8 files changed, 438 insertions, 220 deletions

diff --git a/source/know/concept/einstein-coefficients/index.md b/source/know/concept/einstein-coefficients/index.md
index ad75ce5..2367abc 100644
--- a/source/know/concept/einstein-coefficients/index.md
+++ b/source/know/concept/einstein-coefficients/index.md
@@ -157,10 +157,10 @@ $$\begin{aligned}
     = - q E_0 z \cos(\omega t)
 \end{aligned}$$
 
-This form of $$\hat{H}_1$$ is a well-known case for
-[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
-which tells us that the transition probability
-from $$\ket{a}$$ to $$\ket{b}$$ is (to first order):
+From the derivation of
+[Fermi's golden rule](/know/concept/fermis-golden-rule/),
+we then know that the probability $$P_{ab}$$ of a transition
+from $$\ket{a}$$ to $$\ket{b}$$ is as follows (to first order):
 
 $$\begin{aligned}
     P_{ab}
@@ -210,6 +210,7 @@ However, the result would not be constant in time $$t$$,
 so is not really useful.
 
 
+
 ## Polarized light
 
 To solve this "problem", we generalize to (incoherent) polarized polychromatic light.
@@ -238,8 +239,7 @@ $$\begin{aligned}
     \int_0^\infty \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2} u(\omega) \dd{\omega}
 \end{aligned}$$
 
-From here, the derivation is similar to that of
-[Fermi's golden rule](/know/concept/fermis-golden-rule/),
+From here, the derivation is similar to that of Fermi's golden rule,
 despite the distinction that we are integrating over frequencies rather than states.
 
 At sufficiently large $$t$$, the integrand is sharply peaked at $$\omega = \omega_0$$
@@ -276,6 +276,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## Unpolarized light
 
 We can generalize the above result even further to unpolarized light.
diff --git a/source/know/concept/fermi-gas/index.md b/source/know/concept/fermi-gas/index.md
index 6d316cc..cd3d1a1 100644
--- a/source/know/concept/fermi-gas/index.md
+++ b/source/know/concept/fermi-gas/index.md
@@ -165,47 +165,38 @@ $$\begin{aligned}
     &= \frac{S}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{3/2}
 \end{aligned}$$
 
-The total population $$N = L^3 n$$ can therefore be expressed
-as a function of $$\varepsilon_F$$:
-
-$$\begin{aligned}
-    N(\varepsilon_F)
-    &= \frac{S L^3}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{3/2}
-\end{aligned}$$
-
-And from this we obtain a formula for the
-[density of states](/know/concept/density-of-states/)
-$$g$$ of a 3D Fermi gas:
+And from this we obtain the [density of states](/know/concept/density-of-states/)
+$$g$$ per unit volume of a 3D Fermi gas:
 
 $$\begin{aligned}
     \boxed{
         g(\varepsilon_F)
-        = \dv{N}{\varepsilon_F}
-        = \frac{S L^3}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{1/2}
+        = \dv{n}{\varepsilon_F}
+        = \frac{S}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{1/2}
     }
 \end{aligned}$$
 
 Now, $$\varepsilon_F$$ is the highest energy of a single fermion,
-but what about the total $$N$$-particle energy $$E$$?
+but what about the total $$N$$-particle energy $$E$$ of the cube?
 This is easy to calculate using the density of states:
 
 $$\begin{aligned}
-    E
+    \frac{E}{L^3}
     &= \int_0^{\varepsilon_F} \varepsilon \: g(\varepsilon) \dd{\varepsilon}
     \\
-    &= \frac{S L^3}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2}
+    &= \frac{S}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2}
     \int_0^{\varepsilon_F} \varepsilon^{3/2} \dd{\varepsilon}
     \\
-    &= \frac{3}{2} \frac{S L^3}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \: \frac{2}{5} \varepsilon_F^{5/2}
+    &= \frac{3}{2} \frac{S}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \: \frac{2}{5} \varepsilon_F^{5/2}
 \end{aligned}$$
 
-Here, we recognize $$N(\varepsilon_F)$$ from earlier,
-leading to the following expression for the total $$E$$:
+Here, we recognize $$n(\varepsilon_F)$$ from earlier,
+leading to the following total $$E$$ per unit volume:
 
 $$\begin{aligned}
     \boxed{
-        E
-        = \frac{3}{5} N \varepsilon_F
+        \frac{E}{L^3}
+        = \frac{3}{5} n \varepsilon_F
     }
 \end{aligned}$$
 
diff --git a/source/know/concept/fermis-golden-rule/index.md b/source/know/concept/fermis-golden-rule/index.md
index ea58ee6..cce8894 100644
--- 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.
 
