Fix "Time-independent perturbation theory"

2 files changed, 75 insertions, 77 deletions

diff --git a/latex/know/concept/time-independent-perturbation-theory/source.md b/latex/know/concept/time-independent-perturbation-theory/source.md
index d076457..a3167cd 100644
--- a/latex/know/concept/time-independent-perturbation-theory/source.md
+++ b/latex/know/concept/time-independent-perturbation-theory/source.md
@@ -16,31 +16,31 @@ $$\begin{aligned}
 Where $\hat{H}_0$ is a Hamiltonian for which the time-independent
 Schrödinger equation has a known solution, and $\hat{H}_1$ is a small
 perturbing Hamiltonian. The eigenenergies $E_n$ and eigenstates
-$\ket{\psi_n}$ of the composite problem are expanded accordingly in the
+$\ket{\psi_n}$ of the composite problem are expanded in the
 perturbation "bookkeeping" parameter $\lambda$:
 
 $$\begin{aligned}
     \ket{\psi_n}
-    &= \ket{\psi_n^{(0)}} + \lambda \ket{\psi_n^{(1)}} + \lambda^2 \ket{\psi_n^{(2)}} + ...
+    &= \ket*{\psi_n^{(0)}} + \lambda \ket*{\psi_n^{(1)}} + \lambda^2 \ket*{\psi_n^{(2)}} + ...
     \\
     E_n
     &= E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + ...
 \end{aligned}$$
 
-Where $E_n^{(1)}$ and $\ket{\psi_n^{(1)}}$ are called the *first-order
+Where $E_n^{(1)}$ and $\ket*{\psi_n^{(1)}}$ are called the *first-order
 corrections*, and so on for higher orders. We insert this into the
 Schrödinger equation:
 
 $$\begin{aligned}
     \hat{H} \ket{\psi_n}
-    &= \hat{H}_0 \ket{\psi_n^{(0)}}
-    + \lambda \big( \hat{H}_1 \ket{\psi_n^{(0)}} + \hat{H}_0 \ket{\psi_n^{(1)}} \big) \\
-    &\qquad + \lambda^2 \big( \hat{H}_1 \ket{\psi_n^{(1)}} + \hat{H}_0 \ket{\psi_n^{(2)}} \big) + ...
+    &= \hat{H}_0 \ket*{\psi_n^{(0)}}
+    + \lambda \big( \hat{H}_1 \ket*{\psi_n^{(0)}} + \hat{H}_0 \ket*{\psi_n^{(1)}} \big) \\
+    &\qquad + \lambda^2 \big( \hat{H}_1 \ket*{\psi_n^{(1)}} + \hat{H}_0 \ket*{\psi_n^{(2)}} \big) + ...
     \\
     E_n \ket{\psi_n}
-    &= E_n^{(0)} \ket{\psi_n^{(0)}}
-    + \lambda \big( E_n^{(1)} \ket{\psi_n^{(0)}} + E_n^{(0)} \ket{\psi_n^{(1)}} \big) \\
-    &\qquad + \lambda^2 \big( E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}} \big) + ...
+    &= E_n^{(0)} \ket*{\psi_n^{(0)}}
+    + \lambda \big( E_n^{(1)} \ket*{\psi_n^{(0)}} + E_n^{(0)} \ket*{\psi_n^{(1)}} \big) \\
+    &\qquad + \lambda^2 \big( E_n^{(2)} \ket*{\psi_n^{(0)}} + E_n^{(1)} \ket*{\psi_n^{(1)}} + E_n^{(0)} \ket*{\psi_n^{(2)}} \big) + ...
 \end{aligned}$$
 
