12 files changed, 599 insertions, 280 deletions

diff --git a/source/know/concept/bb84-protocol/index.md b/source/know/concept/bb84-protocol/index.md
index 0f75930..44ea57d 100644
--- a/source/know/concept/bb84-protocol/index.md
+++ b/source/know/concept/bb84-protocol/index.md
@@ -48,7 +48,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 After Alice has sent all her qubits,
-the next step is **basis reconciliation**:
+the next step is *basis reconciliation*:
 over the classical channel, Bob announces, for each bit,
 which basis he chose, and Alice tells him if he was right or wrong.
 Bob discards all bits where he guessed wrongly.
@@ -56,6 +56,7 @@ If their quantum channel did not have any noise or eavesdroppers,
 Alice and Bob now have a perfectly correlated secret string of bits.
 
 
+
 ## Eavesdropper detection
 
 But what if there is actually an eavesdropper?
@@ -99,6 +100,7 @@ that as long as the error rate is below 11%,
 the BB84 protocol is fully secure, i.e. there cannot be any eavesdroppers.
 
 
+
 ## Error correction
 
 In practice, even without Eve, quantum channels are imperfect,
@@ -125,11 +127,14 @@ $$\begin{aligned}
 
 If $$A = B$$, then $$a_{n+1}$$ and $$b_{n+1}$$ are discarded to prevent
 a listener on the classical channel from learning anything about the string.
-If $$A \neq B$$, all of $$a_n$$, $$b_n$$, $$a_{n+1}$$ and $$b_{n+1}$$ are discarded,
-and then Alice and Bob move on to $$n = 3$$, etc.
+If $$A \neq B$$, something went wrong,
+so all of $$a_n$$, $$b_n$$, $$a_{n+1}$$ and $$b_{n+1}$$ are discarded.
+Then Alice and Bob move on to $$n = 3$$, etc.
 
-Given that $$A = B$$, the probability that $$a_n = b_n$$,
-which is what we want, is given by:
+Given that $$A = B$$, the probability that $$a_n = b_n$$ is as shown below.
+Note that there are two possible explanations for $$A = B$$:
+either $$a_{n} = b_{n} \land a_{n+1} = b_{n+1}$$,
+or $$a_{n} \neq b_{n} \land a_{n+1} \neq b_{n+1}$$:
 
 $$\begin{aligned}
     P(a_{n} = b_{n} | A = B)
@@ -164,6 +169,7 @@ $$\begin{aligned}
 More efficient schemes exist, which do not consume so many bits.
 
 
+
 ## Privacy amplification
 
 Suppose that after the error correction step, $$p = 1$$,
@@ -177,10 +183,10 @@ $$\begin{aligned}
     q = P(e_n = a_n) > \frac{1}{2}
 \end{aligned}$$
 
-**Privacy amplification** is an optional final step of the BB84 protocol
-which aims to reduce Eve's $$q$$.
+*Privacy amplification* is an optional final step of the BB84 protocol
+that aims to reduce Eve's $$q$$.
 Alice and Bob use their existing strings to generate a new one
-$$\{a_1', ..., a_M'\}$$:
+$$\{a_1', ..., a_M'\}$$ where:
 
 $$\begin{aligned}
     a_1'
@@ -225,6 +231,7 @@ Eve would only know 50% of the bits,
 which is equivalent to her guessing at random.
 
 
+
 ## References
 1.  N. Brunner,
     *Quantum information theory: lecture notes*,
diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md
index fa289de..d4508e6 100644
--- a/source/know/concept/bell-state/index.md
+++ b/source/know/concept/bell-state/index.md
@@ -24,14 +24,14 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where e.g. $$\ket{0}_A \ket{1}_B = \ket{0}_A \otimes \ket{1}_B$$
+Where e.g. $$\ket{0}_A \ket{1}_B \equiv \ket{0}_A \otimes \ket{1}_B$$
 is the tensor product of qubit $$A$$ in state $$\ket{0}$$ and $$B$$ in $$\ket{1}$$.
 These states form an orthonormal basis for the two-qubit
 [Hilbert space](/know/concept/hilbert-space/).
 
 More importantly, however,
-is that the Bell states are maximally entangled,
-which we prove here for $$\ket{\Phi^{+}}$$.
+is that all four Bell states are *maximally entangled*.
+For brevity, we will only show this for $$\ket{\Phi^{+}}$$ here.
 Consider the following pure [density operator](/know/concept/density-operator/):
 
 $$\begin{aligned}
@@ -40,7 +40,8 @@ $$\begin{aligned}
     &= \frac{1}{2} \Big( \ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big) \Big( \bra{0}_A \bra{0}_B + \bra{1}_A \bra{1}_B \Big)
 \end{aligned}$$
 
-The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated as follows:
+The *reduced* density operator $$\hat{\rho}_A$$ of qubit $$A$$
+is then calculated like so, using a partial trace:
 
 $$\begin{aligned}
     \hat{\rho}_A
@@ -54,12 +55,13 @@ $$\begin{aligned}
     = \frac{1}{2} \hat{I}
 \end{aligned}$$
 
-This result is maximally mixed, therefore $$\ket{\Phi^{+}}$$ is maximally entangled.
-The same holds for the other three Bell states,
-and is equally true for qubit $$B$$.
-
+The same holds for qubit $$B$$. This result is *maximally mixed*,
+therefore $$\ket{\Phi^{+}}$$ is maximally entangled.
 This means that a measurement of qubit $$A$$
-has a 50-50 chance to yield $$\ket{0}$$ or $$\ket{1}$$.
+has a 50-50 chance to yield $$\ket{0}$$ or $$\ket{1}$$,
+or in other words, no useful information can be gathered
+from measuring just one of the qubits.
+
 However, due to the entanglement,
 measuring $$A$$ also has consequences for qubit $$B$$:
 
