Discrete hellmann-feynman theorem ?

1. May 22, 2009

juliette sekx

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. May 22, 2009

VKint

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 $$A(s)$$ be one-parameter smooth family of Hermitian matrices and $$\lambda(s)$$ a smooth parametrization of the eigenvalue of $$A(s)$$ corresponding to the eigenvector $$\mathbf{v}(s)$$. (Assume that $$\mathbf{v}$$ is normalized so that $$\mathbf{v}^{\dagger} \mathbf{v} = 1$$.) Then $$A \mathbf{v} = \lambda \mathbf{v}$$ for all $$s$$. Differentiating this relation, we have
$$A' \mathbf{v} + A \mathbf{v}' = \lambda' \mathbf{v} + \lambda \mathbf{v} \textrm{.}$$
Note that $$\mathbf{v}^{\dagger} A \mathbf{v} = (A^{\dagger} \mathbf{v})^{\dagger} \mathbf{v} = (A \mathbf{v})^{\dagger} \mathbf{v}$$, since $$A$$ is Hermitian; but $$A \mathbf{v} = \lambda \mathbf{v}$$, so $$\mathbf{v}^{\dagger} A \mathbf{v} = \lambda^{*} \mathbf{v}^{\dagger} \mathbf{v} = \lambda^{*}$$ (where stars indicate complex conjugates). However, recall that Hermitian matrices have real eigenvalues; thus, $$\lambda^{*} = \lambda$$. Thus, multiplying the above equation by $$\mathbf{v}^{\dagger}$$ on the left, we have
$$\mathbf{v}^{\dagger} A' \mathbf{v} + \lambda = \lambda' + \lambda \textrm{,}$$
and, canceling $$\lambda$$ from both sides, we obtain the desired result.

3. May 28, 2009

VKint

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:

$$(1) \quad \textrm{WLOG, } \mathbf{v}^{\dagger} \mathbf{v} = 1 \textrm{;}$$

(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*}

(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*}

4. May 28, 2009

juliette sekx

Re: Discrete hellmann-feynman theorem ??

Very nice!

So this works not only for Hermitian matrices, but for real eigenvalues of any matrix!

5. May 28, 2009

VKint

Re: Discrete hellmann-feynman theorem ??

Not quite. In step (2), the first line requires that $$A$$ be Hermitian (otherwise, it would read $$\mathbf{v}^{\dagger} A \mathbf{v}' = (A^{\dagger} \mathbf{v})^{\dagger} \mathbf{v}'$$, which does not give the required cancellation).

6. May 29, 2009

juliette sekx

Re: Discrete hellmann-feynman theorem ??

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