0
$\begingroup$

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?

  • 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...2011-11-21
  • 0
    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.2011-11-22

1 Answers 1