Fix "Pauli exclusion principle"

4 files changed, 32 insertions, 32 deletions

diff --git a/content/know/category/quantum-mechanics.md b/content/know/category/quantum-mechanics.md
index d61f03f..12c62d6 100644
--- a/content/know/category/quantum-mechanics.md
+++ b/content/know/category/quantum-mechanics.md
@@ -13,7 +13,7 @@ Alphabetical list of concepts in this category.
 * [Dirac notation](/know/concept/dirac-notation/)
 
 ## P
-* [Pauli exclusion principle](/known/concept/pauli-exclusion-principle/)
+* [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)
 * [Probability current](/know/concept/probability-current/)
 
 ## T
diff --git a/content/know/concept/index.md b/content/know/concept/index.md
index 43f5928..4171da7 100644
--- a/content/know/concept/index.md
+++ b/content/know/concept/index.md
@@ -13,7 +13,7 @@ Alphabetical list of concepts in this knowledge base.
 * [Dirac notation](/know/concept/dirac-notation/)
 
 ## P
-* [Pauli exclusion principle](/known/concept/pauli-exclusion-principle/)
+* [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)
 * [Probability current](/know/concept/probability-current/)
 
 ## T
diff --git a/latex/know/concept/pauli-exclusion-principle/source.md b/latex/know/concept/pauli-exclusion-principle/source.md
index 0a35869..e9e4d42 100644
--- a/latex/know/concept/pauli-exclusion-principle/source.md
+++ b/latex/know/concept/pauli-exclusion-principle/source.md
@@ -7,24 +7,24 @@ In quantum mechanics, the *Pauli exclusion principle* is a theorem that
 has profound consequences for how the world works.
 
 Suppose we have a composite state
-$\ket*{x_1}\!\ket*{x_2} = \ket*{x_1} \otimes \ket*{x_2}$, where the two
-identical particles $x_1$ and $x_2$ each have the same two allowed
+$\ket*{x_1}\ket*{x_2} = \ket*{x_1} \otimes \ket*{x_2}$, where the two
+identical particles $x_1$ and $x_2$ each can occupy the same two allowed
 states $a$ and $b$. We then define the permutation operator $\hat{P}$ as
 follows:
 
 $$\begin{aligned}
-    \hat{P} \ket{a}\!\ket{b} = \ket{b}\!\ket{a}
+    \hat{P} \ket{a}\ket{b} = \ket{b}\ket{a}
 \end{aligned}$$
 
 That is, it swaps the states of the particles. Obviously, swapping the
 states twice simply gives the original configuration again, so:
 
 $$\begin{aligned}
-    \hat{P}^2 \ket{a}\!\ket{b} = \ket{a}\!\ket{b}
+    \hat{P}^2 \ket{a}\ket{b} = \ket{a}\ket{b}
 \end{aligned}$$
 
-Therefore, $\ket{a}\!\ket{b}$ is an eigenvector of $\hat{P}^2$ with
-eigenvalue $1$. Since $[\hat{P}, \hat{P}^2] = 0$, $\ket{a}\!\ket{b}$
+Therefore, $\ket{a}\ket{b}$ is an eigenvector of $\hat{P}^2$ with
+eigenvalue $1$. Since $[\hat{P}, \hat{P}^2] = 0$, $\ket{a}\ket{b}$
 must also be an eigenket of $\hat{P}$ with eigenvalue $\lambda$,
 satisfying $\lambda^2 = 1$, so we know that $\lambda = 1$ or
 $\lambda = -1$.
@@ -38,14 +38,14 @@ define $\hat{P}_f$ with $\lambda = -1$ and $\hat{P}_b$ with
 $\lambda = 1$, such that:
 
 $$\begin{aligned}
-    \hat{P}_f \ket{a}\!\ket{b} = \ket{b}\!\ket{a} = - \ket{a}\!\ket{b}
+    \hat{P}_f \ket{a}\ket{b} = \ket{b}\ket{a} = - \ket{a}\ket{b}
     \qquad
-    \hat{P}_b \ket{a}\!\ket{b} = \ket{b}\!\ket{a} = \ket{a}\!\ket{b}
+    \hat{P}_b \ket{a}\ket{b} = \ket{b}\ket{a} = \ket{a}\ket{b}
 \end{aligned}$$
 
 Another fundamental fact of nature is that identical particles cannot be
 distinguished by any observation. Therefore it is impossible to tell
-apart $\ket{a}\!\ket{b}$ and the permuted state $\ket{b}\!\ket{a}$,
+apart $\ket{a}\ket{b}$ and the permuted state $\ket{b}\ket{a}$,
 regardless of the eigenvalue $\lambda$. There is no physical difference!
 