 If we collect the terms according to the order of $\lambda$, we arrive
@@ -48,14 +48,14 @@ at the following endless series of equations, of which in practice only
 the first three are typically used:
 
 $$\begin{aligned}
-    \hat{H}_0 \ket{\psi_n^{(0)}}
-    &= E_n^{(0)} \ket{\psi_n^{(0)}}
+    \hat{H}_0 \ket*{\psi_n^{(0)}}
+    &= E_n^{(0)} \ket*{\psi_n^{(0)}}
     \\
-    \hat{H}_1 \ket{\psi_n^{(0)}} + \hat{H}_0 \ket{\psi_n^{(1)}}
-    &= E_n^{(1)} \ket{\psi_n^{(0)}} + E_n^{(0)} \ket{\psi_n^{(1)}}
+    \hat{H}_1 \ket*{\psi_n^{(0)}} + \hat{H}_0 \ket*{\psi_n^{(1)}}
+    &= E_n^{(1)} \ket*{\psi_n^{(0)}} + E_n^{(0)} \ket*{\psi_n^{(1)}}
     \\
-    \hat{H}_1 \ket{\psi_n^{(1)}} + \hat{H}_0 \ket{\psi_n^{(2)}}
-    &= E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}}
+    \hat{H}_1 \ket*{\psi_n^{(1)}} + \hat{H}_0 \ket*{\psi_n^{(2)}}
+    &= E_n^{(2)} \ket*{\psi_n^{(0)}} + E_n^{(1)} \ket*{\psi_n^{(1)}} + E_n^{(0)} \ket*{\psi_n^{(2)}}
     \\
     ...
     &= ...
@@ -63,7 +63,7 @@ $$\begin{aligned}
 
 The first equation is the unperturbed problem, which we assume has
 already been solved, with eigenvalues $E_n^{(0)} = \varepsilon_n$ and
-eigenvectors $\ket{\psi_n^{(0)}} = \ket{n}$:
+eigenvectors $\ket*{\psi_n^{(0)}} = \ket{n}$:
 
 $$\begin{aligned}
     \hat{H}_0 \ket{n} = \varepsilon_n \ket{n}
@@ -79,28 +79,27 @@ $\varepsilon_n$ corresponds to one $\ket{n}$. At order $\lambda^1$, we
 rewrite the equation as follows:
 
 $$\begin{aligned}
-    (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} = 0
+    (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket*{\psi_n^{(1)}} = 0
 \end{aligned}$$
 
 Since $\ket{n}$ form a complete basis, we can express
-$\ket{\psi_n^{(1)}}$ in terms of them:
+$\ket*{\psi_n^{(1)}}$ in terms of them:
 
 $$\begin{aligned}
-    \ket{\psi_n^{(1)}} = \sum_{m \neq n} c_m \ket{m}
+    \ket*{\psi_n^{(1)}} = \sum_{m \neq n} c_m \ket{m}
 \end{aligned}$$
 
 Importantly, $n$ has been removed from the summation to prevent dividing
-by zero later. This is allowed, because
-$\ket{\psi_n^{(1)}} - c_n \ket{n}$ also satisfies the $\lambda^1$-order
+by zero later. We are allowed to do this, because
+$\ket*{\psi_n^{(1)}} - c_n \ket{n}$ also satisfies the order-$\lambda^1$
 equation for any value of $c_n$, as demonstrated here:
 
 $$\begin{aligned}
-    (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} - (\varepsilon_n - \varepsilon_n) c_n \ket{n} = 0
+    (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket*{\psi_n^{(1)}} - (\varepsilon_n - \varepsilon_n) c_n \ket{n} = 0
 \end{aligned}$$
 