diff --git a/source/know/concept/discrete-spectrum-summation/index.md b/source/know/concept/discrete-spectrum-summation/index.md
new file mode 100644
index 0000000..dbbd5f9
--- /dev/null
+++ b/source/know/concept/discrete-spectrum-summation/index.md
@@ -0,0 +1,77 @@
+---
+title: "Discrete spectrum summation"
+sort_title: "Discrete spectrum summation"
+date: 2026-09-02
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+This article is about a trick used in many calculations,
+especially in condensed matter physics and advanced quantum mechanics,
+which, as far as I know, does not have a specific name
+(this is the best I could come up with),
+but is so common and useful that it deserves attention.
+
+Often, we find ourselves doing calculations
+about a $$D$$-dimensional system with periodic boundary conditions.
+Generally, there are two sources of such boundary conditions:
+an inherent periodicity of the system (e.g. crystals),
+and/or a need to chop up an infinite system into finite pieces
+to prevent mathematical problems (e.g. divergences).
+
+In the second case, if studying the whole infinity directly is not possible,
+we restrict ourselves to a hypercube with side $$L$$
+and $$D$$-dimensional volume $$V = L^D$$,
+with the intention to let $$L \to \infty$$ at the end.
+We then often impose periodic boundary conditions on the hypercube,
+in order to be able to use [Fourier transforms](/know/concept/fourier-transform/)
+on such a finite domain, and/or to study transport phenomena.
+This idea is trivial to generalize to "hyperrectangles"
+with different side lengths $$L_x$$, $$L_y$$, etc.
+
+In both cases, we might end up expanding functions
+from a [Hilbert space](/know/concept/hilbert-space/) defined on the hypercube
+in a basis of plane waves $$\ket{\psi_\vb{k}}$$ with wavevectors $$\vb{k}$$,
+often as the result of a Fourier transform.
+But due to the hypercube's finite size and its boundary conditions,
+those plane waves occupy a discrete set of allowed $$\vb{k}$$-values
+(whereas in an infinite system, $$\vb{k}$$ would be a continuous variable).
+
+Hence, those normalized basis waves $$\ket{\psi_\vb{k}}$$ are as follows
+in $$\vb{r}$$-space (modulo a constant phase):
+
+$$\begin{aligned}
+    \inprod{\vb{r}}{\psi_{\vb{k}}}
+    = \psi_{\vb{k}}(\vb{r})
+    = \frac{1}{\sqrt{L^D}} \exp(i \vb{k} \cdot \vb{r})
+    \qquad \qquad
+    \vb{k} = \frac{2 \pi}{L} (n_1, ..., n_D)
+\end{aligned}$$
+
+Where $$n_1, ..., n_D \in \mathbb{Z}$$.
+The discreteness is typically an artifact of our mathematical setup,
+and then disappears into the true continuous spectrum for $$L \to \infty$$.
+Until then, every plane wave occupies a nonzero volume
+$$(2 \pi)^D / L^D$$ in $$\vb{k}$$-space.
+
+Here is the key: as $$L$$ increases, the allowed $$\vb{k}$$-values become denser,
+until any sum over those $$\vb{k}$$ turns into a Riemann integral:
+
+$$\begin{aligned}
+    \lim_{L \to \infty} \frac{(2 \pi)^D}{L^D} \sum_{\vb{k}} f(\vb{k})
+    = \int_{-\infty}^\infty f(\vb{k}) \dd{\vb{k}}
+\end{aligned}$$
+
+Where $$(2 \pi) / L$$ is the spacing between $$\vb{k}$$-values.
+This trick to convert nasty sums to easier integrals
+is used all over physics because it is so powerful.
+We can even get away with postponing taking the limit,
+and doing the conversion as an exact equality in the middle of our calculation,
+on the condition that we take $$L \to \infty$$ at the end.
+
+Actually, this trick is not exclusive to periodic boundary conditions,
+but is also valid for Dirichlet ("particle in a box") boundaries,
+in which case the wavevector spectrum is discrete too,
+also with constant spacing between allowed $$\vb{k}$$-values.
diff --git a/source/know/concept/fermi-gas/index.md b/source/know/concept/fermi-gas/index.md
new file mode 100644
index 0000000..6d316cc
--- /dev/null
+++ b/source/know/concept/fermi-gas/index.md
@@ -0,0 +1,219 @@
+---
+title: "Fermi gas"
+sort_title: "Fermi gas"
+date: 2026-09-02
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+A **Fermi gas** is a system of many fermions
+that do not interact directly, only indirectly through
+the [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/),
+and hence obey [Fermi-Dirac statistics](/know/concept/fermi-dirac-distribution/).
+
+There are several real-life systems for which this model is relevant,
+but most notably it serves as the foundation of the quantum-mechanical study
+of electrons (or electron holes) in materials.
+Obviously, electrons *do* interact strongly via the Coulomb force,
+but it is nevertheless a useful starting point to neglect that fact,
+and to then add the interactions later (see e.g. [jellium](/know/concept/jellium)).
+
+Consider a collection of infinitely many non-interacting fermions.
+For mathematical convenience, we restrict ourselves to a cube with side $$L$$,
+and impose periodic boundary conditions.
+Then, at the end of our calculation,
+we should in theory take the limit $$L \to \infty$$
+to recover the "true" system.
+
+In the absence of any potentials, all the fermions' wavefunctions
+are simply plane waves $$\ket{\psi_\vb{k}}$$ with wavevector $$\vb{k}$$.
+Due to the cube's finite size and its periodic boundary conditions,
+those waves have a discrete spectrum of allowed wavevectors $$\vb{k}$$,
+meaning that each particle's wavefunction $$\ket{\psi_\vb{k}}$$
+is as follows in $$\vb{r}$$-space (modulo a constant phase):
+
+$$\begin{aligned}
+    \psi_{\vb{k}}(\vb{r})
+    = \frac{1}{\sqrt{L^3}} \exp(i \vb{k} \cdot \vb{r})
+    \qquad \qquad
+    \vb{k} = \frac{2 \pi}{L} (n_x, n_y, n_z)
+\end{aligned}$$
+
+Where $$n_x, n_y, n_z \in \mathbb{Z}$$.
+This is a discrete (but infinite) set of independent orbitals,
+so it is natural to use the
+[second quantization](/know/concept/second-quantization/)'s
+operators $$\hat{c}^\dagger$$ and $$\hat{c}$$ in our analysis.
+
+Let the temperature $$T = 0$$,
+then the $$N$$ fermions inside our cube
+fill the $$N$$ lowest-energy orbitals.
+The resulting $$N$$-particle ground state
+is known as the **Fermi sea** or **Fermi sphere** $$\ket{\mathrm{FS}}$$,
+and can be written as follows, where $$S$$ is the spin degeneracy,
+i.e. for each $$\vb{k}$$ there are $$S$$ orbitals
+with the same energy but different spin $$s$$
+(for most relevant fermions $$S = 2$$):
+
+$$\begin{aligned}
+    \ket{\mathrm{FS}}
+    = \prod_{s} \prod_{j = 1}^{N/S} \hat{c}_{s,\vb{k}_j}^\dagger \ket{0}
+\end{aligned}$$
+
+The energy and wavenumber $$|\vb{k}|$$ of the highest filled orbital
+are called the **Fermi energy** $$\varepsilon_F$$ and **Fermi wavenumber** $$k_F$$,
+and obey the expected kinetic energy relation:
+
+$$\begin{aligned}
+    \boxed{
+        \varepsilon_F
+        = \frac{\hbar^2}{2 m} k_F^2
+    }
+\end{aligned}$$
+
+The Fermi sphere can be visualized in $$\vb{k}$$-space
+as a sphere with radius $$k_F$$.
+Because $$\vb{k}$$ is discrete, the sphere's surface is not smooth,
+but in the limit $$L \to \infty$$ that "roughness" disappears.
+
+Now, we would like a relation between the system's parameters,
+e.g. $$N$$ and $$L$$, and the resulting values of $$\varepsilon_F$$ or $$k_F$$.
+The total number $$N$$ of fermions in our cube is given by:
+
+$$\begin{aligned}
+    N
+    = \sum_{s} \sum_{\vb{k}} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}
+    = \sum_{s} \frac{L^3}{(2 \pi)^3} \int_{-\infty}^\infty \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} \dd{\vb{k}}
+\end{aligned}$$
+
+Where the periodic boundary conditions have
+[enabled us](/know/concept/discrete-spectrum-summation/)
+to convert the sum over $$\vb{k}$$ to an integral.
+For $$T = 0$$, the matrix element
+$$\matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}$$
+is either $$0$$ or $$1$$,
+depending on whether $$\vb{k}$$ is outside or inside the Fermi sphere.
+We can write this using
+a [Heaviside step function](/know/concept/heaviside-step-function/):
+
+$$\begin{aligned}
+    N
+    = \sum_{s} \frac{L^3}{(2 \pi)^3} \int_{-\infty}^\infty \Theta(k_F - |\vb{k}|) \dd{\vb{k}}
+    = \frac{S L^3}{(2 \pi)^3} \int_{-\infty}^\infty \Theta(k_F - |\vb{k}|) \dd{\vb{k}}
+\end{aligned}$$
+
+Where we realized that spin does not matter,
+to replace the sum with a factor $$S$$.
+To evaluate this 3D integral, we transition to
+[spherical coordinates](/know/concept/spherical-coordinates/)
+$$(|\vb{k}|, \theta, \varphi)$$:
+
+$$\begin{aligned}
+    N
+    &= \frac{S L^3}{8 \pi^3} \int_0^{2 \pi} \int_0^\pi \int_0^\infty \Theta(k_F - |\vb{k}|) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi}
+    \\
+    &= \frac{S L^3}{8 \pi^3} 4 \pi \int_0^\infty \Theta(k_F - |\vb{k}|) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|}
+    \\
+    &= \frac{S L^3}{2 \pi^2} \int_0^{k_F} |\vb{k}|^2 \dd{|\vb{k}|}
+    \\
+    &= \frac{S L^3}{6 \pi^2} k_F^3
+\end{aligned}$$
+
+Since the particle density $$n = N / L^3$$,
+we can rearrange this result to the following relation:
+
+$$\begin{aligned}
+    \boxed{
+        k_F^3
+        = \frac{6 \pi^2}{S} n
+    }
+    \qquad
+\end{aligned}$$
+
+Consequently, the Fermi energy $$\varepsilon_F$$
+and the corresponding orbital's velocity $$v_F = \hbar k_F / m$$
+can be expressed as a function of the density $$n$$:
+
+$$\begin{aligned}
+    \boxed{
+        \varepsilon_F
+        = \frac{\hbar^2}{2 m} \bigg( \frac{6 \pi^2}{S} \bigg)^{2/3} n^{2/3}
+    }
+    \qquad \qquad
+    \boxed{
+        v_F
+        = \frac{\hbar}{m} \bigg( \frac{6 \pi^2}{S} \bigg)^{1/3} n^{1/3}
+    }
+\end{aligned}$$
+
+This is an important result, especially for electrons in metals.
+We know the electron density $$n$$ for many conductors,
+and then these relations tell us that $$v_F \ll c$$,
+and that the "Fermi temperature" $$T_F = \varepsilon_F / k_B$$
+is very large (e.g. $$T_F \approx 8 \cdot 10^4 \: \mathrm{K}$$ for copper).
+This justifies our implicit assumptions that relativity
+and thermal fluctuations are negligible under normal circumstances.
+
+We now have an expression for $$\varepsilon_F$$ as a function of $$n$$,
+which we can control by adding or removing fermions from the system.
+But it is also useful to isolate this relation for $$n$$ instead:
+
+$$\begin{aligned}
+    n
+    &= \frac{S}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{3/2}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \boxed{
+        g(\varepsilon_F)
+        = \dv{N}{\varepsilon_F}
+        = \frac{S L^3}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{1/2}
+    }
+\end{aligned}$$
+
+Now, $$\varepsilon_F$$ is the highest energy of a single fermion,
+but what about the total $$N$$-particle energy $$E$$?
+This is easy to calculate using the density of states:
+
+$$\begin{aligned}
+    E
+    &= \int_0^{\varepsilon_F} \varepsilon \: g(\varepsilon) \dd{\varepsilon}
+    \\
+    &= \frac{S L^3}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2}
+    \int_0^{\varepsilon_F} \varepsilon^{3/2} \dd{\varepsilon}
+    \\
+    &= \frac{3}{2} \frac{S L^3}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \: \frac{2}{5} \varepsilon_F^{5/2}
+\end{aligned}$$
+
+Here, we recognize $$N(\varepsilon_F)$$ from earlier,
+leading to the following expression for the total $$E$$:
+
+$$\begin{aligned}
+    \boxed{
+        E
+        = \frac{3}{5} N \varepsilon_F
+    }
+\end{aligned}$$
+
+This model is a strong foundation for many more advanced calculations.
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
diff --git a/source/know/concept/hilbert-space/index.md b/source/know/concept/hilbert-space/index.md
index 42b9cb1..2a60896 100644
--- a/source/know/concept/hilbert-space/index.md
+++ b/source/know/concept/hilbert-space/index.md
@@ -18,22 +18,32 @@ is an abstract **vector space** with a notion of length and angle.
 An abstract **vector space** $$\mathbb{V}$$ is a generalization
 of the traditional concept of vectors as "arrows".
 It consists of a set of objects called **vectors**
-which support the following (familiar) operations:
+that support the following (familiar) operations:
 
