Expand knowledge base

6 files changed, 527 insertions, 16 deletions

diff --git a/content/know/category/thermodynamic-ensembles.md b/content/know/category/thermodynamic-ensembles.md
new file mode 100644
index 0000000..4477f23
--- /dev/null
+++ b/content/know/category/thermodynamic-ensembles.md
@@ -0,0 +1,9 @@
+---
+title: "Thermodynamic ensembles"
+firstLetter: "T"
+date: 2021-07-09T16:54:35+02:00
+draft: false
+layout: "category"
+---
+
+This page will fill itself.
diff --git a/content/know/concept/maxwell-relations/index.pdc b/content/know/concept/maxwell-relations/index.pdc
new file mode 100644
index 0000000..7e17a66
--- /dev/null
+++ b/content/know/concept/maxwell-relations/index.pdc
@@ -0,0 +1,296 @@
+---
+title: "Maxwell relations"
+firstLetter: "M"
+publishDate: 2021-07-08
+categories:
+- Physics
+- Thermodynamics
+
+date: 2021-07-08T10:58:37+02:00
+draft: false
+markup: pandoc
+---
+
+# Maxwell relations
+
+The **Maxwell relations** are a useful set of relations in thermodynamics.
+They arise from the fact that the order of differentiation is irrelevant
+for well-behaved functions (sometimes known as the *Schwarz theorem*),
+applied to the [thermodynamic potentials](/know/concept/thermodynamic-potential/).
+
+We start by proving the general "recipe".
+Given that the differential element of some $z$ is defined in terms of
+two constant quantities $A$ and $B$ and two independent variables $x$ and $y$:
+
+$$\begin{aligned}
+    \dd{z} \equiv A \dd{x} + B \dd{y}
+\end{aligned}$$
+
+Then the quantities $A$ and $B$ can be extracted
+by dividing by $\dd{x}$ and $\dd{y}$ respectively:
+
+$$\begin{aligned}
+    A = \Big( \pdv{z}{x} \Big)_y
+    \qquad
+    B = \Big( \pdv{z}{y} \Big)_x
+\end{aligned}$$
+
+By differentiating $A$ and $B$,
+and using that the order of differentiation is irrelevant, we find:
+
+$$\begin{aligned}
+    \pdv{z}{y}{x} =
+    \boxed{
+        \Big( \pdv{A}{y} \Big)_x
+        = \Big( \pdv{B}{x} \Big)_y
+    }
+    = \pdv{z}{x}{y}
+\end{aligned}$$
+
+Using this, all Maxwell relations are derived.
+Each relation also has a complement:
+
+$$\begin{aligned}
+    \Big( \pdv{A}{y} \Big)_x^{-1} =
+    \boxed{
+        \Big( \pdv{y}{A} \Big)_x
+        = \Big( \pdv{x}{B} \Big)_y
+    }
+    = \Big( \pdv{B}{x} \Big)_y^{-1}
+\end{aligned}$$
+
+The following quantities are useful to rewrite some of the Maxwell relations:
+the iso-$P$ thermal expansion coefficient $\alpha$,
+the iso-$T$ combressibility $\kappa_T$,
+the iso-$S$ combressibility $\kappa_S$,
+the iso-$V$ heat capacity $C_V$,
+and the iso-$P$ heat capacity $C_P$:
+
+$$\begin{gathered}
+    \alpha \equiv \frac{1}{V} \Big( \pdv{V}{T} \Big)_{P,N}
+    \\
+    \kappa_T \equiv - \frac{1}{V} \Big( \pdv{V}{P} \Big)_{T,N}
+    \qquad \quad
+    \kappa_S \equiv - \frac{1}{V} \Big( \pdv{V}{P} \Big)_{S,N}
+    \\
+    C_V \equiv T \Big( \pdv{S}{T} \Big)_{V,N}
+    \qquad \qquad
+    C_P \equiv T \Big( \pdv{S}{T} \Big)_{P,N}
+\end{gathered}$$
+
+
+## Internal energy
+
+The following Maxwell relations can be derived
+from the internal energy $U(S, V, N)$:
+
+$$\begin{gathered}
+    \pdv{U}{V}{S} =
+    \boxed{
+        \Big( \pdv{T}{V} \Big)_S = - \Big( \pdv{P}{S} \Big)_V
+    }
+    = \pdv{U}{S}{V}
+    \\
+    \pdv{U}{V}{N} =
+    \boxed{
+        \Big( \pdv{\mu}{V} \Big)_N = - \Big( \pdv{P}{N} \Big)_V
+    }
+    = \pdv{U}{N}{V}
+    \\
+    \pdv{U}{S}{N} =
+    \boxed{
+        \Big( \pdv{\mu}{S} \Big)_N = \Big( \pdv{T}{N} \Big)_S
+    }
+    = \pdv{U}{N}{S}
+\end{gathered}$$
+
+And the corresponding reciprocal relations are then given by:
+
+$$\begin{gathered}
+    \boxed{
+        \Big( \pdv{V}{T} \Big)_S = - \Big( \pdv{S}{P} \Big)_V
+    }
+    \\
+    \boxed{
+        \Big( \pdv{V}{\mu} \Big)_N = - \Big( \pdv{N}{P} \Big)_V
+    }
+    \\
+    \boxed{
+        \Big( \pdv{S}{\mu} \Big)_N = \Big( \pdv{N}{T} \Big)_S
+    }
+\end{gathered}$$
+
+
+## Enthalpy
+
+The following Maxwell relations can be derived
+from the enthalpy $H(S, P, N)$:
+
