LSWT: Goes into more detail about H(q) derivation

docs/developers/design/001_linear_spinwave_theory.md

@@ -57,13 +57,14 @@ to be described in terms of the local spin ordered moment direction, since it de
 We thus define a set of rotation matrices $`R_i^{\alpha}`$ which transforms a spin vector $`\mathbf{S}'_i`$
 in the local coordinate system (where $z$ is along the ordered moment direction)
 to a vector $`\mathbf{S}_i`$ in a Cartesian coordinate system connected to the crystal lattice
-(In SpinW, this Cartesian system is defined by $x \parallel a$, $z \perp  (a, c)$ and $y \perp (x, z$):
+(In SpinW, this Cartesian system is defined by $x \parallel a$, $z \perp  (a, c)$ and $y \perp (x, z$)):
 
 ```math
 \mathbf{S}_i = R_i \mathbf{S}'_i
 ```
 
 where $`R_i`$ is a $3 \times 3$ matrix.
+
 Additionally, to more easily map to the operators $S^z$, $S^-$ ($`b^{\dagger}`$) and $S^+$ ($b$) operators above,
 we will define the following vectors from the columns of $`R_i`$ for each spin:
 
@@ -116,23 +117,59 @@ Thus in the following treatment, $`J_{ij}`$ is a superset and includes both the
 
 Now in the above equation, the sum over site indices $i$ and $j$ extends over all space,
 but in SpinW we deal exclusively with periodic (crystalline) systems, so we will separate out each index $i$ ($j$)
-into components $i'$ ($j'$) within a unit cell and components $m$ ($n$) between the first and $m^{\mathrm{th}}$ ($n^{\mathrm{th}}$) unit cell.
-The sum over sites within the unit cell thus becomes finite, but we still have an infinite sum over unit cells $`\sum_{m,n}`$.
+into components $k$ ($l$) within a unit cell and components $m$ ($n$) between the first and $m^{\mathrm{th}}$ ($n^{\mathrm{th}}$) unit cell.
+The sum over sites within the unit cell thus becomes finite, but we still have an infinite sum over unit cells $`\sum_{m,n}`$:
+
+```math
+\mathcal{H} = \sum_{m,n} \sum_{k, l} \mathbf{S}(\mathbf{r}_{k}+\mathbf{r}_m)
+    J(\mathbf{r}_{k}+\mathbf{r}_m, \mathbf{r}_{l}-\mathbf{r}_n) \mathbf{S}(\mathbf{r}_{l}+\mathbf{r}_n)
+```
+
 As we are looking for wave-like solutions to the Hamiltonian, the next step is thus to perform a Fourier transform to obtain a sum over
-$\mathbf{q}$ vectors in the first Brillouin zone.
+$\mathbf{q}$ vectors in the first Brillouin zone. The Fourier transform of the spin vector at site $k, m$ is:
+
+
+```math
+\mathbf{S}_{k,m}(\mathbf{q}) = \sum_m \left[ \sum_k \mathbf{S} \exp(-i\mathbf{r}_k\cdot\mathbf{q}) \right] \exp(-i\mathbf{r}_m\cdot\mathbf{q})
+```
+
+and similarly for the $n$ indexed terms $`\mathbf{S}_{l}`$. In the next part we will denote the term in square brackets as $`\mathbf{S}_k(\mathbf{q})`$.
+In a similar vein, the Fourier transform of the exchange interaction becomes:
+
+```math
+J_{kl,mn}(\mathbf{q}) = \sum_m \sum_n 
+    \left[ \sum_k \sum_l J \exp(-i(\mathbf{r}_k - \mathbf{r}_l)\cdot\mathbf{q}) \right] \exp(-i(\mathbf{r}_m - \mathbf{r}_n)\cdot\mathbf{q})
+```
+
+And like with the spins the terms in the square brackets will be denoted $`J_{kl}(\mathbf{q})`$.
+
+We can now expressed the Fourier transform of the Heisenberg Hamiltonian:
+
+```math
+\mathcal{H}(\mathbf{q}) = \sum_{m,n} \left[ \sum_{k,l} \mathbf{S}_k(\mathbf{q}) J_{kl}(\mathbf{q}) \mathbf{S}_l(\mathbf{q}') \right] 
+    \exp(-i\mathbf{r}_m\cdot\mathbf{q}) \exp(-i\mathbf{r}_n\cdot\mathbf{q}') \exp(-i(\mathbf{r}_m - \mathbf{r}_n)\cdot\mathbf{q})
+```
+
+Now we can use the identity $`\sum_{r} \exp(i(\mathbf{q}-\mathbf{q}')\cdot\mathbf{r}) = \delta_{\mathbf{qq}'}`$ to cancel the
+$`\mathbf{r}_n`$ terms leaving a single Fourier series in $`\mathbf{r}_m`$.
+If we now take the inverse Fourier transform of this, we can replace the sum $`\sum_{mn}`$ by a sum over $\mathbf{q}$, giving:
+
+```math
+\mathcal{H} = \sum_{\mathbf{q}} \sum_{k,l} \mathbf{S}_k(\mathbf{q}) J_{kl}(\mathbf{q}) \mathbf{S}_l(\mathbf{q})
+```
+
+Expressing the spin vectors in terms of the boson operators using the Holstein-Primakoff transformation above
+(and noting that because of the complex conjugation the Fourier transform of the creation operator $`b^{\dagger}`$ is
+$`b^{\dagger}(-\mathbf{q})`$):
 
