1 files changed, 29 insertions, 4 deletions
diff --git a/latex/know/concept/blochs-theorem/source.md b/latex/know/concept/blochs-theorem/source.md
index 528c218..79ee9a6 100644
--- a/latex/know/concept/blochs-theorem/source.md
+++ b/latex/know/concept/blochs-theorem/source.md
@@ -2,7 +2,7 @@
 
 
 # Bloch's theorem
-In quantum mechanics, *Bloch's theorem* states that,
+In quantum mechanics, **Bloch's theorem** states that,
 given a potential $V(\vec{r})$ which is periodic on a lattice,
 i.e. $V(\vec{r}) = V(\vec{r} + \vec{a})$
 for a primitive lattice vector $\vec{a}$,
@@ -22,13 +22,38 @@ $$
 
 In other words, in a periodic potential,
 the solutions are simply plane waves with a periodic modulation,
-known as *Bloch functions* or *Bloch states*.
+known as **Bloch functions** or **Bloch states**.
 
 This is suprisingly easy to prove:
 if the Hamiltonian $\hat{H}$ is lattice-periodic,
-then it will commute with the unitary translation operator $\hat{T}(\vec{a})$,
+then both $\psi(\vec{r})$ and $\psi(\vec{r} + \vec{a})$
+are eigenstates with the same energy:
+
+$$
+\begin{aligned}
+	\hat{H} \psi(\vec{r}) = E \psi(\vec{r})
+	\qquad
+	\hat{H} \psi(\vec{r} + \vec{a}) = E \psi(\vec{r} + \vec{a})
+\end{aligned}
+$$
+
+Now define the unitary translation operator $\hat{T}(\vec{a})$ such that
+$\psi(\vec{r} + \vec{a}) = \hat{T}(\vec{a}) \psi(\vec{r})$.
+From the previous equation, we then know that:
+
+$$
+\begin{aligned}
+	\hat{H} \hat{T}(\vec{a}) \psi(\vec{r})
+	= E \hat{T}(\vec{a}) \psi(\vec{r})
+	= \hat{T}(\vec{a}) \big(E \psi(\vec{r})\big)
+	= \hat{T}(\vec{a}) \hat{H} \psi(\vec{r})
+\end{aligned}
+$$
+
+In other words, if $\hat{H}$ is lattice-periodic,
+then it will commute with $\hat{T}(\vec{a})$,
 i.e. $[\hat{H}, \hat{T}(\vec{a})] = 0$.
-Therefore $\hat{H}$ and $\hat{T}(\vec{a})$ must share eigenstates $\psi(\vec{r})$:
+Consequently, $\hat{H}$ and $\hat{T}(\vec{a})$ must share eigenstates $\psi(\vec{r})$:
 
 $$
 \begin{aligned}