 
 
diff --git a/source/know/concept/lehmann-representation/index.md b/source/know/concept/lehmann-representation/index.md
index 0dec718..d96eb66 100644
--- a/source/know/concept/lehmann-representation/index.md
+++ b/source/know/concept/lehmann-representation/index.md
@@ -78,7 +78,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-We now go through the same process for the lesser Green's function $$G_{\nu \nu'}^<(t, t')$$:
+We now go through the same process
+for the lesser Green's function $$G_{\nu \nu'}^<(t, t')$$:
 
 $$\begin{aligned}
     G_{\nu \nu'}^<(t - t')
@@ -137,20 +138,22 @@ $$\begin{aligned}
     \\
     &= - \frac{i}{\hbar Z} \Theta(t \!-\! t') \sum_{n n'}
     \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n}
-     \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) e^{i (E_n - E_{n'}) (t - t') / \hbar}
+    \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) e^{i (E_n - E_{n'}) (t - t') / \hbar}
 \end{aligned}$$
 
 We take the Fourier transform, but to ensure convergence,
-we must introduce an infinitesimal positive $$\eta \to 0^+$$ to the exponent
+we introduce an infinitesimal positive $$\eta \to 0^+$$ to the exponent
 (and eventually take the limit):
 
 $$\begin{aligned}
     G_{\nu \nu'}^R(\omega)
-    &= - \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \int_{-\infty}^\infty \Theta(t) e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega + i \eta) t} \dd{t}
+    &= - \frac{i}{\hbar Z} \sum_{n n'} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+     \int_{-\infty}^\infty \Theta(t) e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega + i \eta) t} \dd{t}
     \\
-    &= - \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \int_0^\infty e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega + i \eta) t} \dd{t}
+    &= - \frac{i}{\hbar Z} \sum_{n n'} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+    \int_0^\infty e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega + i \eta) t} \dd{t}
     \\
-    &= - \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big)
+    &= - \frac{i}{\hbar Z} \sum_{n n'} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
     \bigg[ \frac{\hbar e^{i (\hbar \omega + E_n - E_{n'}) t / \hbar} e^{- \eta t}}{i (\hbar \omega + E_n - E_{n'}) - \hbar \eta} \bigg]_0^\infty
 \end{aligned}$$
 
@@ -166,7 +169,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Finally, we go through the same steps for the advanced Green's function $$G_{\nu \nu'}^A(t, t')$$:
+Finally, we go through the same steps
+for the advanced Green's function $$G_{\nu \nu'}^A(t, t')$$:
 
 $$\begin{aligned}
     G_{\nu \nu'}^A(t \!-\! t')
@@ -174,7 +178,7 @@ $$\begin{aligned}
     \\
     &= \frac{i}{\hbar Z} \Theta(t' \!-\! t) \sum_{n n'}
     \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n}
-     \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) e^{i (E_n - E_{n'}) (t - t') / \hbar}
+    \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) e^{i (E_n - E_{n'}) (t - t') / \hbar}
 \end{aligned}$$
 
 For the Fourier transform, we must again introduce $$\eta \to 0^+$$
@@ -182,11 +186,13 @@ For the Fourier transform, we must again introduce $$\eta \to 0^+$$
 
 $$\begin{aligned}
     G_{\nu \nu'}^A(\omega)
-    &= \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \int_{-\infty}^\infty \Theta(-t) e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega - i \eta) t} \dd{t}
+    &= \frac{i}{\hbar Z} \sum_{n n'} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+    \int_{-\infty}^\infty \Theta(-t) e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega - i \eta) t} \dd{t}
     \\
-    &= \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \int_{-\infty}^0 e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega - i \eta) t} \dd{t}
+    &= \frac{i}{\hbar Z} \sum_{n n'} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+    \int_{-\infty}^0 e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega - i \eta) t} \dd{t}
     \\
-    &= \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big)
+    &= \frac{i}{\hbar Z} \sum_{n n'} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
     \bigg[ \frac{\hbar e^{i (\hbar \omega + E_n - E_{n'}) t / \hbar} e^{\eta t}}{i (\hbar \omega + E_n - E_{n'}) + \hbar \eta} \bigg]_{-\infty}^0
 \end{aligned}$$
 