-Where we used $\hat{H}_0 \ket{n} = \varepsilon_n \ket{n}$. Inserting the
-series form of $\ket{\psi_n^{(1)}}$ into the order-$\lambda^1$ equation
-gives us:
+Where we used $\hat{H}_0 \ket{n} = \varepsilon_n \ket{n}$.
+We insert the series form of $\ket*{\psi_n^{(1)}}$ into the $\lambda^1$-equation:
 
 $$\begin{aligned}
     (\hat{H}_1 - E_n^{(1)}) \ket{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \ket{m} = 0
@@ -135,12 +134,12 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We isolate this result for $c_k$ and insert it into the series form of
-$\ket{\psi_n^{(1)}}$ to get the full first-order correction to the wave
+$\ket*{\psi_n^{(1)}}$ to get the full first-order correction to the wave
 function:
 
 $$\begin{aligned}
     \boxed{
-        \ket{\psi_n^{(1)}}
+        \ket*{\psi_n^{(1)}}
         = \sum_{m \neq n} \frac{\matrixel{m}{\hat{H}_1}{n}}{\varepsilon_n - \varepsilon_m} \ket{m}
     }
 \end{aligned}$$
@@ -166,9 +165,9 @@ $$\begin{aligned}
     = \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \braket{n}{\psi_n^{(1)}}
 \end{aligned}$$
 
-We explicitly removed the $\ket{n}$-dependence of $\ket{\psi_n^{(1)}}$,
+We explicitly removed the $\ket{n}$-dependence of $\ket*{\psi_n^{(1)}}$,
 so the last term is zero. By simply inserting our result for
-$\ket{\psi_n^{(1)}}$, we thus arrive at:
+$\ket*{\psi_n^{(1)}}$, we thus arrive at:
 
 $$\begin{aligned}
     \boxed{
@@ -257,9 +256,8 @@ $$\begin{aligned}
 
 This is an eigenvalue problem for $E_n^{(1)}$, where $c_d$ are the
 components of the eigenvectors which represent the "good" states.
-Suppose that this eigenvalue problem has been solved, and that
-$\ket{n, g}$ are the resulting "good" states. Then, as long as
-$E_n^{(1)}$ is a non-degenerate eigenvalue of $M$:
+After solving this, let $\ket{n, g}$ be the resulting "good" states.
+Then, as long as $E_n^{(1)}$ is a non-degenerate eigenvalue of $M$:
 
 $$\begin{aligned}
     \boxed{
@@ -272,36 +270,36 @@ first-order wave function correction is also unchanged:
 
 $$\begin{aligned}
     \boxed{
-        \ket{\psi_{n,g}^{(1)}}
+        \ket*{\psi_{n,g}^{(1)}}
         = \sum_{m \neq (n, g)} \frac{\matrixel{m}{\hat{H}_1}{n, g}}{\varepsilon_n - \varepsilon_m} \ket{m}
     }
 \end{aligned}$$
 
 This works because the matrix $M$ is diagonal in the $\ket{n, g}$-basis,
 such that when $\ket{m}$ is any vector $\ket{n, \gamma}$ in the
-$\ket{n}$-eigenspace (except for $\ket{n,g}$ of course, which is
-explicitly excluded), then conveniently the corresponding numerator
+$\ket{n}$-eigenspace (except for $\ket{n,g}$, which is
+explicitly excluded), then the corresponding numerator
 $\matrixel{n, \gamma}{\hat{H}_1}{n, g} = M_{\gamma, g} = 0$, so the term
 does not contribute.
 
 If any of the eigenvalues $E_n^{(1)}$ of $M$ are degenerate, then there
-is still some information missing about the components $c_d$ of the
-"good" states, in which case we must find these states some other way.
+is still information missing about the components $c_d$ of the
+"good" states, in which case we must find them some other way.
 
-An alternative way of determining these "good" states is also of
-interest if there is no degeneracy in $M$, since such a shortcut would
-allow us use the formulae from non-degenerate perturbation theory
+Such an alternative way of determining these "good" states is also of
+interest even if there is no degeneracy in $M$, since such a shortcut would
+allow us to use the formulae from non-degenerate perturbation theory
 straight away.
 
