Various improvements to knowledge base

1 files changed, 7 insertions, 5 deletions

diff --git a/latex/know/concept/time-independent-perturbation-theory/source.md b/latex/know/concept/time-independent-perturbation-theory/source.md
index a3167cd..48504f4 100644
--- a/latex/know/concept/time-independent-perturbation-theory/source.md
+++ b/latex/know/concept/time-independent-perturbation-theory/source.md
@@ -3,8 +3,8 @@
 
 # Time-independent perturbation theory
 
-*Time-independent perturbation theory*, sometimes also called
-*stationary state perturbation theory*, is a specific application of
+**Time-independent perturbation theory**, sometimes also called
+**stationary state perturbation theory**, is a specific application of
 perturbation theory to the time-independent Schrödinger
 equation in quantum physics, for
 Hamiltonians of the following form:
@@ -27,8 +27,8 @@ $$\begin{aligned}
     &= E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + ...
 \end{aligned}$$
 
-Where $E_n^{(1)}$ and $\ket*{\psi_n^{(1)}}$ are called the *first-order
-corrections*, and so on for higher orders. We insert this into the
+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:
 
 $$\begin{aligned}
@@ -72,6 +72,7 @@ $$\begin{aligned}
 The approach to solving the other two equations varies depending on
 whether this $\hat{H}_0$ has a degenerate spectrum or not.
 
+
 ## Without degeneracy
 
 We start by assuming that there is no degeneracy, in other words, each
@@ -178,6 +179,7 @@ $$\begin{aligned}
 
 In practice, it is not particulary useful to calculate more corrections.
 
+
 ## With degeneracy
 
 If $\varepsilon_n$ is $D$-fold degenerate, then its eigenstate could be
@@ -295,7 +297,7 @@ The trick is to find a Hermitian operator $\hat{L}$ (usually using
 symmetries of the system) which commutes with both $\hat{H}_0$ and $\hat{H}_1$:
 
 $$\begin{aligned}
-    [\hat{L}, \hat{H}_0] = [\hat{L}, \hat{H}_1] = 0
+    \comm*{\hat{L}}{\hat{H}_0} = \comm{\hat{L}}{\hat{H}_1} = 0
 \end{aligned}$$
 
 So that it shares its eigenstates with $\hat{H}_0$ (and $\hat{H}_1$),