Minor rewrites and corrections

10 files changed, 239 insertions, 112 deletions

diff --git a/content/know/concept/electric-dipole-approximation/index.pdc b/content/know/concept/electric-dipole-approximation/index.pdc
index 265babf..67c73ee 100644
--- a/content/know/concept/electric-dipole-approximation/index.pdc
+++ b/content/know/concept/electric-dipole-approximation/index.pdc
@@ -20,120 +20,142 @@ The general Hamiltonian of an electron in such a wave is given by:
 
 $$\begin{aligned}
     \hat{H}
-    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{2 m} (\vec{A} \cdot \vec{P} + \vec{P} \cdot \vec{A}) + \frac{q^2 \vec{A}{}^2}{2m} + V
+    &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \varphi
+    \\
+    &= \frac{\vu{P}{}^2}{2 m} - \frac{q}{2 m} (\vb{A} \cdot \vu{P} + \vu{P} \cdot \vb{A}) + \frac{q^2 \vb{A}^2}{2m} + q \varphi
+\end{aligned}$$
+
+With charge $q = - e$,
+canonical momentum operator $\vu{P} = - i \hbar \nabla$,
+and magnetic vector potential $\vb{A}(\vb{x}, t)$.
+We reduce this by fixing the Coulomb gauge $\nabla \cdot \vb{A} = 0$,
+so that $\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$:
+
+$$\begin{aligned}
+    \comm*{\vb{A}}{\vu{P}} \psi
+    &= -i \hbar \vb{A} \cdot (\nabla \psi) + i \hbar \nabla \cdot (\vb{A} \psi)
+    \\
+    &= i \hbar (\nabla \cdot \vb{A}) \psi
+    = 0
 \end{aligned}$$
 
-With charge $q = - e$
-and electromagnetic vector potential $\vec{A}(\vec{r}, t)$.
-We reduce this by fixing the Coulomb gauge $\nabla \cdot \vec{A} = 0$,
-so that $\vec{A} \cdot \vec{P} = \vec{P} \cdot \vec{A}$,
-and assume that $\vec{A}{}^2$ is negligible:
+Where $\psi$ is an arbitrary test function.
+Assuming $\vb{A}$ is so small that $\vb{A}{}^2$ is negligible, we split $\hat{H}$ as follows,
+where $\hat{H}_1$ can be regarded as a perturbation to $\hat{H}_0$:
 
 $$\begin{aligned}
     \hat{H}
     = \hat{H}_0 + \hat{H}_1
     \qquad \quad
     \hat{H}_0
-    \equiv \frac{\vec{P}{}^2}{2 m} + V
+    \equiv \frac{\vu{P}{}^2}{2 m} + q \varphi
     \qquad \quad
     \hat{H}_1
-    \equiv - \frac{q}{m} \vec{P} \cdot \vec{A}
+    \equiv - \frac{q}{m} \vu{P} \cdot \vb{A}
 \end{aligned}$$
 
-We have split $\hat{H}$ into $\hat{H}_0$
-and a perturbation $\hat{H}_1$, since $\vec{A}$ is small.
-In an electromagnetic wave,
-$\vec{A}$ is oscillating sinusoidally in time and space as follows:
+In an electromagnetic wave, $\vb{A}$ is oscillating sinusoidally in time and space:
 
 $$\begin{aligned}
-    \vec{A}(\vec{r}, t) = - i \vec{A}_0 \exp\!(i \vec{k} \cdot \vec{r} - i \omega t)
+    \vb{A}(\vb{x}, t) = \vb{A}_0 \sin\!(\vb{k} \cdot \vb{x} - \omega t)
 \end{aligned}$$
 
-The corresponding perturbative
-[electric field](/know/concept/electric-field/) $\vec{E}$
-points in the same direction:
+Mathematically, it is more convenient to represent this with a complex exponential,
+whose real part should be taken at the end of the calculation:
 
 $$\begin{aligned}
-    \vec{E}(\vec{r}, t)
-    = - \pdv{\vec{A}}{t}
-    = \vec{E}_0 \exp\!(i \vec{k} \cdot \vec{r} - i \omega t)
+    \vb{A}(\vb{x}, t) = - i \vb{A}_0 \exp\!(i \vb{k} \cdot \vb{x} - i \omega t)
 \end{aligned}$$
 