-The method is to find a Hermitian operator $\hat{L}$ (usually using
-symmetry) which commutes with both $\hat{H}_0$ and $\hat{H}_1$:
+The trick is to find a Hermitian operator $\hat{L}$ (usually using
+symmetries of the system) which commutes with both $\hat{H}_0$ and $\hat{H}_1$:
 
 $$\begin{aligned}
-= [\hat{L}, \hat{H}_1] = 0
+    [\hat{L}, \hat{H}_0] = [\hat{L}, \hat{H}_1] = 0
 \end{aligned}$$
 
 So that it shares its eigenstates with $\hat{H}_0$ (and $\hat{H}_1$),
-meaning at least $D$ of the vectors of the $D$-dimensional
+meaning all the vectors of the $D$-dimensional
 $\ket{n}$-eigenspace are also eigenvectors of $\hat{L}$.
 
 The crucial part, however, is that $\hat{L}$ must be chosen such that
diff --git a/static/know/concept/time-independent-perturbation-theory/index.html b/static/know/concept/time-independent-perturbation-theory/index.html
index eeb53a3..aeaa570 100644
--- a/static/know/concept/time-independent-perturbation-theory/index.html
+++ b/static/know/concept/time-independent-perturbation-theory/index.html
@@ -54,41 +54,41 @@
 <p><span class="math display">\[\begin{aligned}
     \hat{H} = \hat{H}_0 + \lambda \hat{H}_1
 \end{aligned}\]</span></p>
-<p>Where <span class="math inline">\(\hat{H}_0\)</span> is a Hamiltonian for which the time-independent Schrödinger equation has a known solution, and <span class="math inline">\(\hat{H}_1\)</span> is a small perturbing Hamiltonian. The eigenenergies <span class="math inline">\(E_n\)</span> and eigenstates <span class="math inline">\(\ket{\psi_n}\)</span> of the composite problem are expanded accordingly in the perturbation “bookkeeping” parameter <span class="math inline">\(\lambda\)</span>:</p>
+<p>Where <span class="math inline">\(\hat{H}_0\)</span> is a Hamiltonian for which the time-independent Schrödinger equation has a known solution, and <span class="math inline">\(\hat{H}_1\)</span> is a small perturbing Hamiltonian. The eigenenergies <span class="math inline">\(E_n\)</span> and eigenstates <span class="math inline">\(\ket{\psi_n}\)</span> of the composite problem are expanded in the perturbation “bookkeeping” parameter <span class="math inline">\(\lambda\)</span>:</p>
 <p><span class="math display">\[\begin{aligned}
     \ket{\psi_n}
-    &amp;= \ket{\psi_n^{(0)}} + \lambda \ket{\psi_n^{(1)}} + \lambda^2 \ket{\psi_n^{(2)}} + ...
+    &amp;= \ket*{\psi_n^{(0)}} + \lambda \ket*{\psi_n^{(1)}} + \lambda^2 \ket*{\psi_n^{(2)}} + ...
     \\
     E_n
     &amp;= E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + ...
 \end{aligned}\]</span></p>
-<p>Where <span class="math inline">\(E_n^{(1)}\)</span> and <span class="math inline">\(\ket{\psi_n^{(1)}}\)</span> are called the <em>first-order corrections</em>, and so on for higher orders. We insert this into the Schrödinger equation:</p>
+<p>Where <span class="math inline">\(E_n^{(1)}\)</span> and <span class="math inline">\(\ket*{\psi_n^{(1)}}\)</span> are called the <em>first-order corrections</em>, and so on for higher orders. We insert this into the Schrödinger equation:</p>
 <p><span class="math display">\[\begin{aligned}
     \hat{H} \ket{\psi_n}
-    &amp;= \hat{H}_0 \ket{\psi_n^{(0)}}
-    + \lambda \big( \hat{H}_1 \ket{\psi_n^{(0)}} + \hat{H}_0 \ket{\psi_n^{(1)}} \big) \\
