More improvements to knowledge base

15 files changed, 316 insertions, 226 deletions

diff --git a/source/know/concept/amplitude-rate-equations/index.md b/source/know/concept/amplitude-rate-equations/index.md
new file mode 100644
index 0000000..0ca3248
--- /dev/null
+++ b/source/know/concept/amplitude-rate-equations/index.md
@@ -0,0 +1,94 @@
+---
+title: "Amplitude rate equations"
+sort_title: "Amplitude rate equations"
+date: 2023-01-03
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+In quantum mechanics, the **amplitude rate equations** give
+the evolution of a quantum state's superposition coefficients through time.
+They are known as the precursors for
+[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
+but by themselves they are exact and widely applicable.
+
+Let $$\hat{H}_0$$ be a "simple" time-independent part
+of the full Hamiltonian,
+and $$\hat{H}_1$$ a time-varying other part,
+whose contribution need not be small:
+
+$$\begin{aligned}
+    \hat{H}(t) = \hat{H}_0 + \hat{H}_1(t)
+\end{aligned}$$
+
+We assume that the time-independent problem
+$$\hat{H}_0 \Ket{n} = E_n \Ket{n}$$ has already been solved,
+such that its general solution is a superposition as follows:
+
+$$\begin{aligned}
+    \Ket{\Psi_0(t)} = \sum_{n} c_n \Ket{n} e^{- i E_n t / \hbar}
+\end{aligned}$$
+
+Since these $$\Ket{n}$$ form a complete basis,
+the full solution for $$\hat{H}_0 + \hat{H}_1$$ can be written in the same form,
+but now with time-dependent coefficients $$c_n(t)$$:
+
+$$\begin{aligned}
+    \Ket{\Psi(t)} = \sum_{n} c_n(t) \Ket{n} e^{- i E_n t / \hbar}
+\end{aligned}$$
+
+We put this ansatz into the full Schrödinger equation,
+and use the known solution for $$\hat{H}_0$$:
+
+$$\begin{aligned}
+    0
+    &= \hat{H}_0 \Ket{\Psi(t)} + \hat{H}_1 \Ket{\Psi(t)} - i \hbar \dv{}{t}\Ket{\Psi(t)}
+    \\
+    &= \sum_{n}
+    \Big( c_n \hat{H}_0 \Ket{n} + c_n \hat{H}_1 \Ket{n} - c_n E_n \Ket{n} - i \hbar \dv{c_n}{t} \Ket{n} \Big) e^{- i E_n t / \hbar}
+    \\
+    &= \sum_{n} \Big( c_n \hat{H}_1 \Ket{n} - i \hbar \dv{c_n}{t} \Ket{n} \Big) e^{- i E_n t / \hbar}
+\end{aligned}$$
+
+We then take the inner product with an arbitrary stationary basis state $$\Ket{m}$$:
+
+$$\begin{aligned}
+    0
+    &= \sum_{n} \Big( c_n \matrixel{m}{\hat{H}_1}{n} - i \hbar \dv{c_n}{t} \inprod{m}{n} \Big) e^{- i E_n t / \hbar}
+\end{aligned}$$
+
+Thanks to orthonormality, this moves the latter term outside the summation:
+
+$$\begin{aligned}
+    i \hbar \dv{c_m}{t} e^{- i E_m t / \hbar}
+    &= \sum_{n} c_n \matrixel{m}{\hat{H}_1}{n} e^{- i E_n t / \hbar}
+\end{aligned}$$
+
+We divide by the left-hand exponential and define
+$$\omega_{mn} \equiv (E_m - E_n) / \hbar$$ to arrive at
+the desired set of amplitude rate equations,
+one for each basis state $$\ket{m}$$:
+
+$$\begin{aligned}
+    \boxed{
+        i \hbar \dv{c_m}{t}
+        = \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1(t)}{n} e^{i \omega_{mn} t}
+    }
+\end{aligned}$$
+
+We have not made any approximations,
+so it is possible to exactly solve for $$c_n(t)$$ in some simple systems.
+This is worth pointing out, because these equations' most famous uses
+are for deriving time-dependent-perturbation theory
+(by making a truncated power series approximation)
+and [Rabi oscillation](/know/concept/rabi-oscillation/)
+(by making the [rotating wave approximation](/know/concept/rotating-wave-approximation/)).
+
+
+
+## References
+1.  D.J. Griffiths, D.F. Schroeter,
+    *Introduction to quantum mechanics*, 3rd edition,
+    Cambridge.
diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md
index f454264..fa289de 100644
--- a/source/know/concept/bell-state/index.md
+++ b/source/know/concept/bell-state/index.md
@@ -16,16 +16,16 @@ $$\begin{aligned}
     \boxed{
         \begin{aligned}
             \ket{\Phi^{\pm}}
-            &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{0}_B \pm \Ket{1}_A \Ket{1}_B \Big)
+            &= \frac{1}{\sqrt{2}} \Big( \ket{0}_A \ket{0}_B \pm \ket{1}_A \ket{1}_B \Big)
             \\
             \ket{\Psi^{\pm}}
-            &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{1}_B \pm \Ket{1}_A \Ket{0}_B \Big)
+            &= \frac{1}{\sqrt{2}} \Big( \ket{0}_A \ket{1}_B \pm \ket{1}_A \ket{0}_B \Big)
         \end{aligned}
     }
 \end{aligned}$$
 
-Where e.g. $$\Ket{0}_A \Ket{1}_B = \Ket{0}_A \otimes \Ket{1}_B$$
-is the tensor product of qubit $$A$$ in state $$\Ket{0}$$ and $$B$$ in $$\Ket{1}$$.
+Where e.g. $$\ket{0}_A \ket{1}_B = \ket{0}_A \otimes \ket{1}_B$$
+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/).
 
