QFI from charge excitations

In this analyze the Quantum Fisher Information (QFI) from the charge excitations. In the following analysis we make the following assumptions:

• The RIXS intensity of the charge excitations is directly proportional to the imaginary part of the charge susceptibility.

In a certain unit system, we can directly use $\mathrm{Im}\chi$ to represent the RIXS intensity of the charge excitations.

We model the measured intensity as a sum of charge and phonon contributions:

$$I_{\mathrm{meas}}(T, q, \omega)=\mathrm{Im}\chi(T, q, \omega) + I_\text{phonons}(T, q, \omega),$$ where $\mathrm{Im}\chi$ is the charge intensity and $I_\text{phonons}$ denotes the phonon intensity.

The Quantum Fisher Information (QFI) is defined (see this paper) as:

$$F_Q(T,q)=\int_{0}^{\omega_{\max}} \left[\mathrm{Im}\chi(T,q,\omega)\right]\, \tanh\left(\frac{\omega}{2T}\right)\, d\omega.$$

Since we currently do not filter out the phonon contributions, in this report we calculate:

$$F_Q(T,q)=\int_{0}^{\omega_{\max}} \left[\mathrm{Im}\chi(T,q,\omega)+I_\text{phonons}(T,q,\omega)\right]\, \tanh\left(\frac{\omega}{2T}\right)\, d\omega.$$

instead.

Here $T$ is the temperature expressed in meV. Therefore the implemented weighting is $\tanh(\omega/(2T))$ over $\omega \in [0,\omega_{\max}]$. In our analysis we set $\omega_{\max} = 90$ meV. Beyond that, other contributions kick in.

Absolute intensity maps at 21K, 62K, 104K, and 155K.

1D cuts of absolute intensity versus energy loss, color-coded by $(H - 0.24)$. On top we overlay the QFI weighting factor $\tanh(\omega/(2T))$.

Energy-integrated QFI quantity computed with the weighted integrand: $\tanh(\omega/(2T_{\mathrm{meV}}))$ for $\omega \in [0,\omega_{\max}]$. The selected H-value trends are shown in the right panel.

To suppress phonon background, we also show the reference-subtracted quantity using the momentum point $H=0.15$:

$$\Delta F_Q(H,T)=F_Q(H,T)-F_Q(0.15,T).$$

There are something I would like to mention here:

From RIXS intensity to susceptibility?

The assumption we made (RIXS intensity is proportional to the imaginary part of the charge susceptibility) is not necessarily correct as in RIXS the intensity is more complicated than normal X-ray scattering.

Difference between charge and magnetic excitations.

Even if the assumption (RIXS intensity is proportional to the imaginary part of the charge susceptibility) is valid, we still don't get the absolute value of the susceptibility. Assuming that for charge excitations:

$$I_\text{charge} = C_\text{charge} \mathrm{Im}\chi_\text{charge},$$ where $C_\text{charge}$ is a coefficient for the charge excitations. It is possibly (this I am not sure) that the coefficient for magnetic excitations $C_\text{mag}$ is different from $C_\text{charge}$. Therefore, the differencen between the RIXS intensity of the charge and magnetic excitations

$$\Delta I \equiv I_\text{charge} - I_\text{mag} = C_\text{charge} \mathrm{Im}\chi_\text{charge} - C_\text{mag} \mathrm{Im}\chi_\text{mag},$$

is not necessarily proportional to the difference between the susceptibility of the charge and magnetic excitations.

$$\Delta \mathrm{Im}\chi \equiv \mathrm{Im}\chi_\text{charge} - \mathrm{Im}\chi_\text{mag},$$

Phonon contributions not taken out.

Keep in mind that in our analysis the phonon contributions are not taken out. Therefore, the QFI is contaminated by the phonon contributions.

Therefore, in the last analysis, I made an attempt to subtract the phonon contributions by

$$\Delta F_Q(H,T)=F_Q(H,T)-F_Q(0.15,T).$$

assuming that (1) at H=0.15 r.l.u. the phonons dominate and (2) the phonon intensity is temperature independent and (3) The phonon intensity is also momentum-independent.