-Where $\vec{E}_0 = \omega \vec{A}_0$.
+The corresponding perturbative [electric field](/know/concept/electric-field/) $\vb{E}$ is then given by:
+
+$$\begin{aligned}
+    \vb{E}(\vb{x}, t)
+    = - \pdv{\vb{A}}{t}
+    = \vb{E}_0 \exp\!(i \vb{k} \cdot \vb{x} - i \omega t)
+\end{aligned}$$
+
+Where $\vb{E}_0 = \omega \vb{A}_0$.
 Let us restrict ourselves to visible light,
-whose wavelength $2 \pi / k \approx 10^{-6} \:\mathrm{m}$.
-Meanwhile, an atomic orbital is on the order of $10^{-10} \:\mathrm{m}$,
-so $\vec{k} \cdot \vec{r}$ is negligible:
+whose wavelength $2 \pi / |\vb{k}| \sim 10^{-6} \:\mathrm{m}$.
+Meanwhile, an atomic orbital is several Bohr $\sim 10^{-10} \:\mathrm{m}$,
+so $\vb{k} \cdot \vb{x}$ is negligible:
 
 $$\begin{aligned}
     \boxed{
-        \vec{E}(\vec{r}, t)
-        \approx \vec{E}_0 \exp\!(- i \omega t)
+        \vb{E}(\vb{x}, t)
+        \approx \vb{E}_0 \exp\!(- i \omega t)
     }
 \end{aligned}$$
 
 This is the **electric dipole approximation**:
-we ignore all spatial variation of $\vec{E}$,
+we ignore all spatial variation of $\vb{E}$,
 and only consider its temporal oscillation.
 Also, since we have not used the word "photon",
 we are implicitly treating the radiation classically,
 and the electron quantum-mechanically.
 
-Next, we want to convert $\hat{H}_1$
-to use the electric field $\vec{E}$ instead of the potential $\vec{A}$.
-To do so, we rewrite the momemtum $\vec{P} = m \: \dv*{\vec{r}}{t}$
+Next, we want to rewrite $\hat{H}_1$
+to use the electric field $\vb{E}$ instead of the potential $\vb{A}$.
+To do so, we use that $\vu{P} = m \: \dv*{\vu{x}}{t}$
 and evaluate this in the [interaction picture](/know/concept/interaction-picture/):
 
 $$\begin{aligned}
-    \matrixel{2}{\dv*{\vec{r}}{t}}{1}
-    &= \frac{i}{\hbar} \matrixel{2}{[\hat{H}_0, \vec{r}]}{1}
-    = \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vec{r} - \vec{r} \hat{H}_0}{1}
-    \\
-    &= \frac{i}{\hbar} (E_2 - E_1) \matrixel{2}{\vec{r}}{1}
-    = i \omega_0 \matrixel{2}{\vec{r}}{1}
+    \vu{P}
+    = m \dv*{\vu{x}}{t}
+    = m \frac{i}{\hbar} \comm*{\hat{H}_0}{\vu{x}}
+    = m \frac{i}{\hbar} (\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0)
+\end{aligned}$$
+
+Taking the off-diagonal inner product with
+the two-level system's states $\ket{1}$ and $\ket{2}$ gives:
+
+$$\begin{aligned}
+    \matrixel{2}{\vu{P}}{1}
+    = m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1}
+    = m i \omega_0 \matrixel{2}{\vu{x}}{1}
 \end{aligned}$$
 
-Therefore, $\vec{P} / m = i \omega_0 \vec{r}$,
-where $\omega_0 \equiv (E_2 - E_1) / \hbar$ is the resonance frequency of the transition,
-close to which we assume that $\vec{A}$ and $\vec{E}$ are oscillating, i.e. $\omega \approx \omega_0$.
+Therefore, $\vu{P} / m = i \omega_0 \vu{x}$,
+where $\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$ is the resonance of the energy gap,
+close to which we assume that $\vb{A}$ and $\vb{E}$ are oscillating, i.e. $\omega \approx \omega_0$.
 We thus get:
 
 $$\begin{aligned}
     \hat{H}_1(t)
-    &= - \frac{q}{m} \vec{P} \cdot \vec{A}
-    = - (- i i) q \omega_0 \vec{r} \cdot \vec{A}_0 \exp\!(- i \omega t)
+    &= - \frac{q}{m} \vu{P} \cdot \vb{A}
+    = - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp\!(- i \omega t)
     \\
-    &\approx - q \vec{r} \cdot \vec{E}_0 \exp\!(- i \omega t)
-    = - \vec{d} \cdot \vec{E}_0 \exp\!(- i \omega t)
+    &\approx - q \vu{x} \cdot \vb{E}_0 \exp\!(- i \omega t)
+    = - \vu{d} \cdot \vb{E}_0 \exp\!(- i \omega t)
 \end{aligned}$$
 
