1 files changed, 22 insertions, 22 deletions
diff --git a/source/know/concept/fermis-golden-rule/index.md b/source/know/concept/fermis-golden-rule/index.md
index 18fcfd8..021c8e4 100644
--- a/source/know/concept/fermis-golden-rule/index.md
+++ b/source/know/concept/fermis-golden-rule/index.md
@@ -13,29 +13,29 @@ layout: "concept"
 In quantum mechanics, **Fermi's golden rule** expresses
 the transition rate between two states of a system,
 when a sinusoidal perturbation is applied
-at the resonance frequency $\omega = E_g / \hbar$ of the
-energy gap $E_g$. The main conclusion is that the rate is independent of
+at the resonance frequency $$\omega = E_g / \hbar$$ of the
+energy gap $$E_g$$. The main conclusion is that the rate is independent of
 time.
 
 From [time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
 we know that the transition probability
-for a particle in state $\Ket{a}$ to go to $\Ket{b}$
-is as follows for a periodic perturbation at frequency $\omega$:
+for a particle in state $$\Ket{a}$$ to go to $$\Ket{b}$$
+is as follows for a periodic perturbation at frequency $$\omega$$:
 
 $$\begin{aligned}
     P_{ab}
     = \frac{|V_{ba}|^2}{\hbar^2} \frac{\sin^2\!\big((\omega_{ba} - \omega) t / 2\big)}{(\omega_{ba} - \omega)^2}
 \end{aligned}$$
 
-Where $\omega_{ba} \equiv (E_b - E_a) / \hbar$.
-If we assume that $\Ket{b}$ irreversibly absorbs an unlimited number of particles,
-then we can interpret $P_{ab}$ as the "amount" of the current particle
-that has transitioned since the last period $2 \pi n / (\omega_{ba} \!-\! \omega)$.
+Where $$\omega_{ba} \equiv (E_b - E_a) / \hbar$$.
+If we assume that $$\Ket{b}$$ irreversibly absorbs an unlimited number of particles,
+then we can interpret $$P_{ab}$$ as the "amount" of the current particle
+that has transitioned since the last period $$2 \pi n / (\omega_{ba} \!-\! \omega)$$.
 
-For generality, let $E_b$ be the center
-of a state continuum with width $\Delta E$.
-In that case, $P_{ab}$ must be modified as follows,
-where $\rho(E_x)$ is the destination's
+For generality, let $$E_b$$ be the center
+of a state continuum with width $$\Delta E$$.
+In that case, $$P_{ab}$$ must be modified as follows,
+where $$\rho(E_x)$$ is the destination's
 [density of states](/know/concept/density-of-states/):
 
 $$\begin{aligned}
@@ -44,26 +44,26 @@ $$\begin{aligned}
     \frac{\sin^2\!\big((\omega_{xa} - \omega) t / 2\big)}{(\omega_{xa} - \omega)^2} \:\rho(E_x) \dd{E_x}
 \end{aligned}$$
 
-If $E_b$ is not in a continuum, then $\rho(E_x) = \delta(E_x - E_b)$.
-The integrand is a sharp sinc-function around $E_x$.
-For large $t$, it is so sharp that we can take out $\rho(E_x)$.
+If $$E_b$$ is not in a continuum, then $$\rho(E_x) = \delta(E_x - E_b)$$.
+The integrand is a sharp sinc-function around $$E_x$$.
+For large $$t$$, it is so sharp that we can take out $$\rho(E_x)$$.
 In that case, we also simplify the integration limits.
-Then we substitute $x \equiv (\omega_{xa}\!-\!\omega) / 2$ to get:
+Then we substitute $$x \equiv (\omega_{xa}\!-\!\omega) / 2$$ to get:
 
 $$\begin{aligned}
     P_{ab}
     &\approx \frac{2}{\hbar} |V_{ba}|^2 \rho(E_b) \int_{-\infty}^\infty \frac{\sin^2(x t)}{x^2} \:dx
 \end{aligned}$$
 
-This definite integral turns out to be $\pi |t|$,
-so we find, because clearly $t > 0$:
+This definite integral turns out to be $$\pi |t|$$,
+so we find, because clearly $$t > 0$$:
 
 $$\begin{aligned}
     P_{ab}
     &= \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b) \: t
 \end{aligned}$$
 
-The transition rate $R_{ab}$,
+The transition rate $$R_{ab}$$,
 i.e. the number of particles per unit time,
 then takes this form:
 
@@ -75,9 +75,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Note that the $t$-dependence has disappeared,
-and all that remains is a constant factor involving $E_b = E_a \!+\! \hbar \omega$,
-where $\omega$ is the resonance frequency.
+Note that the $$t$$-dependence has disappeared,
+and all that remains is a constant factor involving $$E_b = E_a \!+\! \hbar \omega$$,
+where $$\omega$$ is the resonance frequency.