+$$\begin{gathered}
+    \pdv{H}{P}{S} =
+    \boxed{
+        \Big( \pdv{T}{P} \Big)_S = \Big( \pdv{V}{S} \Big)_P
+    }
+    = \pdv{H}{S}{P}
+    \\
+    \pdv{H}{P}{N} =
+    \boxed{
+        \Big( \pdv{\mu}{P} \Big)_N = \Big( \pdv{V}{N} \Big)_P
+    }
+    = \pdv{H}{N}{P}
+    \\
+    \pdv{H}{N}{S} =
+    \boxed{
+        \Big( \pdv{T}{N} \Big)_S = \Big( \pdv{\mu}{S} \Big)_N
+    }
+    = \pdv{H}{S}{N}
+\end{gathered}$$
+
+And the corresponding reciprocal relations are then given by:
+
+$$\begin{gathered}
+    \boxed{
+        \Big( \pdv{P}{T} \Big)_S = \Big( \pdv{S}{V} \Big)_P
+    }
+    \\
+    \boxed{
+        \Big( \pdv{P}{\mu} \Big)_N = \Big( \pdv{N}{V} \Big)_P
+    }
+    \\
+    \boxed{
+        \Big( \pdv{N}{T} \Big)_S = \Big( \pdv{S}{\mu} \Big)_N
+    }
+\end{gathered}$$
+
+
+## Helmholtz free energy
+
+The following Maxwell relations can be derived
+from the Helmholtz free energy $F(T, V, N)$:
+
+$$\begin{gathered}
+    - \pdv{F}{V}{T} =
+    \boxed{
+        \Big( \pdv{S}{V} \Big)_T = \Big( \pdv{P}{T} \Big)_V
+    }
+    = - \pdv{F}{T}{V}
+    \\
+    \pdv{F}{V}{N} =
+    \boxed{
+        \Big( \pdv{\mu}{V} \Big)_N = - \Big( \pdv{P}{N} \Big)_V
+    }
+    = \pdv{F}{N}{V}
+    \\
+    \pdv{F}{T}{N} =
+    \boxed{
+        \Big( \pdv{\mu}{T} \Big)_N = - \Big( \pdv{S}{N} \Big)_T
+    }
+    = \pdv{F}{N}{T}
+\end{gathered}$$
+
+And the corresponding reciprocal relations are then given by:
+
+$$\begin{gathered}
+    \boxed{
+        \Big( \pdv{V}{S} \Big)_T = \Big( \pdv{T}{P} \Big)_V
+    }
+    \\
+    \boxed{
+        \Big( \pdv{V}{\mu} \Big)_N = - \Big( \pdv{N}{P} \Big)_V
+    }
+    \\
+    \boxed{
+        \Big( \pdv{T}{\mu} \Big)_N = - \Big( \pdv{N}{S} \Big)_T
+    }
+\end{gathered}$$
+
+
+## Gibbs free energy
+
+The following Maxwell relations can be derived
+from the Gibbs free energy $G(T, P, N)$:
+
+$$\begin{gathered}
+    \pdv{G}{T}{P} =
+    \boxed{
+        \Big( \pdv{V}{T} \Big)_P = - \Big( \pdv{S}{P} \Big)_T
+    }
+    = \pdv{G}{P}{T}
+    \\
+    \pdv{G}{N}{P} =
+    \boxed{
+        \Big( \pdv{V}{N} \Big)_P = \Big( \pdv{\mu}{P} \Big)_N
+    }
+    = \pdv{G}{P}{N}
+    \\
+    \pdv{G}{T}{N} =
+    \boxed{
+        \Big( \pdv{\mu}{T} \Big)_N = - \Big( \pdv{S}{N} \Big)_T
+    }
+    = \pdv{G}{N}{T}
+\end{gathered}$$
+
+And the corresponding reciprocal relations are then given by:
+
+$$\begin{gathered}
+    \boxed{
+        \Big( \pdv{T}{V} \Big)_P = - \Big( \pdv{P}{S} \Big)_T
+    }
+    \\
+    \boxed{
+        \Big( \pdv{N}{V} \Big)_P = \Big( \pdv{P}{\mu} \Big)_N
+    }
+    \\
+    \boxed{
+        \Big( \pdv{T}{\mu} \Big)_N = - \Big( \pdv{N}{S} \Big)_T
+    }
+\end{gathered}$$
+
+
+## Landau potential
+
+The following Maxwell relations can be derived
+from the Gibbs free energy $\Omega(T, V, \mu)$:
+
+$$\begin{gathered}
+    - \pdv{\Omega}{V}{T} =
+    \boxed{
+        \Big( \pdv{S}{V} \Big)_T = \Big( \pdv{P}{T} \Big)_V
+    }
+    = - \pdv{\Omega}{T}{V}
+    \\
+    - \pdv{\Omega}{\mu}{V} =
+    \boxed{
+        \Big( \pdv{P}{\mu} \Big)_V = \Big( \pdv{N}{V} \Big)_\mu
+    }
+    = - \pdv{\Omega}{V}{\mu}
+    \\
+    - \pdv{\Omega}{T}{\mu} =
+    \boxed{
+        \Big( \pdv{N}{T} \Big)_\mu = \Big( \pdv{S}{\mu} \Big)_T
+    }
+    = - \pdv{\Omega}{\mu}{T}
+\end{gathered}$$
+
+And the corresponding reciprocal relations are then given by:
+
+$$\begin{gathered}
+    \boxed{
+        \Big( \pdv{V}{S} \Big)_T = \Big( \pdv{T}{P} \Big)_V
+    }
+    \\
+    \boxed{
+        \Big( \pdv{\mu}{P} \Big)_V = \Big( \pdv{V}{N} \Big)_\mu
+    }
+    \\
+    \boxed{
+        \Big( \pdv{T}{N} \Big)_\mu = \Big( \pdv{\mu}{S} \Big)_T
+    }
+\end{gathered}$$
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/content/know/concept/microcanonical-ensemble/index.pdc b/content/know/concept/microcanonical-ensemble/index.pdc
new file mode 100644
index 0000000..89d114b
--- /dev/null
+++ b/content/know/concept/microcanonical-ensemble/index.pdc
@@ -0,0 +1,123 @@
+---
+title: "Microcanonical ensemble"
+firstLetter: "M"