-Where $\vec{d} \equiv q \vec{r} = - e \vec{r}$ is
+Where $\vu{d} \equiv q \vu{x} = - e \vu{x}$ is
 the **transition dipole moment operator** of the electron,
 hence the name **electric dipole approximation**.
-Finally, since electric fields are actually real
-(we let it be complex for mathematical convenience),
-we take the real part, yielding:
+Finally, we take the real part, yielding:
 
 $$\begin{aligned}
     \boxed{
         \hat{H}_1(t)
-        = - q \vec{r} \cdot \vec{E}_0 \cos\!(\omega t)
+        = - \vu{d} \cdot \vb{E}(t)
+        = - q \vu{x} \cdot \vb{E}_0 \cos\!(\omega t)
     }
 \end{aligned}$$
 
 If this approximation is too rough,
-$\vec{E}$ can always be Taylor-expanded in $(i \vec{k} \cdot \vec{r})$:
+$\vb{E}$ can always be Taylor-expanded in $(i \vb{k} \cdot \vb{x})$:
 
 $$\begin{aligned}
-    \vec{E}(\vec{r}, t)
-    = \vec{E}_0 \Big( 1 + (i \vec{k} \cdot \vec{r}) + \frac{1}{2} (i \vec{k} \cdot \vec{r})^2 + \: ... \Big) \exp\!(- i \omega t)
+    \vb{E}(\vb{x}, t)
+    = \vb{E}_0 \Big( 1 + (i \vb{k} \cdot \vb{x}) + \frac{1}{2} (i \vb{k} \cdot \vb{x})^2 + \: ... \Big) \exp\!(- i \omega t)
 \end{aligned}$$
 
 Taking the real part then yields the following series of higher-order correction terms:
 
 $$\begin{aligned}
-    \vec{E}(\vec{r}, t)
-    = \vec{E}_0 \Big( \cos\!(\omega t) + (\vec{k} \cdot \vec{r}) \sin\!(\omega t) - \frac{1}{2} (\vec{k} \cdot \vec{r})^2 \cos\!(\omega t) + \: ... \Big)
+    \vb{E}(\vb{x}, t)
+    = \vb{E}_0 \Big( \cos\!(\omega t) + (\vb{k} \cdot \vb{x}) \sin\!(\omega t) - \frac{1}{2} (\vb{k} \cdot \vb{x})^2 \cos\!(\omega t) + \: ... \Big)
 \end{aligned}$$
 
 
diff --git a/content/know/concept/fabry-perot-cavity/cavity.png b/content/know/concept/fabry-perot-cavity/cavity.png
new file mode 100644
index 0000000..547e3e9
--- /dev/null
+++ b/content/know/concept/fabry-perot-cavity/cavity.png
diff --git a/content/know/concept/fabry-perot-cavity/index.pdc b/content/know/concept/fabry-perot-cavity/index.pdc
index d40852f..50b7c62 100644
--- a/content/know/concept/fabry-perot-cavity/index.pdc
+++ b/content/know/concept/fabry-perot-cavity/index.pdc
@@ -14,45 +14,154 @@ markup: pandoc
 # Fabry-Pérot cavity
 
 In its simplest form, a **Fabry-Pérot cavity**
-is a region of light-transmitting medium
-surrounded by two mirrors,
+is a region of light-transmitting medium surrounded by two mirrors,
 which may transmit some of the incoming light.
 Such a setup can be used as e.g. an interferometer or a laser cavity.
 
+Below, we calculate its quasinormal modes in 1D.
+We divide the $x$-axis into three domains: left $L$, center $C$, and right $R$.
+The cavity $C$ has length $\ell$ and is centered on $x = 0$.
+Let $n_L$, $n_C$ and $n_R$ be the respective domains' refractive indices:
 
-## Modes of macroscopic cavity
+<a href="cavity.png">
+<img src="cavity.png" style="width:70%;display:block;margin:auto;">
+</a>
 
-Consider a Fabry-Pérot cavity large enough
-that we can neglect the mirrors' thicknesses,
-which have reflection coefficients $r_L$ and $r_R$.
-Let $\tilde{n}_C$ be the complex refractive index inside,
-and $\tilde{n}_L$ and $\tilde{n}_R$ be the indices outside.
-The cavity has length $L$, centered on $x = 0$.
 