@@ -37,7 +37,7 @@ Consider the following pure [density operator](/know/concept/density-operator/):
 $$\begin{aligned}
     \hat{\rho}
     = \ket{\Phi^{+}} \bra{\Phi^{+}}
-    &= \frac{1}{2} \Big( \Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big) \Big( \Bra{0}_A \Bra{0}_B + \Bra{1}_A \Bra{1}_B \Big)
+    &= \frac{1}{2} \Big( \ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big) \Big( \bra{0}_A \bra{0}_B + \bra{1}_A \bra{1}_B \Big)
 \end{aligned}$$
 
 The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated as follows:
@@ -45,12 +45,12 @@ The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated
 $$\begin{aligned}
     \hat{\rho}_A
     &= \Tr_B(\hat{\rho})
-    = \sum_{b = 0, 1} \Bra{b}_B \Big( \ket{\Phi^{+}} \bra{\Phi^{+}} \Big) \Ket{b}_B
+    = \sum_{b = 0, 1} \bra{b}_B \Big( \ket{\Phi^{+}} \bra{\Phi^{+}} \Big) \ket{b}_B
     \\
-    &= \sum_{b = 0, 1} \Big( \Ket{0}_A \Inprod{b}{0}_B + \Ket{1}_A \Inprod{b}{1}_B \Big)
-    \Big( \Bra{0}_A \Inprod{0}{b}_B + \Bra{1}_A \Inprod{1}{b}_B \Big)
+    &= \sum_{b = 0, 1} \Big( \ket{0}_A \inprod{b}{0}_B + \ket{1}_A \inprod{b}{1}_B \Big)
+    \Big( \bra{0}_A \inprod{0}{b}_B + \bra{1}_A \inprod{1}{b}_B \Big)
     \\
-    &= \frac{1}{2} \Big( \Ket{0}_A \Bra{0}_A + \Ket{1}_A \Bra{1}_A \Big)
+    &= \frac{1}{2} \Big( \ket{0}_A \bra{0}_A + \ket{1}_A \bra{1}_A \Big)
     = \frac{1}{2} \hat{I}
 \end{aligned}$$
 
@@ -59,35 +59,36 @@ The same holds for the other three Bell states,
 and is equally true for qubit $$B$$.
 
 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 $$\ket{1}$$.
 However, due to the entanglement,
 measuring $$A$$ also has consequences for qubit $$B$$:
 
 $$\begin{aligned}
-    \big| \Bra{0}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
-    &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{0}{0}_B + \Inprod{0}{1}_A \Inprod{0}{1}_B \Big)^2
+    \big| \bra{0}_A \! \bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \inprod{0}{0}_A \inprod{0}{0}_B + \inprod{0}{1}_A \inprod{0}{1}_B \Big)^2
     = \frac{1}{2}
     \\
-    \big| \Bra{0}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
-    &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{1}{0}_B + \Inprod{0}{1}_A \Inprod{1}{1}_B \Big)^2
+    \big| \bra{0}_A \! \bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \inprod{0}{0}_A \inprod{1}{0}_B + \inprod{0}{1}_A \inprod{1}{1}_B \Big)^2
     = 0
     \\
-    \big| \Bra{1}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
-    &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{0}{0}_B + \Inprod{1}{1}_A \Inprod{0}{1}_B \Big)^2
+    \big| \bra{1}_A \! \bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \inprod{1}{0}_A \inprod{0}{0}_B + \inprod{1}{1}_A \inprod{0}{1}_B \Big)^2
     = 0
     \\
-    \big| \Bra{1}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
-    &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{1}{0}_B + \Inprod{1}{1}_A \Inprod{1}{1}_B \Big)^2
+    \big| \bra{1}_A \! \bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \inprod{1}{0}_A \inprod{1}{0}_B + \inprod{1}{1}_A \inprod{1}{1}_B \Big)^2
     = \frac{1}{2}
 \end{aligned}$$
 
-As an example, if $$A$$ collapses into $$\Ket{0}$$ due to a measurement,
-then $$B$$ instantly also collapses into $$\Ket{0}$$, never $$\Ket{1}$$,
+As an example, if $$A$$ collapses into $$\ket{0}$$ due to a measurement,
+then $$B$$ instantly also collapses into $$\ket{0}$$, never $$\ket{1}$$,
 even if it was not measured.
 This was a specific example for $$\ket{\Phi^{+}}$$,
 but analogous results can be found for the other Bell states.
 