-    &amp;\qquad + \lambda^2 \big( \hat{H}_1 \ket{\psi_n^{(1)}} + \hat{H}_0 \ket{\psi_n^{(2)}} \big) + ...
+    &amp;= \hat{H}_0 \ket*{\psi_n^{(0)}}
+    + \lambda \big( \hat{H}_1 \ket*{\psi_n^{(0)}} + \hat{H}_0 \ket*{\psi_n^{(1)}} \big) \\
+    &amp;\qquad + \lambda^2 \big( \hat{H}_1 \ket*{\psi_n^{(1)}} + \hat{H}_0 \ket*{\psi_n^{(2)}} \big) + ...
     \\
     E_n \ket{\psi_n}
-    &amp;= E_n^{(0)} \ket{\psi_n^{(0)}}
-    + \lambda \big( E_n^{(1)} \ket{\psi_n^{(0)}} + E_n^{(0)} \ket{\psi_n^{(1)}} \big) \\
-    &amp;\qquad + \lambda^2 \big( E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}} \big) + ...
+    &amp;= E_n^{(0)} \ket*{\psi_n^{(0)}}
+    + \lambda \big( E_n^{(1)} \ket*{\psi_n^{(0)}} + E_n^{(0)} \ket*{\psi_n^{(1)}} \big) \\
+    &amp;\qquad + \lambda^2 \big( E_n^{(2)} \ket*{\psi_n^{(0)}} + E_n^{(1)} \ket*{\psi_n^{(1)}} + E_n^{(0)} \ket*{\psi_n^{(2)}} \big) + ...
 \end{aligned}\]</span></p>
 <p>If we collect the terms according to the order of <span class="math inline">\(\lambda\)</span>, we arrive at the following endless series of equations, of which in practice only the first three are typically used:</p>
 <p><span class="math display">\[\begin{aligned}
-    \hat{H}_0 \ket{\psi_n^{(0)}}
-    &amp;= E_n^{(0)} \ket{\psi_n^{(0)}}
+    \hat{H}_0 \ket*{\psi_n^{(0)}}
+    &amp;= E_n^{(0)} \ket*{\psi_n^{(0)}}
     \\
-    \hat{H}_1 \ket{\psi_n^{(0)}} + \hat{H}_0 \ket{\psi_n^{(1)}}
-    &amp;= E_n^{(1)} \ket{\psi_n^{(0)}} + E_n^{(0)} \ket{\psi_n^{(1)}}
+    \hat{H}_1 \ket*{\psi_n^{(0)}} + \hat{H}_0 \ket*{\psi_n^{(1)}}
+    &amp;= E_n^{(1)} \ket*{\psi_n^{(0)}} + E_n^{(0)} \ket*{\psi_n^{(1)}}
     \\
-    \hat{H}_1 \ket{\psi_n^{(1)}} + \hat{H}_0 \ket{\psi_n^{(2)}}
-    &amp;= E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}}
+    \hat{H}_1 \ket*{\psi_n^{(1)}} + \hat{H}_0 \ket*{\psi_n^{(2)}}
+    &amp;= E_n^{(2)} \ket*{\psi_n^{(0)}} + E_n^{(1)} \ket*{\psi_n^{(1)}} + E_n^{(0)} \ket*{\psi_n^{(2)}}
     \\
     ...
     &amp;= ...
 \end{aligned}\]</span></p>
-<p>The first equation is the unperturbed problem, which we assume has already been solved, with eigenvalues <span class="math inline">\(E_n^{(0)} = \varepsilon_n\)</span> and eigenvectors <span class="math inline">\(\ket{\psi_n^{(0)}} = \ket{n}\)</span>:</p>
+<p>The first equation is the unperturbed problem, which we assume has already been solved, with eigenvalues <span class="math inline">\(E_n^{(0)} = \varepsilon_n\)</span> and eigenvectors <span class="math inline">\(\ket*{\psi_n^{(0)}} = \ket{n}\)</span>:</p>
 <p><span class="math display">\[\begin{aligned}
     \hat{H}_0 \ket{n} = \varepsilon_n \ket{n}
 \end{aligned}\]</span></p>
@@ -96,17 +96,17 @@
 <h2 id="without-degeneracy">Without degeneracy</h2>
 <p>We start by assuming that there is no degeneracy, in other words, each <span class="math inline">\(\varepsilon_n\)</span> corresponds to one <span class="math inline">\(\ket{n}\)</span>. At order <span class="math inline">\(\lambda^1\)</span>, we rewrite the equation as follows:</p>
 <p><span class="math display">\[\begin{aligned}
-    (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} = 0
+    (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket*{\psi_n^{(1)}} = 0
 \end{aligned}\]</span></p>