-+ **Vector addition**: the sum of two vectors $$V$$ and $$W$$, denoted by $$V + W$$.
-+ **Scalar multiplication**: product of a vector $$V$$ with a scalar $$a$$, denoted by $$a V$$.
++ **Vector addition**:
+  the sum of two vectors $$V$$ and $$W$$, denoted by $$V + W$$.
++ **Scalar multiplication**:
+  product of a vector $$V$$ with a scalar $$a$$, denoted by $$a V$$.
 
-In addition, for a given $$\mathbb{V}$$ to qualify as a proper vector
-space, these operations must obey the following axioms:
+In addition, for a given $$\mathbb{V}$$ to qualify as a proper vector space,
+these operations must have the following (again familiar) properties:
 
-+ **Addition is associative**: $$U + (V + W) = (U + V) + W$$
-+ **Addition is commutative**: $$U + V = V + U$$
-+ **Addition has an identity**: there exists a $$\mathbf{0}$$ such that $$V + 0 = V$$
-+ **Addition has an inverse**: for every $$V$$ there exists $$-V$$ so that $$V + (-V) = 0$$
-+ **Multiplication is associative**: $$a (b V) = (a b) V$$
-+ **Multiplication has an identity**: There exists a $$1$$ such that $$1 V = V$$
-+ **Multiplication is distributive over scalars**: $$(a + b)V = aV + bV$$
-+ **Multiplication is distributive over vectors**: $$a (U + V) = a U + a V$$
++ **Addition is associative**:
+  $$U + (V + W) = (U + V) + W$$
++ **Addition is commutative**:
+  $$U + V = V + U$$
++ **Addition has an identity**:
+  there exists a $$\mathbf{0}$$ such that $$V + 0 = V$$
++ **Addition has an inverse**:
+  for every $$V$$ there exists $$-V$$ so that $$V + (-V) = 0$$
++ **Multiplication is associative**:
+  $$a (b V) = (a b) V$$
++ **Multiplication has an identity**:
+  There exists a $$1$$ such that $$1 V = V$$
++ **Multiplication is distributive over scalars**:
+  $$(a + b)V = aV + bV$$
++ **Multiplication is distributive over vectors**:
+  $$a (U + V) = a U + a V$$
 
 A set of $$N$$ vectors $$V_1, V_2, ..., V_N$$ is **linearly independent**
 if the only way to satisfy the following relation
@@ -46,25 +56,28 @@ $$\begin{aligned}
 In other words, these vectors cannot be expressed in terms of each other.
 Otherwise, they would be **linearly dependent**.
 
-A vector space $$\mathbb{V}$$ has **dimension** $$N$$
-if only up to $$N$$ of its vectors can be linearly indepedent.
+$$\mathbb{V}$$ has **dimension** $$N$$
+if only up to $$N$$ of its vectors can be linearly independent.
 All other vectors in $$\mathbb{V}$$ can then be written
 as a **linear combination** of these $$N$$ **basis vectors**.
 
-Let $$\vu{e}_1, ..., \vu{e}_N$$ be the basis vectors,
-then any vector $$V$$ in the same space can be **expanded**
-in the basis according to the unique weights $$v_n$$,
-known as the **components** of $$V$$ in that basis:
+Let $$\vu{e}_1, ..., \vu{e}_N$$ be a (generally not unique)
+valid set of basis vectors of $$\mathbb{V}$$,
+then any vector $$V$$ in that space can be **expanded**
+in that basis according to unique weights $$v_n$$,
+called the **components** of $$V$$ in that basis:
 
 $$\begin{aligned}
     V = \sum_{n = 1}^N v_n \vu{e}_n
 \end{aligned}$$
 
-Using these, the vector space operations can then be implemented as follows:
+Using these components,
+the operations of vector addition and scalar multiplication
+can then be implemented as follows:
 
 $$\begin{gathered}
     V = \sum_{n = 1} v_n \vu{e}_n
-    \quad
+    \qquad
     W = \sum_{n = 1} w_n \vu{e}_n
     \\
     \quad \implies \quad
@@ -73,18 +86,24 @@ $$\begin{gathered}
     a V = \sum_{n = 1}^N a v_n \vu{e}_n
 \end{gathered}$$
 
+It is straightforward to see that this implementation satisfies the properties above.
+
 
 
 ## Inner product
 
-A given vector space $$\mathbb{V}$$ can be promoted to a **Hilbert space** or **inner product space**
+A given vector space $$\mathbb{V}$$ can be promoted
+to a **Hilbert space** or **inner product space**
 if it supports an operation $$\Inprod{U}{V}$$ called the **inner product**,
 which takes two vectors and returns a scalar,
 and has the following properties:
 
-+ **Skew symmetry**: $$\Inprod{U}{V} = (\Inprod{V}{U})^*$$, where $${}^*$$ is the complex conjugate.
-+ **Positive semidefiniteness**: $$\Inprod{V}{V} \ge 0$$, and $$\Inprod{V}{V} = 0$$ if $$V = \mathbf{0}$$.
-+ **Linearity in second operand**: $$\Inprod{U}{(a V + b W)} = a \Inprod{U}{V} + b \Inprod{U}{W}$$.
++ **Skew symmetry**:
+  $$\Inprod{U}{V} = (\Inprod{V}{U})^*$$, where $${}^*$$ is the complex conjugate.
++ **Positive semidefiniteness**:
+  $$\Inprod{V}{V} \ge 0$$, and $$\Inprod{V}{V} = 0$$ if $$V = \mathbf{0}$$.
++ **Linearity in second operand**:
+  $$\Inprod{U}{(a V + b W)} = a \Inprod{U}{V} + b \Inprod{U}{W}$$.
 
 The inner product describes the lengths and angles of vectors,
 and in Euclidean space it is implemented by the dot product.
@@ -93,34 +112,39 @@ The **magnitude** or **norm** $$|V|$$ of a vector $$V$$ is given by
 $$|V| = \sqrt{\Inprod{V}{V}}$$ and represents the real positive length of $$V$$.
 A **unit vector** has a norm of 1.
 
-Two vectors $$U$$ and $$V$$ are **orthogonal** if their inner product
-$$\Inprod{U}{V} = 0$$. If in addition to being orthogonal, $$|U| = 1$$ and
-$$|V| = 1$$, then $$U$$ and $$V$$ are known as **orthonormal** vectors.
+Two vectors $$U$$ and $$V$$ are **orthogonal**
+if their inner product $$\Inprod{U}{V} = 0$$.
+If, in addition to being orthogonal, $$|U| = 1$$ and $$|V| = 1$$,
+then $$U$$ and $$V$$ are known as **orthonormal** vectors.
 
-Orthonormality is desirable for basis vectors, so if they are
-not already like that, it is common to manually turn them into a new
-orthonormal basis using e.g. the [Gram-Schmidt method](/know/concept/gram-schmidt-method).
+Orthonormality is desirable for basis vectors,
+so if they are not already like that,
+it is common to manually turn them into a new orthonormal basis,
+using e.g. the [Gram-Schmidt method](/know/concept/gram-schmidt-method).
 
-As for the implementation of the inner product, it is given by:
+The implementation of the inner product in terms of components and basis vectors
+is as follows, which can easily be shown to satisfy the properties above:
 
 $$\begin{gathered}
     V = \sum_{n = 1}^N v_n \vu{e}_n
-    \quad
+    \qquad
     W = \sum_{n = 1}^N w_n \vu{e}_n
     \\
     \quad \implies \quad
     \Inprod{V}{W} = \sum_{n = 1}^N \sum_{m = 1}^N v_n^* w_m \Inprod{\vu{e}_n}{\vu{e}_j}
 \end{gathered}$$
 
-If the basis vectors $$\vu{e}_1, ..., \vu{e}_N$$ are already
-orthonormal, this reduces to:
+If the basis vectors $$\vu{e}_1, ..., \vu{e}_N$$ are already orthonormal,
+this reduces to:
 
 $$\begin{aligned}
     \Inprod{V}{W} = \sum_{n = 1}^N v_n^* w_n
 \end{aligned}$$
 
-As it turns out, the components $$v_n$$ are given by the inner product
-with $$\vu{e}_n$$, where $$\delta_{nm}$$ is the Kronecker delta:
+This suggests a way to calculate the components $$v_n$$:
+taking the inner product of $$V$$ with a basis vector $$\vu{e}_n$$
+"picks out" the corresponding component $$v_n$$.
+Let $$\delta_{nm}$$ be the Kronecker delta:
 
 $$\begin{aligned}
     \Inprod{\vu{e}_n}{V} = \sum_{m = 1}^N \delta_{nm} v_m = v_n
@@ -134,40 +158,46 @@ As the dimensionality $$N$$ tends to infinity, things may or may not
 change significantly, depending on whether $$N$$ is **countably** or
 **uncountably** infinite.
 