-To do this we first need to re-express the sum over unit cells as a sum over distances
-$`\mathbf{r}_m`$ and $`\boldsymbol{\delta}=\mathbf{r}_m - \mathbf{r}_n`$.
-After Fourier transforming both the operators $b$, $b^{\dagger}$ and the exchange interactions $J$,
-we can use the identity $`\sum_{r} \exp(i(\mathbf{q}-\mathbf{q}')\cdot\mathbf{r}) = \delta_{\mathbf{qq}'}`$ to obtain:
 
 ```math
-\mathcal{H} = \sum_{\mathbf{q}} \sum_{i',j'} \left[
-    \sqrt{\frac{S_{i'}}{2}}\left( \mathbf{z}_{i'}^* b_{i'}(\mathbf{q}) + \mathbf{z}_{i'}b_{i'}^{\dagger}(-\mathbf{q}) \right)
-                           + \boldsymbol{\eta}_i \left( S_{i'} - b_{i'}(\mathbf{q}) b_{i'}^{\dagger}(-\mathbf{q}) \right)
-    \right]^{\intercal} J(\mathbf{q}) \left[
-    \sqrt{\frac{S_{j'}}{2}}\left( \mathbf{z}_{j'}^* b_{j'}(\mathbf{q}) + \mathbf{z}_{j'}b_{j'}^{\dagger}(-\mathbf{q}) \right)
-                           + \boldsymbol{\eta}_j \left( S_{j'} - b_{j'}(\mathbf{q}) b_{j'}^{\dagger}(-\mathbf{q}) \right)
+\mathcal{H} = \sum_{\mathbf{q}} \sum_{k,l} \left[
+    \sqrt{\frac{S_{k}}{2}}\left( \mathbf{z}_{k}^* b_{k}(\mathbf{q}) + \mathbf{z}_{k}b_{k}^{\dagger}(-\mathbf{q}) \right)
+                           + \boldsymbol{\eta}_i \left( S_{k} - b_{k}(\mathbf{q}) b_{k}^{\dagger}(-\mathbf{q}) \right)
+    \right]^{\intercal} J_{kl}(\mathbf{q}) \left[
+    \sqrt{\frac{S_{l}}{2}}\left( \mathbf{z}_{l}^* b_{l}(\mathbf{q}) + \mathbf{z}_{l}b_{l}^{\dagger}(-\mathbf{q}) \right)
+                           + \boldsymbol{\eta}_j \left( S_{l} - b_{l}(\mathbf{q}) b_{l}^{\dagger}(-\mathbf{q}) \right)
     \right]
 ```
 
@@ -142,7 +179,7 @@ where the sum over $`\mathbf{q}`$ extends over both positive and negative vector
 Noting that the boson operators obey the commutation relation
 
 ```math
-[b_{i'}, b_{j'}^{\dagger}] = \delta_{i', j'}
+[b_{k}, b_{l}^{\dagger}] = \delta_{k, l}
 ```
 
 and that the $`\mathbf{z}`$ and $`\boldsymbol{\eta}`$ vectors are perpendicular, we can rewrite the Hamiltonian as a matrix equation