-<p>Since <span class="math inline">\(\ket{n}\)</span> form a complete basis, we can express <span class="math inline">\(\ket{\psi_n^{(1)}}\)</span> in terms of them:</p>
+<p>Since <span class="math inline">\(\ket{n}\)</span> form a complete basis, we can express <span class="math inline">\(\ket*{\psi_n^{(1)}}\)</span> in terms of them:</p>
 <p><span class="math display">\[\begin{aligned}
-    \ket{\psi_n^{(1)}} = \sum_{m \neq n} c_m \ket{m}
+    \ket*{\psi_n^{(1)}} = \sum_{m \neq n} c_m \ket{m}
 \end{aligned}\]</span></p>
-<p>Importantly, <span class="math inline">\(n\)</span> has been removed from the summation to prevent dividing by zero later. This is allowed, because <span class="math inline">\(\ket{\psi_n^{(1)}} - c_n \ket{n}\)</span> also satisfies the <span class="math inline">\(\lambda^1\)</span>-order equation for any value of <span class="math inline">\(c_n\)</span>, as demonstrated here:</p>
+<p>Importantly, <span class="math inline">\(n\)</span> has been removed from the summation to prevent dividing by zero later. We are allowed to do this, because <span class="math inline">\(\ket*{\psi_n^{(1)}} - c_n \ket{n}\)</span> also satisfies the order-<span class="math inline">\(\lambda^1\)</span> equation for any value of <span class="math inline">\(c_n\)</span>, as demonstrated here:</p>
 <p><span class="math display">\[\begin{aligned}
-    (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} - (\varepsilon_n - \varepsilon_n) c_n \ket{n} = 0
+    (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket*{\psi_n^{(1)}} - (\varepsilon_n - \varepsilon_n) c_n \ket{n} = 0
 \end{aligned}\]</span></p>
-<p>Where we used <span class="math inline">\(\hat{H}_0 \ket{n} = \varepsilon_n \ket{n}\)</span>. Inserting the series form of <span class="math inline">\(\ket{\psi_n^{(1)}}\)</span> into the order-<span class="math inline">\(\lambda^1\)</span> equation gives us:</p>
+<p>Where we used <span class="math inline">\(\hat{H}_0 \ket{n} = \varepsilon_n \ket{n}\)</span>. We insert the series form of <span class="math inline">\(\ket*{\psi_n^{(1)}}\)</span> into the <span class="math inline">\(\lambda^1\)</span>-equation:</p>
 <p><span class="math display">\[\begin{aligned}
     (\hat{H}_1 - E_n^{(1)}) \ket{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \ket{m} = 0
 \end{aligned}\]</span></p>
@@ -125,10 +125,10 @@
 <p><span class="math display">\[\begin{aligned}
     \matrixel{k}{\hat{H}_1}{n} + c_k (\varepsilon_k - \varepsilon_n) = 0
 \end{aligned}\]</span></p>
-<p>We isolate this result for <span class="math inline">\(c_k\)</span> and insert it into the series form of <span class="math inline">\(\ket{\psi_n^{(1)}}\)</span> to get the full first-order correction to the wave function:</p>
+<p>We isolate this result for <span class="math inline">\(c_k\)</span> and insert it into the series form of <span class="math inline">\(\ket*{\psi_n^{(1)}}\)</span> to get the full first-order correction to the wave function:</p>
 <p><span class="math display">\[\begin{aligned}
     \boxed{
-        \ket{\psi_n^{(1)}}
+        \ket*{\psi_n^{(1)}}
         = \sum_{m \neq n} \frac{\matrixel{m}{\hat{H}_1}{n}}{\varepsilon_n - \varepsilon_m} \ket{m}
     }
 \end{aligned}\]</span></p>
@@ -143,7 +143,7 @@
     E_n^{(2)}
     = \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \braket{n}{\psi_n^{(1)}}