-In the former case, not much changes: the infinitely many **discrete**
-basis vectors $$\vu{e}_n$$ can all still be made orthonormal as usual,
-and as before:
+In the former case, not much changes:
+the infinitely many **discrete** basis vectors $$\vu{e}_n$$
+can all still be made orthonormal as usual, and as before:
 
 $$\begin{aligned}
     V = \sum_{n = 1}^\infty v_n \vu{e}_n
 \end{aligned}$$
 
-A good example of such a countably-infinitely-dimensional basis are the
-solution eigenfunctions of a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/).
+A good example of such a countably-infinitely-dimensional basis
+are the solution eigenfunctions of
+a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/).
 
-However, if the dimensionality is uncountably infinite, the basis
-vectors are **continuous** and cannot be labeled by $$n$$. For example, all
-complex functions $$f(x)$$ defined for $$x \in [a, b]$$ which
-satisfy $$f(a) = f(b) = 0$$ form such a vector space.
-In this case $$f(x)$$ is expanded as follows, where $$x$$ is a basis vector:
+However, if the dimensionality is uncountably infinite,
+the basis vectors are **continuous** and cannot be labeled by $$n$$.
+For example, all complex functions $$f(x)$$ defined on the interval $$x \in [a, b]$$
+satisfying the boundary condition $$f(a) = f(b) = 0$$, form such a vector space.
+In this case, every value of $$f(x)$$ is the component of
+an abstract vector $$\Ket{f}$$ with respect to a basis vector $$\Ket{x}$$:
 
 $$\begin{aligned}
-    f(x) = \int_a^b \Inprod{x}{f} \dd{x}
+    f(x) = \Inprod{x}{f}
 \end{aligned}$$
 
-Similarly, the inner product $$\Inprod{f}{g}$$ must also be redefined as
-follows:
+The inner product $$\Inprod{f}{g}$$ must be redefined as follows,
+effectively turning the sum over a discrete basis
+into an integral over a continuous basis:
 
 $$\begin{aligned}
     \Inprod{f}{g} = \int_a^b f^*(x) \: g(x) \dd{x}
 \end{aligned}$$
 
-The concept of orthonormality must be also weakened. A finite function
-$$f(x)$$ can be normalized as usual, but the basis vectors $$x$$ themselves
-cannot, since each represents an infinitesimal section of the real line.
+The concept of orthonormality must be also weakened.
+A finite function $$f(x)$$ can be normalized as usual,
+but the basis vectors $$x$$ themselves cannot,
+since each represents an infinitesimal section of the real line.
+So how to proceed?
 
-The rationale in this case is that action of the identity operator $$\hat{I}$$ must
-be preserved, which is given here in [Dirac notation](/know/concept/dirac-notation/):
+The rationale in this case is that the action
+of the identity operator $$\hat{I}$$ must be preserved,
+which is given here in [Dirac notation](/know/concept/dirac-notation/):
 
 $$\begin{aligned}
     \hat{I} = \int_a^b \Ket{\xi} \Bra{\xi} \dd{\xi}
@@ -181,8 +211,9 @@ $$\begin{aligned}
     = \int_a^b \Inprod{x}{\xi} f(\xi) \dd{\xi}
 \end{aligned}$$
 
-Since we want the latter integral to reduce to $$f(x)$$, it is plain to see that
-$$\Inprod{x}{\xi}$$ can only be a [Dirac delta function](/know/concept/dirac-delta-function/),
+Since we want the latter integral to reduce to $$f(x)$$,
+it is plain to see that $$\Inprod{x}{\xi}$$ can only be
+a [Dirac delta function](/know/concept/dirac-delta-function/),
 i.e $$\Inprod{x}{\xi} = \delta(x - \xi)$$:
 
 $$\begin{aligned}
@@ -191,12 +222,13 @@ $$\begin{aligned}
     = f(x)
 \end{aligned}$$
 
-Consequently, $$\Inprod{x}{\xi} = 0$$ if $$x \neq \xi$$ as expected for an
-orthogonal set of vectors, but if $$x = \xi$$ the inner product
-$$\Inprod{x}{\xi}$$ is infinite, unlike earlier.
+Consequently, $$\Inprod{x}{\xi} = 0$$ if $$x \neq \xi$$
+as expected for an orthogonal set of vectors,
+but if $$x = \xi$$ then the inner product $$\Inprod{x}{\xi}$$ is infinite,
+unlike earlier.
 
-Technically, because the basis vectors $$x$$ cannot be normalized, they
-are not members of a Hilbert space, but rather of a superset called a
-**rigged Hilbert space**. Such vectors have no finite inner product with
-themselves, but do have one with all vectors from the actual Hilbert
-space.
+Technically, because the basis vectors $$x$$ cannot be normalized,
+they are not members of a Hilbert space,
+but rather of a superset called a **rigged Hilbert space**.
+Such vectors have no finite inner product with themselves,
+but do have one with all vectors from the actual Hilbert space.
diff --git a/source/know/concept/jellium/index.md b/source/know/concept/jellium/index.md
index 5c50f80..5cd8483 100644
--- a/source/know/concept/jellium/index.md
+++ b/source/know/concept/jellium/index.md
@@ -12,32 +12,24 @@ layout: "concept"
 **Jellium**, also called the **uniform** or **homogeneous electron gas**,
 is a theoretical material where all electrons are free,
 and the ions' positive charge is smeared into a uniform background "jelly".
-This simple model lets us study electron interactions easily.
+This is a version of the [Fermi gas](/know/concept/fermi-gas/) model,
+which we extend by including electron-electron interactions using
+[time-independent perturbation theory](/know/concept/time-independent-perturbation-theory/).
 
 
-## Without interactions
 
-Let us start by neglecting electron-electron interactions.
-This is clearly a dubious assumption, but we will stick with it for now.
-For an infinitely large sample of jellium,
-the single-electron states are simply plane waves.
-We consider an arbitrary cube of volume $$V$$,
-and impose periodic boundary conditions on it,
-such that the single-particle orbitals are (suppressing spin):
+## 0th order
 
-$$\begin{aligned}
-    \Inprod{\vb{r}}{\psi_{\vb{k}}}
-    = \psi_{\vb{k}}(\vb{r})
-    = \frac{1}{\sqrt{V}} \exp(i \vb{k} \cdot \vb{r})
-    \qquad \quad
-    \vb{k} = \frac{2 \pi}{V^{1/3}} (n_x, n_y, n_z)
-\end{aligned}$$
+Let us start with the 0th order of the perturbation expansion.
+Without interactions or potentials, this is simply a Fermi gas,
+so the single-electron wavefunctions are just plane waves.
+For mathematical convenience, we consider these waves
+in a cube of volume $$V$$ with periodic boundaries,
+leading to a discrete spectrum of allowed wavevectors $$\vb{k}$$,
+which becomes continuous for $$V \to \infty$$.
 
-Where $$n_x, n_y, n_z \in \mathbb{Z}$$.
-This is a discrete (but infinite) set of independent orbitals,
-so it is natural to use the
-[second quantization](/know/concept/second-quantization/)
-to write the non-interacting Hamiltonian $$\hat{H}_0$$,
+The unperturbed many-particle Hamiltonian $$\hat{H}_0$$ is given below in the
+[second quantization](/know/concept/second-quantization/),
 where $$\hbar^2 |\vb{k}|^2 / (2 m)$$ is the kinetic energy
 of the orbital with wavevector $$\vb{k}$$, and $$s$$ is the spin:
 
@@ -46,10 +38,9 @@ $$\begin{aligned}
     = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}
 \end{aligned}$$
 
-Assuming that the temperature $$T = 0$$,
-the $$N$$-electron ground state of this Hamiltonian
-is known as the **Fermi sea** or **Fermi sphere** $$\Ket{\mathrm{FS}}$$,
-and is constructed by filling up the single-electron states
+Which, at absolute zero $$T = 0$$, has an $$N$$-electron ground state 
+known as the *Fermi sphere* $$\Ket{\mathrm{FS}}$$
+that is constructed by filling up the single-electron states
 starting from the lowest energy:
 
 $$\begin{aligned}
@@ -57,63 +48,9 @@ $$\begin{aligned}
     = \prod_{s} \prod_{j = 1}^{N/2} \hat{c}_{s,\vb{k}_j}^\dagger \Ket{0}
 \end{aligned}$$
 
