Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Two body harmonic oscillator in 3D, and 2D diamond lattice vibrations

Posted by peeterjoot on January 7, 2014

Abridged harmonic oscillator notes

[This is an abbreviation of more extensive PDF notes associated with the latter part of this post.]

Motivation and summary of harmonic oscillator background

After having had some trouble on a non-1D harmonic oscillator lattice problem on the exam, I attempted such a problem with enough time available to consider it properly. I found it helpful to consider first just two masses interacting harmonically in 3D, each displaced from an equilibrium position.

The Lagrangian that described this most naturally was found to be

\begin{aligned}\mathcal{L} = \frac{1}{2} m_1 \left( \dot{\mathbf{r}}_1 \right)^2+\frac{1}{2} m_2 \left( \dot{\mathbf{r}}_2 \right)^2- \frac{K}{2} \left( \left\lvert {\mathbf{r}_2 - \mathbf{r}_1} \right\rvert - a \right)^2.\end{aligned} \hspace{\stretch{1}}(2.1)

This was solved in absolute and displacement coordinates, and then I moved on to consider a linear expansion of the harmonic potential about the equilibrium point, a problem closer to the exam problem (albeit still considering only two masses). The equilibrium points were described with vectors \mathbf{a}_1, \mathbf{a}_2 as in fig. 2.1, where \Delta \mathbf{a} = \left\lvert {\Delta \mathbf{a}} \right\rvert (\cos \theta_1, \cos\theta_2, \cos\theta_3).

fig 2.1: Direction cosines relative to equilibrium position difference vector

 

Using such coordinates, and generalizing, it was found that the complete Lagrangian, to second order about the equilibrium positions, is

\begin{aligned}\mathcal{L} = \sum_j \frac{m_i}{2} \dot{u}_{ij}^2 -\frac{K}{2} \sum_{i j} \cos\theta_i \cos\theta_j \left( u_{2 i} - u_{1 i} \right)\left( u_{2 j} - u_{1 j} \right).\end{aligned} \hspace{\stretch{1}}(2.2)

Evaluating the Euler-Lagrange equations, the equations of motion for the displacements were found to be

\begin{aligned}\begin{aligned}m_1 \ddot{\mathbf{u}}_1 &= K \widehat{\Delta \mathbf{a}} \left( \widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u} \right) \\ m_2 \ddot{\mathbf{u}}_2 &= -K \widehat{\Delta \mathbf{a}} \left( \widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u} \right),\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.3)

or

\begin{aligned}\boxed{\begin{aligned}\mu \Delta \ddot{\mathbf{u}} &= -K \widehat{\Delta \mathbf{a}} \left( \widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u} \right) \\ m_1 \ddot{\mathbf{u}}_1 + m_2 \ddot{\mathbf{u}}_2 &= 0.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(2.4)

Observe that on the RHS above we have a projection operator, so we could also write

\begin{aligned}\mu \Delta \ddot{\mathbf{u}} = -K \text{Proj}_{\widehat{\Delta \mathbf{a}}} \Delta \mathbf{u}.\end{aligned} \hspace{\stretch{1}}(2.5)

We see that the equations of motion for the displacements of a system of harmonic oscillators has a rather pleasant expression in terms of projection operators, where we have projections onto the unit vectors between each pair of equilibrium position.

A number of harmonically coupled masses

Now let’s consider masses at lattice points indexed by a lattice vector \mathbf{n}, as illustrated in fig. 2.2.

fig 2.2: Masses harmonically coupled in a lattice

 

With a coupling constant of K_{\mathbf{n} \mathbf{m}} between lattice points indexed \mathbf{n} and \mathbf{m} (located at \mathbf{a}_\mathbf{n} and \mathbf{a}_\mathbf{m} respectively), and direction cosines for the equilibrium direction vector between those points given by

\begin{aligned}\mathbf{a}_\mathbf{n} - \mathbf{a}_\mathbf{m} = \Delta \mathbf{a}_{\mathbf{n} \mathbf{m}}= \left\lvert {\Delta \mathbf{a}_{\mathbf{n} \mathbf{m}}} \right\rvert (\cos \theta_{\mathbf{n} \mathbf{m} 1},\cos \theta_{\mathbf{n} \mathbf{m} 2},\cos \theta_{\mathbf{n} \mathbf{m} 3}),\end{aligned} \hspace{\stretch{1}}(2.6)

