermitian operator A has the spectral decomposition

    A = &Sigma; an|n><n| (summation in n)

    where A|n> = an|n> (the an's are eigenvalues of A, and |n>'s the eigenstates).

    So, how can I find the spectral decomposition of the inverse of A so that
    AA-1 = A-1A = 1?


    My intuition would be A = &Sigma; (1/an)|n><n| (summation in n), since 1 = &Sigma; |k><k| (summation in k), but I don't think it's that easy.

    Thanks in advance!
     
  2. jcsd
  3. Oct 26, 2003 #2

    Hurkyl

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    Try multiplying the given sum for A and your sum for A-1 and see if you get 1 as the answer.
     
  4. Oct 26, 2003 #3
    Thanks. I have a question about multiplying the 2 sums though. Should I make n --> n' for one of them, then sum over n and n'? Like this:

    A = &Sigma; an|n><n| (summation in n)
    A-1 = &Sigma; (1/an')|n'><n'| (summation in n')

    then AA-1 = &Sigma;&Sigma; an(1/an') |n><n|n'><n'| (summation in n and n') ?

    And what is the condition imposed on A so that the inverse exists? I have no idea...
     
    Last edited: Oct 26, 2003
  5. Oct 26, 2003 #4

    Hurkyl

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    Right, that's how you do the summation. (it will simplify to 1)

    As for the condition imposed on A, your formula for A-1 contains a strong hint as to what that condition might me...
     
  6. Oct 26, 2003 #5
    Thank you!

    *scratches head*

    Hmm......is it that A and A-1 must have the same eigenstates, so that <n|n'> gives a delta function? And eigenvalues of A must be none zero otherwise the term in the inverse diverges?
     
  7. Oct 26, 2003 #6

    Hurkyl

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    It is true that A and A-1 have the same eigenstates. For example, if |1> is an eigenstate of A with nonzero eigenvalue &lambda;, then:


    |1> = A-1A|1> = A-1&lambda;|1>
    and thus A-1|1> = &lambda;-1|1>

    So any eigenstate of A is an eigenstate of A-1 (and vice-versa, by symmetry).


    More importantly, we can always choose an orthonormal eigenbasis, in which <n|n'> would indeed be a delta function. (I would presume the basis used by spectral decomposition would be orthonormal, but don't quote me on that!)


    And yes, the right condition here is that all of the eigenvalues of A be nonzero. However, there is still work to do! At the moment, your formula only proves that if all of the eigenvalues of A are nonzero then there exists an inverse! You still need to prove that if A has a zero eigenvalue then it cannot have an inverse.
     
  8. Oct 26, 2003 #7
    That's a very good explanation, thank you.

    Yes, they're orthonormal basis since it's the summation over the eigenstates.

    Yikes, I'm not too sure if I understand this part... I mean, if A has a zero eigenvalue, then its inverse would have an eigenstate with eigenvalue 1/0 which diverges....couldn't that prove A cannot have an inverse?
     
  9. Oct 26, 2003 #8

    Hurkyl

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    I'm not sure if you're off the right track, or if you're on the right track but are missing a detail or two... so I'll give an example of a wrong proof and a right proof. :smile:


    Wrong proof:

    A-1 is given by that formula, and if we plug in a zero for one of the eigenvalues, the sum diverges, so A-1 doesn't exist.

    This is wrong because you haven't proved that A-1 must have the form given by your formula; you can only prove that the formula works when the eigenvalues are nonzero.

    It might be the case that there's another formula that will give you the inverse when one of the eigenvalues is zero. (It turns out that this is not the case, but from just the information I mentioned above, we can't determine this fact!)

    Right proof:

    Suppose |1> is an eigenstate of A with eigenvalue 0. Then:

    |1> = A-1A|1> = A-10|1> = A-1 0 = 0

    But since |1> is not the zero vector, we have a contradiction in assuming A-1 exists.


    This is kinda like what you were saying, because we have:

    |1> = 0 A-1 |1>

    Implying A-1 has an "infinite" eigenvalue which is impossible... I just wasn't sure if you were aiming at this ponit or not.
     
  10. Oct 26, 2003 #9
    Oh, I see! Now I understand... I've always been bad at proving things :smile:. Thank you again for your help!
     
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