-To find the quasinormal modes,
-we make the following ansatz, with mode number $m$:
+## Microscopic cavity
+
+In its simplest "microscopic" form, the reflection at the boundaries
+is simply caused by the index differences there.
+Consider this ansatz for the [electric field](/know/concept/electric-field/) $E_m(x)$,
+where $m$ is the mode:
+
+$$\begin{aligned}
+    E_m(x)
+    = \begin{cases}
+        A_1 e^{- i k_m n_L x} & \mathrm{for}\; x < -\ell/2 \\
+        A_2 e^{- i k_m n_C x} + A_3 e^{i k_m n_C x} & \mathrm{for}\; \!-\!\ell/2 < x < \ell/2 \\
+        A_4 e^{i k_m n_R x} & \mathrm{for}\; x > \ell/2
+    \end{cases}
+\end{aligned}$$
+
+The goal is to find the modes' wavenumbers $k_m$.
+First, we demand that $E_m$ and its derivative $\dv*{E_m}{x}$
+are continuous at the boundaries $x = \pm \ell/2$:
+
+$$\begin{aligned}
+    A_1 e^{i k_m n_L \ell/2}
+    &= A_2 e^{i k_m n_C \ell/2} + A_3 e^{- i k_m n_C \ell/2}
+    \\
+    A_4 e^{i k_m n_R \ell/2}
+    &= A_2 e^{- i k_m n_C \ell/2} + A_3 e^{i k_m n_C \ell/2}
+\end{aligned}$$
+$$\begin{aligned}
+    - i k_m n_L A_1 e^{i k_m n_L \ell/2}
+    &= - i k_m n_C A_2 e^{i k_m n_C \ell/2} + i k_m n_C A_3 e^{- i k_m n_C \ell/2}
+    \\
+    i k_m n_R A_4 e^{i k_m n_R \ell/2}
+    &= - i k_m n_C A_2 e^{- i k_m n_C \ell/2} + i k_m n_C A_3 e^{i k_m n_C \ell/2}
+\end{aligned}$$
+
+Rearranging the four equations above yields the following linear system:
+
+$$\begin{aligned}
+    0
+    &= A_1 - A_2 e^{i k_m (n_C - n_L) \ell/2} - A_3 e^{- i k_m (n_C + n_L) \ell/2}
+    \\
+    0
+    &= A_2 e^{- i k_m (n_C + n_R) \ell/2} + A_3 e^{i k_m (n_C - n_R) \ell/2} - A_4
+    \\
+    0
+    &= n_L A_1 + n_C \big( A_3 e^{- i k_m (n_C + n_L) \ell/2} - A_2 e^{i k_m (n_C - n_L) \ell/2} \big)
+    \\
+    0
+    &= n_C \big( A_3 e^{i k_m (n_C - n_R) \ell/2} - A_2 e^{- i k_m (n_C + n_R) \ell/2} \big) - n_R A_4
+\end{aligned}$$
+
+Which can be rewritten in matrix form as follows, with the system matrix on the left:
+
+$$\begin{aligned}
+    \begin{bmatrix}
+        1 & -e^{i k_m (n_C - n_L) \ell/2} & -e^{- i k_m (n_C + n_L) \ell/2} & 0 \\
+        0 & e^{- i k_m (n_C + n_R) \ell/2} & e^{i k_m (n_C - n_R) \ell/2} & -1 \\
+        n_L & -n_C e^{i k_m (n_C - n_L) \ell/2} & n_C e^{- i k_m (n_C + n_L) \ell/2} & 0 \\
+        0 & -n_C e^{- i k_m (n_C + n_R) \ell/2} & n_C e^{i k_m (n_C - n_R) \ell/2} & -n_R
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        A_1 \\ A_2 \\ A_3 \\ A_4
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        0 \\ 0 \\ 0 \\ 0
+    \end{bmatrix}
+\end{aligned}$$
+
+We want non-trivial solutions, where we
+cannot simply satisfy the system by setting $A_1$, $A_2$, $A_3$ and
+$A_4$; this constraint will give us an equation for $k_m$. Therefore, we
+demand that the system matrix is singular, i.e. its determinant is zero:
+
+$$\begin{aligned}
+    0 =
+    &- n_C (n_L + n_R) \big( e^{i k_m (2 n_C - n_L - n_R) \ell/2} + e^{- i k_m (2 n_C + n_L + n_R) \ell/2} \big)
+    \\
+    &+ (n_C^2 + n_L n_R) \big( e^{i k_m (2 n_C - n_L - n_R) \ell/2} - e^{- i k_m (2 n_C + n_L + n_R) \ell/2} \big)
+\end{aligned}$$
+
+We multiply by $e^{i k_m (n_L + n_R) \ell / 2}$ and
+decompose the exponentials into sines and cosines:
+
+$$\begin{aligned}
+    0
+    = i 2 (n_C^2 + n_L n_R) \sin\!(k_m n_C \ell)
+    - 2 n_C (n_L + n_R) \cos\!(k_m n_C \ell)
+\end{aligned}$$
+
+Finally, some further rearranging gives a convenient transcendental equation:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \tan\!(k_m n_C \ell) + i \frac{n_C (n_L + n_R)}{n_C^2 + n_L n_R}
+    }
+\end{aligned}$$
+
+Thanks to linearity, we can choose one of the amplitudes
+$A_1$, $A_2$, $A_3$ or $A_4$ freely,
+and then the others are determined by $k_m$ and the field's continuity.
+
+
+## Macroscopic cavity
+
+Next, consider a "macroscopic" Fabry-Pérot cavity
+with complex mirror structures at boundaries, e.g. Bragg reflectors.
+If the cavity is large enough, we can neglect the mirrors' thicknesses,
+and just use their reflection coefficients $r_L$ and $r_R$.
+We use the same ansatz:
 