+publishDate: 2021-07-09
+categories:
+- Physics
+- Thermodynamics
+- Thermodynamic ensembles
+
+date: 2021-07-08T11:00:59+02:00
+draft: false
+markup: pandoc
+---
+
+# Microcanonical ensemble
+
+The **microcanonical** or **NVE ensemble** is a statistical model
+of a theoretical system with constant internal energy $U$,
+volume $V$, and particle count $N$.
+
+Consider a box with those properties.
+We now put an imaginary rigid wall inside the box,
+thus dividing it into two subsystems $A$ and $B$,
+which can exchange energy (i.e. heat), but no particles.
+At any time, $A$ has energy $U_A$, and $B$ has $U_B$,
+so that in total $U = U_A \!+\! U_B$.
+
+The particles in each subsystem are in a certain **microstate** (configuration).
+For a given $U$, there is a certain number $c$
+of possible whole-box microstates with that energy, given by:
+
+$$\begin{aligned}
+    c(U)
+    = \sum_{U_A \le U} c_A(U_A) \: c_B(U - U_A)
+\end{aligned}$$
+
+Where $c_A$ and $c_B$ are the number of microstates of
+the subsystems at the given energy levels.
+
+The core assumption of the microcanonical ensemble
+is that each of these microstates has the same probability $1 / c$.
+Consequently, the probability of finding an energy $U_A$ in $A$ is:
+
+$$\begin{aligned}
+    p_A(U_A)
+    = \frac{c_A(U_A) \:c_B(U - U_A)}{c(U)}
+\end{aligned}$$
+
+If a certain $U_A$ has a higher probability,
+then there are more $A$-microstates with that energy,
+meaning that $U_A$ is "easier to reach" or "more comfortable" for the system.
+Note that $c(U)$ is a constant, because $U$ is given.
+
+After some time, the system will reach equilibrium,
+where both $A$ and $B$ have settled into a "comfortable" position.
+In other words, the subsystem microstates at equilibrium
+must be maxima of their probability distributions $p_A$ and $p_B$.
+
+We only need to look at $p_A$.
+Clearly, a maximum of $p_A$ is also a maximum of $\ln p_A$:
+
+$$\begin{aligned}
+    \ln p_A(U_A)
+    = \ln{c_A(U_A)} + \ln{c_B(U - U_A)} - \ln{c(U)}
+\end{aligned}$$
+
+Here, in the quantity $\ln{c_A}$,
+we recognize the definition of
+the entropy $S_A \equiv k_B \ln{c_A}$,
+where $k_B$ is Boltzmann's constant.
+We thus multiply by $k_B$:
+
+$$\begin{aligned}
+    k_B \ln p_A(U_A)
+    = S_A(U_A) + S_B(U - U_A) - k_B \ln{c(U)}
+\end{aligned}$$
+
+Since entropy is additive over subsystems,
+the total is $S = S_A + S_B$.
+To reach equilibrium, we are thus
+**maximizing the total entropy**,
+meaning that $S$ is the [thermodynamic potential](/know/concept/thermodynamic-potential/)
+that corresponds to the microcanonical ensemble.
+
+For our example, maximizing gives the following,
+more concrete, equilibrium condition:
+
+$$\begin{aligned}
+    0
+    = k_B \dv{(\ln{p_A})}{U_A}
+    = \pdv{S_A}{U_A} + \pdv{S_B}{U_A}
+    = \pdv{S_A}{U_A} - \pdv{S_B}{U_B}
+\end{aligned}$$
+
+By definition, the energy-derivative of the entropy
+is the reciprocal temperature $1 / T$.
+In other words,
+equilibrium is reached when both subsystems
+are at the same temperature:
+
+$$\begin{aligned}
+    \frac{1}{T_A}
+    = \pdv{S_A}{U_A}
+    = \pdv{S_B}{U_B}
+    = \frac{1}{T_B}
+\end{aligned}$$
+
+Recall that our partitioning into $A$ and $B$ was arbitrary,
+meaning that, in fact, the temperature $T$ must be uniform in the whole box.
+
+We get this specific result because
+heat was the only thing that $A$ and $B$ could exchange.
+The key point, however,
+is that the total entropy $S$ must be maximized.
+We also would have reached that conclusion if our imaginary wall
+allowed changes in volume $V_A$ and particle count $N_A$.
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/content/know/concept/thermodynamic-potential/index.pdc b/content/know/concept/thermodynamic-potential/index.pdc
index 5d154d5..b65891f 100644
--- a/content/know/concept/thermodynamic-potential/index.pdc
+++ b/content/know/concept/thermodynamic-potential/index.pdc
@@ -226,7 +226,7 @@ with respect to the natural variables $T$, $V$, and $\mu$:
 