diff --git a/source/know/concept/lindhard-function/index.md b/source/know/concept/lindhard-function/index.md
index 1e3c675..67af408 100644
--- a/source/know/concept/lindhard-function/index.md
+++ b/source/know/concept/lindhard-function/index.md
@@ -376,30 +376,46 @@ $$\begin{aligned}
     - \int_{-\infty}^{\infty} \frac{n_F(\xi_{- \vb{k}'})}{\xi_{- \vb{k}' - \vb{q}} - \xi_{- \vb{k}'} + \hbar (\omega + i \eta)} \dd{\vb{k}'}
     \bigg)
     \\
-    &= \frac{1}{4 \pi^3} \int_{-\infty}^{\infty} n_F(\xi_{\vb{k}}) \bigg(
+    &= \frac{1}{4 \pi^3} \int_{-\infty}^{\infty} \bigg(
     \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} + \hbar (\omega + i \eta)}
     + \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} - \hbar (\omega + i \eta)}
-    \bigg) \dd{\vb{k}}
+    \bigg) n_F(\xi_{\vb{k}}) \dd{\vb{k}}
+\end{aligned}$$
+
+Considering the system's symmetry, it is natural
+to switch to [spherical coordinates](/know/concept/spherical-coordinates/)
+$$(|\vb{k}|, \theta, \varphi)$$:
+
+$$\begin{aligned}
+    \chi_0
+    &= \frac{1}{4 \pi^3} \int_0^\infty \!\! \int_0^\pi \! \int_{-\pi}^\pi \! \bigg(
+    \frac{1}{\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}} + \hbar (\omega \!+\! i \eta)}
+    + \frac{1}{\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}} - \hbar (\omega \!+\! i \eta)}
+    \bigg) n_F(\xi_{\vb{k}}) \, |\vb{k}|^2 \sin{\theta} \dd{\varphi} \dd{\theta} \dd{|\vb{k}|}
+    \\
+    &= \frac{2 \pi}{4 \pi^3} \int_0^\infty \!\! \int_0^\pi \bigg(
+    \frac{1}{\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}} + \hbar (\omega \!+\! i \eta)}
+    + \frac{1}{\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}} - \hbar (\omega \!+\! i \eta)}
+    \bigg) n_F(\xi_{\vb{k}}) \, |\vb{k}|^2 \sin{\theta} \dd{\theta} \dd{|\vb{k}|}
 \end{aligned}$$
 
 At absolute zero $$T = 0$$, the Fermi gas has a well-known ground state,
 in which the occupied single-particle plane wave states
 form a sphere in $$\vb{k}$$-space with radius $$k_F$$.
-In that case, $$n_F(\xi_\vb{k})$$ reduces to
-a Heaviside step function $$\Theta(k_F - |\vb{k}|)$$.
-We switch to [spherical coordinates](/know/concept/spherical-coordinates/)
-$$(|\vb{k}|, \theta, \varphi)$$ and adjust $$|\vb{k}|$$'s
-integration limit based on the step function's demand that $$|\vb{k}| < k_F$$:
+This is a good approximation for nonzero $$T$$ too, because in practice,
+the electrons' Fermi level $$\xi_F \equiv \hbar^2 k_F^2 / (2 m)$$
+is huge compared to thermal excitations.
+Formally, $$n_F(\xi_\vb{k}) \approx \Theta(k_F - |\vb{k}|)$$,
+so we adjust the integration limit:
 
 $$\begin{aligned}
-    \chi_0(\vb{q}, \omega)
-    &= \frac{2 \pi}{4 \pi^3} \int_0^{k_F} \! \int_0^\pi
-    \bigg( \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} + \hbar (\omega \!+\! i \eta)}
-    + \frac{1}{\xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}} - \hbar (\omega \!+\! i \eta)} \bigg) |\vb{k}|^2 \sin{\theta} \dd{\theta} \dd{|\vb{k}|}
+    \chi_0
+    &\approx \frac{1}{2 \pi^2} \int_0^{k_F} \! \int_0^\pi \bigg(
+    \frac{1}{\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}} + \hbar (\omega \!+\! i \eta)}
+    + \frac{1}{\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}} - \hbar (\omega \!+\! i \eta)}
+    \bigg) |\vb{k}|^2 \sin{\theta} \dd{\theta} \dd{|\vb{k}|}
 \end{aligned}$$
 
-Where we have already used that $$\varphi$$ does not appear
-in the integrand to get a factor $$2 \pi$$.
 Next, let us take a closer look at the excitation energy in the denominator.
 Thanks to symmetry, we can always choose our coordinate system
 such that $$\vb{q}$$ points in the $$\theta = 0$$ direction, yielding:
@@ -426,14 +442,20 @@ $$y \equiv \hbar (\omega + i \eta) / (4 \xi_F)$$
 and $$c \equiv \cos{\theta}$$, giving:
 
 $$\begin{aligned}
-    \chi_0(\vb{q}, \omega)
+    \chi_0
     &= \frac{k_F^3}{8 \pi^2 \xi_F} \int_0^1 \! \int_{-1}^1 \bigg(
     \frac{1}{-x^2 - x \kappa c + y} + \frac{1}{-x^2 - x \kappa c - y}
     \bigg) \kappa^2 \dd{c} \dd{\kappa}
 \end{aligned}$$
 
