Fix "Pauli exclusion principle"

1 files changed, 16 insertions, 16 deletions

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:
 
 $$\begin{aligned}
     \ket{\Psi(a, a)}_f
-    = C \big( \ket{a}\!\ket{a} - \ket{a}\!\ket{a} \!\big)
+    = C \big( \ket{a}\ket{a} - \ket{a}\ket{a} \big)
     = 0
 \end{aligned}$$