 But this does not mean that $\hat{P}$ is useless: despite not having any
@@ -55,7 +55,7 @@ where $\alpha$ and $\beta$ are unknown:
 
 $$\begin{aligned}
     \ket{\Psi(a, b)}
-    = \alpha \ket{a}\!\ket{b} + \beta \ket{b}\!\ket{a}
+    = \alpha \ket{a}\ket{b} + \beta \ket{b}\ket{a}
 \end{aligned}$$
 
 When we apply $\hat{P}$, we can "choose" between two "intepretations" of
@@ -64,10 +64,10 @@ equal, the right-hand sides must be equal too:
 
 $$\begin{aligned}
     \hat{P} \ket{\Psi(a, b)}
-    &= \lambda \alpha \ket{a}\!\ket{b} + \lambda \beta \ket{b}\!\ket{a}
+    &= \lambda \alpha \ket{a}\ket{b} + \lambda \beta \ket{b}\ket{a}
     \\
     \hat{P} \ket{\Psi(a, b)}
-    = \alpha \ket{b}\!\ket{a} + \beta \ket{a}\!\ket{b}
+    &= \alpha \ket{b}\ket{a} + \beta \ket{a}\ket{b}
 \end{aligned}$$
 
 This gives us the equations $\lambda \alpha = \beta$ and
@@ -76,7 +76,7 @@ that $\lambda$ can be either $-1$ or $1$. In any case, for bosons
 ($\lambda = 1$), we thus find that $\alpha = \beta$:
 
 $$\begin{aligned}
-    \ket{\Psi(a, b)}_b = C \big( \ket{a}\!\ket{b} + \ket{b}\!\ket{a} \!\big)
+    \ket{\Psi(a, b)}_b = C \big( \ket{a}\ket{b} + \ket{b}\ket{a} \big)
 \end{aligned}$$
 
 Where $C$ is a normalization constant. As expected, this state is
@@ -84,7 +84,7 @@ Where $C$ is a normalization constant. As expected, this state is
 fermions ($\lambda = -1$), we find that $\alpha = -\beta$:
 
 $$\begin{aligned}
-    \ket{\Psi(a, b)}_f = C \big( \ket{a}\!\ket{b} - \ket{b}\!\ket{a} \!\big)
+    \ket{\Psi(a, b)}_f = C \big( \ket{a}\ket{b} - \ket{b}\ket{a} \big)
 \end{aligned}$$
 
 This state called *antisymmetric* under exchange: switching $a$ and $b$
@@ -95,14 +95,14 @@ For bosons, we just need to update the normalization constant $C$:
 
 $$\begin{aligned}
     \ket{\Psi(a, a)}_b
-    = C \ket{a}\!\ket{a}
+    = C \ket{a}\ket{a}
 \end{aligned}$$
 
 However, for fermions, the state is unnormalizable and thus unphysical:
 
 $$\begin{aligned}
     \ket{\Psi(a, a)}_f
-    = C \big( \ket{a}\!\ket{a} - \ket{a}\!\ket{a} \!\big)
+    = C \big( \ket{a}\ket{a} - \ket{a}\ket{a} \big)
     = 0
 \end{aligned}$$
 