 $$\begin{aligned}
     \boxed{
-        S = \Big( \pdv{\Omega}{T} \Big)_{V,\mu}
+        S = - \Big( \pdv{\Omega}{T} \Big)_{V,\mu}
     \qquad
         P = - \Big( \pdv{\Omega}{V} \Big)_{T,\mu}
     \qquad
diff --git a/content/know/concept/time-dependent-perturbation-theory/index.pdc b/content/know/concept/time-dependent-perturbation-theory/index.pdc
index fbb71b2..b16e3ee 100644
--- a/content/know/concept/time-dependent-perturbation-theory/index.pdc
+++ b/content/know/concept/time-dependent-perturbation-theory/index.pdc
@@ -17,7 +17,7 @@ markup: pandoc
 In quantum mechanics, **time-dependent perturbation theory** exists to deal
 with time-varying perturbations to the Schrödinger equation.
 This is in contrast to [time-independent perturbation theory](/know/concept/time-independent-perturbation-theory/),
-where the perturbation is is stationary.
+where the perturbation is stationary.
 
 Let $\hat{H}_0$ be the base time-independent
 Hamiltonian, and $\hat{H}_1$ be a time-varying perturbation, with
@@ -32,14 +32,14 @@ $\hat{H}_0 \ket{n} = E_n \ket{n}$ has already been solved, such that the
 full solution is:
 