-Evaluating the integral over $$c$$ gives
-a result that involves the complex natural logarithm $$\ln$$:
+From now on, we need to treat $$\chi_0$$'s real and imaginary parts separately
+using the [Sokhotski-Plemelj theorem](/know/concept/sokhotski-plemelj-theorem/).
+The following is valid for the real part,
+but is (for now) still written as a complex expression.
+The point is that, to the best of my knowledge,
+it is not practical to extract the imaginary part
+from the "real part's complex result" so to speak.
+So, hiding the [principal value](/know/concept/cauchy-principal-value/):
 
 $$\begin{aligned}
     \chi_0
@@ -494,13 +516,23 @@ $$\begin{aligned}
     \bigg)
 \end{aligned}$$
 
-We thus arrive at this expression
+Finally, we also know that the [density of states](/know/concept/density-of-states/)
+$$g$$ per unit volume of a Fermi gas is given by:
+
+$$\begin{aligned}
+    g(\xi_F)
+    = \frac{2}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \xi_F^{1/2}
+    = \frac{m k_F}{\pi^2 \hbar^2}
+\end{aligned}$$
+
+Now we officially take the real part,
+and arrive at the following expression
 for the response function of non-interacting electrons:
 
 $$\begin{aligned}
     \boxed{
-        \chi_0(\vb{q}, \omega)
-        = - \frac{m k_F}{\pi^2 \hbar^2} \: \mathcal{F}\bigg(\frac{|\vb{q}|}{2 k_F}, \: \frac{\hbar (\omega \!+\! i \eta)}{4 \xi_F}\bigg)
+        \Real{\chi_0(\vb{q}, \omega)}
+        = - g(\xi_F) \: \mathcal{F}\bigg(\frac{|\vb{q}|}{2 k_F}, \: \frac{\hbar \omega}{4 \xi_F}\bigg)
     }
 \end{aligned}$$
 
@@ -511,17 +543,144 @@ $$\begin{aligned}
     \boxed{
         \mathcal{F}(x, y)
         \equiv \frac{1}{2}
-        + \frac{1}{8 x} \bigg( 1 - \Big(x \!+\! \frac{y}{x}\Big)^{\!2} \bigg) \ln\!\bigg( \frac{x + y/x + 1}{x + y/x - 1} \bigg)
-        + \frac{1}{8 x} \bigg( 1 - \Big(x \!-\! \frac{y}{x}\Big)^{\!2} \bigg) \ln\!\bigg( \frac{x - y/x + 1}{x - y/x - 1} \bigg)
+        + \frac{1}{8 x} \bigg( 1 - \Big(x + \frac{y}{x}\Big)^{\!2} \bigg) \ln\!\bigg| \frac{x + y/x + 1}{x + y/x - 1} \bigg|
+        + \frac{1}{8 x} \bigg( 1 - \Big(x - \frac{y}{x}\Big)^{\!2} \bigg) \ln\!\bigg| \frac{x - y/x + 1}{x - y/x - 1} \bigg|
     }
 \end{aligned}$$
 