 $$\begin{aligned}
     E_m(x)
     =
     \begin{cases}
-        A_m \exp\!(-i \tilde{n}_L \tilde{k}_m x) & \mathrm{if}\; x < -L/2 \\
-        B_m \exp\!(i \tilde{n}_C \tilde{k}_m x) + C_m \exp\!(-i \tilde{n}_C \tilde{k}_m x) & \mathrm{if}\; -\!L/2 < x < L/2 \\
-        D_m \exp\!(i \tilde{n}_R \tilde{k}_m x) & \mathrm{if}\; L/2 < x
+        A_1 e^{-i k_m n_L x} & \mathrm{for}\; x < -\ell/2 \\
+        A_2 e^{-i k_m n_C x} + A_3 e^{i k_m n_C x} & \mathrm{for}\; \!-\!\ell/2 < x < \ell/2 \\
+        A_4 e^{i k_m n_R x} & \mathrm{for}\; \ell/2 < x
     \end{cases}
 \end{aligned}$$
 
-On the left, $B_m$ is the reflection of $C_m$,
-and on the right, $C_m$ is the reflection of $B_m$,
-where the reflected amplitude is determined
-by the coefficients $r_L$ and $r_L$, respectively:
+On the left, $A_3$ is the reflection of $A_2$,
+and on the right, $A_2$ is the reflection of $A_3$,
+where the reflected amplitudes are determined
+by the coefficients $r_L$ and $r_R$, respectively:
 
 $$\begin{aligned}
-    B_m \exp\!(-i \tilde{n}_C \tilde{k}_m L/2)
-    &= r_L C_m \exp\!(i \tilde{n}_C \tilde{k}_m L/2)
+    A_3 e^{- i k_m n_C \ell/2}
+    &= r_L A_2 e^{i k_m n_C \ell/2}
     \\
-    C_m \exp\!(-i \tilde{n}_C \tilde{k}_m L/2)
-    &= r_R B_m \exp\!(i \tilde{n}_C \tilde{k}_m L/2)
+    A_2 e^{-i k_m n_C \ell/2}
+    &= r_R A_3 e^{i k_m n_C \ell/2}
 \end{aligned}$$
 