-Because $$T = 0$$, all the electrons stay in their assigned state.
-The energy and wavenumber $$|\vb{k}|$$ of the highest filled orbital
-are called the **Fermi energy** $$\epsilon_F$$ and **Fermi wavenumber** $$k_F$$,
-and obey the expected kinetic energy relation:
-
-$$\begin{aligned}
-    \boxed{
-        \epsilon_F
-        = \frac{\hbar^2}{2 m} k_F^2
-    }
-\end{aligned}$$
-
-The Fermi sea can be visualized in $$\vb{k}$$-space as a sphere with radius $$k_F$$.
-Because $$\vb{k}$$ is discrete, the sphere's surface is not smooth,
-but in the limit $$V \to \infty$$ it becomes perfect.
-
-Now, we would like a relation between the system's parameters,
-e.g. $$N$$ and $$V$$, and the resulting values of $$\epsilon_F$$ or $$k_F$$.
-The total population $$N$$ must be given by:
-
-$$\begin{aligned}
-    N
-    = \sum_{s} \sum_{\vb{k}} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}
-    = \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} \dd{\vb{k}}
-\end{aligned}$$
-
-Where we have turned the sum over $$\vb{k}$$ into an integral with a constant factor,
-by using that each orbital exclusively occupies a volume $$(2 \pi)^3 / V$$ in $$\vb{k}$$-space.
-
-At zero temperature, this inner product can only be $$0$$ or $$1$$,
-depending on whether $$\vb{k}$$ is outside or inside the Fermi sphere.
-We can therefore rewrite using a
-[Heaviside step function](/know/concept/heaviside-step-function/):
-
-$$\begin{aligned}
-    N
-    = \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \Theta(k_F - |\vb{k}|) \dd{\vb{k}}
-    = 2 \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \Theta(k_F - |\vb{k}|) \dd{\vb{k}}
-\end{aligned}$$
-
-Where we realized that spin does not matter,
-and replaced the sum over $$s$$ by a factor $$2$$.
-In order to evaluate this 3D integral,
-we go to [spherical coordinates](/know/concept/spherical-coordinates/)
-$$(|\vb{k}|, \theta, \varphi)$$:
-
-$$\begin{aligned}
-    N
-    &= \frac{V}{4 \pi^3} \int_0^{2 \pi} \int_0^\pi \int_0^\infty \Theta(k_F - |\vb{k}|) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi}
-    \\
-    &= \frac{V}{4 \pi^3} 4 \pi \int_0^{k_F} |\vb{k}|^2 \dd{|\vb{k}|}
-    = \frac{V}{\pi^2} \bigg[ \frac{|\vb{k}|^3}{3} \bigg]_0^{k_F}
-    = \frac{V}{3 \pi^2} k_F^3
-\end{aligned}$$
-
-Using that the electron density $$n = N/V$$,
-we thus arrive at the following relation:
+From our analysis of the Fermi gas, we have an important result
+for the wavenumber $$k_F = |\vb{k}_{N/2}|$$ of the highest filled orbital,
+as a function of the particle density $$n = N / V$$:
 
 $$\begin{aligned}
     \boxed{
@@ -122,103 +59,137 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-This result also justifies our assumption that $$T = 0$$:
-we can accurately calculate the density $$n$$ for many conducting materials,
-and this relation then gives $$k_F$$ and $$\epsilon_F$$.
-It turns out that $$\epsilon_F$$ is usually very large
-compared to the thermal energy $$k_B T$$ at reasonable temperatures,
-so we can conclude that thermal fluctuations are negligible.
-
-Now, $$\epsilon_F$$ is the highest single-electron energy,
-but about the total $$N$$-particle energy $$E^{(0)}$$?
+Now, let us calculate the total $$N$$-particle ground state
+energy $$E^{(0)}$$ of the unperturbed system:
 
 $$\begin{aligned}
     E^{(0)}
     = \matrixel{\mathrm{FS}}{\hat{H}_0}{\mathrm{FS}}
-    = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}
+    = 2 \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}
 \end{aligned}$$
 
-Once again, we turn the sum over $$\vb{k}$$ into an integral,
-and recognize the spin's irrelevance:
+Where we have recognized the spin's irrelevance,
+by replacing the sum over $$s$$ with a factor $$2$$.
+Next, we turn the sum over the allowed $$\vb{k}$$-values into an integral,
+which is a [common trick](/know/concept/discrete-spectrum-summation/)
+enabled by our periodic boundary conditions, yielding:
 
 $$\begin{aligned}
     E^{(0)}
-    &= \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \frac{\hbar^2 |\vb{k}|^2}{2 m}
+    &= \frac{2 V}{(2 \pi)^3} \int_{-\infty}^\infty \frac{\hbar^2 |\vb{k}|^2}{2 m}
     \matrixel{\mathrm{FS}}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}}}{\mathrm{FS}} \dd{\vb{k}}
-    \\
+\end{aligned}$$
+
+The matrix element
+$$\matrixel{\mathrm{FS}}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}}}{\mathrm{FS}} \dd{\vb{k}}$$
+is either $$0$$ or $$1$$, depending on whether $$\vb{k}$$
+is outside the Fermi sphere, or, equivalently,
+whether $$|\vb{k}|$$ is above or below $$k_F$$.
+We can write this fact by introducing a
+[Heaviside step function](/know/concept/heaviside-step-function/) $$\Theta(k)$$:
+
+$$\begin{aligned}
+    E^{(0)}
     &= \frac{\hbar^2 V}{8 \pi^3 m} \int_{-\infty}^\infty |\vb{k}|^2 \: \Theta(k_F - |\vb{k}|) \dd{\vb{k}}
 \end{aligned}$$
 
-In spherical coordinates,
-we evaluate the integral and find that $$E^{(0)}$$ is proportional to $$k_F^5$$:
+We evaluate this in
+[spherical coordinates](/know/concept/spherical-coordinates/)
+and find that $$E^{(0)}$$ is proportional to $$k_F^5$$:
 
 $$\begin{aligned}
     E^{(0)}
     &= \frac{\hbar^2 V}{8 \pi^3 m} \int_0^{2 \pi}
     \int_0^\pi \int_0^\infty \Big( |\vb{k}|^2 \: \Theta(k_F - |\vb{k}|) \Big) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi}
     \\
-    &= \frac{\hbar^2 V}{8 \pi^3 m} 4 \pi \int_0^{k_F} |\vb{k}|^4 \dd{|\vb{k}|}
-    = \frac{\hbar^2 V}{2 \pi^2 m} \bigg[ \frac{|\vb{k}|^5}{5} \bigg]_0^{k_F}
-    = \frac{\hbar^2 V}{10 \pi^2 m} k_F^5
+    &= \frac{\hbar^2 V}{8 \pi^3 m} \: 4 \pi \int_0^{k_F} |\vb{k}|^4 \dd{|\vb{k}|}
+    \\
+    &= \frac{\hbar^2 V}{10 \pi^2 m} k_F^5
 \end{aligned}$$
 
 In general, it is more useful to consider
 the average kinetic energy per electron $$E^{(0)} / N$$,
-which we find to be as follows, using that $$k_F^3 = 3 \pi^2 n$$:
+which we find to be as follows,
+using that $$k_F^3 = 3 \pi^2 N / V$$:
 
 $$\begin{aligned}
     \boxed{
         \frac{E^{(0)}}{N}
         = \frac{3 \hbar^2}{10 m} k_F^2
-        = \frac{3}{5} \epsilon_F
     }
-    \:\sim\: n^{2/3}
+    \:\:\propto\: n^{2/3}
 \end{aligned}$$
 
-Traditionally, this is expressed using a dimensionless parameter $$r_s$$,
+Traditionally, this is rewritten using the **Wigner-Seitz radius** $$r_s$$,
 defined as the radius of a sphere containing a single electron,
-measured in Bohr radii $$a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m)$$:
+measured in Bohr radii $$a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e_0^2 m)$$:
 
 $$\begin{aligned}
         \frac{4 \pi}{3} (a_0 r_s)^3
-        = \frac{1}{n}
-        = \frac{3 \pi^2}{k_F^3}
-        \quad \implies \quad
+        \equiv \frac{1}{n}
+        \qquad \implies \qquad
         r_s
-        = \Big( \frac{3}{4 \pi a_0^3 n} \Big)^{1/3}
-        = \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 k_F}
+        = \bigg( \frac{3}{4 \pi a_0^3 n} \bigg)^{1/3}
 \end{aligned}$$
 
-Such that the ground state energy can be rewritten in Rydberg units of energy like so:
+Note that this is dimensionless due to our choice of $$a_0$$ as a unit.
+In the Fermi gas, we have:
+
+$$\begin{aligned}
+        r_s
+        = \bigg( \frac{9 \pi}{4} \bigg)^{1/3} \frac{1}{a_0 k_F}
+        \qquad \implies \qquad
+        k_F
+        = \bigg( \frac{9 \pi}{4} \bigg)^{1/3} \frac{1}{a_0 r_s}
+\end{aligned}$$
+
+By inserting this into the ground state energy
+and using the definition of $$a_0$$, we can write:
 
 $$\begin{aligned}
     \frac{E^{(0)}}{N}
-    = \frac{3 \hbar^2}{10 m} \frac{4 \pi \varepsilon_0 e^2}{4 \pi \varepsilon_0 e^2} \frac{a_0^2 k_F^2}{a_0^2}
-    = \frac{3 e^2}{40 \pi \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{2/3} \frac{1}{a_0 r_s^2}
-    \approx \frac{2.21}{r_s^2} \; \mathrm{Ry}
+    &= \frac{3 \hbar^2}{10 m} \bigg( \frac{9 \pi}{4} \bigg)^{2/3} \frac{1}{a_0^2 r_s^2}
+    \\
+    &= \frac{3}{5} \bigg( \frac{9 \pi}{4} \bigg)^{2/3} \bigg( \frac{e_0^2}{8 \pi \varepsilon_0 a_0} \bigg) \frac{1}{r_s^2}
 \end{aligned}$$
 
+Where the last parenthesized expression is
+the Rydberg unit of energy $$\mathrm{Ry} \approx 13.6 \:\mathrm{eV}$$, so:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{E^{(0)}}{N}
+        \approx \frac{2.21}{r_s^2} \; \mathrm{Ry}
+    }
+\end{aligned}$$
+
+This result is found in a lot of literature.
+The choice of Rydberg units is simply a tradition.
+
+
 
-## With interactions
+## 1st order
 
-To include Coulomb interactions, let us try
-[time-independent pertubation theory](/know/concept/time-independent-perturbation-theory/).
-Clearly, this will give better results when the interaction is relatively weak, if ever.
+In the next term of the perturbation expansion,
+we start to include Coulomb interactions.
+Clearly, this will give better results when the interaction is relatively weak,
+but is that ever the case?
 
