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diff --git a/source/know/concept/fermis-golden-rule/index.md b/source/know/concept/fermis-golden-rule/index.md
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--- a/source/know/concept/fermis-golden-rule/index.md
+++ b/source/know/concept/fermis-golden-rule/index.md
@@ -10,74 +10,210 @@ categories:
 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
-time.
+Consider a general quantum system that is being affected by a sinusoidal potential.
+Specifically, the system's time-independent base Hamiltonian $$\hat{H}_0$$
+is perturbed by an $$\hat{H}_1(t)$$ given by:
 
-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 sinusoidal perturbation at frequency $$\omega$$:
+$$\begin{aligned}
+    \hat{H}_1(\vb{r}, t)
+    \equiv \hat{V}(\vb{r}) \cos(\omega t)
+    = \frac{1}{2} \hat{V}(\vb{r}) \: (e^{i \omega t} + e^{-i \omega t})
+\end{aligned}$$
+
+This potential is switched on at $$t = 0$$.
+Let us study the system's response using
+[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
+by expressing the perturbed wavefunction $$\ket{\Psi(t)}$$
+as a superposition of known eigenstates $$\ket{n}$$ of $$\hat{H}_0$$
+with time-dependent coefficient $$c_n(t)$$:
 
 $$\begin{aligned}
-    P_{ab}
-    = \frac{|V_{ba}|^2}{\hbar^2} \frac{\sin^2\!\big((\omega_{ba} - \omega) t / 2\big)}{(\omega_{ba} - \omega)^2}
+    \ket{\Psi(t)}
+    = \sum_n c_n(t) \ket{n} e^{-i E_n t / h}
 \end{aligned}$$
 
-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 $$2 \pi n / (\omega_{ba} \!-\! \omega)$$.
+In the absence of $$\hat{H}_1$$,
+the coefficients would simply be constants $$c_n^{(0)}$$.
+Perturbation theory then states that the first-order correction
+$$c_n^{(1)}(t) \approx c_n(t) - c_n^{(0)}$$ is as follows,
+with $$V_{mn} \equiv \matrixel{m}{\hat{V}}{n}$$:
 
-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/):
+$$\begin{aligned}
+    c_m^{(1)}(t)
+    &= - \frac{i}{2 \hbar} \sum_{n} c_n^{(0)} \: V_{mn}
+    \int_0^t e^{i \tau (\omega_{mn} + \omega)} + e^{i \tau (\omega_{mn} - \omega)} \dd{\tau}
+    \\
+    &= - \frac{1}{2 \hbar} \sum_{n} c_n^{(0)} \: V_{mn}
+    \bigg( \frac{e^{i t (\omega_{mn} + \omega)} - 1}{\omega_{mn} + \omega}
+    + \frac{e^{i t (\omega_{mn} - \omega)} - 1}{\omega_{mn} - \omega} \bigg)
+\end{aligned}$$
+
+For simplicity, we let the system start in a known state $$\ket{a}$$,
+such that $$c_n^{(0)} = \delta_{na}$$,
+and we focus on a single destination state $$m = b$$,
+effectively turning this into a two-level system.
+Let us assume that the driving frequency is close
+to that transition's resonance $$\omega \approx \omega_{ba}$$,
+such that the second term is dominant over the first, which can then be neglected.
+We therefore get:
 
 $$\begin{aligned}
-    P_{ab}
-    &= \frac{|V_{ba}|^2}{\hbar^2} \int_{E_b - \Delta E / 2}^{E_b + \Delta E / 2}
-    \frac{\sin^2\!\big((\omega_{xa} - \omega) t / 2\big)}{(\omega_{xa} - \omega)^2} \:\rho(E_x) \dd{E_x}
+    c_b^{(1)}(t)
+    &= - \frac{V_{ba}}{2 \hbar} \frac{e^{i (\omega_{ba} - \omega) t} - 1}{\omega_{ba} - \omega}
+    \\
+    &= - i \frac{V_{ba}}{2 i \hbar} \frac{e^{i (\omega_{ba} - \omega) t / 2} - e^{- i (\omega_{ba} - \omega) t / 2}}{\omega_{ba} - \omega} \: e^{i (\omega_{ba} - \omega) t / 2}
+    \\
+    &= - i \frac{V_{ba}}{\hbar} \frac{\sin\!\big( (\omega_{ba} - \omega) t / 2 \big)}{\omega_{ba} - \omega} \: e^{i t (\omega_{ba} - \omega) / 2}
+\end{aligned}$$
+
+Taking the norm squared yields the **transition probability**:
+the probability that a particle starting in state $$\ket{a}$$
+will be found in $$\ket{b}$$ at time $$t > 0$$:
+
+$$\begin{aligned}
+    \boxed{
+        P_{a \to b}(t)
+        = |c_b^{(1)}(t)|^2
+        = \frac{|V_{ba}|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_{ba} - \omega) t / 2 \big)}{(\omega_{ba} - \omega)^2}
+    }
 \end{aligned}$$
 
