Deployed 69966fb with MkDocs version: 1.6.0

Author: github-actions[bot]

index.html

@@ -1301,7 +1301,10 @@ <h3 id="subproblems">Subproblems</h3>
 {\mathbf{a}_1} = (0,a),
 \]</div>
 <div class="arithmatex">\[
-{\mathbf{a}_2} = (\sqrt 3 a/2, - a/2),{\mathbf{a}_3} = ( - \sqrt 3 a/2, - a/2)
+{\mathbf{a}_2} = (\sqrt 3 a/2, - a/2),
+\]</div>
+<div class="arithmatex">\[
+{\mathbf{a}_3} = ( - \sqrt 3 a/2, - a/2)
 \]</div>
 <div class="arithmatex">\[
 {\mathbf{b}_1} = {\mathbf{a}_2} - {\mathbf{a}_3} = (\sqrt 3 a,0),
@@ -1312,14 +1315,21 @@ <h3 id="subproblems">Subproblems</h3>
 <div class="arithmatex">\[
 {\mathbf{b}_3} = {\mathbf{a}_1} - {\mathbf{a}_2} = ( - \sqrt 3 a/2,3a/2)
 \]</div>
-<p>Then the Haldane model on a hexagonal lattice can be written as
-<span class="arithmatex">\(<span class="arithmatex">\(H(k) = {d_0}I + {d_1}{\sigma _1} + {d_2}{\sigma _2} + {d_3}{\sigma _3}\)</span>\)</span>
-<span class="arithmatex">\(<span class="arithmatex">\({d_0} = 2{t_2}\cos \phi \sum\nolimits_i {\cos (\mathbf{k} \cdot {\mathbf{b}_i})} = 2{t_2}\cos \phi \left[ {\cos \left( {\sqrt 3 {k_x}a} \right) + \cos \left( { - \sqrt 3 {k_x}a/2 + 3{k_y}a/2} \right) + \cos \left( { - \sqrt 3 {k_x}a/2 - 3{k_y}a/2} \right)} \right]\)</span>\)</span>
-$$
-{d_1} = {t_1}\sum\nolimits_i {\cos (\mathbf{k} \cdot {\mathbf{a}_i})}  = {t_1}\left[ {\cos \left( {{k_y}a} \right) + \cos \left( {\sqrt 3 {k_x}a/2 - {k_y}a/2} \right) + \cos \left( { - \sqrt 3 {k_x}a/2 - {k_y}a/2} \right)} \right]\
-{d_2} = {t_1}\sum\nolimits_i {\sin (\mathbf{k} \cdot {\mathbf{a}_i})}  = {t_1}\left[ {\sin \left( {{k_y}a} \right) + \sin \left( {\sqrt 3 {k_x}a/2 - {k_y}a/2} \right) + \sin \left( { - \sqrt 3 {k_x}a/2 - {k_y}a/2} \right)} \right] \
-{d_3} = m - 2{t_2}\sin \phi \sum\nolimits_i {\sin (\mathbf{k} \cdot {\mathbf{b}_i})}  = m - 2{t_2}\sin \phi \left[ {\sin \left( {\sqrt 3 {k_x}a} \right) + \sin \left( { - \sqrt 3 {k_x}a/2 + 3{k_y}a/2} \right) + \sin \left( { - \sqrt 3 {k_x}a/2 - 3{k_y}a/2} \right)} \right] \
-$$</p>
+<p>Then the Haldane model on a hexagonal lattice can be written as</p>
+<div class="arithmatex">\[
+H(k) = {d_0}I + {d_1}{\sigma _1} + {d_2}{\sigma _2} + {d_3}{\sigma _3}
+\]</div>
+<div class="arithmatex">\[{d_0} = 2{t_2}\cos \phi \sum\nolimits_i {\cos (\mathbf{k} \cdot {\mathbf{b}_i})} = 2{t_2}\cos \phi \left[ {\cos \left( {\sqrt 3 {k_x}a} \right) + \cos \left( { - \sqrt 3 {k_x}a/2 + 3{k_y}a/2} \right) + \cos \left( { - \sqrt 3 {k_x}a/2 - 3{k_y}a/2} \right)} \right]$$
+\]</div>
+<div class="arithmatex">\[
+{d_1} = {t_1}\sum\nolimits_i {\cos (\mathbf{k} \cdot {\mathbf{a}_i})}  = {t_1}\left[ {\cos \left( {{k_y}a} \right) + \cos \left( {\sqrt 3 {k_x}a/2 - {k_y}a/2} \right) + \cos \left( { - \sqrt 3 {k_x}a/2 - {k_y}a/2} \right)} \right]\\
+\]</div>
+<div class="arithmatex">\[
+{d_2} = {t_1}\sum\nolimits_i {\sin (\mathbf{k} \cdot {\mathbf{a}_i})}  = {t_1}\left[ {\sin \left( {{k_y}a} \right) + \sin \left( {\sqrt 3 {k_x}a/2 - {k_y}a/2} \right) + \sin \left( { - \sqrt 3 {k_x}a/2 - {k_y}a/2} \right)} \right] \\
+\]</div>
+<div class="arithmatex">\[
+{d_3} = m - 2{t_2}\sin \phi \sum\nolimits_i {\sin (\mathbf{k} \cdot {\mathbf{b}_i})}  = m - 2{t_2}\sin \phi \left[ {\sin \left( {\sqrt 3 {k_x}a} \right) + \sin \left( { - \sqrt 3 {k_x}a/2 + 3{k_y}a/2} \right) + \sin \left( { - \sqrt 3 {k_x}a/2 - 3{k_y}a/2} \right)} \right] \\
+\]</div>
 <p>where <span class="arithmatex">\(\sigma_i\)</span> are the Pauli matrices and <span class="arithmatex">\(I\)</span> is the identity matrix.
 <div class="highlight"><pre><span></span><code><span class="k">def</span> <span class="nf">calc_hamiltonian</span><span class="p">(</span><span class="n">kx</span><span class="p">,</span> <span class="n">ky</span><span class="p">,</span> <span class="n">a</span><span class="p">,</span> <span class="n">t1</span><span class="p">,</span> <span class="n">t2</span><span class="p">,</span> <span class="n">phi</span><span class="p">,</span> <span class="n">m</span><span class="p">):</span>
 <span class="w">    </span><span class="sd">&quot;&quot;&quot;</span>