 \end{aligned}\]</span></p>
-<p>We explicitly removed the <span class="math inline">\(\ket{n}\)</span>-dependence of <span class="math inline">\(\ket{\psi_n^{(1)}}\)</span>, so the last term is zero. By simply inserting our result for <span class="math inline">\(\ket{\psi_n^{(1)}}\)</span>, we thus arrive at:</p>
+<p>We explicitly removed the <span class="math inline">\(\ket{n}\)</span>-dependence of <span class="math inline">\(\ket*{\psi_n^{(1)}}\)</span>, so the last term is zero. By simply inserting our result for <span class="math inline">\(\ket*{\psi_n^{(1)}}\)</span>, we thus arrive at:</p>
 <p><span class="math display">\[\begin{aligned}
     \boxed{
         E_n^{(2)}
@@ -196,7 +196,7 @@
         c_1 \\ \vdots \\ c_D
     \end{bmatrix}
 \end{aligned}\]</span></p>
-<p>This is an eigenvalue problem for <span class="math inline">\(E_n^{(1)}\)</span>, where <span class="math inline">\(c_d\)</span> are the components of the eigenvectors which represent the “good” states. Suppose that this eigenvalue problem has been solved, and that <span class="math inline">\(\ket{n, g}\)</span> are the resulting “good” states. Then, as long as <span class="math inline">\(E_n^{(1)}\)</span> is a non-degenerate eigenvalue of <span class="math inline">\(M\)</span>:</p>
+<p>This is an eigenvalue problem for <span class="math inline">\(E_n^{(1)}\)</span>, where <span class="math inline">\(c_d\)</span> are the components of the eigenvectors which represent the “good” states. After solving this, let <span class="math inline">\(\ket{n, g}\)</span> be the resulting “good” states. Then, as long as <span class="math inline">\(E_n^{(1)}\)</span> is a non-degenerate eigenvalue of <span class="math inline">\(M\)</span>:</p>
 <p><span class="math display">\[\begin{aligned}
     \boxed{
         E_{n, g}^{(1)} = \matrixel{n, g}{\hat{H}_1}{n, g}
@@ -205,18 +205,18 @@
 <p>Which is the same as in the non-degenerate case! Even better, the first-order wave function correction is also unchanged:</p>
 <p><span class="math display">\[\begin{aligned}
     \boxed{
-        \ket{\psi_{n,g}^{(1)}}
+        \ket*{\psi_{n,g}^{(1)}}
         = \sum_{m \neq (n, g)} \frac{\matrixel{m}{\hat{H}_1}{n, g}}{\varepsilon_n - \varepsilon_m} \ket{m}
     }
 \end{aligned}\]</span></p>
-<p>This works because the matrix <span class="math inline">\(M\)</span> is diagonal in the <span class="math inline">\(\ket{n, g}\)</span>-basis, such that when <span class="math inline">\(\ket{m}\)</span> is any vector <span class="math inline">\(\ket{n, \gamma}\)</span> in the <span class="math inline">\(\ket{n}\)</span>-eigenspace (except for <span class="math inline">\(\ket{n,g}\)</span> of course, which is explicitly excluded), then conveniently the corresponding numerator <span class="math inline">\(\matrixel{n, \gamma}{\hat{H}_1}{n, g} = M_{\gamma, g} = 0\)</span>, so the term does not contribute.</p>
-<p>If any of the eigenvalues <span class="math inline">\(E_n^{(1)}\)</span> of <span class="math inline">\(M\)</span> are degenerate, then there is still some information missing about the components <span class="math inline">\(c_d\)</span> of the “good” states, in which case we must find these states some other way.</p>