-Keep in mind that these are complex logarithms,
-whose branch cut should lie along the negative real axis,
-such that the results' imaginary parts are always between $$-\pi$$ and $$\pi$$.
-That way, it is consistent with alternative derivations
-that use the [Sokhotski-Plemelj theorem](/know/concept/sokhotski-plemelj-theorem/)
-(e.g. Mihaila's paper).
+Now we move on to $$\chi_0$$'s imaginary part.
+Recall the following expression from earlier:
+
+$$\begin{aligned}
+    \chi_0
+    &= \frac{- k_F^3}{8 \pi^2 \xi_F} \frac{1}{x} \int_0^1 \! \int_{-1}^1 \bigg(
+    \frac{1}{x + y/x + \kappa c} + \frac{1}{x - y/x + \kappa c}
+    \bigg) \kappa^2 \dd{c} \dd{\kappa}
+\end{aligned}$$
+
+The Sokhotski-Plemelj theorem then states that the imaginary part is as shown below.
+There is a nasty detail here regarding the positive infinitesimal $$\eta \to 0^{+}$$:
+the first input term's denominator contains $$+ i \eta$$,
+so the theorem can be applied normally,
+but the second term has $$-i \eta$$, meaning the theorem's output must be negated!
+
+$$\begin{aligned}
+    \Imag{\chi_0}
+    &= \frac{-2 m k_F^3}{8 \pi^2 \hbar^2 k_F^2} \frac{1}{x} \int_0^1 \! \int_{-1}^1 \bigg(
+    \!-\! \pi \delta(z_{+} + \kappa c) + \pi \delta(z_{-} + \kappa c)
+    \bigg) \kappa^2 \dd{c} \dd{\kappa}
+    \\
+    &= \frac{\pi m k_F}{4 \pi^2 \hbar^2} \frac{1}{x} \int_0^1 \bigg[
+    \frac{1}{\kappa} \Theta(z_{+} + \kappa c) - \frac{1}{\kappa} \Theta(z_{-} + \kappa c)
+    \bigg]_{-1}^1 \kappa^2 \dd{\kappa}
+    \\
+    &= \frac{\pi}{4 x} \: g(\xi_F) \int_0^1 \bigg(
+    \Theta(z_{+} + \kappa) - \Theta(z_{+} - \kappa) - \Theta(z_{-} + \kappa) + \Theta(z_{-} - \kappa)
+    \bigg) \kappa \dd{\kappa}
+\end{aligned}$$
+
+These integrals are easily evaluated
+according to the following straightforward relations:
+
+$$\begin{aligned}
+    \int_0^1 \Theta(\alpha - \kappa) \: \kappa \dd{\kappa}
+    &= \frac{1}{2} \big( \alpha^2 - (\alpha^2 - 1) \: \Theta(\alpha - 1) \big) \: \Theta(\alpha)
+    \\
+    \int_0^1 \Theta(\alpha + \kappa) \: \kappa \dd{\kappa}
+    &= \frac{1}{2} \big( \alpha^2 \: \Theta(\alpha) - \alpha^2 + 1 \big) \: \Theta(\alpha + 1)
+\end{aligned}$$
+
+Let us assume that $$y > 0$$ for now,
+such that $$z_{+} > 0$$ and $$z_{+} > z_{-}$$.
+We can then naively distinguish several options to be considered:
+
+1.  $$z_{+} > z_{-} > 1 > 0 > -1$$.
+2.  $$z_{+} > 1 > z_{-} > 0 > -1$$.
+3.  $$z_{+} > 1 > 0 > z_{-} > -1$$.
+4.  $$z_{+} > 1 > 0 > -1 > z_{-}$$.
+5.  $$1 > z_{+} > z_{-} > 0 > -1$$.
+6.  $$1 > z_{+} > 0 > z_{-} > -1$$.
+7.  $$1 > z_{+} > 0 > -1 > z_{-}$$.
+
+It can be shown that #7 is impossible
+due to the definitions of $$z_{+}$$ and $$z_{-}$$.
+A tedious investigation of the six remaining cases then yields:
+
+1.  In cases #1 or #4,
+    or, equivalently, if both $$|z_{+}| > 1$$ and $$|z_{-}| > 1$$,
+    the result vanishes:
+
+    $$\begin{aligned}
+        \Imag{\chi_0}
+        = 0
+    \end{aligned}$$
+
+    An alternative statement of those conditions can be found, namely:
+
+    $$\begin{aligned}
+        \begin{cases}
+            y > x^2 + x
+            &\mathrm{if} \: x < 1
+            \\
+            x^2 - x > y > 0
+            &\mathrm{if} \: x > 1
+        \end{cases}
+    \end{aligned}$$
+
+2.  In cases #2 or #3,
+    or, equivalently, if both $$|z_{+}| > 1$$ and $$|z_{-}| < 1$$,
+    the result is parabolic in $$y$$.
+    These conditions can be restated as $$x^2 + x > y > |x^2 - x|$$.
+
+    $$\begin{aligned}
+        \Imag{\chi_0}
+        &= -\frac{\pi}{8 x} \: g(\xi_F) \: (1 - z_{-}^2)
+    \end{aligned}$$
+
+3.  In cases #5 or #6,
+    or, equivalently, if both $$|z_{+}| < 1$$ and $$|z_{-}| < 1$$
+    the result is linear in $$y$$.
+    These conditions can be restated as $$-(x^2 - x) > y > 0$$.
+
+    $$\begin{aligned}
+        \Imag{\chi_0}
+        &= \frac{\pi}{8 x} \: g(\xi_F) \: (z_{-}^2 - z_{+}^2)
+        = - \frac{\pi}{2 x} \: g(\xi_F) \: y
+    \end{aligned}$$
+
+Repeating the analysis for $$y < 0$$ and putting everything together,
+we arrive at the following formula for the imaginary part
+of the non-interacting response function:
+
+$$\begin{gathered}
+    \boxed{
+        \Imag{\chi_0(\vb{q}, \omega)}
+        = - g(\xi_F) \: \mathcal{G}\bigg( \frac{|\vb{q}|}{2 k_F}, \: \frac{\hbar \omega}{4 \xi_F} \bigg)
+    }
+    \\
+    \boxed{
+        \mathcal{G}(x, y)
+        \equiv
+        \begin{cases}
+            0
+            &\mathrm{if} \: |y| > x^2 + x > 0
+            \\
+            \sgn(y) \displaystyle\frac{\pi}{8 x} \bigg( 1 - \Big( x - \displaystyle\frac{|y|}{x} \Big)^2 \bigg)
+            &\mathrm{if} \: x^2 + x > |y| > |x^2 - x|
+            \\
+            \displaystyle\frac{\pi}{2 x} y
+            &\mathrm{if} \: \!-\!(x^2 - x) > |y| > 0
+            \\
+            0
+            &\mathrm{if} \: x^2 - x > |y| > 0
+        \end{cases}
+    }
+\end{gathered}$$
+
+We are being a bit sloppy here by ignoring the boundaries,
+e.g. $$|y| = x^2 - x$$.
+Fortunately, they are all naturally continuous,
+so either of the two neighboring formulas can be used.
 