 $$\begin{aligned}
-    \ket{\Psi_0(t)} = \sum_{n} c_n \ket{n} \exp(- i E_n t / \hbar)
+    \ket{\Psi_0(t)} = \sum_{n} c_n \ket{n} \exp\!(- i E_n t / \hbar)
 \end{aligned}$$
 
 Since these $\ket{n}$ form a complete basis, the perturbed wave function
 can be written in the same form, but with time-dependent coefficients $c_n(t)$:
 
 $$\begin{aligned}
-    \ket{\Psi(t)} = \sum_{n} c_n(t) \ket{n} \exp(- i E_n t / \hbar)
+    \ket{\Psi(t)} = \sum_{n} c_n(t) \ket{n} \exp\!(- i E_n t / \hbar)
 \end{aligned}$$
 
 We insert this ansatz in the time-dependent Schrödinger equation, and
@@ -50,32 +50,32 @@ $$\begin{aligned}
     &= \hat{H}_0 \ket{\Psi(t)} + \lambda \hat{H}_1 \ket{\Psi(t)} - i \hbar \dv{t} \ket{\Psi(t)}
     \\
     &= \sum_{n}
-    \Big( c_n \hat{H}_0 \ket{n} + \lambda c_n \hat{H}_1 \ket{n} - c_n E_n \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp(- i E_n t / \hbar)
+    \Big( c_n \hat{H}_0 \ket{n} + \lambda c_n \hat{H}_1 \ket{n} - c_n E_n \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp\!(- i E_n t / \hbar)
     \\
-    &= \sum_{n} \Big( \lambda c_n \hat{H}_1 \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp(- i E_n t / \hbar)
+    &= \sum_{n} \Big( \lambda c_n \hat{H}_1 \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp\!(- i E_n t / \hbar)
 \end{aligned}$$
 
 We then take the inner product with an arbitrary stationary basis state $\ket{m}$:
 
 $$\begin{aligned}
     0
-    &= \sum_{n} \Big( \lambda c_n \matrixel{m}{\hat{H}_1}{n} - i \hbar \frac{d c_n}{dt} \braket{m}{n} \Big) \exp(- i E_n t / \hbar)
+    &= \sum_{n} \Big( \lambda c_n \matrixel{m}{\hat{H}_1}{n} - i \hbar \frac{d c_n}{dt} \braket{m}{n} \Big) \exp\!(- i E_n t / \hbar)
 \end{aligned}$$
 
 Thanks to orthonormality, this removes the latter term from the summation:
 
 $$\begin{aligned}
-    i \hbar \frac{d c_m}{dt} \exp(- i E_m t / \hbar)
-    &= \lambda \sum_{n} c_n \matrixel{m}{\hat{H}_1}{n} \exp(- i E_n t / \hbar)
+    i \hbar \frac{d c_m}{dt} \exp\!(- i E_m t / \hbar)
+    &= \lambda \sum_{n} c_n \matrixel{m}{\hat{H}_1}{n} \exp\!(- i E_n t / \hbar)
 \end{aligned}$$
 