-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 $$\rho(E_x)$$.
-In that case, we also simplify the integration limits.
-Then we substitute $$x \equiv (\omega_{xa}\!-\!\omega) / 2$$ to get:
+This would be the same if $$\hat{H}_1 \equiv \hat{V} \sin(\omega t)$$.
+However, many authors define $$\hat{H}_1 \equiv \hat{V} e^{- i \omega t}$$,
+and then $$P_{a \to b}$$ is larger by a factor of $$4$$,
+which can cause confusion when comparing literature.
+
+In any case, the probability oscillates as a function of $$t$$
+with period $$T = 2 \pi / (\omega_{ba} - \omega)$$,
+so after one full period the particle is back in $$\ket{a}$$,
+and after $$T/2$$ the particle is fully in $$\ket{b}$$.
+See [Rabi oscillation](/know/concept/rabi-oscillation/)
+for a more accurate treatment of this "flopping" behaviour.
+
+However, when regarded as a function of $$\omega$$, the probability
+takes the form of a sinc-function centred around $$(\omega_{ba} - \omega)$$,
+so it is maximized for transitions with energy
+$$\hbar \omega = E_b - E_a$$.
+Also notice that the sinc-function becomes narrower over time:
+this has no effect if $$\omega = \omega_{ba}$$,
+however, if we are slightly off-resonance,
+then this means that $$P_{a \to b}$$ decreases over time.
+
+In fact, for $$t \to \infty$$, the sinc-function
+becomes so sharp that it can be related to
+the [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta$$.
+Consider this general identity (yes, the notation is sloppy, but it is clear):
 
 $$\begin{aligned}
-    P_{ab}
-    &\approx \frac{2}{\hbar} |V_{ba}|^2 \rho(E_b) \int_{-\infty}^\infty \frac{\sin^2(x t)}{x^2} \:dx
+    \lim_{t \to \infty} \frac{\sin^2(x t / 2)}{x^2}
+    = \frac{\pi}{2} \: \delta(x) \: t
 \end{aligned}$$
 
-This definite integral turns out to be $$\pi |t|$$,
-so we find, because clearly $$t > 0$$:
+
+{% include proof/start.html id="proof-relation" -%}
+First, observe that we can rewrite the fraction using an integral:
 
 $$\begin{aligned}
-    P_{ab}
-    &= \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b) \: t
+    \frac{\sin(x t / 2)}{x}
+    = \frac{e^{i x t / 2} - e^{-i x t / 2}}{2 i x}
+    = \frac{i}{2 i} \int_{-t/2}^{t/2} e^{i x \tau} \dd{\tau}
 \end{aligned}$$
 