the Lagrangian is

\begin{aligned}\mathcal{L} = \sum_{\mathbf{n}, i} \frac{m_\mathbf{n}}{2} \dot{u}_{\mathbf{n} i}^2-\frac{1}{2} \sum_{\mathbf{n} \ne \mathbf{m}, i, j} \frac{K_{\mathbf{n} \mathbf{m}}}{2} \cos\theta_{\mathbf{n} \mathbf{m} i}\cos\theta_{\mathbf{n} \mathbf{m} j}\left( u_{\mathbf{n} i} - u_{\mathbf{m} i} \right)\left( u_{\mathbf{n} j} - u_{\mathbf{m} j} \right)\end{aligned} \hspace{\stretch{1}}(2.7)

Evaluating the Euler-Lagrange equations for the mass at index \mathbf{n} we have

\begin{aligned}\frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{u}_{\mathbf{n} k}}} =m_\mathbf{n} \ddot{u}_{\mathbf{n} k},\end{aligned} \hspace{\stretch{1}}(2.8)

and

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {u_{\mathbf{n} k}}} &= -\sum_{\mathbf{m}, i, j}\frac{K_{\mathbf{n} \mathbf{m}}}{2} \cos\theta_{\mathbf{n} \mathbf{m} i}\cos\theta_{\mathbf{n} \mathbf{m} j}\left(\delta_{i k}\left( u_{\mathbf{n} j} - u_{\mathbf{m} j} \right)+\left( u_{\mathbf{n} i} - u_{\mathbf{m} i} \right)\delta_{j k}\right) \\ &= -\sum_{\mathbf{m}, i}K_{\mathbf{n} \mathbf{m}}\cos\theta_{\mathbf{n} \mathbf{m} k}\cos\theta_{\mathbf{n} \mathbf{m} i}\left( u_{\mathbf{n} i} - u_{\mathbf{m} i} \right) \\ &= -\sum_{\mathbf{m}}K_{\mathbf{n} \mathbf{m}}\cos\theta_{\mathbf{n} \mathbf{m} k}\widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u}_{\mathbf{n} \mathbf{m}},\end{aligned} \hspace{\stretch{1}}(2.9)

where \Delta \mathbf{u}_{\mathbf{n} \mathbf{m}} = \mathbf{u}_\mathbf{n} - \mathbf{u}_\mathbf{m}. Equating both, we have in vector form

\begin{aligned}m_\mathbf{n} \ddot{\mathbf{u}}_\mathbf{n} = -\sum_{\mathbf{m}}K_{\mathbf{n} \mathbf{m}}\widehat{\Delta \mathbf{a}}\left( \widehat{\Delta \mathbf{a}} \cdot \Delta \mathbf{u}_{\mathbf{n} \mathbf{m}} \right),\end{aligned} \hspace{\stretch{1}}(2.10)

or

\begin{aligned}\boxed{m_\mathbf{n} \ddot{\mathbf{u}}_\mathbf{n} = -\sum_{\mathbf{m}}K_{\mathbf{n} \mathbf{m}}\text{Proj}_{ \widehat{\Delta \mathbf{a}} } \Delta \mathbf{u}_{\mathbf{n} \mathbf{m}},}\end{aligned} \hspace{\stretch{1}}(2.11)

This is an intuitively pleasing result. We have displacement and the direction of the lattice separations in the mix, but not the magnitude of the lattice separation itself.

Two atom basis, 2D diamond lattice

As a concrete application of the previously calculated equilibrium harmonic oscillator result, let’s consider a two atom basis diamond lattice where the horizontal length is a and vertical height is b.

