Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to tackle the following exercise in a quantum chemistry textbook:

Show that: If $\mathbf{G}(\omega) = (\omega \mathbf{1}-\mathbf{A})^{-1}$, and $\mathbf{A}$ is Hermitian (i.e. $\mathbf{U}^{\dagger}\mathbf{A}\mathbf{U} =\mathbf{a}$ where $(\mathbf{a})_{ij} = a_{i}\delta_{ij}$ ) that $[\mathbf{G}(\omega)]_{ij} = \sum_{\alpha = 1}^{N} \frac{U_{i\alpha}U^{*}_{j\alpha}}{\omega-a_{\alpha}}$ (where $a_{\alpha}$ are the eigenvalues of $\mathbf{A}$).

I think that if $\mathbf{A}$ is Hermitian, then $(\omega\mathbf{1}-\mathbf{A})$ is also Hermitian, since the property $M_{ij} = M^{*}_{ji}$ is conserved when $\mathbf{A}$ is multiplied by $-1$ and when it's diagonal matrix elements are modified (assuming $\omega$ is real).

How should I approach this problem?

share|improve this question
1  
The idea is that if you have the eigendecomposition of $\mathbf A$, it is a simple matter to evaluate $f(\mathbf A)$: if $\mathbf A=\mathbf U\mathbf D\mathbf U^\dagger$, then $f(\mathbf A)=\mathbf U f(\mathbf D)\mathbf U^\dagger$, and it is a simple matter to evaluate the matrix function of a diagonal matrix... –  J. M. Nov 21 '11 at 12:07
    
Ah thanks. I knew this, but I think what I missed that not only is $(\omega\mathbf{1}-\mathbf{A})^{1}$ a function of $(\omega\mathbf{1}-\mathbf{A})$, it is also a function of $\mathbf{A}$. I'll work though the problem and write up the solution as an answer if I get there. –  James Womack Nov 22 '11 at 9:46

1 Answer 1

The comments on my question helped me make the necessary logical jump to solve this problem.

First, it is important to realize that if $\mathbf{A} = \mathbf{U}\mathbf{D}\mathbf{U}^{\dagger}$, then $f(\mathbf{A}) = \mathbf{U}f(\mathbf{D})\mathbf{U}^{\dagger}$ (where $\mathbf{D}$ is a diagonal matrix and $f(\mathbf{D})$ is a diagonal matrix with elements $f(D_{ii})$). The jump I needed to make was to realize that $(\omega\mathbf{1}-\mathbf{A})^{-1}$ is a function of $\mathbf{A}$, not just of $(\omega\mathbf{1} - \mathbf{A})$, suggesting

$$\mathbf{G}(\omega) = (\omega\mathbf{1}-\mathbf{A})^{-1} = \mathbf{U}(\omega\mathbf{1}-\mathbf{a})^{-1}\mathbf{U}^{\dagger}.$$

$(\omega\mathbf{1}-\mathbf{a})^{-1}$ is a diagonal matrix and the eigenvalues of $(\omega\mathbf{1}-\mathbf{A})^{-1}$ are therefore $(\omega-a_{\alpha})^{-1}$ an

$$[\mathbf{G}(\omega)]_{ij} = \sum_{\alpha} U_{i\alpha}(\omega-a_{\alpha})^{-1}U_{j\alpha}^{*} = \sum_{\alpha} \frac{U_{i\alpha}U_{j\alpha}^{*}}{\omega-a_{\alpha}},$$

as required by the original problem.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.