 
 
@@ -594,7 +753,7 @@ $$\begin{aligned}
     \boxed{
         \mathcal{F}(x, 0)
         = \frac{1}{2}
-        + \frac{1}{4 x} (1 - x^2) \ln\!\bigg( \frac{x + 1}{x - 1} \bigg)
+        + \frac{1}{4 x} (1 - x^2) \ln\!\bigg| \frac{x + 1}{x - 1} \bigg|
     }
 \end{aligned}$$
 
diff --git a/source/know/concept/parsevals-theorem/index.md b/source/know/concept/parsevals-theorem/index.md
index b2db490..06e9406 100644
--- a/source/know/concept/parsevals-theorem/index.md
+++ b/source/know/concept/parsevals-theorem/index.md
@@ -45,7 +45,8 @@ $$\begin{aligned}
     &= 2 \pi B^2 \iint \tilde{f}^*(k') \: \tilde{g}(k) \: \delta\big(s (k' \!-\! k)\big) \dd{k'} \dd{k}
     \\
     &= \frac{2 \pi B^2}{|s|} \int_{-\infty}^\infty \tilde{f}^*(k) \: \tilde{g}(k) \dd{k}
-    = \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}}{\tilde{g}}
+    \\
+    &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}}{\tilde{g}}
 \end{aligned}$$
 
 Where $$\delta(k)$$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
@@ -67,7 +68,8 @@ $$\begin{aligned}
     &= 2 \pi A^2 \iint f^*(x') \: g(x) \: \delta\big(s (x \!-\! x')\big) \dd{x'} \dd{x}
     \\
     &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
-    = \frac{2 \pi A^2}{|s|} \inprod{f}{g}
+    \\
+    &= \frac{2 \pi A^2}{|s|} \inprod{f}{g}
 \end{aligned}$$
 {% include proof/end.html id="proof-fourier" %}
 
diff --git a/source/know/concept/schwartz-distribution/index.md b/source/know/concept/schwartz-distribution/index.md
index 2492d91..3d5186f 100644
--- a/source/know/concept/schwartz-distribution/index.md
+++ b/source/know/concept/schwartz-distribution/index.md
@@ -11,7 +11,7 @@ A **Schwartz distribution**, also known as a **generalized function**,
 is a generalization of a function,
 allowing us to work with otherwise pathological definitions.
 
-Notable examples of distributions are
+Notable examples of such distributions are
 the [Dirac delta function](/know/concept/dirac-delta-function/)
 and the [Heaviside step function](/know/concept/heaviside-step-function/),
 whose unusual properties are justified by this generalization.
@@ -31,13 +31,13 @@ Furthermore, all test functions must be infinitely differentiable.
 These are quite strict requirements.
 
 The **space of distributions** $$\mathcal{S}'$$ (note the prime)
-is then said to consist of *functionals* $$f[\phi]$$
-which map a test function $$\phi$$ from $$\mathcal{S}$$,
+is then said to consist of *functionals* $$\delta[\phi]$$
+that map a test function $$\phi$$ from $$\mathcal{S}$$
 to a number from $$\mathbb{C}$$;
-this is often written as $$\Inprod{f}{\phi}$$.
+this is often written as $$\Inprod{\delta}{\phi}$$.
 This notation looks like the inner product of
 a [Hilbert space](/know/concept/hilbert-space/),
-for good reason: any well-behaved function $$f(x)$$ can be embedded
+and there is a good reason for that: any well-behaved function $$f(x)$$ can be embedded
 into $$\mathcal{S}'$$ by defining the corresponding functional $$f[\phi]$$ as follows:
 
 $$\begin{aligned}
@@ -49,10 +49,11 @@ $$\begin{aligned}
 Not all functionals qualify for $$\mathcal{S}'$$:
 they also need to be linear in $$\phi$$, and **continuous**,
 which in this context means: if a series $$\phi_n$$
-converges to $$\phi$$, then $$\Inprod{f}{\phi_n}$$
-converges to $$\Inprod{f}{\phi}$$ for all $$f$$.
+converges to $$\phi$$, then $$\Inprod{\delta}{\phi_n}$$
+converges to $$\Inprod{\delta}{\phi}$$ for all $$\delta$$.
 