diff --git a/static/know/concept/pauli-exclusion-principle/index.html b/static/know/concept/pauli-exclusion-principle/index.html
index 74a0954..5389c02 100644
--- a/static/know/concept/pauli-exclusion-principle/index.html
+++ b/static/know/concept/pauli-exclusion-principle/index.html
@@ -51,53 +51,53 @@
 <hr>
 <h1 id="pauli-exclusion-principle">Pauli exclusion principle</h1>
 <p>In quantum mechanics, the <em>Pauli exclusion principle</em> is a theorem that has profound consequences for how the world works.</p>
-<p>Suppose we have a composite state <span class="math inline">\(\ket*{x_1}\!\ket*{x_2} = \ket*{x_1} \otimes \ket*{x_2}\)</span>, where the two identical particles <span class="math inline">\(x_1\)</span> and <span class="math inline">\(x_2\)</span> each have the same two allowed states <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span>. We then define the permutation operator <span class="math inline">\(\hat{P}\)</span> as follows:</p>
+<p>Suppose we have a composite state <span class="math inline">\(\ket*{x_1}\ket*{x_2} = \ket*{x_1} \otimes \ket*{x_2}\)</span>, where the two identical particles <span class="math inline">\(x_1\)</span> and <span class="math inline">\(x_2\)</span> each can occupy the same two allowed states <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span>. We then define the permutation operator <span class="math inline">\(\hat{P}\)</span> as follows:</p>
 <p><span class="math display">\[\begin{aligned}
-    \hat{P} \ket{a}\!\ket{b} = \ket{b}\!\ket{a}
+    \hat{P} \ket{a}\ket{b} = \ket{b}\ket{a}
 \end{aligned}\]</span></p>
 <p>That is, it swaps the states of the particles. Obviously, swapping the states twice simply gives the original configuration again, so:</p>
 <p><span class="math display">\[\begin{aligned}
-    \hat{P}^2 \ket{a}\!\ket{b} = \ket{a}\!\ket{b}
+    \hat{P}^2 \ket{a}\ket{b} = \ket{a}\ket{b}
 \end{aligned}\]</span></p>
-<p>Therefore, <span class="math inline">\(\ket{a}\!\ket{b}\)</span> is an eigenvector of <span class="math inline">\(\hat{P}^2\)</span> with eigenvalue <span class="math inline">\(1\)</span>. Since <span class="math inline">\([\hat{P}, \hat{P}^2] = 0\)</span>, <span class="math inline">\(\ket{a}\!\ket{b}\)</span> must also be an eigenket of <span class="math inline">\(\hat{P}\)</span> with eigenvalue <span class="math inline">\(\lambda\)</span>, satisfying <span class="math inline">\(\lambda^2 = 1\)</span>, so we know that <span class="math inline">\(\lambda = 1\)</span> or <span class="math inline">\(\lambda = -1\)</span>.</p>
+<p>Therefore, <span class="math inline">\(\ket{a}\ket{b}\)</span> is an eigenvector of <span class="math inline">\(\hat{P}^2\)</span> with eigenvalue <span class="math inline">\(1\)</span>. Since <span class="math inline">\([\hat{P}, \hat{P}^2] = 0\)</span>, <span class="math inline">\(\ket{a}\ket{b}\)</span> must also be an eigenket of <span class="math inline">\(\hat{P}\)</span> with eigenvalue <span class="math inline">\(\lambda\)</span>, satisfying <span class="math inline">\(\lambda^2 = 1\)</span>, so we know that <span class="math inline">\(\lambda = 1\)</span> or <span class="math inline">\(\lambda = -1\)</span>.</p>
 <p>As it turns out, in nature, each class of particle has a single associated permutation eigenvalue <span class="math inline">\(\lambda\)</span>, or in other words: whether <span class="math inline">\(\lambda\)</span> is <span class="math inline">\(-1\)</span> or <span class="math inline">\(1\)</span> depends on the species of particle that <span class="math inline">\(x_1\)</span> and <span class="math inline">\(x_2\)</span> represent. Particles with <span class="math inline">\(\lambda = -1\)</span> are called <em>fermions</em>, and those with <span class="math inline">\(\lambda = 1\)</span> are known as <em>bosons</em>. We define <span class="math inline">\(\hat{P}_f\)</span> with <span class="math inline">\(\lambda = -1\)</span> and <span class="math inline">\(\hat{P}_b\)</span> with <span class="math inline">\(\lambda = 1\)</span>, such that:</p>
 <p><span class="math display">\[\begin{aligned}
-    \hat{P}_f \ket{a}\!\ket{b} = \ket{b}\!\ket{a} = - \ket{a}\!\ket{b}
+    \hat{P}_f \ket{a}\ket{b} = \ket{b}\ket{a} = - \ket{a}\ket{b}
     \qquad
-    \hat{P}_b \ket{a}\!\ket{b} = \ket{b}\!\ket{a} = \ket{a}\!\ket{b}
+    \hat{P}_b \ket{a}\ket{b} = \ket{b}\ket{a} = \ket{a}\ket{b}
 \end{aligned}\]</span></p>
