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diff --git a/source/know/concept/pauli-exclusion-principle/index.md b/source/know/concept/pauli-exclusion-principle/index.md
index 5b83b69..15130d9 100644
--- a/source/know/concept/pauli-exclusion-principle/index.md
+++ b/source/know/concept/pauli-exclusion-principle/index.md
@@ -8,57 +8,69 @@ categories:
 layout: "concept"
 ---
 
-In quantum mechanics, the **Pauli exclusion principle** is a theorem with
-profound consequences for how the world works.
+In quantum mechanics, the **Pauli exclusion principle** is a theorem
+with 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 can occupy the same two allowed
-states $$a$$ and $$b$$. We then define the permutation operator $$\hat{P}$$ as
-follows:
+$$\ket{x_1}\ket{x_2} = \ket{x_1} \otimes \ket{x_2}$$,
+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:
+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 $$[\hat{P}, \hat{P}^2] = 0$$, $$\Ket{a}\Ket{b}$$
-must also be an eigenket of $$\hat{P}$$ with eigenvalue $$\lambda$$,
-satisfying $$\lambda^2 = 1$$, so we know that $$\lambda = 1$$ or $$\lambda = -1$$:
+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 $$\lambda = -1$$:
 
 $$\begin{aligned}
-    \hat{P} \Ket{a}\Ket{b} = \lambda \Ket{a}\Ket{b}
+    \hat{P} \Ket{a}\Ket{b}
+    = \lambda \Ket{a}\Ket{b}
 \end{aligned}$$
 
-As it turns out, in nature, each class of particle has a single
-associated permutation eigenvalue $$\lambda$$, or in other words: whether
-$$\lambda$$ is $$-1$$ or $$1$$ depends on the type of particle that $$x_1$$
-and $$x_2$$ are. Particles with $$\lambda = -1$$ are called
-**fermions**, and those with $$\lambda = 1$$ are known as **bosons**. We
-define $$\hat{P}_f$$ with $$\lambda = -1$$ and $$\hat{P}_b$$ with
-$$\lambda = 1$$, such that:
+As it turns out, in nature, each type of particle has a single
+associated permutation eigenvalue $$\lambda$$, or in other words:
+whether $$\lambda$$ is $$-1$$ or $$1$$ depends on
+the type of particle that $$x_1$$ and $$x_2$$ are.
+Particles with $$\lambda = -1$$ are called **fermions**,
+and those with $$\lambda = 1$$ are known as **bosons**.
+We define $$\hat{P}_f$$ with $$\lambda = -1$$ and $$\hat{P}_b$$ with $$\lambda = 1$$,
+such that:
 
 $$\begin{aligned}
-    \hat{P}_f \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = - \Ket{a}\Ket{b}
-    \qquad
-    \hat{P}_b \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = \Ket{a}\Ket{b}
+    \hat{P}_f \Ket{a}\Ket{b}
+    &= \Ket{b}\Ket{a}
+    = - \Ket{a}\Ket{b}
+    \\
+    \hat{P}_b \Ket{a}\Ket{b}
+    &= \Ket{b}\Ket{a}
+    = \Ket{a}\Ket{b}
 \end{aligned}$$
 
-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 $$\Ket{b}\Ket{a}$$,
-regardless of the eigenvalue $$\lambda$$. There is no physical difference!
+Another fundamental fact 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 $$\Ket{b}\Ket{a}$$,
+regardless of the eigenvalue $$\lambda$$.
+There is no physical difference!
 
-But this does not mean that $$\hat{P}$$ is useless: despite not having any
-observable effect, the resulting difference between fermions and bosons
-is absolutely fundamental. Consider the following superposition state,
+But this does not mean that $$\hat{P}$$ is useless:
+despite not having any observable effect,
+the resulting difference between fermions and bosons is absolutely profound.
+Consider the following superposition state,
 where $$\alpha$$ and $$\beta$$ are unknown:
 
 $$\begin{aligned}
@@ -66,9 +78,10 @@ $$\begin{aligned}
     = \alpha \Ket{a}\Ket{b} + \beta \Ket{b}\Ket{a}
 \end{aligned}$$
 
-When we apply $$\hat{P}$$, 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:
+When we apply $$\hat{P}$$, 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:
 
 $$\begin{aligned}
     \hat{P} \Ket{\Psi(a, b)}
@@ -78,25 +91,28 @@ $$\begin{aligned}
     &= \alpha \Ket{b}\Ket{a} + \beta \Ket{a}\Ket{b}
 \end{aligned}$$
 
-This gives us the equations $$\lambda \alpha = \beta$$ and
-$$\lambda \beta = \alpha$$. In fact, just from this we could have deduced
-that $$\lambda$$ can be either $$-1$$ or $$1$$. In any case, for bosons
-($$\lambda = 1$$), we thus find that $$\alpha = \beta$$:
+This gives us the equations $$\lambda \alpha = \beta$$ and $$\lambda \beta = \alpha$$.
+In fact, just from this we could have deduced
+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
-**symmetric**: switching $$a$$ and $$b$$ gives the same result. Meanwhile, for
-fermions ($$\lambda = -1$$), we find that $$\alpha = -\beta$$:
+Where $$C$$ is a normalization constant.
+As expected, this state is **symmetric**:
+switching $$a$$ and $$b$$ gives the same result.
+Meanwhile, for 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 is called **antisymmetric** under exchange: switching $$a$$ and $$b$$
-causes a sign change, as we would expect for fermions.
+This state is called **antisymmetric** under exchange:
+switching $$a$$ and $$b$$ causes a sign change, as we would expect for fermions.
 
 Now, what if the particles $$x_1$$ and $$x_2$$ are in the same state $$a$$?
 For bosons, we just need to update the normalization constant $$C$$:
@@ -114,7 +130,8 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-And 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 atoms only fit a limited number of
-electrons (which are fermions), since each must have a different quantum number.
+And 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 atoms only fit a limited number of electrons (which are fermions),
+since each must have a different quantum number.