 
+
 ## References
 1.  J.B. Brask,
     *Quantum information: lecture notes*,
diff --git a/source/know/concept/dielectric-function/index.md b/source/know/concept/dielectric-function/index.md
index 529ce2a..d55cc91 100644
--- a/source/know/concept/dielectric-function/index.md
+++ b/source/know/concept/dielectric-function/index.md
@@ -13,7 +13,8 @@ The **dielectric function** or **relative permittivity** $$\varepsilon_r$$
 is a measure of how strongly a given medium counteracts
 [electric fields](/know/concept/electric-field/) compared to a vacuum.
 Let $$\vb{D}$$ be the applied external field,
-and $$\vb{E}$$ the effective field inside the material:
+and $$\vb{E}$$ the effective field inside the material,
+then $$\varepsilon_r$$ is defined such that:
 
 $$\begin{aligned}
     \boxed{
@@ -23,7 +24,7 @@ $$\begin{aligned}
 
 If $$\varepsilon_r$$ is large, then $$\vb{D}$$ is strongly suppressed,
 because the material's electrons and nuclei move to create an opposing field.
-In order for $$\varepsilon_r$$ to be well defined, we only consider linear media,
+In order for $$\varepsilon_r$$ to be well-defined, we only consider *linear* media,
 where the induced polarization $$\vb{P}$$ is proportional to $$\vb{E}$$.
 
 We would like to find an alternative definition of $$\varepsilon_r$$.
@@ -54,13 +55,10 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-
-## From induced charge density
-
-A common way to calculate $$\varepsilon_r$$ is from
+In practice, a common way to calculate $$\varepsilon_r$$ is from
 the induced charge density $$\rho_\mathrm{ind}$$,
 i.e. the offset caused by the material's particles responding to the field.
-We start from [Gauss' law](/know/concept/maxwells-equations/) for $$\vb{P}$$:
+Starting from [Gauss' law](/know/concept/maxwells-equations/) for $$\vb{P}$$:
 
 $$\begin{aligned}
     \nabla \cdot \vb{P}
@@ -68,27 +66,27 @@ $$\begin{aligned}
     = - \rho_\mathrm{ind}(\vb{r})
 \end{aligned}$$
 
-This is Poisson's equation, which has the following well-known
-[Fourier transform](/know/concept/fourier-transform/):
+This is Poisson's equation, which has a well-known solution
+via [Fourier transformation](/know/concept/fourier-transform/):
 
 $$\begin{aligned}
     \Phi_\mathrm{ind}(\vb{q})
     = \frac{\rho_\mathrm{ind}(\vb{q})}{\varepsilon_0 |\vb{q}|^2}
-    = V(\vb{q}) \: \rho_\mathrm{ind}(\vb{q})
+    \equiv V(\vb{q}) \: \rho_\mathrm{ind}(\vb{q})
 \end{aligned}$$
 
 Where $$V(\vb{q})$$ represents Coulomb interactions,
-and $$V(0) = 0$$ to ensure overall neutrality:
+and $$V(0) \equiv 0$$ to ensure overall neutrality:
 
 $$\begin{aligned}
     V(\vb{q})
-    = \frac{1}{\varepsilon_0 |\vb{q}|^2}
+    \equiv \frac{1}{\varepsilon_0 |\vb{q}|^2}
     \qquad \implies \qquad
     V(\vb{r} - \vb{r}')
     = \frac{1}{4 \pi \varepsilon_0 |\vb{r} - \vb{r}'|}
 \end{aligned}$$
 
-The [convolution theorem](/know/concept/convolution-theorem/)
+Note that the [convolution theorem](/know/concept/convolution-theorem/)
 then gives us the solution $$\Phi_\mathrm{ind}$$ in the $$\vb{r}$$-domain:
 
 $$\begin{aligned}
@@ -97,37 +95,56 @@ $$\begin{aligned}
     = \int_{-\infty}^\infty V(\vb{r} - \vb{r}') \: \rho_\mathrm{ind}(\vb{r}') \dd{\vb{r}'}
 \end{aligned}$$
 
-To proceed, we need to find an expression for $$\rho_\mathrm{ind}$$
-that is proportional to $$\Phi_\mathrm{tot}$$ or $$\Phi_\mathrm{ext}$$,
+To proceed to calculate $$\varepsilon_r$$ from $$\rho_\mathrm{ind}$$,
+one needs an expression for $$\rho_\mathrm{ind}$$
+that is proportional to $$\Phi_\mathrm{tot}$$ or $$\Phi_\mathrm{ext}$$
 or some linear combination thereof.
-Such an expression must exist for a linear material.
+Such an expression must exist for a linear medium,
+but the details depend on the physics being considered
+and are thus beyond our current scope;
+we will just show the general form of $$\varepsilon_r$$
+once such an expression has been found.
 
