# Out-of-plane and in-plane correlation function for single-q-magnetic structures

I want to derive the statical correlation function of a general Heisenberg system (e.g. antiferromagnetic) with the Hamiltonian

$$H= J S \sum_{\mathbf{k}} \omega_{\mathbf{k}} b^{\dagger}_{\mathbf{k}} b_{\mathbf{k}}$$

whereas $$\omega_{\mathbf{k}} = \sqrt{A_{\mathbf{k}} - B_{\mathbf{k}}}$$ is the dispersion. The spin-spin correlation function is defined as \begin{align} S^{\alpha \beta} (\omega, \mathbf{k}) = \int_{-\infty}^{\infty} \sum_{ij} \mathrm{d}t \exp (-i \omega t) \langle S_i^{\alpha}(0) S_j^{\beta}(t) \rangle \end{align} with $\alpha,\beta \in \{x,y,z\}$. We use now Holstein-Primakoff-Transformation: \begin{align} S_i^{+} &\approx \sqrt{2S}b_i \\ S_i^{-} &\approx b_i^{\dagger} \sqrt{2S} \\ S_i^{z} &= S-\widehat{n}_i \, . \end{align} with large $1/S$-approximation. Now I found a solution to my problem here: https://indico.psi.ch/conferenceDisplay.py/getPic?picId=56&confId=3407 (my $b_{\mathbf{k}}$ is his $a_{\mathbf{k}}$) on page 33 and 34. But didn't understand all the steps.

