Discrete hellmann-Feynman theorem ?

[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!

 

[VKint]

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
<br /> A&#039; \mathbf{v} + A \mathbf{v}&#039; = \lambda&#039; \mathbf{v} + \lambda \mathbf{v} \textrm{.}<br />
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
<br /> \mathbf{v}^{\dagger} A&#039; \mathbf{v} + \lambda = \lambda&#039; + \lambda \textrm{,}<br />
and, canceling \lambda from both sides, we obtain the desired result.

 

[VKint]

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

<br /> (1) \quad \textrm{WLOG, } \mathbf{v}^{\dagger} \mathbf{v} = 1 \textrm{;}<br />

<br /> (2) \begin{align*} \mathbf{v}^{\dagger} A \mathbf{v}&#039; &amp;= (A \mathbf{v})^{\dagger} \mathbf{v}&#039;\\<br /> &amp;= \lambda^{*} \mathbf{v}^{\dagger} \mathbf{v}&#039;\\<br /> &amp;= \lambda \mathbf{v}^{\dagger} \mathbf{v}&#039; \quad \textrm{(since $\lambda$ is real);}<br /> \end{align*}<br />

<br /> (3) \begin{align*} A \mathbf{v} &amp;= \lambda \mathbf{v}\\<br /> \Rightarrow A&#039; \mathbf{v} + A \mathbf{v}&#039; &amp;= \lambda&#039; \mathbf{v} + \lambda \mathbf{v}&#039;\\<br /> \Rightarrow \mathbf{v}^{\dagger} A&#039; \mathbf{v} + \mathbf{v}^{\dagger} A \mathbf{v}&#039; &amp;= \lambda&#039; \mathbf{v}^{\dagger} \mathbf{v} + \lambda \mathbf{v}^{\dagger} \mathbf{v}&#039;\\<br /> \Rightarrow \mathbf{v}^{\dagger} A&#039; \mathbf{v} &amp;= \lambda&#039; \quad \textrm{(by (1) and (2)).}<br /> \end{align*}<br />

 

[juliette sekx]

Very nice!

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

 

[VKint]

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}&#039; = (A^{\dagger} \mathbf{v})^{\dagger} \mathbf{v}&#039;, which does not give the required cancellation).

 

[juliette sekx]

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