-Suppose we can show that $$\rho_\mathrm{ind} = C_\mathrm{ext} \Phi_\mathrm{ext}$$,
-for some $$C_\mathrm{ext}$$, which may depend on $$\vb{q}$$. Then:
+Suppose we know that $$\rho_\mathrm{ind} = c_\mathrm{ext} \Phi_\mathrm{ext}$$
+for some factor $$c_\mathrm{ext}$$, which may depend on $$\vb{q}$$.
+Then, since $$\Phi_\mathrm{tot} = \Phi_\mathrm{ext} \!+\! \Phi_\mathrm{ind}$$,
+we find in the $$\vb{q}$$-domain:
 
 $$\begin{aligned}
     \Phi_\mathrm{tot}
-    = (1 + C_\mathrm{ext} V) \Phi_\mathrm{ext}
+    = (1 + c_\mathrm{ext} V) \Phi_\mathrm{ext}
     \quad \implies \quad
     \boxed{
         \varepsilon_r(\vb{q})
-        = \frac{1}{1 + C_\mathrm{ext}(\vb{q}) V(\vb{q})}
+        = \frac{1}{1 + c_\mathrm{ext}(\vb{q}) V(\vb{q})}
     }
 \end{aligned}$$
 
-Similarly, suppose we can show that $$\rho_\mathrm{ind} = C_\mathrm{tot} \Phi_\mathrm{tot}$$,
-for some quantity $$C_\mathrm{tot}$$, then:
+Likewise, suppose we can instead show that
+$$\rho_\mathrm{ind} = c_\mathrm{tot} \Phi_\mathrm{tot}$$
+for some quantity $$c_\mathrm{tot}$$, then:
 
 $$\begin{aligned}
     \Phi_\mathrm{ext}
-    = (1 - C_\mathrm{tot} V) \Phi_\mathrm{tot}
+    = (1 - c_\mathrm{tot} V) \Phi_\mathrm{tot}
     \quad \implies \quad
     \boxed{
         \varepsilon_r(\vb{q})
-        = 1 - C_\mathrm{tot}(\vb{q}) V(\vb{q})
+        = 1 - c_\mathrm{tot}(\vb{q}) V(\vb{q})
     }
 \end{aligned}$$
 
+And in the unlikely event that an expression of the form
+$$\rho_\mathrm{ind} = c_\mathrm{ext} \Phi_\mathrm{ext} \!+\! c_\mathrm{tot} \Phi_\mathrm{tot}$$ is found:
+
+$$\begin{aligned}
+    (1 - c_\mathrm{tot} V) \Phi_\mathrm{tot}
+    = (1 + c_\mathrm{ext} V) \Phi_\mathrm{ext}
+    \quad \implies \quad
+    \varepsilon_r(\vb{q})
+    = \frac{1 - c_\mathrm{tot}(\vb{q}) V(\vb{q})}{1 + c_\mathrm{ext}(\vb{q}) V(\vb{q})}
+\end{aligned}$$
+
 
 
 ## References
diff --git a/source/know/concept/einstein-coefficients/index.md b/source/know/concept/einstein-coefficients/index.md
index 179d866..ad75ce5 100644
--- a/source/know/concept/einstein-coefficients/index.md
+++ b/source/know/concept/einstein-coefficients/index.md
@@ -159,7 +159,8 @@ $$\begin{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:
+which tells us that the transition probability
+from $$\ket{a}$$ to $$\ket{b}$$ is (to first order):
 
 $$\begin{aligned}
     P_{ab}
@@ -167,7 +168,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 If the nucleus is at $$z = 0$$,
-then generally $$\Ket{1}$$ and $$\Ket{2}$$ will be even or odd functions of $$z$$,
+then generally $$\ket{1}$$ and $$\ket{2}$$ will be even or odd functions of $$z$$,
 meaning that $$\matrixel{1}{z}{1} = \matrixel{2}{z}{2} = 0$$
 (see also [Laporte's selection rule](/know/concept/selection-rules/)),
 leading to:
diff --git a/source/know/concept/fermis-golden-rule/index.md b/source/know/concept/fermis-golden-rule/index.md
index 021c8e4..ea58ee6 100644
--- a/source/know/concept/fermis-golden-rule/index.md
+++ b/source/know/concept/fermis-golden-rule/index.md
@@ -20,7 +20,7 @@ time.
 From [time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
 we know that the transition probability
 for a particle in state $$\Ket{a}$$ to go to $$\Ket{b}$$
-is as follows for a periodic perturbation at frequency $$\omega$$:
+is as follows for a sinusoidal perturbation at frequency $$\omega$$:
 
 $$\begin{aligned}
     P_{ab}
diff --git a/source/know/concept/ghz-paradox/index.md b/source/know/concept/ghz-paradox/index.md
index a59ccfe..9951883 100644
--- a/source/know/concept/ghz-paradox/index.md
+++ b/source/know/concept/ghz-paradox/index.md
@@ -12,43 +12,46 @@ layout: "concept"
 The **Greenberger-Horne-Zeilinger** or **GHZ paradox**
 is an alternative proof of [Bell's theorem](/know/concept/bells-theorem/)
 that does not use inequalities,
-but the three-particle entangled **GHZ state** $$\Ket{\mathrm{GHZ}}$$ instead,
+but the three-particle entangled **GHZ state** $$\ket{\mathrm{GHZ}}$$ instead,
 
 $$\begin{aligned}
     \boxed{
-        \Ket{\mathrm{GHZ}}
-        = \frac{1}{\sqrt{2}} \Big( \Ket{000} + \Ket{111} \Big)
+        \ket{\mathrm{GHZ}}
+        = \frac{1}{\sqrt{2}} \Big( \ket{000} + \ket{111} \Big)
     }
 \end{aligned}$$
 
-Where $$\Ket{0}$$ and $$\Ket{1}$$ are qubit states,
-for example, the eigenvalues of the Pauli matrix $$\hat{\sigma}_z$$.
+Where $$\ket{0}$$ and $$\ket{1}$$ are qubit states,
+specifically the eigenvalues of the Pauli matrix $$\hat{\sigma}_z$$.
 
