It looks like many papers (maybe all papers containing "static spin structure factor") use the terminology, static spin structure factor, to refer to the equal-time spin structure factor defined as: $\mathcal{S}_{0}(q)=\frac{1}{N}\sum_{ij} e^{iq(r_i-r_j)} \langle \hat{S}_i \cdot \hat{S}_j\rangle$, which can also be viewed as the integral over the dynamical spin structure factor $\mathcal{S}(q,\omega)$.

I thought the equal-time spin structure factor $\mathcal{S}_{0}(q)$ is very different from the true static spin structure factor which is $\mathcal{S}(q,\omega=0)$ just like the equal-time (simultaneous) spin susceptibility is different from the static spin susceptibility.

Is there any reason for this unusual and maybe misleading way to name the spin structure factor?

Another related question is which spin structure factor is related to the experiment, $\mathcal{S}_{0}(q)$ or $\mathcal{S}(q,\omega=0)$? As far as I know for spin susceptibility, the static spin susceptibility is more related to the experiment rather than the equal-time spin susceptibility. But why do people seem only care about $\mathcal{S}_{0}(q)$ and no one discuss $\mathcal{S}(q,\omega=0)$?


1 Answer 1


As you say, the term static spin structure factor is often used for $\mathcal{S}(q)$. Another word for $\mathcal{S}(q)$ is the total scattering function. The elastic part of the dynamical spin structure factor, $\mathcal{S}(q,\omega=0)$, is sometimes referred to as the elastic scattering function.

The physical interpretations of the two functions are slightly different. $\mathcal{S}(q)$ can be viewed as the Fourier transform of the real-space correlation function at a snapshot in time (i.e. the instantaneous or equal-time correlation function). Thus you can assign a time variable to the state in which the correlation functions are measured, and label the total scattering function accordingly, i.e. $\mathcal{S}_t(q)=\frac{1}{N}\sum_{ij} e^{iq(r_i-r_j)} \langle \psi(t)| \hat{S}_i \cdot \hat{S}_j | \psi(t) \rangle$. This labeling matters if the system is time-dependent or out-of-equilibrium. However, mostly we are concerned with states $|\psi(t)\rangle\equiv |\psi\rangle$ representing thermal equilibria or ground states, which are stationary and removes the time-dependence from $\mathcal{S}(q)$. In such cases it is justified to call $\mathcal{S}(q)$ a static spin structure factor. The elastic scattering function $\mathcal{S}(q,\omega=0)$ can be viewed as the Fourier transform of a long-time average correlation function, which is by definition static.

In practice, one often finds that $\mathcal{S}(q,\omega=0)$ dominates $\mathcal{S}(q)$, at least in magnetically ordered systems at thermal equilibrium, where the scattering is expected to be dominated by Bragg peaks. So while the two quantities are defined differently, often they are approximately equal and can be treated more or less interchangeably, which I think adds to the confusion.

Traditionally, $\mathcal{S}(q)$ is the quantity that is experimentally more accessible. As is implied by "total scattering", it can be measured through diffraction experiments with wide-frequency ("white") beams, a widely used experiment type. For this reason, in the context of scattering experiments static structure factors are typically understood as scattering measured without resolving the energy. However, at least in inelastic neutron scattering experiments there are also ways to exploit cross correlations in order to measure the elastic scattering $\mathcal{S}(q,\omega=0)$, see for example this paper.

  • $\begingroup$ It's really helpful, thanks! I also dig a little bit yesterday and try to connect $\mathcal{S}(q)$ with the static spin susceptibility $\chi(q,\omega=0)$ which has a more direct interpretation in terms of linear response theory (fluctuation-dissipation theorem) but it seems there is no obvious way to include $\mathcal{S}$ in that framework. Do you have any idea about this point? I notice that Sachdev made a simple relation $\mathcal{S}(q)=T\chi(q,\omega=0)$ in his book (Quantum Phase transition Eq.(10.76)) for the special cases but I still don't think I fully understand the problem. $\endgroup$
    – Yang
    Sep 1, 2022 at 21:44
  • $\begingroup$ @Yang I don't know of a general relation between those quantities. Generically by the fluctuation-dissipation theorem $\mathcal{S}(q,\omega)=\left( 1- e^{-\omega/T}\right)^{-1} \mathrm{Im}\left[ \chi \left( q, \hbar\omega, T\right) \right]$ (maybe up to a constant factor in Sachdev's notation?), which you can try to play with in different limits. Sachdev assumes $\hbar\omega \ll k_B T$, which certainly is not generically true for quantum problems. $\endgroup$
    – Anyon
    Sep 2, 2022 at 14:04
  • $\begingroup$ Yeah, and at the zero frequency, this relation becomes $\mathcal{S}(q,\omega=0)=\infty \times 0$ which seems to be not well defined. Let's see if other people can give more details... Thanks. $\endgroup$
    – Yang
    Sep 2, 2022 at 14:30

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