 The Coulomb potential is proportional to the inverse distance,
 and the average electron spacing is roughly $$n^{-1/3}$$,
-so the interaction energy $$E_\mathrm{int}$$ should scale as $$n^{1/3}$$.
-We already know that the kinetic energy $$E_\mathrm{kin} = E^{(0)}$$ scales as $$n^{2/3}$$,
-meaning perturbation theory should be reasonable
-if $$1 \gg E_\mathrm{int} / E_\mathrm{kin} \sim n^{-1/3}$$,
-so in the limit of high density $$n \to \infty$$.
+so the interaction energy $$E_\mathrm{int}$$ scales as $$n^{1/3}$$.
+We also know that the kinetic energy $$E_\mathrm{kin} = E^{(0)}$$
+is proportional to $$n^{2/3}$$,
+meaning that it is reasonable to use perturbation theory
+as long as $$1 \gg E_\mathrm{int} / E_\mathrm{kin} \propto n^{-1/3}$$,
+i.e. in the limit of high density $$n \to \infty$$.
 
 The two-body Coulomb interaction operator $$\hat{W}$$
 is as follows in second-quantized form:
 
 $$\begin{aligned}
     \hat{W}
-    = \frac{1}{2 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{e^2}{\varepsilon_0 |\vb{q}|^2}
+    = \frac{1}{2 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{e_0^2}{\varepsilon_0 |\vb{q}|^2}
     \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1}
 \end{aligned}$$
 
@@ -228,7 +199,7 @@ is then given by:
 $$\begin{aligned}
     E^{(1)}
     = \matrixel{\mathrm{FS}}{\hat{W}}{\mathrm{FS}}
-    = \frac{e^2}{2 \varepsilon_0 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
+    = \frac{e_0^2}{2 \varepsilon_0 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
     \matrixel{\mathrm{FS}}{
         \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1}
     }{\mathrm{FS}}
@@ -246,17 +217,17 @@ Let $$s = s_1$$ and $$\vb{k} = \vb{k}_1$$:
 
 $$\begin{aligned}
     E^{(1)}
-    &= \frac{e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
+    &= \frac{e_0^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
     \matrixel{\mathrm{FS}}{
         \hat{c}_{s, \vb{k} + \vb{q}}^\dagger \hat{c}_{s, \vb{k}}^\dagger \hat{c}_{s, \vb{k} + \vb{q}} \hat{c}_{s, \vb{k}}
     }{\mathrm{FS}}
     \\
-    &= \frac{- e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
+    &= \frac{- e_0^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
     \matrixel{\mathrm{FS}}{
         \big( \hat{c}_{s, \vb{k} + \vb{q}}^\dagger \hat{c}_{s, \vb{k} + \vb{q}}\big) \big(\hat{c}_{s, \vb{k}}^\dagger \hat{c}_{s, \vb{k}}\big)
     }{\mathrm{FS}}
     \\
-    &= \frac{- e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
+    &= \frac{- e_0^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
     \Theta(k_F - |\vb{k}|) \:\Theta(k_F - |\vb{k} \!+\! \vb{q}|)
 \end{aligned}$$
 
@@ -269,11 +240,11 @@ This yields the integration limit, and therefore leads to:
 
 $$\begin{aligned}
     E^{(1)}
-    &= \frac{- e^2}{(2 \pi)^3 \varepsilon_0} \sum_{\vb{k}}
+    &= \frac{- e_0^2}{(2 \pi)^3 \varepsilon_0} \sum_{\vb{k}}
     \int_0^{2 \pi} \!\!\int_0^\pi \!\!\int_0^\infty \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \frac{|\vb{q}|^2}{|\vb{q}|^2}
     \sin(\theta_q) \dd{|\vb{q}|} \dd{\theta_q} \dd{\varphi_q}
     \\
-    &= \frac{- e^2}{2 \pi^2 \varepsilon_0} \sum_{\vb{k}}
+    &= \frac{- e_0^2}{2 \pi^2 \varepsilon_0} \sum_{\vb{k}}
     \int_0^{2 k_F} \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \dd{|\vb{q}|}
 \end{aligned}$$
 
@@ -285,11 +256,11 @@ when we go to spherical coordinates $$(|\vb{k}|, \theta_k, \varphi_k)$$ for $$\v
 
 $$\begin{aligned}
     E^{(1)}
-    &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^\infty
+    &= \frac{- e_0^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^\infty
     \!\Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|)
     \: |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|}
     \\
-    &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^{k_F}
+    &= \frac{- e_0^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^{k_F}
     \!\Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|)
     \: |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|}
 \end{aligned}$$
@@ -335,13 +306,13 @@ substituting $$\xi \equiv \cos(\theta_k)$$:
 
 $$\begin{aligned}
     E^{(1)}
-    &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} 2 \int_0^{2 k_F} \!\!\!\int_0^{2 \pi} \!\!\int_0^{\arccos{|\vb{q}| / (2 k_F)}}
+    &= \frac{- e_0^2 V}{16 \pi^5 \varepsilon_0} 2 \int_0^{2 k_F} \!\!\!\int_0^{2 \pi} \!\!\int_0^{\arccos{|\vb{q}| / (2 k_F)}}
     \!\!\int_{|\vb{q}|/(2 \cos{\theta_k})}^{k_F} |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|}
     \\
-    &= \frac{e^2 V}{8 \pi^5 \varepsilon_0} 2 \pi \int_0^{2 k_F} \!\!\!\int_1^{|\vb{q}| / (2 k_F)}
+    &= \frac{e_0^2 V}{8 \pi^5 \varepsilon_0} 2 \pi \int_0^{2 k_F} \!\!\!\int_1^{|\vb{q}| / (2 k_F)}
     \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \frac{\sin(\theta_k)}{\sin(\theta_k)} \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|}
     \\
-    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
+    &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
     \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|}
 \end{aligned}$$
 
@@ -350,23 +321,23 @@ Evaluating these integrals:
 
 $$\begin{aligned}
     E^{(1)}
-    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
+    &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
     \bigg[ \frac{|\vb{k}|^3}{3} \bigg]_{|\vb{q}|/(2 \xi)}^{k_F} \dd{\xi} \dd{|\vb{q}|}
     \\
-    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
+    &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
     \bigg( \frac{k_F^3}{3} - \frac{|\vb{q}|^3}{24 \xi^3} \bigg) \dd{\xi} \dd{|\vb{q}|}
     \\
-    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
+    &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
     \bigg[ \frac{k_F^3}{3} x + \frac{|\vb{q}|^3}{48 \xi^2} \bigg]_{|\vb{q}| / (2 k_F)}^1 \dd{|\vb{q}|}
     \\
-    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
+    &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
     \bigg( \frac{k_F^3}{3} + \frac{|\vb{q}|^3}{48} - \frac{k_F^2 |\vb{q}|}{4} \bigg) \dd{|\vb{q}|}
     \\
-    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \bigg[ \frac{k_F^3 |\vb{q}|}{3} + \frac{|\vb{q}|^4}{192} - \frac{k_F^2 |\vb{q}|^2}{8} \bigg]_0^{2 k_F}
+    &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \bigg[ \frac{k_F^3 |\vb{q}|}{3} + \frac{|\vb{q}|^4}{192} - \frac{k_F^2 |\vb{q}|^2}{8} \bigg]_0^{2 k_F}
     \\
-    &= \frac{- e^2 V}{16 \pi^4 \varepsilon_0} k_F^4
-    = \frac{- e^2 N}{16 \pi^4 \varepsilon_0 n} k_F^4
-    = -\frac{3 e^2 N}{16 \pi^2 \varepsilon_0} k_F
+    &= \frac{- e_0^2 V}{16 \pi^4 \varepsilon_0} k_F^4
+    = \frac{- e_0^2 N}{16 \pi^4 \varepsilon_0 n} k_F^4
+    = -\frac{3 e_0^2 N}{16 \pi^2 \varepsilon_0} k_F
 \end{aligned}$$
 