 If we now apply certain products of the Pauli matrices $$\hat{\sigma}_x$$ and $$\hat{\sigma}_y$$
-to the three particles, we find:
+as [quantum gates](/know/concept/quantum-gate/)
+to this three-particle state, we find:
 
 
 $$\begin{aligned}
-    \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \Ket{\mathrm{GHZ}}
-    &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_x \Ket{0}
-    + \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_x \Ket{1} \Big)
+    \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \ket{\mathrm{GHZ}}
+    &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_x \ket{0}
+    + \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_x \ket{1} \Big)
     \\
-    &= \frac{1}{\sqrt{2}} \Big( \Ket{1} \otimes \Ket{1} \otimes \Ket{1} + \Ket{0} \otimes \Ket{0} \otimes \Ket{0} \Big)
-    = \Ket{\mathrm{GHZ}}
+    &= \frac{1}{\sqrt{2}} \Big( \ket{1} \otimes \ket{1} \otimes \ket{1} + \ket{0} \otimes \ket{0} \otimes \ket{0} \Big)
     \\
-    \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \Ket{\mathrm{GHZ}}
-    &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_y \Ket{0} \otimes \hat{\sigma}_y \Ket{0}
-    + \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_y \Ket{1} \otimes \hat{\sigma}_y \Ket{1} \Big)
+    &= \ket{\mathrm{GHZ}}
     \\
-    &= \frac{1}{\sqrt{2}} \Big( \Ket{1} \otimes i \Ket{1} \otimes i \Ket{1} + \Ket{0} \otimes i \Ket{0} \otimes i \Ket{0} \Big)
-    = - \Ket{\mathrm{GHZ}}
+    \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \ket{\mathrm{GHZ}}
+    &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_y \ket{0} \otimes \hat{\sigma}_y \ket{0}
+    + \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_y \ket{1} \otimes \hat{\sigma}_y \ket{1} \Big)
+    \\
+    &= \frac{1}{\sqrt{2}} \Big( \ket{1} \otimes i \ket{1} \otimes i \ket{1} + \ket{0} \otimes i \ket{0} \otimes i \ket{0} \Big)
+    \\
+    &= - \ket{\mathrm{GHZ}}
 \end{aligned}$$
 
 In other words, the GHZ state is a simultaneous eigenstate of these composite operators,
 with eigenvalues $$+1$$ and $$-1$$, respectively.
-Let us introduce two other product operators,
-such that we have a set of four observables,
-for which $$\Ket{\mathrm{GHZ}}$$ gives these eigenvalues:
+Let us introduce two more operators in the same way,
+so that we have a set of four observables,
+for which $$\ket{\mathrm{GHZ}}$$ gives these eigenvalues:
 
 $$\begin{aligned}
     \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x
diff --git a/source/know/concept/hellmann-feynman-theorem/index.md b/source/know/concept/hellmann-feynman-theorem/index.md
index e18acc2..c6bf720 100644
--- a/source/know/concept/hellmann-feynman-theorem/index.md
+++ b/source/know/concept/hellmann-feynman-theorem/index.md
@@ -9,20 +9,20 @@ layout: "concept"
 ---
 
 Consider the time-independent Schrödinger equation,
-where the Hamiltonian $$\hat{H}$$ depends on a general parameter $$\lambda$$,
-whose meaning or type we will not specify:
+where the Hamiltonian $$\hat{H}$$ depends on some parameter $$\lambda$$
+whose meaning we will not specify:
 
 $$\begin{aligned}
-    \hat{H}(\lambda) \Ket{\psi_n(\lambda)}
-    = E_n(\lambda) \Ket{\psi_n(\lambda)}
+    \hat{H}(\lambda) \ket{\psi_n(\lambda)}
+    = E_n(\lambda) \ket{\psi_n(\lambda)}
 \end{aligned}$$
 
-Assuming all eigenstates $$\Ket{\psi_n}$$ are normalized,
+Assuming all eigenstates $$\ket{\psi_n}$$ are normalized,
 this gives us the following basic relation:
 
 $$\begin{aligned}
     \matrixel{\psi_m}{\hat{H}}{\psi_n}
-    = E_n \Inprod{\psi_m}{\psi_n}
+    = E_n \inprod{\psi_m}{\psi_n}
     = \delta_{mn} E_n
 \end{aligned}$$
 
@@ -38,33 +38,32 @@ $$\begin{aligned}
     + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
     + \matrixel{\psi_m}{\hat{H}}{\nabla_\lambda \psi_n}
     \\
-    &= E_m \Inprod{\psi_m}{\nabla_\lambda \psi_n} + E_n \Inprod{\nabla_\lambda \psi_m}{\psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
+    &= E_m \inprod{\psi_m}{\nabla_\lambda \psi_n} + E_n \inprod{\nabla_\lambda \psi_m}{\psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
 \end{aligned}$$
 
