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Operator with strictly positive eigenvalues

  1. Oct 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider a Hilbert space with a (not necessarily orthogonal) basis [tex]\{f_i\}[/tex] Show that [tex]G=\sum_i |f_i\rangle\langle f_i|[/tex] has strictly positive eigenvalues.


    2. Relevant equations



    3. The attempt at a solution

    I know that [tex]G=\sum_i |f_i\rangle\langle f_i|[/tex] is hermitian. Therefore, the eigenvalues are real and the Hilbert space has an orthonormal basis of eigenvectors of G. However, a general hermitian matrix does not need to be positive definite. Therefore, I need another approach.

    I consider an eigenvector [tex]|a\rangle = \sum_i a_i |f_i\rangle[/tex] with [tex]G|a\rangle = \lambda|a\rangle[/tex]
    [tex]\Rightarrow \lambda\sum_j a_j |f_j\rangle = \lambda |a\rangle = G |a\rangle = \sum_{ij}a_i |f_j\rangle\langle f_j|f_i\rangle[/tex]
    [tex]\Rightarrow \lambda = \frac{1}{a_j}\sum_{i}a_i \langle f_j|f_i\rangle[/tex]
    Unfortunately, I cannot conclude [tex]\lambda>0[/tex] from that.

    Can anybody help me?
     
  2. jcsd
  3. Oct 26, 2012 #2

    dextercioby

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    Can you show that G and G2 are related ?
     
  4. Oct 26, 2012 #3
    Unfortunately, I don't know how to relate G² and G.

    [tex]G^2=\sum_{ij}|f_i\rangle\langle f_i|f_j\rangle\langle f_j|[/tex]
    I would need an expression for [tex]\langle f_i|f_j\rangle[/tex] But since the basis set is not orthonormal I don't see what this could be.
     
  5. Oct 27, 2012 #4

    dextercioby

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    Alright. Different approach. Let psi be an eigenvector of G with eigenvalue a. Then from the spectral equation you have

    a=1/normsq(psi) * <psi|G|psi> = 1/normsq(psi) * sum (i) <psi|f_i><f_i|psi> = ...

    Can you now show the dots hide a number > 0 ?
     
  6. Oct 27, 2012 #5
    [tex]\ldots = \frac{1}{||\,|\psi\rangle||^2} \sum_i |\langle \psi | f_i\rangle|^2[/tex]

    Since psi ist not the zero vector (as it is an eigenvector) its overlap with at least one of the basis vectors is non-zero. Therefore, the sum of squares is strictly greater than 0.

    Thank you very much.
     
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