 These equations might seem to contradict each other.
@@ -60,12 +169,12 @@ We recast them into matrix form:
 
 $$\begin{aligned}
     \begin{bmatrix}
-        1 & - r_L \exp\!(i \tilde{n}_C \tilde{k}_m L) \\
-        - r_R \exp\!(i \tilde{n}_C \tilde{k}_m L) & 1
+        1 & - r_R e^{i k_m n_C \ell} \\
+        - r_L e^{i k_m n_C \ell} & 1
     \end{bmatrix}
     \cdot
     \begin{bmatrix}
-        B_m \\ C_m
+        A_2 \\ A_3
     \end{bmatrix}
     =
     \begin{bmatrix}
@@ -73,58 +182,53 @@ $$\begin{aligned}
     \end{bmatrix}
 \end{aligned}$$
 
-Now, we do not want to be able to find values for $B_m$ and $C_m$
-that satisfy this for a given $\tilde{k}_m$.
-Instead, we only want specific values of $\tilde{k}_m$ to be allowed,
-corresponding to the cavity's modes.
-We thus demand that the determinant to zero:
+Again, we demand that the determinant is zero, in order to get non-trivial solutions:
 
 $$\begin{aligned}
     0
-    &= 1 - r_L r_R \exp\!(i 2 \tilde{n}_C \tilde{k}_m L)
+    &= 1 - r_L r_R e^{i 2 k_m n_C \ell}
 \end{aligned}$$
 
-Isolating this for $\tilde{k}_m$ yields the following modes,
+Isolating this for $k_m$ yields the following modes,
 where $m$ is an arbitrary integer:
 
 $$\begin{aligned}
     \boxed{
-        \tilde{k}_m
-        = - \frac{\ln\!(r_L r_R) + i 2 \pi m}{i 2 \tilde{n}_C L}
+        k_m
+        = - \frac{\ln\!(r_L r_R) + i 2 \pi m}{i 2 n_C \ell}
     }
 \end{aligned}$$
 
-These $\tilde{k}_m$ satisfy the matrix equation above.
-Thanks to linearity, we can choose one of $B_m$ or $C_m$,
-and then the other is determined by the corresponding equation.
+These $k_m$ satisfy the matrix equation above.
+Thanks to linearity, we can choose one of $A_2$ or $A_3$,
+and then the other is determined by the corresponding reflection equation.
 
 Finally, we look at the light transmitted through the mirrors,
 according to $1 \!-\! r_L$ and $1 \!-\! r_R$:
 
 $$\begin{aligned}
-    A_m \exp\!(i \tilde{n}_L \tilde{k}_m L/2)
-    &= (1 - r_L) C_m \exp\!(i \tilde{n}_C \tilde{k}_m L/2)
+    A_1 e^{i k_m n_L \ell/2}
+    &= (1 - r_L) A_2 e^{i k_m n_C \ell/2}
     \\
-    D_m \exp\!(i \tilde{n}_R \tilde{k}_m L/2)
-    &= (1 - r_R) B_m \exp\!(i \tilde{n}_C \tilde{k}_m L/2)
+    A_4 e^{i k_m n_R \ell/2}
+    &= (1 - r_R) A_3 e^{i k_m n_C \ell/2}
 \end{aligned}$$
 
-We simply isolate for $A_m$ and $D_m$ respectively,
+We simply isolate for $A_1$ and $A_4$ respectively,
 yielding the following amplitudes:
 
 $$\begin{aligned}
-    A_m
-    &= (1 - r_L) C_m \exp\!\big( i (\tilde{n}_C \!-\! \tilde{n}_L) \tilde{k}_m L/2 \big)
+    A_1
+    &= (1 - r_L) A_2 e^{i k_m (n_C - n_L) \ell/2}
     \\
-    D_m
-    &= (1 - r_R) B_m \exp\!\big( i (\tilde{n}_C \!-\! \tilde{n}_R) \tilde{k}_m L/2 \big)
+    A_4
+    &= (1 - r_R) A_3 e^{i k_m (n_C - n_R) \ell/2}
 \end{aligned}$$
 
 Note that we have not demanded continuity of the electric field.
 This is because the mirrors are infinitely thin "magic" planes;
-if we had instead used a full physical mirror structure,
-then the we would have demanded continuity,
-as you might have expected.
+had we instead used the full mirror structure,
+then we would have demanded continuity, as you maybe expected.
 