 In order to simplify this,
 we differentiate the orthogonality relation
-$$\Inprod{\psi_m}{\psi_n} = \delta_{mn}$$,
-which ends up telling us that
-$$\Inprod{\nabla_\lambda \psi_m}{\psi_n} = - \Inprod{\psi_m}{\nabla_\lambda \psi_n}$$:
+$$\inprod{\psi_m}{\psi_n} = \delta_{mn}$$:
 
 $$\begin{aligned}
     0
     = \nabla_\lambda \delta_{mn}
-    = \nabla_\lambda \Inprod{\psi_m}{\psi_n}
-    = \Inprod{\nabla_\lambda \psi_m}{\psi_n} + \Inprod{\psi_m}{\nabla_\lambda \psi_n}
+    = \nabla_\lambda \inprod{\psi_m}{\psi_n}
+    = \inprod{\nabla_\lambda \psi_m}{\psi_n} + \inprod{\psi_m}{\nabla_\lambda \psi_n}
 \end{aligned}$$
 
-Using this result to replace $$\Inprod{\nabla_\lambda \psi_m}{\psi_n}$$
+Meaning that $$\inprod{\nabla_\lambda \psi_m}{\psi_n} = - \inprod{\psi_m}{\nabla_\lambda \psi_n}$$.
+Using this result to replace $$\inprod{\nabla_\lambda \psi_m}{\psi_n}$$
 in the previous equation leads to:
 
 $$\begin{aligned}
     \delta_{mn} \nabla_\lambda E_n
-    &= (E_m - E_n) \Inprod{\psi_m}{\nabla_\lambda \psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
+    &= (E_m - E_n) \inprod{\psi_m}{\nabla_\lambda \psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
 \end{aligned}$$
 
 For $$m = n$$, we therefore arrive at the **Hellmann-Feynman theorem**,
 which is useful when doing numerical calculations
-to minimize energies with respect to $$\lambda$$:
+that often involve minimizing energies with respect to $$\lambda$$:
 
 $$\begin{aligned}
     \boxed{
@@ -79,7 +78,7 @@ the [Berry phase](/know/concept/berry-phase/):
 
 $$\begin{aligned}
     \boxed{
-        (E_n - E_m) \Inprod{\psi_m}{\nabla_\lambda \psi_n}
+        (E_n - E_m) \inprod{\psi_m}{\nabla_\lambda \psi_n}
         = \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
     }
 \end{aligned}$$
diff --git a/source/know/concept/maxwell-bloch-equations/index.md b/source/know/concept/maxwell-bloch-equations/index.md
index ba8a677..0252b5c 100644
--- a/source/know/concept/maxwell-bloch-equations/index.md
+++ b/source/know/concept/maxwell-bloch-equations/index.md
@@ -12,13 +12,13 @@ layout: "concept"
 ---
 
 For an electron in a two-level system with time-independent states
-$$\Ket{g}$$ (ground) and $$\Ket{e}$$ (excited),
+$$\ket{g}$$ (ground) and $$\ket{e}$$ (excited),
 consider the following general solution
-to the full Schrödinger equation:
+to the time-dependent Schrödinger equation:
 
 $$\begin{aligned}
-    \Ket{\Psi}
-    &= c_g \: \Ket{g} \exp(-i E_g t / \hbar) + c_e \: \Ket{e} \exp(-i E_e t / \hbar)
+    \ket{\Psi}
+    &= c_g \ket{g} \exp(-i E_g t / \hbar) + c_e \ket{e} \exp(-i E_e t / \hbar)
 \end{aligned}$$
 
 Perturbing this system with
@@ -87,15 +87,16 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## Optical Bloch equations
 
-For $$\Ket{\Psi}$$ as defined above,
+For $$\ket{\Psi}$$ as defined above,
 the corresponding pure [density operator](/know/concept/density-operator/)
 $$\hat{\rho}$$ is as follows:
 
 $$\begin{aligned}
     \hat{\rho}
-    = \Ket{\Psi} \Bra{\Psi}
+    = \ket{\Psi} \bra{\Psi}
     =
     \begin{bmatrix}
         c_e c_e^* & c_e c_g^* \exp(-i \omega_0 t) \\
@@ -159,11 +160,10 @@ $$\begin{aligned}
 
 These equations are correct if nothing else is affecting $$\hat{\rho}$$.
 But in practice, these quantities decay due to various processes,
-e.g. spontaneous emission (see [Einstein coefficients](/know/concept/einstein-coefficients/)).
+e.g. [spontaneous emission](/know/concept/einstein-coefficients/).
 
-Let $$\rho_{ee}$$ decays with rate $$\gamma_e$$.
-Since the total probability $$\rho_{ee} + \rho_{gg} = 1$$,
-we thus have:
+Suppose $$\rho_{ee}$$ decays with rate $$\gamma_e$$.
+Because the total probability $$\rho_{ee} + \rho_{gg} = 1$$, we have:
 
 $$\begin{aligned}
     \Big( \dv{\rho_{ee}}{t} \Big)_{e}
@@ -220,10 +220,11 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Many authors simplify these equations a bit by choosing
+Some authors simplify these equations a bit by choosing
 $$\gamma_g = 0$$ and $$\gamma_\perp = \gamma_e / 2$$.
 