 Per particle, the first-order energy correction $$E^{(1)}$$
@@ -375,7 +346,7 @@ is therefore found to be as follows:
 $$\begin{aligned}
     \boxed{
         \frac{E^{(1)}}{N}
-        = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} k_F
+        = -\frac{3 e_0^2}{16 \pi^2 \varepsilon_0} k_F
     }
 \end{aligned}$$
 
@@ -383,8 +354,8 @@ This can also be written using the parameter $$r_s$$ introduced above, leading t
 
 $$\begin{aligned}
     \frac{E^{(1)}}{N}
-    = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \frac{a_0 k_F}{a_0}
-    = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 r_s}
+    = -\frac{3 e_0^2}{16 \pi^2 \varepsilon_0} \frac{a_0 k_F}{a_0}
+    = -\frac{3 e_0^2}{16 \pi^2 \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 r_s}
 \end{aligned}$$
 
 Consequently, for sufficiently high densities $$n$$,
diff --git a/source/know/concept/matsubara-sum/index.md b/source/know/concept/matsubara-summation/index.md
index 3347476..de08024 100644
--- a/source/know/concept/matsubara-sum/index.md
+++ b/source/know/concept/matsubara-summation/index.md
@@ -1,6 +1,6 @@
 ---
-title: "Matsubara sum"
-sort_title: "Matsubara sum"
+title: "Matsubara summation"
+sort_title: "Matsubara summation"
 date: 2021-11-13
 categories:
 - Physics
@@ -8,7 +8,7 @@ categories:
 layout: "concept"
 ---
 
-A **Matsubara sum** is a summation of the following form,
+**Matsubara summation** is a technique for evaluating sums of the following form,
 which notably appears as the inverse
 [Fourier transform](/know/concept/fourier-transform/) of the
 [Matsubara Green's function](/know/concept/matsubara-greens-function/):
diff --git a/source/know/concept/quantum-teleportation/index.md b/source/know/concept/quantum-teleportation/index.md
index 095c2c6..f57f981 100644
--- a/source/know/concept/quantum-teleportation/index.md
+++ b/source/know/concept/quantum-teleportation/index.md
@@ -22,7 +22,7 @@ $$\begin{aligned}
 
 She can only directly communicate with Bob over a classical channel.
 This is not enough: even if Alice did know $$\alpha$$ and $$\beta$$ exactly
-(which would need her having infinitely many copies to measure),
+(for which she would need infinitely many copies to measure),
 sending an arbitrary real number requires an infinite amount of classical data.
 
 However, between them, she and Bob also have an entangled [Bell state](/know/concept/bell-state/),
@@ -32,7 +32,7 @@ with $$A'$$ being Alice' qubit, $$A$$ her side of the Bell state, and $$B$$ Bob'
 
 $$\begin{aligned}
     \Ket{q}_{A'} \otimes \ket{\Phi^+}_{AB}
-    &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{0} + \beta \Ket{1} \Big)_{A'} \Big( \Ket{00} + \Ket{11} \Big)_{AB}
+    &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{0} + \beta \Ket{1} \Big)_{A'} \otimes \Big( \Ket{00} + \Ket{11} \Big)_{AB}
     \\
     &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{000} + \beta \Ket{100}
     + \alpha \Ket{011} + \beta \Ket{111} \Big)_{A'AB}
@@ -78,8 +78,10 @@ $$\begin{aligned}
     + \ket{\Psi^{-}}_{A'A} \Big( \alpha \Ket{1} - \beta \Ket{0} \Big)_{B} \bigg)
 \end{aligned}$$
 
-Thus, purely due to entanglement,
-Bob's qubit $$B$$ is in a superposition of the following states:
+Therefore, thanks to entanglement,
+Bob's qubit $$B$$ is in a superposition of the following states,
+where $$\hat{\sigma}_x$$ and $$\hat{\sigma}_z$$ are Pauli matrices
+(see [quantum gate](/know/concept/quantum-gate/)):
 
 $$\begin{aligned}
     \Ket{q}
@@ -95,8 +97,8 @@ $$\begin{aligned}
     = \alpha \Ket{1} - \beta \Ket{0}
 \end{aligned}$$
 
-Consequently, Alice and Bob are sharing (or, to be precise, seeing different sides of)
-the following entangled three-qubit state:
+Consequently, Alice and Bob are seeing different sides of
+this entangled three-qubit state:
 
 $$\begin{aligned}
     \Ket{q}_{A'} \ket{\Phi^+}_{AB}
@@ -123,7 +125,7 @@ who then either does nothing (for $$\Ket{q}$$),
 applies $$\hat{\sigma}_z$$ (for $$\hat{\sigma}_z \Ket{q}$$),
 applies $$\hat{\sigma}_x$$ (for $$\hat{\sigma}_x \Ket{q}$$),
 or applies $$\hat{\sigma}_z \hat{\sigma}_x$$ (for $$\hat{\sigma}_x \hat{\sigma}_z \Ket{q}$$).
-Then, due to the fact that $$\hat{\sigma}_x^2 = \hat{\sigma}_z^2 = \hat{I}$$,
+Then, thanks to the fact that $$\hat{\sigma}_x^2 = \hat{\sigma}_z^2 = \hat{I}$$,
 he recovers $$\Ket{q}$$ in his local qubit $$B$$.
 
 This is not violating the [no-cloning theorem](/know/concept/no-cloning-theorem)
@@ -140,6 +142,7 @@ Before receiving that, Bob only sees his side of the maximally entangled
 Bell state $$\ket{\Phi^{+}}_{AB}$$, which contains nothing of $$\Ket{q}$$.
 
 
+
 ## References
 1.  J.B. Brask,
     *Quantum information: lecture notes*,
diff --git a/source/know/concept/random-phase-approximation/index.md b/source/know/concept/random-phase-approximation/index.md
index 03fd302..ab2681f 100644
--- a/source/know/concept/random-phase-approximation/index.md
+++ b/source/know/concept/random-phase-approximation/index.md
@@ -127,7 +127,7 @@ $$\begin{aligned}
     \frac{1}{i \hbar \omega_n^B + i \hbar \omega_m^F - \varepsilon_{\vb{k}+\vb{q}}} \: \frac{1}{i \hbar \omega_m^F - \varepsilon_{\vb{q}}} \dd{\vb{q}}
 \end{aligned}$$
 
-Here we recognize a [Matsubara sum](/know/concept/matsubara-sum/),
+Here we recognize a [Matsubara sum](/know/concept/matsubara-summation/),
 and rewrite accordingly.
 Note that the residues of $$n_F$$ are $$1 / (\hbar \beta)$$
 when it is a function of frequency,
diff --git a/source/know/concept/runge-kutta-method/index.md b/source/know/concept/runge-kutta-method/index.md
index 4c3dacf..f0e54ba 100644
--- a/source/know/concept/runge-kutta-method/index.md
+++ b/source/know/concept/runge-kutta-method/index.md
@@ -88,12 +88,13 @@ since this is not a practical way to describe RKMs,
 but it is helpful to understand how they work.
 
 
+
 ## Example derivation
 
 For example, let us truncate at $$n = 3$$,
 such that $$N_1 = 3$$, $$N_2 = 3$$ and $$N_3 = 1$$.
 The following derivation is very general,
-except it requires all $$\alpha_j \neq 0$$.
+only requiring all $$\alpha_j \neq 0$$.
 Renaming $$\omega_{mj}$$, we start from:
 
 $$\begin{aligned}
@@ -177,6 +178,7 @@ there is an enormous freedom of choice here,
 all leading to valid RKMs, although not necessarily good ones.
 
 
+
 ## General form
 
 A more practical description goes as follows:
diff --git a/source/know/concept/rutherford-scattering/index.md b/source/know/concept/rutherford-scattering/index.md
index edf391c..7a2a1f2 100644
--- a/source/know/concept/rutherford-scattering/index.md
+++ b/source/know/concept/rutherford-scattering/index.md
@@ -27,13 +27,13 @@ Intuitively, we expect $$\theta$$ to be larger for smaller $$b$$.
 