-<p>An alternative way of determining these “good” states is also of interest if there is no degeneracy in <span class="math inline">\(M\)</span>, since such a shortcut would allow us use the formulae from non-degenerate perturbation theory straight away.</p>
-<p>The method is to find a Hermitian operator <span class="math inline">\(\hat{L}\)</span> (usually using symmetry) which commutes with both <span class="math inline">\(\hat{H}_0\)</span> and <span class="math inline">\(\hat{H}_1\)</span>:</p>
+<p>This works because the matrix <span class="math inline">\(M\)</span> is diagonal in the <span class="math inline">\(\ket{n, g}\)</span>-basis, such that when <span class="math inline">\(\ket{m}\)</span> is any vector <span class="math inline">\(\ket{n, \gamma}\)</span> in the <span class="math inline">\(\ket{n}\)</span>-eigenspace (except for <span class="math inline">\(\ket{n,g}\)</span>, which is explicitly excluded), then the corresponding numerator <span class="math inline">\(\matrixel{n, \gamma}{\hat{H}_1}{n, g} = M_{\gamma, g} = 0\)</span>, so the term does not contribute.</p>
+<p>If any of the eigenvalues <span class="math inline">\(E_n^{(1)}\)</span> of <span class="math inline">\(M\)</span> are degenerate, then there is still information missing about the components <span class="math inline">\(c_d\)</span> of the “good” states, in which case we must find them some other way.</p>
+<p>Such an alternative way of determining these “good” states is also of interest even if there is no degeneracy in <span class="math inline">\(M\)</span>, since such a shortcut would allow us to use the formulae from non-degenerate perturbation theory straight away.</p>
+<p>The trick is to find a Hermitian operator <span class="math inline">\(\hat{L}\)</span> (usually using symmetries of the system) which commutes with both <span class="math inline">\(\hat{H}_0\)</span> and <span class="math inline">\(\hat{H}_1\)</span>:</p>
 <p><span class="math display">\[\begin{aligned}
-= [\hat{L}, \hat{H}_1] = 0
+    [\hat{L}, \hat{H}_0] = [\hat{L}, \hat{H}_1] = 0
 \end{aligned}\]</span></p>
-<p>So that it shares its eigenstates with <span class="math inline">\(\hat{H}_0\)</span> (and <span class="math inline">\(\hat{H}_1\)</span>), meaning at least <span class="math inline">\(D\)</span> of the vectors of the <span class="math inline">\(D\)</span>-dimensional <span class="math inline">\(\ket{n}\)</span>-eigenspace are also eigenvectors of <span class="math inline">\(\hat{L}\)</span>.</p>
+<p>So that it shares its eigenstates with <span class="math inline">\(\hat{H}_0\)</span> (and <span class="math inline">\(\hat{H}_1\)</span>), meaning all the vectors of the <span class="math inline">\(D\)</span>-dimensional <span class="math inline">\(\ket{n}\)</span>-eigenspace are also eigenvectors of <span class="math inline">\(\hat{L}\)</span>.</p>
 <p>The crucial part, however, is that <span class="math inline">\(\hat{L}\)</span> must be chosen such that <span class="math inline">\(\ket{n, d_1}\)</span> and <span class="math inline">\(\ket{n, d_2}\)</span> have distinct eigenvalues <span class="math inline">\(\ell_1 \neq \ell_2\)</span> for <span class="math inline">\(d_1 \neq d_2\)</span>:</p>
 <p><span class="math display">\[\begin{aligned}
     \hat{L} \ket{n, b_1} = \ell_1 \ket{n, b_1}