 
+
 ## Including Maxwell's equations
 
 This two-level system has a dipole moment $$\vb{p}$$ as follows,
@@ -286,7 +287,7 @@ $$\begin{aligned}
 We can rewrite the first two terms in the following intuitive form,
 which describes a decay with
 rate $$\gamma_\parallel \equiv \gamma_g + \gamma_e$$
-towards an equilbrium $$d_0$$:
+towards an equilibrium $$d_0$$:
 
 $$\begin{aligned}
     2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
diff --git a/source/know/concept/no-cloning-theorem/index.md b/source/know/concept/no-cloning-theorem/index.md
index a91ae6f..840a598 100644
--- a/source/know/concept/no-cloning-theorem/index.md
+++ b/source/know/concept/no-cloning-theorem/index.md
@@ -62,6 +62,7 @@ This is clearly not the same as before: we have a contradiction,
 which implies that such a general cloning machine cannot ever exist.
 
 
+
 ## References
 1.  N. Brunner,
     *Quantum information theory: lecture notes*,
diff --git a/source/know/concept/quantum-gate/index.md b/source/know/concept/quantum-gate/index.md
index 9704e53..dd198f2 100644
--- a/source/know/concept/quantum-gate/index.md
+++ b/source/know/concept/quantum-gate/index.md
@@ -17,15 +17,15 @@ so we only consider the most important examples here.
 
 ## One-qubit gates
 
-As an example, consider the following must general single-qubit state $$\Ket{\psi}$$:
+As an example, consider the following most general single-qubit state $$\ket{\psi}$$:
 
 $$\begin{aligned}
-    \Ket{\psi}
-    = \alpha \Ket{0} + \beta \Ket{1}
+    \ket{\psi}
+    = \alpha \ket{0} + \beta \ket{1}
     = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}
 \end{aligned}$$
 
-Arguably the most famous and/or most fundamental quantum gates are the **Pauli matrices**:
+Arguably the most famous and most fundamental quantum gates are the **Pauli matrices**:
 
 $$\begin{aligned}
     \boxed{
@@ -53,19 +53,19 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-They have the following effect on $$\Ket{\psi}$$.
+They have the following effect on $$\ket{\psi}$$.
 Note that $$X$$ is equivalent to the classical $$\mathrm{NOT}$$ gate
 (and is often given that name),
 and $$Z$$ is sometimes called the **phase-flip gate**:
 
 $$\begin{aligned}
-    X \Ket{\psi}
+    X \ket{\psi}
     = \begin{bmatrix} \beta \\ \alpha \end{bmatrix}
     \qquad
-    Y \Ket{\psi}
+    Y \ket{\psi}
     = \begin{bmatrix} -i \beta \\ i \alpha \end{bmatrix}
     \qquad
-    Z \Ket{\psi}
+    Z \ket{\psi}
     = \begin{bmatrix} \alpha \\ -\beta \end{bmatrix}
 \end{aligned}$$
 
@@ -87,7 +87,7 @@ For $$\phi = \pi$$, we recover the Pauli-$$Z$$ gate.
 In general, the action of $$R_\phi$$ is as follows:
 
 $$\begin{aligned}
-    R_\phi \Ket{\psi}
+    R_\phi \ket{\psi}
     = \begin{bmatrix} \alpha \\ e^{i \phi} \beta \end{bmatrix}
 \end{aligned}$$
 
@@ -128,10 +128,11 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Its action consists of rotating the qubit
-by $$\pi$$ around the axis $$(X + Z) / \sqrt{2}$$ of the Bloch sphere:
+by $$\pi$$ around the axis $$(X + Z) / \sqrt{2}$$ of
+the [Bloch sphere](/know/concept/bloch-sphere/):
 
 $$\begin{aligned}
-    H \Ket{\psi}
+    H \ket{\psi}
     = \frac{1}{\sqrt{2}} \begin{bmatrix} \alpha + \beta \\ \alpha - \beta \end{bmatrix}
 \end{aligned}$$
 
@@ -139,20 +140,20 @@ Notably, it maps the eigenstates of $$X$$ and $$Z$$ to each other,
 and is its own inverse (i.e. unitary):
 
 $$\begin{aligned}
-    H \Ket{0} = \Ket{+}
+    H \ket{0} = \ket{+}
     \qquad
-    H \Ket{1} = \Ket{-}
+    H \ket{1} = \ket{-}
     \qquad
-    H \Ket{+} = \Ket{0}
+    H \ket{+} = \ket{0}
     \qquad
-    H \Ket{-} = \Ket{1}
+    H \ket{-} = \ket{1}
 \end{aligned}$$
 
 The **Clifford gates** are a set including $$X$$, $$Y$$, $$Z$$, $$H$$ and $$S$$,
 or more generally any gates that rotate
 by multiples of $$\pi/2$$ around the Bloch sphere.
-This set is **not universal**, meaning that if we start from $$\Ket{0}$$,
-we can only reach $$\Ket{0}$$, $$\Ket{1}$$, $$\Ket{+}$$, $$\Ket{-}$$, $$\Ket{+i}$$ $$\Ket{-i}$$ using these gates.
+This set is **not universal**, meaning that if we start from $$\ket{0}$$,
+we can only reach $$\ket{0}$$, $$\ket{1}$$, $$\ket{+}$$, $$\ket{-}$$, $$\ket{+i}$$ $$\ket{-i}$$ using these gates.
 