 By combining Coulomb's law with Newton's laws,
 these particles' equations of motion are found to be as follows,
-where $$r = |\vb{r}_1 - \vb{r}_2|$$ is the distance between 1 and 2:
+where $$r \equiv |\vb{r}_1 \!-\! \vb{r}_2|$$ is the distance between 1 and 2:
 
 $$\begin{aligned}
     m_1 \dv{\vb{v}_1}{t}
     = \vb{F}_1
     = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}_1 - \vb{r}_2}{r^3}
-    \qquad \quad
+    \qquad \qquad
     m_2 \dv{\vb{v}_2}{t}
     = \vb{F}_2
     = - \vb{F}_1
@@ -56,8 +56,9 @@ $$(r, \varphi, z)$$:
 
 $$\begin{aligned}
     \vb{r}
-    = r \cos{\varphi} \:\vu{e}_x + r \sin{\varphi} \:\vu{e}_y + z \:\vu{e}_z
-    = r \:\vu{e}_r + z \:\vu{e}_z
+    &= r \cos{\varphi} \:\vu{e}_x + r \sin{\varphi} \:\vu{e}_y + z \:\vu{e}_z
+    \\
+    &= r \:\vu{e}_r + z \:\vu{e}_z
 \end{aligned}$$
 
 These new coordinates are sketched below,
@@ -76,6 +77,7 @@ we can find $$\vb{v}$$ by differentiating with respect to time:
 
 $$\begin{aligned}
     \vb{v}
+    = \vb{r}'
     &= \big( r' \cos{\varphi} - r \varphi' \sin{\varphi} \big) \:\vu{e}_x
     + \big( r' \sin{\varphi} + r \varphi' \cos{\varphi} \big) \:\vu{e}_y + z' \:\vu{e}_z
     \\
@@ -107,32 +109,34 @@ $$\begin{aligned}
     = \mu r^2 \varphi' \:\vu{e}_z
 \end{aligned}$$
 
-Now, from the figure above,
-we can argue geometrically that at infinity $$t = \pm \infty$$,
-the ratio $$b/r$$ is related to the angle $$\chi$$ between $$\vb{v}$$ and $$\vb{r}$$ like so:
+Now, in the figure above, imagine a right-angled triangle
+with hypotenuse $$\vb{r}$$ and short side $$b$$.
+When $$t \to +\infty$$, trigonometry tells us the following,
+where $$\chi$$ is the final angle between $$\vb{v}$$ and $$\vb{r}$$:
 
 $$\begin{aligned}
-    \frac{b}{r(\pm \infty)}
-    = \sin{\chi(\pm \infty)}
-    \qquad \quad
-    \chi(t)
-    \equiv \measuredangle(\vb{r}, \vb{v})
+    \lim_{t \to +\infty} \frac{b}{r(t)}
+    = \sin{\chi}
+    \qquad \qquad
+    \chi
+    \equiv
+    \lim_{t \to +\infty} \measuredangle(\vb{r}(t), \vb{v}(t))
 \end{aligned}$$
 
-With this, we can rewrite
-the magnitude of the angular momentum $$\vb{L}$$ as follows,
-where the total velocity $$|\vb{v}|$$ is a constant,
-thanks to conservation of energy:
+With this, we can rewrite the magnitude of the angular momentum $$\vb{L}$$ as follows,
+where the relative speed $$|\vb{v}|$$ is a constant thanks to energy conservation:
 
 $$\begin{aligned}
-    \big| \vb{L}(\pm \infty) \big|
-    = \mu \big| \vb{r} \cross \vb{v} \big|
+    \lim_{t \to +\infty}
+    \big| \vb{L}(t) \big|
     = \mu r |\vb{v}| \sin{\chi}
     = \mu b |\vb{v}|
 \end{aligned}$$
 
-However, conveniently,
-angular momentum is also conserved, i.e. $$\vb{L}$$ is constant in time:
+This is useful, because angular momentum is conserved,
+i.e. $$\vb{L}$$ is constant in time.
+We prove this by using the product rule of differentiation,
+and replacing $$\mu \vb{v}'$$ with the reduced equation of motion:
 
 $$\begin{aligned}
     \vb{L}'(t)
@@ -142,8 +146,8 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-Where we have replaced $$\mu \vb{v}'$$ with the equation of motion.
-Thanks to this, we can equate the two preceding expressions for $$\vb{L}$$,
+Thanks to this, we can equate the two preceding expressions
+for the magnitude $$|\vb{L}|$$,
 leading to the relation below.
 Note the appearance of a new minus,
 because the sketch shows that $$\varphi' < 0$$,
@@ -178,8 +182,8 @@ $$\begin{aligned}
     = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \dd{(\cos{\varphi})}
 \end{aligned}$$
 
-Integrating this from the initial state $$i$$ at $$t = -\infty$$
-to the final state $$f$$ at $$t = \infty$$ yields:
+Integrating this from the initial state $$i$$ at $$t \to -\infty$$
+to the final state $$f$$ at $$t \to +\infty$$ yields:
 
 $$\begin{aligned}
     \Delta v_y
@@ -187,18 +191,18 @@ $$\begin{aligned}
     = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos{\varphi_i} \big)
 \end{aligned}$$
 
-From symmetry, we see that $$\varphi_i = \pi \!-\! \varphi_f$$,
-and that $$\Delta v_y = v_{y,f} \!-\! v_{y,i} = 2 v_{y,f}$$, such that:
+From symmetry, we see that $$\Delta v_y = v_{y,f} \!-\! v_{y,i} = 2 v_{y,f}$$,
+and that $$\varphi_i = \pi \!-\! \varphi_f$$, such that:
 
 $$\begin{aligned}
-    2 v_{y,f}
+    \Delta v_y
+    = 2 v_{y,f}
     = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos(\pi \!-\! \varphi_f) \big)
     = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( 2 \cos{\varphi_f} \big)
 \end{aligned}$$
 
-Furthermore, geometrically, at $$t = \infty$$
-we notice that $$v_{y,f} = |\vb{v}| \sin{\varphi_f}$$,
-leading to:
+Furthermore, geometrically for $$t \to +\infty$$
+we notice that $$v_{y,f} = |\vb{v}| \sin{\varphi_f}$$, leading to:
 
 $$\begin{aligned}
     2 |\vb{v}| \sin{\varphi_f}
@@ -206,7 +210,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Rearranging this yields the following equation
-for the final polar angle $$\varphi_f \equiv \varphi(\infty)$$:
+for the final polar angle $$\varphi_f$$:
 
 $$\begin{aligned}
     \tan{\varphi_f}
@@ -214,14 +218,14 @@ $$\begin{aligned}
     = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu}
 \end{aligned}$$
 
-However, we want $$\theta$$, not $$\varphi_f$$.
+However, we want the deflection angle $$\theta$$, not $$\varphi_f$$.
 One last use of symmetry and geometry
 tells us that $$\theta = 2 \varphi_f$$,
 and we thus arrive at the celebrated **Rutherford scattering formula**:
 
 $$\begin{aligned}
     \boxed{
-        \tan\!\Big( \frac{\theta}{2} \Big)
+        \tan\!\bigg( \frac{\theta}{2} \bigg)
         = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu}
     }
 \end{aligned}$$
diff --git a/source/know/concept/superdense-coding/index.md b/source/know/concept/superdense-coding/index.md
index 4338205..0ad8e9e 100644
--- a/source/know/concept/superdense-coding/index.md
+++ b/source/know/concept/superdense-coding/index.md
@@ -25,16 +25,17 @@ where $$A$$ and $$B$$ are qubits belonging to Alice and Bob, respectively.
 
 Based on the values of the two classical bits $$(a_1, a_2)$$,
 Alice performs the following operations on her side $$A$$
-of the Bell state:
+of the Bell state, where $$\hat{\sigma}_x$$ and $$\hat{\sigma}_z$$
+are Pauli matrices (see [quantum gate](/know/concept/quantum-gate/)):
 
 | $$(a_1, a_2)$$ | **Operator** | **Result** |
 | :-: | :-: | :-: |
-| $$00$$ | $$\hat{I}$$ | $$\displaystyle \ket{\Phi^{+}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big)$$ |
-| $$01$$ | $$\hat{\sigma}_z$$ | $$\displaystyle \ket{\Phi^{-}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{0}_B - \Ket{1}_A \Ket{1}_B \Big)$$ |
-| $$10$$ | $$\hat{\sigma}_x$$ | $$\displaystyle \ket{\Psi^{+}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{1}_B + \Ket{1}_A \Ket{0}_B \Big)$$ |
-| $$11$$ | $$\hat{\sigma}_x \hat{\sigma}_z$$ | $$\displaystyle \ket{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{1}_B - \Ket{1}_A \Ket{0}_B \Big)$$ |
+| $$00$$ | $$\hat{I}$$ | $$\displaystyle \ket{\Phi^{+}} = \frac{1}{\sqrt{2}} \Big( \:\:\: \Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big)$$ |
+| $$01$$ | $$\hat{\sigma}_x$$ | $$\displaystyle \ket{\Psi^{+}} = \frac{1}{\sqrt{2}} \Big( \:\:\: \Ket{1}_A \Ket{0}_B + \Ket{0}_A \Ket{1}_B \Big)$$ |
+| $$10$$ | $$\hat{\sigma}_z$$ | $$\displaystyle \ket{\Phi^{-}} = \frac{1}{\sqrt{2}} \Big( \:\:\: \Ket{0}_A \Ket{0}_B - \Ket{1}_A \Ket{1}_B \Big)$$ |
+| $$11$$ | $$\hat{\sigma}_x \hat{\sigma}_z$$ | $$\displaystyle \ket{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big( \!-\! \Ket{1}_A \Ket{0}_B + \Ket{0}_A \Ket{1}_B \Big)$$ |
 
-Her actions affect the state on Bob's side $$B$$ due to entanglement.
+Her actions indirectly affect the state on Bob's side $$B$$ due to entanglement.
 Alice then sends her qubit $$A$$ to Bob over the quantum channel,
 so he has both sides of the entangled pair.
 
@@ -45,6 +46,7 @@ In the end, Alice only sent a single qubit,
 and the rest of the information transfer was via entanglement.
 
 
+
 ## References
 1.  J.B. Brask,
     *Quantum information: lecture notes*,