# X-Ray Absorption Spectroscopy (XAS)¶

Schematic illustration of XAS (from [Nil04]):

The oscillator strengths are proportional to $$|\langle \phi_{1s}| \mathbf{r} | \psi_n \rangle|^2$$, where the one-center expansion of $$\psi_n$$ for the core-hole atom can be used.

## Introduction¶

The pseudo wave-functions are solutions to this generalized eigenvalue problem:

$H \tilde{\psi}_n = \epsilon_n S \tilde{\psi}_n.$

This can be transformed into a standard eigenvalue problem:

$S^{-1/2} H S^{-1/2} \psi_n = \epsilon_n \psi_n,$

where $$\psi_n = S^{1/2} \tilde{\psi}_n$$ is an all-electron wave function.

## XAS cross section¶

For the cross section, we need this quantity:

$\langle \psi_n | x | \phi^a \rangle = \sum_i \langle \tilde{\psi}_n | \tilde{p}_i^a \rangle \langle \phi_i^a | x | \phi^a \rangle = \langle \tilde{\psi}_n | \tilde{\phi}^a \rangle,$

where $$\phi^a$$ is the core state localized on atom $$a$$ and $$\tilde{\phi}^a = \sum_i \langle \phi_i^a | x | \phi^a \rangle \tilde{p}_i^a$$. Now, the cross section is:

$\sum_n |\langle \tilde{\psi}_n | \tilde{\phi}^a \rangle|^2 \delta(\epsilon_n - E) = \sum_n \langle \tilde{\phi}^a | S^{-1/2} | \psi_n \rangle \delta(\epsilon_n - E) \langle \psi_n | S^{-1/2} | \tilde{\phi}^a \rangle.$

By introducing $$G(E) = (E - S^{-1/2} H S^{-1/2} + i \gamma)^{-1}$$, we get:

$\text{Im}[\langle S^{-1/2} \tilde{\phi}^a | G(E) | S^{-1/2} \tilde{\phi}^a \rangle].$

## Recursion method¶

Instead of working with the $$u_i$$ functions from the Taillefumier paper, we introduce $$w_i=S^{1/2}u_i$$ which are the actual functions that we need to find. We now define $$y_i$$ and $$z_i$$ as:

$w_i = S z_i,$
$y_i = H z_i.$

With these definitions, the recursion formula reads:

$y_i = a_i w_i + b_{i+1} w_{i+1} + b_i w_{i-1},$

where:

$a_i = \langle z_i | y_i \rangle,$

and

$b_i = \langle z_i | y_{i-1} \rangle = \langle z_{i-1} | y_i \rangle.$

The $$w_i$$ functions should be normalized as:

$\langle w_i | S^{-1} | w_i \rangle = \langle w_i | z_i \rangle = 1,$

and the recursion is started with $$w_0 \propto \tilde{\phi}^a$$.

## Inverting the S matrix¶

The S (or O) operator is defined as:

$\hat O = 1 + \sum_a \sum_{i_1 i_2} |\tilde p^a_{i_1}> O^a_{i_1 i_2}< \tilde p^q_{i_2}|$

Where $$O^a_{i_1 i_2} = <\phi ^a_{i_1}| \phi ^a_{i_2}> - <\tilde \phi ^a_{i_1}| \tilde \phi ^a_{i_2}>$$

Assume that $$\hat O^{-1}$$ can be written as

$\hat O^{-1} = 1 + \sum_a \sum_{i_1 i_2} |\tilde p^a_{i_1}> P^a_{i_1 i_2}< \tilde p^a_{i_2}|$

Then according to [P.J. Hasnip et al, Comp. Phys. Comm. 174 (2006) 24-29 ] the coefficients $$P^a_{i_1 i_2}$$ are given by

$P^a_{i_1 i_2} = -O^a_{i_1 j} ( 1 + B^a_{kl} O^a_{lm} )^{-1}_{j i_2}$
$B^a_{kl} = < \tilde p^a_{k}| \tilde p^a_{l}>$

With summation over equal indices (except a). These formulas ignore overlap between projectors on different atoms. The accuracy of the $$\hat O^{-1}$$ operator can be checked for example by doing:

$<\tilde \phi_{i_1}| \hat O \hat O^{-1} \hat O |\tilde \phi_{i_2}> - \delta_{i_1 i_2}$

which should be zero for all normalized, orthogonalized $$\tilde \phi$$