 If we add *any* non-Clifford gate, for example $$T$$,
 then we can reach any point on the Bloch sphere,
@@ -170,15 +171,15 @@ any state can be approximated.
 ## Two-qubit gates
 
 As an example, let us consider
-the following two pure one-qubit states $$\Ket{\psi_1}$$ and $$\Ket{\psi_2}$$:
+the following two pure one-qubit states $$\ket{\psi_1}$$ and $$\ket{\psi_2}$$:
 
 $$\begin{aligned}
-    \Ket{\psi_1}
-    = \alpha_1 \Ket{0} + \beta_1 \Ket{1}
+    \ket{\psi_1}
+    = \alpha_1 \ket{0} + \beta_1 \ket{1}
     = \begin{bmatrix} \alpha_1 \\ \beta_1 \end{bmatrix}
     \qquad \quad
-    \Ket{\psi_2}
-    = \alpha_2 \Ket{0} + \beta_2 \Ket{1}
+    \ket{\psi_2}
+    = \alpha_2 \ket{0} + \beta_2 \ket{1}
     = \begin{bmatrix} \alpha_2 \\ \beta_2 \end{bmatrix}
 \end{aligned}$$
 
@@ -186,23 +187,22 @@ The composite state of both qubits, assuming they are pure,
 is then their tensor product $$\otimes$$:
 
 $$\begin{aligned}
-    \Ket{\psi_1 \psi_2}
-    = \Ket{\psi_1} \otimes \Ket{\psi_2}
-    &= \alpha_1 \alpha_2 \Ket{00} + \alpha_1 \beta_2 \Ket{01} + \beta_1 \alpha_2 \Ket{10} + \beta_1 \beta_2 \Ket{11}
+    \ket{\psi_1 \psi_2}
+    = \ket{\psi_1} \otimes \ket{\psi_2}
+    &= \alpha_1 \alpha_2 \ket{00} + \alpha_1 \beta_2 \ket{01} + \beta_1 \alpha_2 \ket{10} + \beta_1 \beta_2 \ket{11}
     \\
-    &= c_{00} \Ket{00} + c_{01} \Ket{01} + c_{10} \Ket{10} + c_{11} \Ket{11}
+    &= c_{00} \ket{00} + c_{01} \ket{01} + c_{10} \ket{10} + c_{11} \ket{11}
 \end{aligned}$$
 
 Note that a two-qubit system may be [entangled](/know/concept/quantum-entanglement/),
 in which case the coefficients $$c_{00}$$ etc. cannot be written as products,
-i.e. $$\Ket{\psi_2}$$ cannot be expressed separately from $$\Ket{\psi_1}$$, and vice versa.
+i.e. $$\ket{\psi_2}$$ cannot be expressed separately from $$\ket{\psi_1}$$, and vice versa.
+In other words, the action of a two-qubit gate
+can be expressed in the basis of $$\ket{00}$$, $$\ket{01}$$, $$\ket{10}$$ and $$\ket{11}$$,
+but not always in the basis of $$\ket{0}_1$$, $$\ket{1}_1$$, $$\ket{0}_2$$ and $$\ket{1}_2$$.
 
-In other words, the general action of a two-qubit quantum gate
-can be expressed in the basis of $$\Ket{00}$$, $$\Ket{01}$$, $$\Ket{10}$$ and $$\Ket{11}$$,
-but not always in the basis of $$\Ket{0}_1$$, $$\Ket{1}_1$$, $$\Ket{0}_2$$ and $$\Ket{1}_2$$.
-
-With that said, the first two-qubit gate is $$\mathrm{SWAP}$$,
-which simply swaps $$\Ket{\psi_1}$$ and $$\Ket{\psi_2}$$:
+With this noted, the first two-qubit gate is $$\mathrm{SWAP}$$,
+which simply swaps $$\ket{\psi_1}$$ and $$\ket{\psi_2}$$:
 
 {% include image.html file="swap.png" width="22%"
     alt="SWAP gate diagram" %}
@@ -219,18 +219,18 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-This matrix is given in the basis of $$\Ket{00}$$, $$\Ket{01}$$, $$\Ket{10}$$ and $$\Ket{11}$$.
+This matrix is given in the basis of $$\ket{00}$$, $$\ket{01}$$, $$\ket{10}$$ and $$\ket{11}$$.
 Note that $$\mathrm{SWAP}$$ cannot generate entanglement,
 so if its input is separable, its output is too.
 In any case, its effect is clear:
 
 $$\begin{aligned}
-    \mathrm{SWAP} \Ket{\psi_1 \psi_2}
-    &= c_{00} \Ket{00} + c_{10} \Ket{01} + c_{01} \Ket{10} + c_{11} \Ket{11}
+    \mathrm{SWAP} \ket{\psi_1 \psi_2}
+    &= c_{00} \ket{00} + c_{10} \ket{01} + c_{01} \ket{10} + c_{11} \ket{11}
 \end{aligned}$$
 
 Next, there is the **controlled NOT gate** $$\mathrm{CNOT}$$,
-which "flips" (applies $$X$$ to) $$\Ket{\psi_2}$$