Indexing for the primitive unit cells is illustrated in fig. 2.3.

fig 2.3: Primitive unit cells for rectangular lattice

 

Let’s write

\begin{aligned}\begin{aligned}\mathbf{r} &= a (\cos\theta, \sin\theta) = a \hat{\mathbf{r}} \\ \mathbf{s} &= a (-\cos\theta, \sin\theta) = a \hat{\mathbf{s}} \\ \mathbf{n} &= (n_1, n_2) \\ \mathbf{r}_\mathbf{n} &= n_1 \mathbf{r} + n_2 \mathbf{s},\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.12)

For mass m_\alpha, \alpha \in \{1, 2\} assume a trial solution of the form

\begin{aligned}\mathbf{u}_{\mathbf{n},\alpha} = \frac{\boldsymbol{\epsilon}_\alpha(\mathbf{q})}{\sqrt{m_\alpha}} e^{i \mathbf{r}_n \cdot \mathbf{q} - \omega t}.\end{aligned} \hspace{\stretch{1}}(2.13)

The equations of motion for the two particles are

\begin{aligned}\begin{aligned}m_1 \ddot{\mathbf{u}}_{\mathbf{n}, 1} &= - K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} - (0,1), 2}  \right)- K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} - (1,0), 2}  \right) \\  & \quad- K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n}, 2}  \right)- K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} - (1,1), 2}  \right) \\  & \quad- K_3 \sum_\pm\text{Proj}_{\hat{\mathbf{r}}} \left(  \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} \pm (1,0), 1}  \right)- K_4 \sum_\pm\text{Proj}_{\hat{\mathbf{s}}} \left(  \mathbf{u}_{\mathbf{n}, 1} - \mathbf{u}_{\mathbf{n} \pm (0,1), 1}  \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.14.14)

\begin{aligned}\begin{aligned}m_2 \ddot{\mathbf{u}}_{\mathbf{n}, 2} &= - K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} + (1,0), 1}  \right)- K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} + (0,1), 1}  \right)\\  &\quad- K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n}, 1}  \right)- K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} + (1,1), 1}  \right)\\  &\quad- K_3 \sum_\pm\text{Proj}_{\hat{\mathbf{r}}} \left(  \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} \pm (1,0), 2}  \right)- K_4 \sum_\pm\text{Proj}_{\hat{\mathbf{s}}} \left(  \mathbf{u}_{\mathbf{n}, 2} - \mathbf{u}_{\mathbf{n} \pm (0,1), 2}  \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.14.14)

Insertion of the trial solution gives

\begin{aligned}\begin{aligned} \omega^2 \sqrt{m_1} \boldsymbol{\epsilon}_1&= K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} - \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} e^{ - i \mathbf{s} \cdot \mathbf{q} }  \right)+ K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} - \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} e^{ - i \mathbf{r} \cdot \mathbf{q} }  \right) \\  &\quad+ K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} - \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}}  \right)+ K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} - \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} e^{ - i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q} }  \right) \\  &\quad+ K_3 \left(  \text{Proj}_{\hat{\mathbf{r}}} \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}}  \right)\sum_\pm\left(  1 - e^{ \pm i \mathbf{r} \cdot \mathbf{q} }  \right)+ K_4 \left(  \text{Proj}_{\hat{\mathbf{s}}} \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}}  \right)\sum_\pm\left(  1 - e^{ \pm i \mathbf{s} \cdot \mathbf{q} }  \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.15.15)

\begin{aligned}\begin{aligned}\omega^2 \sqrt{m_2} \boldsymbol{\epsilon}_2&=K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} - \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} e^{ + i \mathbf{r} \cdot \mathbf{q} }  \right)+ K_1 \text{Proj}_{\hat{\mathbf{x}}} \left(  \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} - \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} e^{ + i \mathbf{s} \cdot \mathbf{q} }  \right)\\  &\quad+ K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} - \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}}  \right)+ K_2 \text{Proj}_{\hat{\mathbf{y}}} \left(  \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}} - \frac{\boldsymbol{\epsilon}_1}{\sqrt{m_1}} e^{ + i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q} }  \right) \\  &\quad+ K_3 \left(  \text{Proj}_{\hat{\mathbf{r}}} \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}}  \right)\sum_\pm\left(  1 - e^{ \pm i \mathbf{r} \cdot \mathbf{q} }  \right)+ K_4 \left(  \text{Proj}_{\hat{\mathbf{s}}} \frac{\boldsymbol{\epsilon}_2}{\sqrt{m_2}}  \right)\sum_\pm\left(  1 - e^{ \pm i \mathbf{s} \cdot \mathbf{q} }  \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.15.15)

