# Discrete hellmann-feynman theorem ?

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!

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.

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*}

Very nice!

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

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

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).

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