 
 
diff --git a/content/know/concept/feynman-diagram/index.pdc b/content/know/concept/feynman-diagram/index.pdc
index dfb63c1..600be61 100644
--- a/content/know/concept/feynman-diagram/index.pdc
+++ b/content/know/concept/feynman-diagram/index.pdc
@@ -284,12 +284,12 @@ involving the [Matsubara Green's function](/know/concept/matsubara-greens-functi
 $$\begin{aligned}
     i \hbar G_{s_2 s_1}^0(\vb{r}_2, t_2; \vb{r}_1, t_1)
     \:\: &\longrightarrow \:\:
-    -\!\hbar G_{s_2 s_1}^0(\vb{r}_2, \tau_2; \vb{r}_1, \tau_1)
+    \hbar G_{s_2 s_1}^0(\vb{r}_2, \tau_2; \vb{r}_1, \tau_1)
     = \expval{\mathcal{T} \Big\{ \hat{\Psi}_I(\vb{r}_2, \tau_2) \hat{\Psi}_I^\dagger(\vb{r}_1, \tau_1) \Big\}}
     \\
     i \hbar G_{s_2 s_1}(\vb{r}_2, t_2; \vb{r}_1, t_1)
     \:\: &\longrightarrow \:\:
-    -\!\hbar G_{s_2 s_1}(\vb{r}_2, \tau_2; \vb{r}_1, \tau_1)
+    \hbar G_{s_2 s_1}(\vb{r}_2, \tau_2; \vb{r}_1, \tau_1)
     = \expval{\mathcal{T} \Big\{ \hat{\Psi}_H(\vb{r}_2, \tau_2) \hat{\Psi}_H^\dagger(\vb{r}_1, \tau_1) \Big\}}
 \end{aligned}$$
 
@@ -312,7 +312,8 @@ and a distinction must be made between
 fermionic Matsubara frequencies $i \omega_n^f$ (for $G$ and $G^0$)
 and bosonic Matsubara ones $i \omega_n^b$ (for $W$).
 This distinction is compatible with frequency conservation,
-since a sum of two fermionic frequencies is always bosonic:
+since a sum of two fermionic frequencies is always bosonic.
+We have:
 
 $$\begin{aligned}
     G_{s_2 s_1}^0(\vb{r}_2, \tau_2; \vb{r}_1, \tau_1)
diff --git a/content/know/concept/random-phase-approximation/index.pdc b/content/know/concept/random-phase-approximation/index.pdc
index 970a884..ed85106 100644
--- a/content/know/concept/random-phase-approximation/index.pdc
+++ b/content/know/concept/random-phase-approximation/index.pdc
@@ -77,7 +77,7 @@ i.e. the ones where all $n$ interaction lines
 carry the same momentum and energy:
 
 <a href="rpasigma.png">
-<img src="rpasigma.png" style="width:91%;display:block;margin:auto;">
+<img src="rpasigma.png" style="width:92%;display:block;margin:auto;">
 </a>
 
 Where we have defined the **screened interaction** $W^\mathrm{RPA}$,
diff --git a/content/know/concept/random-phase-approximation/rpasigma.png b/content/know/concept/random-phase-approximation/rpasigma.png
index 87ba3cc..c9587b8 100644
--- a/content/know/concept/random-phase-approximation/rpasigma.png
+++ b/content/know/concept/random-phase-approximation/rpasigma.png
diff --git a/content/know/concept/self-energy/dyson.png b/content/know/concept/self-energy/dyson.png
index f576632..168505c 100644
--- a/content/know/concept/self-energy/dyson.png
+++ b/content/know/concept/self-energy/dyson.png
diff --git a/content/know/concept/self-energy/fullgf.png b/content/know/concept/self-energy/fullgf.png
index 5767dba..0ea6958 100644
--- a/content/know/concept/self-energy/fullgf.png
+++ b/content/know/concept/self-energy/fullgf.png
diff --git a/content/know/concept/self-energy/index.pdc b/content/know/concept/self-energy/index.pdc
index 7e67143..935cca8 100644
--- a/content/know/concept/self-energy/index.pdc
+++ b/content/know/concept/self-energy/index.pdc
@@ -27,7 +27,7 @@ and $\beta = 1 / (k_B T)$:
 
 $$\begin{aligned}
     G_{s_b s_a}(\vb{r}_b, \tau_b; \vb{r}_a, \tau_a)
-    = - \frac{\expval{\mathcal{T}\Big\{ \hat{K}(\hbar \beta, 0) \hat{\Psi}_{s_b}(\vb{r}_b, \tau_b) \hat{\Psi}_{s_a}^\dagger(\vb{r}_a, \tau_a) \Big\}}}
+    = \frac{\expval{\mathcal{T}\Big\{ \hat{K}(\hbar \beta, 0) \hat{\Psi}_{s_b}(\vb{r}_b, \tau_b) \hat{\Psi}_{s_a}^\dagger(\vb{r}_a, \tau_a) \Big\}}}
     {\hbar \expval{\hat{K}(\hbar \beta, 0)}}
 \end{aligned}$$
 