Regrouping, and using the matrix form \text{Proj}_{\hat{\mathbf{u}}} = \hat{\mathbf{u}} \hat{\mathbf{u}}^\text{T} for the projection operators, this is

\begin{aligned}\left(\omega^2 - \frac{2}{m_1} \left(  K_1 \hat{\mathbf{x}} \hat{\mathbf{x}}^T + K_2 \hat{\mathbf{y}} \hat{\mathbf{y}}^T + 2 K_3 \hat{\mathbf{r}} \hat{\mathbf{r}}^T \sin^2 (\mathbf{r} \cdot \mathbf{q}/2) + 2 K_4 \hat{\mathbf{s}} \hat{\mathbf{s}}^T \sin^2 (\mathbf{s} \cdot \mathbf{q}/2)  \right)\right)\boldsymbol{\epsilon}_1 = -\left(  K_1 \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T} \left(  e^{ - i \mathbf{s} \cdot \mathbf{q} } + e^{ - i \mathbf{r} \cdot \mathbf{q} }  \right) + K_2 \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T} \left(  1 + e^{ - i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q} }  \right)  \right)\frac{\boldsymbol{\epsilon}_2}{\sqrt{ m_1 m_2 }}\end{aligned} \hspace{\stretch{1}}(2.0.16.16)

\begin{aligned}\left( \omega^2 - \frac{2}{m_2} \left(  K_1 \hat{\mathbf{x}} \hat{\mathbf{x}}^T + K_2 \hat{\mathbf{y}} \hat{\mathbf{y}}^T + 2 K_3 \hat{\mathbf{r}} \hat{\mathbf{r}}^T \sin^2 (\mathbf{r} \cdot \mathbf{q}/2)+ 2 K_4 \hat{\mathbf{s}} \hat{\mathbf{s}}^T \sin^2 (\mathbf{s} \cdot \mathbf{q}/2) \right) \right)\boldsymbol{\epsilon}_2 = -\left( K_1    \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T}   \left(    e^{ i \mathbf{s} \cdot \mathbf{q} }   +   e^{ i \mathbf{r} \cdot \mathbf{q} }    \right) +   K_2    \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T}   \left(    1   +   e^{ i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q} }    \right) \right)\frac{\boldsymbol{\epsilon}_1}{\sqrt{ m_1 m_2 }}\end{aligned} \hspace{\stretch{1}}(2.0.16.16)

As a single matrix equation, this is

\begin{aligned}A = K_1 \hat{\mathbf{x}} \hat{\mathbf{x}}^T + K_2 \hat{\mathbf{y}} \hat{\mathbf{y}}^T + 2 K_3 \hat{\mathbf{r}} \hat{\mathbf{r}}^T \sin^2 (\mathbf{r} \cdot \mathbf{q}/2)+ 2 K_4 \hat{\mathbf{s}} \hat{\mathbf{s}}^T \sin^2 (\mathbf{s} \cdot \mathbf{q}/2)\end{aligned} \hspace{\stretch{1}}(2.0.17.17)

\begin{aligned}B = e^{ i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q}/2 }\left( {   K_1    \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T}   \cos\left(  (\mathbf{r} - \mathbf{s}) \cdot \mathbf{q}/2  \right)+   K_2    \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T}   \cos\left(  (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q}/2  \right)} \right)\end{aligned} \hspace{\stretch{1}}(2.0.17.17)

\begin{aligned}0 =\begin{bmatrix}\omega^2 - \frac{2 A}{m_1} & \frac{B^{*}}{\sqrt{m_1 m_2}} \\ \frac{B}{\sqrt{m_1 m_2}} & \omega^2 - \frac{2 A}{m_2} \end{bmatrix}\begin{bmatrix}\boldsymbol{\epsilon}_1 \\ \boldsymbol{\epsilon}_2\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.17.17)

Observe that this is an eigenvalue problem E \mathbf{e} = \omega^2 \mathbf{e} for matrix

\begin{aligned}E = \begin{bmatrix}\frac{2 A}{m_1} & -\frac{B^{*}}{\sqrt{m_1 m_2}} \\ -\frac{B}{\sqrt{m_1 m_2}} & \frac{2 A}{m_2} \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.0.17.17)

and eigenvalues \omega^2.