-The power of this generalization is that $$f(x)$$ does not need to be well-behaved:
+The power of the generalization is that $$f(x)$$
+in fact does not *need* to be well-behaved:
 for example, the Dirac delta function can also be used,
 whose definition is nonsensical *outside* of an integral,
 but perfectly reasonable *inside* one.
@@ -60,7 +61,7 @@ By treating it as a distribution,
 we gain the ability to sanely define e.g. its derivatives.
 
 Using the example of embedding a well-behaved function $$f(x)$$ into $$\mathcal{S}$$,
-we can work out what the derivative of a distribution is:
+we can work out what the derivative of a distribution is using partial integration:
 
 $$\begin{aligned}
     \Inprod{f'}{\phi}
@@ -68,20 +69,21 @@ $$\begin{aligned}
     = \Big[ f(x) \: \phi(x) \Big]_{-\infty}^\infty - \int_{-\infty}^\infty f(x) \: \phi'(x) \dd{x}
 \end{aligned}$$
 
-The test function removes the boundary term, yielding the result
-$$- \Inprod{f}{\phi'}$$. Although this was an example for a specific $$f(x)$$,
-we use it to define the derivative of any distribution:
+The test function removes the first term,
+yielding the result $$- \Inprod{f}{\phi'}$$.
+Although this was an example for a well-behaved $$f(x)$$,
+we use it to define the derivative of any distribution $$\delta \in \mathcal{S}'$$:
 
 $$\begin{aligned}
     \boxed{
-        \Inprod{f'}{\phi} = - \Inprod{f}{\phi'}
+        \Inprod{\delta'}{\phi}
+        \equiv - \Inprod{\delta}{\phi'}
     }
 \end{aligned}$$
 
 Using the same trick, we can find the
-[Fourier transform](/know/concept/fourier-transform/) (FT)
-of a generalized function.
-We define the FT as follows,
+[Fourier transform](/know/concept/fourier-transform/) (FT) of a distribution.
+Let us define the FT like so,
 but be prepared for some switching of the names $$k$$ and $$x$$:
 
 $$\begin{aligned}
@@ -89,30 +91,34 @@ $$\begin{aligned}
     = \int_{-\infty}^\infty \phi(k) \exp(- i k x) \dd{k}
 \end{aligned}$$
 
-The FT of a Schwartz distribution $$f$$ then turns out to be as follows:
+The FT of a well-behaved $$f(x)$$ is then as follows
+(this is basically [Parseval's theorem](/know/concept/parsevals-theorem/)):
 
 $$\begin{aligned}
     \inprod{\tilde{f}}{\phi}
     &= \int_{-\infty}^\infty \tilde{f}(k) \: \phi(k) \dd{k}
-    = \iint_{-\infty}^\infty f(x) \exp(- i k x) \: \phi(k) \dd{x} \dd{k}
+    \\
+    &= \iint_{-\infty}^\infty f(x) \exp(- i k x) \: \phi(k) \dd{x} \dd{k}
     \\
     &= \int_{-\infty}^\infty f(x) \: \tilde{\phi}(x) \dd{x}
-    = \inprod{f}{\tilde{\phi}}
+    \\
+    &= \inprod{f}{\tilde{\phi}}
 \end{aligned}$$
 
-Note that the ordinary FT $$\tilde{f}(k) = \hat{\mathcal{F}}\{f(x)\}$$ is
-already a 1:1 mapping of test functions $$\phi \leftrightarrow \tilde{\phi}$$.
-As it turns out,
-in this generalization it is also a 1:1 mapping of distributions in $$\mathcal{S}'$$,
-defined as:
+So we defined the FT and derived one of its properties,
+but to generalize to Schwartz distributions $$\delta$$, we swap the roles:
+the property is an axiom, and serves as the definition of the FT $$\tilde{\delta}$$:
 
 $$\begin{aligned}
     \boxed{
-        \inprod{\tilde{f}}{\phi}
-        = \inprod{f}{\tilde{\phi}}
+        \inprod{\tilde{\delta}}{\phi}
+        \equiv \inprod{\delta}{\tilde{\phi}}
     }
 \end{aligned}$$
 
+It can be shown that this generalization is a 1:1 mapping
+between distributions in $$\mathcal{S}'$$.
+
 
 
 ## References
diff --git a/source/know/concept/time-dependent-perturbation-theory/index.md b/source/know/concept/time-dependent-perturbation-theory/index.md
index 841c219..b0ffa42 100644
--- a/source/know/concept/time-dependent-perturbation-theory/index.md
+++ b/source/know/concept/time-dependent-perturbation-theory/index.md
@@ -78,89 +78,6 @@ and we cannot go any further without considering a specific $$\hat{H}_1$$.
 
