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Discrete hellmann-feynman theorem ?

  1. May 22, 2009 #1
    Discrete hellmann-feynman theorem ??

    The Helmann-Feynman theorem states that :
    derivative of eigenvalue with respect to a parameter = eigenfunction dagger * derivative of operator with respect to parameter * eigenfunction

    (assuming that the eigenfuncitons are normalized, otherwise the theorem would also include a normalization integration).

    Does anyone know if this works for the DISCRETE CASE ??

    ie:

    derivative of eigenvalue with respect to parameter = eigenVECTOR dagger * derivative of MATRIX with respect to parameter * eigenVECTOR ??

    I can't find a proof of it anywhere!
     
  2. jcsd
  3. May 22, 2009 #2
    Re: Discrete hellmann-feynman theorem ??

    In order for the theorem to be true, the matrix has to be Hermitian.

    The proof would go something like this. Let [tex] A(s) [/tex] be one-parameter smooth family of Hermitian matrices and [tex] \lambda(s) [/tex] a smooth parametrization of the eigenvalue of [tex] A(s) [/tex] corresponding to the eigenvector [tex] \mathbf{v}(s) [/tex]. (Assume that [tex] \mathbf{v} [/tex] is normalized so that [tex] \mathbf{v}^{\dagger} \mathbf{v} = 1 [/tex].) Then [tex] A \mathbf{v} = \lambda \mathbf{v} [/tex] for all [tex] s [/tex]. Differentiating this relation, we have
    [tex]
    A' \mathbf{v} + A \mathbf{v}' = \lambda' \mathbf{v} + \lambda \mathbf{v} \textrm{.}
    [/tex]
    Note that [tex] \mathbf{v}^{\dagger} A \mathbf{v} = (A^{\dagger} \mathbf{v})^{\dagger} \mathbf{v} = (A \mathbf{v})^{\dagger} \mathbf{v} [/tex], since [tex] A [/tex] is Hermitian; but [tex] A \mathbf{v} = \lambda \mathbf{v} [/tex], so [tex] \mathbf{v}^{\dagger} A \mathbf{v} = \lambda^{*} \mathbf{v}^{\dagger} \mathbf{v} = \lambda^{*}[/tex] (where stars indicate complex conjugates). However, recall that Hermitian matrices have real eigenvalues; thus, [tex] \lambda^{*} = \lambda [/tex]. Thus, multiplying the above equation by [tex] \mathbf{v}^{\dagger} [/tex] on the left, we have
    [tex]
    \mathbf{v}^{\dagger} A' \mathbf{v} + \lambda = \lambda' + \lambda \textrm{,}
    [/tex]
    and, canceling [tex] \lambda [/tex] from both sides, we obtain the desired result.
     
  4. May 28, 2009 #3
    Re: Discrete hellmann-feynman theorem ??

    I was just rereading my work, and I discovered that there are a few typos in the previous post. The proof should have read:

    [tex]
    (1) \quad \textrm{WLOG, } \mathbf{v}^{\dagger} \mathbf{v} = 1 \textrm{;}
    [/tex]

    [tex]
    (2) \begin{align*} \mathbf{v}^{\dagger} A \mathbf{v}' &= (A \mathbf{v})^{\dagger} \mathbf{v}'\\
    &= \lambda^{*} \mathbf{v}^{\dagger} \mathbf{v}'\\
    &= \lambda \mathbf{v}^{\dagger} \mathbf{v}' \quad \textrm{(since $\lambda$ is real);}
    \end{align*}
    [/tex]

    [tex]
    (3) \begin{align*} A \mathbf{v} &= \lambda \mathbf{v}\\
    \Rightarrow A' \mathbf{v} + A \mathbf{v}' &= \lambda' \mathbf{v} + \lambda \mathbf{v}'\\
    \Rightarrow \mathbf{v}^{\dagger} A' \mathbf{v} + \mathbf{v}^{\dagger} A \mathbf{v}' &= \lambda' \mathbf{v}^{\dagger} \mathbf{v} + \lambda \mathbf{v}^{\dagger} \mathbf{v}'\\
    \Rightarrow \mathbf{v}^{\dagger} A' \mathbf{v} &= \lambda' \quad \textrm{(by (1) and (2)).}
    \end{align*}
    [/tex]
     
  5. May 28, 2009 #4
    Re: Discrete hellmann-feynman theorem ??

    Very nice!

    So this works not only for Hermitian matrices, but for real eigenvalues of any matrix!
     
  6. May 28, 2009 #5
    Re: Discrete hellmann-feynman theorem ??

    Not quite. In step (2), the first line requires that [tex] A [/tex] be Hermitian (otherwise, it would read [tex] \mathbf{v}^{\dagger} A \mathbf{v}' = (A^{\dagger} \mathbf{v})^{\dagger} \mathbf{v}' [/tex], which does not give the required cancellation).
     
  7. May 29, 2009 #6
    Re: Discrete hellmann-feynman theorem ??

    Right .. I totally missed that! Thanks! =)
     
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