@@ -50,7 +50,7 @@ and $\hat{\Psi}_a \equiv \hat{\Psi}_{s_a}(\vb{r}_a, \tau_a)$:
 
 $$\begin{aligned}
     G_{ba}
-    &= - \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \idotsint_0^{\hbar \beta}
+    &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \idotsint_0^{\hbar \beta}
     \expval{\mathcal{T}\Big\{ \hat{W}(\tau_1) \cdots \hat{W}(\tau_n) \hat{\Psi}_b \hat{\Psi}_a^\dagger \Big\}} \dd{\tau_1} \cdots \dd{\tau_n}}
     {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \idotsint_0^{\hbar \beta}
     \expval{\mathcal{T}\Big\{ \hat{W}(\tau_1) \cdots \hat{W}(\tau_n) \Big\}} \dd{\tau_1} \cdots \dd{\tau_n}}
@@ -84,7 +84,7 @@ The full $G_{ba}$ thus becomes:
 
 $$\begin{aligned}
     G_{ba}
-    &= - \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{1}{2 \hbar} \Big)^n (-\hbar)^{2n+1}
+    &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{1}{2 \hbar} \Big)^n (-\hbar)^{2n+1}
     \idotsint W_{1'1} \cdots W_{n'n} \: \Big( G^0_\mathrm{num} \Big) \dd{1'} \dd{1} \cdots \dd{n'} \dd{n}}
     {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{1}{2 \hbar} \Big)^n (-\hbar)^{2n}
     \idotsint W_{1'1} \cdots W_{n'n} \: \Big( G^0_\mathrm{den} \Big) \dd{1'} \dd{1} \cdots \dd{n'} \dd{n}}
@@ -131,7 +131,7 @@ times $(-1)^p$ to account for swaps of fermionic operators:
 
 $$\begin{aligned}
     G_{ba}
-    &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+    &= -\frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
     \idotsint W_{1'1} \cdots W_{n'n} \: \Big( \sum_{p} (-1)^p \prod_{m = 1}^{2 n + 1} G^0_{(p,m)} \Big) \dd{1}' \dd{1} \cdots \dd{n'} \dd{n}}
     {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
     \idotsint W_{1'1} \cdots W_{n'n} \: \Big( \sum_{p} (-1)^p \prod_{m = 1}^{2 n} G^0_{(p,m)} \Big) \dd{1'} \dd{1} \cdots \dd{n'} \dd{n}}
@@ -172,7 +172,7 @@ $$\begin{aligned}
     &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!}
     \bigg[ \sum_{m = 0}^{n} \frac{n!}{m! (n \!-\! m)!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
     \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)}_{\!\Sigma\mathrm{all}} \bigg]}
-    {-\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
+    {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
 \end{aligned}$$
 
 Where the total order is the sum of the orders of all considered diagrams,
@@ -186,7 +186,7 @@ $$\begin{aligned}
     &= \frac{\displaystyle\sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
     \bigg[ \sum_{n = 0}^\infty \frac{1}{2^{n-m} (n \!-\! m)!}
     \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)}_{\!\Sigma\mathrm{all}} \bigg]}
-    {-\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
+    {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
 \end{aligned}$$
 
 Since both $n$ and $m$ start at zero,
@@ -194,7 +194,7 @@ and the sums include all possible diagrams,
 we see that the second sum in the numerator does not actually depend on $m$:
 
 $$\begin{aligned}
-    -\hbar G_{ba}
+    \hbar G_{ba}
     &= \frac{\displaystyle\sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
     \bigg[ \sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}} \bigg]}
     {\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
diff --git a/content/know/concept/self-energy/selfenergy.png b/content/know/concept/self-energy/selfenergy.png
index 55e182e..84d3651 100644
--- a/content/know/concept/self-energy/selfenergy.png
+++ b/content/know/concept/self-energy/selfenergy.png