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When calculating the differential cross-section for magnetic scattering you have to compute the spin-spin correlation function

$$ S^{\alpha\beta}(\boldsymbol{k},\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty} dt e^{i\omega t}\sum_{l}e^{i\boldsymbol{k}\cdot \boldsymbol{r}_l} \langle S_0^\alpha(0) S_l^\beta(t)\rangle $$ where $\langle ...\rangle$ denotes an average over configurations, $\alpha,\beta \in \{x,y,z\}$, and $l\in \{1,2,3 \}$. Using linear spin-wave theory which involves mapping spin vectors at site l into bosonic creation/annihilation operators and then expanding them using Fourier transforms we obtain $$ S_l^-(t) = \sqrt{2S}b_l^\dagger(t) \mapsto \sqrt{\frac{2S}{N}}\sum_{\boldsymbol{k}} e^{-i(\boldsymbol{k}\cdot\boldsymbol{r}_l - \omega_{\boldsymbol{k}}t)}b_{\boldsymbol{k}}^\dagger \\ S_l^z(t) = S - b_l^\dagger(t) b_l(t) \mapsto S - \frac{1}{N}\sum_{\boldsymbol{k},\boldsymbol{k'}} e^{i[(\boldsymbol{k'}-\boldsymbol{k})\cdot\boldsymbol{r}_l - (\omega_{\boldsymbol{k'}}-\omega_{\boldsymbol{k}})t]}b_{\boldsymbol{k}}^\dagger b_{\boldsymbol{k}} $$ My question is whether certain averages will always vanish given that I have a linear spin-wave Hamiltonian with spectrum $\omega_{\boldsymbol{k}}$. $$ H = \sum_{\boldsymbol{k}} \begin{pmatrix} b^\dagger_{\boldsymbol{k}}&b_{-\boldsymbol{k}} \end{pmatrix} \begin{pmatrix} h(\boldsymbol{k})&\Delta^*(\boldsymbol{k})\\ \Delta(\boldsymbol{k}) & h(-\boldsymbol{k}) \end{pmatrix} \begin{pmatrix} b_{\boldsymbol{k}} \\ b_{-\boldsymbol{k}}^\dagger \end{pmatrix} \\ \omega_{\boldsymbol{k}} = \sqrt{|h(\boldsymbol{k})|^2 - |\Delta(\boldsymbol{k})}|^2 $$ I currently am referencing the book Introduction to the Theory of Thermal Neutron Scattering by G. Squires which states that terms like $\langle S_0^+(0) S_l^+(t)\rangle$ and $\langle S_0^+(0) S_l^z(t)\rangle$ will always vanish, but when talking with my research advisor he states that we should have the first of those two terms contributing to the correlation function. At this point, I am confused about if we could have those terms contributing and why/why not? If someone could help me out that would be greatly appreciated.

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  • $\begingroup$ In what context does Squires state that? It largely depends on the symmetry of the spin Hamiltonian. $\endgroup$
    – Anyon
    Commented Mar 28, 2021 at 21:46
  • $\begingroup$ @Anyon Looking at Squires on page 162, he is assuming a Hamiltonian of the form H = H^0 + \sum_k H_k so for eigenfunctions of H we must have an equal number of creation/annihilation operators for the correlation function to not vanish. $\endgroup$ Commented Mar 29, 2021 at 4:18
  • $\begingroup$ Squires is concluding that due to using a Heisenberg model (in Eq. (8.26)) as a starting point, such that $H_k$ is diagonalized by just Fourier transforming (i.e. no Bogoliubov transformation required). It's not intended as a more general statement than that. And with other spin Hamiltonians you can have e.g. non-zero <S^x(0)S^z(t)> correlations. $\endgroup$
    – Anyon
    Commented Mar 30, 2021 at 1:12

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