To be explicit lets put the A and B functions in explicit matrix form. The orthogonal projectors have a simple form

\begin{aligned}\text{Proj}_{\hat{\mathbf{x}}} = \hat{\mathbf{x}} \hat{\mathbf{x}}^\text{T}= \begin{bmatrix}1 \\ 0\end{bmatrix}\begin{bmatrix}1 & 0\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.19a)

\begin{aligned}\text{Proj}_{\hat{\mathbf{y}}} = \hat{\mathbf{y}} \hat{\mathbf{y}}^\text{T}= \begin{bmatrix}0 \\ 1\end{bmatrix}\begin{bmatrix}0 & 1\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.19b)

For the \hat{\mathbf{r}} and \hat{\mathbf{s}} projection operators, we can use half angle formulations

\begin{aligned}\text{Proj}_{\hat{\mathbf{r}}} = \hat{\mathbf{r}} \hat{\mathbf{r}}^\text{T}= \begin{bmatrix}\cos\theta \\ \sin\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta\end{bmatrix}=\begin{bmatrix}\cos^2\theta & \cos\theta \sin\theta \\ \cos\theta \sin\theta & \sin^2 \theta\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1 + \cos \left(  2 \theta  \right) & \sin \left(  2 \theta  \right) \\ \sin \left(  2 \theta  \right) & 1 - \cos \left(  2 \theta  \right)\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.20.20)

\begin{aligned}\text{Proj}_{\hat{\mathbf{s}}} = \hat{\mathbf{s}} \hat{\mathbf{s}}^\text{T}= \begin{bmatrix}-\cos\theta \\ \sin\theta\end{bmatrix}\begin{bmatrix}-\cos\theta & \sin\theta\end{bmatrix}=\begin{bmatrix}\cos^2\theta & -\cos\theta \sin\theta \\ -\cos\theta \sin\theta & \sin^2 \theta\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1 + \cos \left(  2 \theta  \right) & -\sin \left(  2 \theta  \right) \\ -\sin \left(  2 \theta  \right) & 1 - \cos \left(  2 \theta  \right)\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.20.20)

After some manipulation, and the following helper functions

\begin{aligned}\begin{aligned}\alpha_\pm &= K_3 \sin^2 (\mathbf{r} \cdot \mathbf{q}/2) \pm K_4 \sin^2 (\mathbf{s} \cdot \mathbf{q}/2) \\ \beta_\pm &= K_1 \cos\left(  (\mathbf{r} - \mathbf{s}) \cdot \mathbf{q}/2  \right) \pm K_2 \cos\left(  (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q}/2  \right),\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.0.20.20)

the block matrices of eq. 2.0.17.17 take the form

\begin{aligned}A = \begin{bmatrix}K_1 + \alpha_+ (1 + \cos\left(  2 \theta  \right)) & \alpha_- \sin\left(  2 \theta  \right) \\ \alpha_- \sin\left(  2 \theta  \right) & K_2 + \alpha_+ (1 - \cos\left(  2 \theta  \right))\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.22.22)

\begin{aligned}B =    e^{ i (\mathbf{r} + \mathbf{s}) \cdot \mathbf{q}/2 }\begin{bmatrix}   \beta_+ (1 + \cos \left(  2 \theta  \right)) & \beta_- \sin \left(  2 \theta  \right) \\    \beta_- \sin \left(  2 \theta  \right) & \beta_+( 1 -\cos \left(  2 \theta  \right))\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.0.22.22)

A final bit of simplification for B possible, noting that \mathbf{r} + \mathbf{s} = 2 a (0, \sin\theta ), and \mathbf{r} - \mathbf{s} = 2 a(\cos\theta, 0), so

\begin{aligned}\beta_\pm = K_1 \cos\left(  a \cos\theta q_x  \right) \pm K_2 \cos\left(  a \sin\theta q_y  \right),\end{aligned} \hspace{\stretch{1}}(2.0.22.22)

and

\begin{aligned}B =    e^{ i a \sin\theta q_y }\begin{bmatrix}   \beta_+ (1 + \cos \left(  2 \theta  \right)) & \beta_- \sin \left(  2 \theta  \right) \\    \beta_- \sin \left(  2 \theta  \right) & \beta_+( 1 -\cos \left(  2 \theta  \right))\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.0.22.22)

It isn’t particularly illuminating to expand out the determinant for such a system, even though it can be done symbolically without too much programming. However, what is easy after formulating the matrix for this system, is actually solving it. This is done, and animated, in twoAtomBasisRectangularLatticeDispersionRelation.cdf

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