-The transition rate $$R_{ab}$$,
+By taking the limit $$t \to \infty$$,
+it can be turned into a nascent Dirac delta function:
+
+$$\begin{aligned}
+    \lim_{t \to \infty} \frac{\sin(x t / 2)}{x}
+    = \lim_{t \to \infty} \frac{\pi}{2 \pi} \int_{-\infty}^{\infty} e^{i x \tau} \dd{\tau}
+    = \lim_{t \to \infty} \pi \delta(x)
+\end{aligned}$$
+
+On the topic of the delta function, we also show that its square is given by:
+
+$$\begin{aligned}
+    \delta^2(x)
+    = \delta(x) \lim_{t \to \infty} \frac{1}{2 \pi} \int_{-t/2}^{t/2} e^{i x \tau} \dd{\tau}
+    = \delta(x) \lim_{t \to \infty} \frac{t}{2 \pi}
+\end{aligned}$$
+
+Where we have set $$x = 0$$ according to the first delta function.
+This gives the target:
+
+$$\begin{aligned}
+    \lim_{t \to \infty} \frac{\sin^2(x t / 2)}{x^2}
+    = \pi^2 \delta^2(x)
+    = \frac{\pi}{2} \: \delta(x) \: t
+\end{aligned}$$
+{% include proof/end.html id="proof-relation" %}
+
+
+Inserting this into $$P_{a \to b}$$
+and using the general property that $$\delta(x) / c = \delta(c x)$$:
+
+$$\begin{aligned}
+    P_{a \to b}
+    &= \frac{\pi t}{2 \hbar^2} |V_{ba}|^2 \: \delta(\omega_{ba} - \omega)
+    \\
+    &= \frac{\pi t}{2 \hbar} |V_{ba}|^2 \: \delta(E_b - E_a - \hbar \omega)
+\end{aligned}$$
+
+By taking the $$t$$-derivative, we obtain the **transition rate** $$R_{a \to b}$$,
 i.e. the number of particles per unit time,
-then takes this form:
+which is given by **Fermi's golden rule**:
+
+$$\begin{aligned}
+    \pdv{P_{a \to b}}{t} =
+    \boxed{
+        R_{a \to b}
+        = \frac{\pi}{2 \hbar} |V_{ba}|^2 \: \delta(E_b - E_a - \hbar \omega)
+    }
+\end{aligned}$$
+
+Once again, beware that many authors define
+$$\hat{H}_1 \equiv \hat{V} e^{- i \omega t}$$,
+in which case the result is larger by a factor of $$4$$,
+which can cause confusion when comparing literature.
+For that alternative derivation, see the one-photon section of
+[multi-photon absorption](/know/concept/multi-photon-absorption/).
+
+What is most noteworthy about this result is
+that the $$t$$-dependence has disappeared,
+and that the rate is proportional to $$|V_{ba}|^2$$,
+i.e. the overlap between $$\ket{a}$$ and $$\ket{b}$$ with respect to $$\hat{V}$$.
+In practice, this result is usually used inside an integral over $$\omega$$,
+hence that "naked" Dirac delta function.
+
+For generality, let us take $$E_b$$ to be the center
+of a state continuum with width $$2 W$$.
+In that case, $$P_{a \to b}$$ must be modified as follows,
+where $$\rho(E_x)$$ is the destination's
+[density of states](/know/concept/density-of-states/):
+
+$$\begin{aligned}
+    P_{a \to b}
+    &= \frac{|V_{ba}|^2}{\hbar^2} \int_{E_b - W}^{E_b + W}
+    \frac{\sin^2\!\big((\omega_{xa} - \omega) t / 2\big)}{(\omega_{xa} - \omega)^2} \:\rho(E_x) \dd{E_x}
+    \\
+    &= \frac{\pi t}{2 \hbar} |V_{ba}|^2
+    \int_{E_b - W}^{E_b + W} \delta(E_x - E_a - \hbar \omega) \: \rho(E_x) \dd{E_x}
+    \\
+    &= \frac{\pi t}{2 \hbar} |V_{ba}|^2 \rho(E_a + \hbar \omega)
+\end{aligned}$$
+
+Where we used the aforementioned relation for $$t \to \infty$$ again.
+The transition rate is therefore:
 
 $$\begin{aligned}
+    \pdv{P_{a \to b}}{t} =
     \boxed{
-        R_{ab}
-        = \pdv{P_{ab}}{t}
-        = \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b)
+        R_{a \to b}
+        = \frac{\pi}{2 \hbar} |V_{ba}|^2 \, \rho(E_b)
     }
 \end{aligned}$$
 
-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.
+This is an alternative version of Fermi's golden rule.
+Unsurprisingly, if $$E_b$$ is not in a continuum,
+then we simply have $$\rho(E_x) = \delta(E_b - E_x)$$ and recover the previous version.