-<p>Another fundamental fact of nature is that identical particles cannot be distinguished by any observation. Therefore it is impossible to tell apart <span class="math inline">\(\ket{a}\!\ket{b}\)</span> and the permuted state <span class="math inline">\(\ket{b}\!\ket{a}\)</span>, regardless of the eigenvalue <span class="math inline">\(\lambda\)</span>. There is no physical difference!</p>
+<p>Another fundamental fact of nature is that identical particles cannot be distinguished by any observation. Therefore it is impossible to tell apart <span class="math inline">\(\ket{a}\ket{b}\)</span> and the permuted state <span class="math inline">\(\ket{b}\ket{a}\)</span>, regardless of the eigenvalue <span class="math inline">\(\lambda\)</span>. There is no physical difference!</p>
 <p>But this does not mean that <span class="math inline">\(\hat{P}\)</span> is useless: despite not having any observable effect, the resulting difference between fermions and bosons is absolutely fundamental. Consider the following superposition state, where <span class="math inline">\(\alpha\)</span> and <span class="math inline">\(\beta\)</span> are unknown:</p>
 <p><span class="math display">\[\begin{aligned}
     \ket{\Psi(a, b)}
-    = \alpha \ket{a}\!\ket{b} + \beta \ket{b}\!\ket{a}
+    = \alpha \ket{a}\ket{b} + \beta \ket{b}\ket{a}
 \end{aligned}\]</span></p>
 <p>When we apply <span class="math inline">\(\hat{P}\)</span>, we can “choose” between two “intepretations” of its action, both shown below. Obviously, since the left-hand sides are equal, the right-hand sides must be equal too:</p>
 <p><span class="math display">\[\begin{aligned}
     \hat{P} \ket{\Psi(a, b)}
-    &amp;= \lambda \alpha \ket{a}\!\ket{b} + \lambda \beta \ket{b}\!\ket{a}
+    &amp;= \lambda \alpha \ket{a}\ket{b} + \lambda \beta \ket{b}\ket{a}
     \\
     \hat{P} \ket{\Psi(a, b)}
-    = \alpha \ket{b}\!\ket{a} + \beta \ket{a}\!\ket{b}
+    &amp;= \alpha \ket{b}\ket{a} + \beta \ket{a}\ket{b}
 \end{aligned}\]</span></p>
 <p>This gives us the equations <span class="math inline">\(\lambda \alpha = \beta\)</span> and <span class="math inline">\(\lambda \beta = \alpha\)</span>. In fact, just from this we could have deduced that <span class="math inline">\(\lambda\)</span> can be either <span class="math inline">\(-1\)</span> or <span class="math inline">\(1\)</span>. In any case, for bosons (<span class="math inline">\(\lambda = 1\)</span>), we thus find that <span class="math inline">\(\alpha = \beta\)</span>:</p>
 <p><span class="math display">\[\begin{aligned}
-    \ket{\Psi(a, b)}_b = C \big( \ket{a}\!\ket{b} + \ket{b}\!\ket{a} \!\big)
+    \ket{\Psi(a, b)}_b = C \big( \ket{a}\ket{b} + \ket{b}\ket{a} \big)
 \end{aligned}\]</span></p>
 <p>Where <span class="math inline">\(C\)</span> is a normalization constant. As expected, this state is <em>symmetric</em>: switching <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span> gives the same result. Meanwhile, for fermions (<span class="math inline">\(\lambda = -1\)</span>), we find that <span class="math inline">\(\alpha = -\beta\)</span>:</p>
 <p><span class="math display">\[\begin{aligned}
-    \ket{\Psi(a, b)}_f = C \big( \ket{a}\!\ket{b} - \ket{b}\!\ket{a} \!\big)
+    \ket{\Psi(a, b)}_f = C \big( \ket{a}\ket{b} - \ket{b}\ket{a} \big)
 \end{aligned}\]</span></p>
 <p>This state called <em>antisymmetric</em> under exchange: switching <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span> causes a sign change, as we would expect for fermions.</p>
 <p>Now, what if the particles <span class="math inline">\(x_1\)</span> and <span class="math inline">\(x_2\)</span> are in the same state <span class="math inline">\(a\)</span>? For bosons, we just need to update the normalization constant <span class="math inline">\(C\)</span>:</p>
 <p><span class="math display">\[\begin{aligned}
     \ket{\Psi(a, a)}_b
-    = C \ket{a}\!\ket{a}
+    = C \ket{a}\ket{a}
 \end{aligned}\]</span></p>
 <p>However, for fermions, the state is unnormalizable and thus unphysical:</p>
 <p><span class="math display">\[\begin{aligned}
     \ket{\Psi(a, a)}_f
-    = C \big( \ket{a}\!\ket{a} - \ket{a}\!\ket{a} \!\big)
+    = C \big( \ket{a}\ket{a} - \ket{a}\ket{a} \big)
     = 0
 \end{aligned}\]</span></p>
 <p>At last, this is the Pauli exclusion principle: fermions may never occupy the same quantum state. One of the many notable consequences of this is that the shells of an atom only fit a limited number of electrons, since each must have a different quantum number.</p>