Out-of-plane

After the Fourier-Transform one gets the out-of-plane correlation function \begin{align} S^{xx}(t, \mathbf{k}) &= \frac{1}{N^{2}} \sum_{ij} \langle S_i^{x}(0) S_j^{x}(t) \rangle \mathrm{e}^{i \mathbf{k} \mathbf{r}_{ij}} \\ &= \frac{S}{2N^{2}} \sum_{ij} \langle \left( b_i + b_i^{\dagger} \right) \left( b_j(t) + b_j^{\dagger}(t) \right) \rangle \mathrm{e}^{i \mathbf{k} \mathbf{r}_{ij}} \\ &= \frac{S}{2N^{2}} \sum_{ij} \langle b_i b_j(t) + b_i^{\dagger} b_j(t) + b_i b_j^{\dagger}(t) + b_i^{\dagger}b_j^{\dagger}(t) \rangle \mathrm{e}^{i \mathbf{k} \mathbf{r}_{ij}} \\ &= \frac{S}{2} \sum_{\mathbf{k} \mathbf{k}'} \sum_{ij} \langle b_{\mathbf{k}'} b_{\mathbf{k}}(t) \mathrm{e}^{i\mathbf{k}' \mathbf{R}_i + i \mathbf{k} \mathbf{R}_j} + b_{\mathbf{k}'}^{\dagger} b_{\mathbf{k}}(t) \mathrm{e}^{-i \mathbf{k}' \mathbf{R}_i + i \mathbf{k} \mathbf{R}_j} + b_{\mathbf{k}'} b_{\mathbf{k}}^{\dagger} \mathrm{e}^{ i \mathbf{k}' \mathbf{R}_i - i \mathbf{k} \mathbf{R}_j} + b_{\mathbf{k}'}^{\dagger} b_{\mathbf{k}}^{\dagger} \mathrm{e}^{-i\mathbf{k}' \mathbf{R}_i - i\mathbf{k} \mathbf{R}_j} \rangle \mathrm{e}^{i \mathbf{k} \mathbf{r}_{ij}} \\ &= \frac{S}{2N^{2}} \sum_{\mathbf{k} } \sum_{ij} \langle b_{-\mathbf{k}} b_{\mathbf{k}}(t) \mathrm{e}^{-i\mathbf{k} \mathbf{R}_i + i \mathbf{k} \mathbf{R}_j} + b_{\mathbf{k}}^{\dagger} b_{\mathbf{k}}(t) \mathrm{e}^{-i \mathbf{k}\mathbf{R}_i + i \mathbf{k} \mathbf{R}_j} + b_{\mathbf{k}} b_{\mathbf{k}}^{\dagger} \mathrm{e}^{ i \mathbf{k} \mathbf{R}_i - i \mathbf{k} \mathbf{R}_j} + b_{-\mathbf{k}}^{\dagger} b_{\mathbf{k}}^{\dagger} \mathrm{e}^{i\mathbf{k} \mathbf{R}_i - i\mathbf{k} \mathbf{R}_j} \rangle \mathrm{e}^{i \mathbf{k} \mathbf{r}_{ij}} \\ &\overset{?}=\frac{S}{2} \sum_{\mathbf{k} } \langle b_{-\mathbf{k}} b_{\mathbf{k}}(t) + b_{\mathbf{k}}^{\dagger} b_{\mathbf{k}}(t) + b_{\mathbf{k}} b_{\mathbf{k}}^{\dagger} + b_{-\mathbf{k}}^{\dagger} b_{\mathbf{k}}^{\dagger} \rangle \end{align} In the last step he somehow uses \begin{align} \sum_i \mathrm{e}^{i (\mathbf{k}+\mathbf{k}') \mathbf{R}_i} = N \delta_{k,-k'} , \quad \sum_i \mathrm{e}^{i (\mathbf{k}-\mathbf{k}') \mathbf{R}_i} = N \delta_{k,k'} \end{align} This one point I didn't understand, but the second point is more difficult. Now he uses some kind of transformation: \begin{align} a = u b + v b^{\dagger} , \quad a^{\dagger} = u b^{\dagger} + v b, \quad u^2 + v^2 =1 \end{align} and Fourier transformation in time to get \begin{align} S^{xx} (\omega, \mathbf{k}) \overset{??}= S \frac{A_{\mathbf{k}} - B_{\mathbf{k}}}{\omega_{\mathbf{k}}} \delta(\omega - \omega_{\mathbf{k}} ) (n_{\omega} +1) \end{align} and by using \begin{align} \langle a_{\mathbf{k}}^{\dagger} a_{\mathbf{k}} \rangle = n_{\omega} \mathrm{e}^{-i\omega_{\mathbf k}t} \end{align} I tried every combination of these products but didn't find a combination in which all the mixed terms cancel and only $a$ and $a^{\dagger}$ terms stay. For example: \begin{align} b b^{\dagger} + b^{\dagger} b + b^{\dagger} b^{\dagger} + b b = \frac{1}{uv} \left[ ( a - v b^{\dagger})(a - u b) + (a^{\dagger} - vb) ( a^{\dagger} - u b^{\dagger}) + (a^{\dagger} - vb)(a - u b) +( a - v b^{\dagger}) ( a^{\dagger} - u b^{\dagger}) \right] \end{align} In-Plane For the in-plane correlation function he even starts with a function that comes from nowhere: \begin{align} S_i^{y} S_j^{y} (t) \overset{???}= - \frac{S}{2} \sum_{ij} \cos (\mathbf{Q} \cdot \mathbf{r}_{ij} ) \langle b_i^{\dagger} b_j^{\dagger} (t) + b_i b_j(t) - b_i b_j^{\dagger}(t) - b_i^{\dagger} b_j(t) \rangle \end{align} I mean the expression in the brakets $\langle \rangle$ follows directly from the definition of $S_i^{y}$ but where comes the $\cos (\mathbf{Q} \cdot \mathbf{r}_{ij} )= \cos (\varphi_{ij})$ (with $\mathbf{Q}$ being the magnetic ordering wave vector, $\varphi_{ij}$ being a rotation angle, because we are working in rotating coordinate system to go into a ferromagnetic ground state) comes from is uncertain.

I looked up all the references he mentioned, also look up in many books about Quantum Theory of Magnetism, but didn't find a fullfilling answer of how to calculate these correlation functions. In the paper https://arxiv.org/abs/1402.6069 that he has written, he goes a more general way of solving the problem, but says nothing to this derivation.

It would be far too much that everything in this post will be answered. I would be very glad if someone has an idea of how to derive one of these correlation function or has an different approach to a problem like this.