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Interval of convergence of a linear operator

  1. Nov 6, 2015 #1
    1. The problem statement, all variables and given/known data
    A function of a hermitian operator H can be written as f(H)=Σ (H)n with n=0 to n=∞.
    When is (1-H)-1 defined?

    2. Relevant equations
    (1-x)-1 = Σ(-x)n= 1-x+x2-x3+...

    3. The attempt at a solution
    (1-H)-1 converges if each element of H converges in this series, that is (1-hi)-1 converges with hi being the diagonal elements of the hermitian operator H.
    So Σ(-hi)n should converge. By the ratio test, (-hi)n+1/(-hi)n = |hi|. So it converges if |hi|<1. Is this correct?
     
  2. jcsd
  3. Nov 6, 2015 #2

    andrewkirk

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    A linear operator doesn't have diagonal elements. The matrix of a linear operator with respect to a specified basis has diagonal elements.

    If the basis is the eigenbasis of the operator then I think the statement will be correct. One can see that by expressing the vector to which it is applied as a linear sum of eigenvectors and looking at the effect of the operator on those.

    Whether it is true for matrices of the operator in other bases, I don't know. The trace of a matrix is invariant between bases but AFAIK individual diagonal elements are not.
     
  4. Nov 6, 2015 #3
    Sorry for the lazy mistake, "diagonal elements of the matrix corresponding to the hermitian operator". Also, the basis that I'm considering is the eigenbasis of the hermitian operator. So is it correct to say that the function (1-H)-1 is defined/meaningful if each element hi of H converges for the given series, that is, in this case (1-hi)-1=Σ (-hi)n converges if |hi|<1?
     
  5. Nov 6, 2015 #4

    andrewkirk

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    Yes I think so.
    Let ##B## be the set of all eigenvectors ##e## of ##H##. They form a basis for the vector space so for any vector ##v## we can write ##v=\sum_{e\in B}v_ee##.

    Then $$(1-H)^{-1}v=(1-H)^{-1}\sum_{e\in B} v_ee=
    \sum_{k=0}^\infty (-H)^k\sum_{e\in B} v_ee=
    \sum_{e\in B} v_e\sum_{k=0}^\infty (-H)^ke
    =\sum_{e\in B} v_e\sum_{k=0}^\infty (-\lambda_e)^ke
    =\sum_{e\in B} v_ee\sum_{k=0}^\infty (-\lambda_e)^k$$
    where ##\lambda_e## is the eigenvalue of ##H## corresponding to eigenvector ##e##.

    This converges if the inner sums ##\sum_{k=0}^\infty (-\lambda_e)^k## converge, which will happen where ##|\lambda_e|<1##.
     
    Last edited: Nov 6, 2015
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