Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

1 files changed, 28 insertions, 28 deletions

diff --git a/source/know/concept/pauli-exclusion-principle/index.md b/source/know/concept/pauli-exclusion-principle/index.md
index 58a3f69..9821718 100644
--- a/source/know/concept/pauli-exclusion-principle/index.md
+++ b/source/know/concept/pauli-exclusion-principle/index.md
@@ -12,9 +12,9 @@ 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
+$$\ket{x_1}\ket{x_2} = \ket{x_1} \otimes \ket{x_2}$$, 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:
 
 $$\begin{aligned}
@@ -28,22 +28,22 @@ $$\begin{aligned}
     \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 $$\lambda = -1$$:
 
 $$\begin{aligned}
     \hat{P} \Ket{a}\Ket{b} = \lambda \Ket{a}\Ket{b}
 \end{aligned}$$
 
 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:
+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:
 
 $$\begin{aligned}
     \hat{P}_f \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = - \Ket{a}\Ket{b}
@@ -53,20 +53,20 @@ $$\begin{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!
+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
+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,
-where $\alpha$ and $\beta$ are unknown:
+where $$\alpha$$ and $$\beta$$ are unknown:
 
 $$\begin{aligned}
     \Ket{\Psi(a, b)}
     = \alpha \Ket{a}\Ket{b} + \beta \Ket{b}\Ket{a}
 \end{aligned}$$
 
-When we apply $\hat{P}$, we can "choose" between two "intepretations" of
+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:
 
@@ -78,28 +78,28 @@ $$\begin{aligned}
     &= \alpha \Ket{b}\Ket{a} + \beta \Ket{a}\Ket{b}
 \end{aligned}$$
 
-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 $$\alpha = \beta$$:
 
 $$\begin{aligned}
     \Ket{\Psi(a, b)}_b = C \big( \Ket{a}\Ket{b} + \Ket{b}\Ket{a} \big)
 \end{aligned}$$
 
-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 $$\alpha = -\beta$$:
 
 $$\begin{aligned}
     \Ket{\Psi(a, b)}_f = C \big( \Ket{a}\Ket{b} - \Ket{b}\Ket{a} \big)
 \end{aligned}$$
 
-This state is called **antisymmetric** under exchange: switching $a$ and $b$
+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 $C$:
+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 $$C$$:
 
 $$\begin{aligned}
     \Ket{\Psi(a, a)}_b