 
 
-## Sinusoidal perturbation
-
-Arguably the most important perturbation
-is a sinusoidally-varying potential, which represents
-e.g. incoming electromagnetic waves,
-or an AC voltage being applied to the system.
-In this case, $$\hat{H}_1$$ has the following form:
-
-$$\begin{aligned}
-    \hat{H}_1(\vec{r}, t)
-    \equiv V(\vec{r}) \sin(\omega t)
-    = \frac{1}{2 i} V(\vec{r}) \: \big( \exp(i \omega t) - \exp(-i \omega t) \big)
-\end{aligned}$$
-
-We abbreviate $$V_{mn} = \matrixel{m}{V}{n}$$,
-and take the first-order correction formula:
-
-$$\begin{aligned}
-    c_m^{(1)}(t)
-    &= - \frac{1}{2 \hbar} \sum_{n} V_{mn} c_n^{(0)}
-    \int_0^t \exp\!\big(i \tau (\omega_{mn} \!+\! \omega)\big) - \exp\!\big(i \tau (\omega_{mn} \!-\! \omega)\big) \dd{\tau}
-    \\
-    &= \frac{i}{2 \hbar} \sum_{n} V_{mn} c_n^{(0)}
-    \bigg( \frac{\exp\!\big(i t (\omega_{mn} \!+\! \omega) \big) - 1}{\omega_{mn} + \omega}
-    + \frac{\exp\!\big(i t (\omega_{mn} \!-\! \omega) \big) - 1}{\omega_{mn} - \omega} \bigg)
-\end{aligned}$$
-
-For simplicity, we let the system start in a known state $$\Ket{a}$$,
-such that $$c_n^{(0)} = \delta_{na}$$,
-and we assume that the driving frequency is close to resonance $$\omega \approx \omega_{ma}$$,
-such that the second term dominates the first, which can then be neglected.
-We thus get:
-
-$$\begin{aligned}
-    c_m^{(1)}(t)
-    &= i \frac{V_{ma}}{2 \hbar} \frac{\exp\!\big(i t (\omega_{ma} \!-\! \omega) \big) - 1}{\omega_{ma} - \omega}
-    \\
-    &= i \frac{V_{ma}}{2 \hbar}
-    \frac{\exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big) - \exp\!\big(\!-\! i t (\omega_{ma} \!-\! \omega) / 2 \big)}{\omega_{ma} - \omega}
-    \: \exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big)
-    \\
-    &= - \frac{V_{ma}}{\hbar}
-    \frac{\sin\!\big( t (\omega_{ma} \!-\! \omega) / 2 \big)}{\omega_{ma} - \omega}
-    \: \exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big)
-\end{aligned}$$
-
-Taking the norm squared yields the **transition probability**:
-the probability that a particle that started in state $$\Ket{a}$$
-will be found in $$\Ket{m}$$ at time $$t$$:
-
-$$\begin{aligned}
-    \boxed{
-        P_{a \to m}
-        = |c_m^{(1)}(t)|^2
-        = \frac{|V_{ma}|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_{ma} - \omega) t / 2 \big)}{(\omega_{ma} - \omega)^2}
-    }
-\end{aligned}$$
-
-The result would be the same if $$\hat{H}_1 \equiv V \cos(\omega t)$$.
-However, if instead $$\hat{H}_1 \equiv V \exp(- i \omega t)$$,
-the result 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 $$T = 2 \pi / (\omega_{ma} \!-\! \omega)$$,
-so after one period the particle is back in $$\Ket{a}$$,
-and after $$T/2$$ the particle is 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 $$(\omega_{ma} \!-\! \omega)$$,
-so it is highest for transitions with energy $$\hbar \omega = E_m \!-\! E_a$$.
-
-Also note that the sinc-distribution becomes narrower over time,
-which roughly means that it takes some time
-for the system to "notice" that
-it is being driven periodically.
-In other words, there is some "inertia" to it.
-
-
-
 ## References
 1.  D.J. Griffiths, D.F. Schroeter,
     *Introduction to quantum mechanics*, 3rd edition,