 We divide by the left-hand exponential and define
-$\omega_{mn} = (E_m - E_n) / \hbar$ to get:
+$\omega_{mn} \equiv (E_m - E_n) / \hbar$ to get:
 
 $$\begin{aligned}
     \boxed{
         i \hbar \frac{d c_m}{dt}
-        = \lambda \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1(t)}{n} \exp(i \omega_{mn} t)
+        = \lambda \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1(t)}{n} \exp\!(i \omega_{mn} t)
     }
 \end{aligned}$$
 
@@ -85,7 +85,7 @@ Furthermore, it is useful to write this equation in integral form instead:
 
 $$\begin{aligned}
     c_m(t)
-    = c_m(0) - \lambda \frac{i}{\hbar} \sum_{n} \int_0^t c_n(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+    = c_m(0) - \lambda \frac{i}{\hbar} \sum_{n} \int_0^t c_n(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp\!(i \omega_{mn} \tau) \dd{\tau}
 \end{aligned}$$
 
 If this cannot be solved exactly, we must approximate it. We expand
@@ -100,15 +100,15 @@ We then insert this into the integral and collect the non-zero orders of $\lambd
 
 $$\begin{aligned}
     c_m^{(1)}(t)
-    &= - \frac{i}{\hbar} \sum_{n} \int_0^t c_n^{(0)} \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+    &= - \frac{i}{\hbar} \sum_{n} \int_0^t c_n^{(0)} \matrixel{m}{\hat{H}_1(\tau)}{n} \exp\!(i \omega_{mn} \tau) \dd{\tau}
     \\
     c_m^{(2)}(t)
     &= - \frac{i}{\hbar} \sum_{n}
-    \int_0^t c_n^{(1)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+    \int_0^t c_n^{(1)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp\!(i \omega_{mn} \tau) \dd{\tau}
     \\
     c_m^{(3)}(t)
     &= - \frac{i}{\hbar} \sum_{n}
-    \int_0^t c_n^{(2)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+    \int_0^t c_n^{(2)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp\!(i \omega_{mn} \tau) \dd{\tau}
 \end{aligned}$$
 
 And so forth. The pattern here is clear: we can calculate the $(j\!+\!1)$th
@@ -116,7 +116,90 @@ correction using only our previous result for the $j$th correction.
 We cannot go any further than this without considering a specific perturbation $\hat{H}_1(t)$.
 
 
+## Sinusoidal perturbation
+
+Arguably the most important perturbation
+is a sinusoidally-varying potential, which represents
+e.g. incoming electromagnetic waves,
+or an AC voltage being applied to the system.
+In this case, $\hat{H}_1$ has the following form:
+
+$$\begin{aligned}
+    \hat{H}_1(\vec{r}, t)
+    \equiv V(\vec{r}) \sin\!(\omega t)
+    = \frac{1}{2 i} V(\vec{r}) \: \big( \exp\!(i \omega t) - \exp\!(-i \omega t) \big)
+\end{aligned}$$
+
+We abbreviate $V_{mn} = \matrixel{m}{V}{n}$,
+and take the first-order correction formula:
+
+$$\begin{aligned}
+    c_m^{(1)}(t)
+    &= - \frac{1}{2 \hbar} \sum_{n} V_{mn} c_n^{(0)}
+    \int_0^t \exp\!\big(i \tau (\omega_{mn} \!+\! \omega)\big) - \exp\big(i \tau (\omega_{mn} \!-\! \omega)\big) \dd{\tau}
+    \\
+    &= \frac{i}{2 \hbar} \sum_{n} V_{mn} c_n^{(0)}
+    \bigg( \frac{\exp\!\big(i t (\omega_{mn} \!+\! \omega) \big) - 1}{\omega_{mn} + \omega}
+    + \frac{\exp\!\big(i t (\omega_{mn} \!-\! \omega) \big) - 1}{\omega_{mn} - \omega} \bigg)
+\end{aligned}$$
+
+For simplicity, we let the system start in a known state $\ket{a}$,
+such that $c_n^{(0)} = \delta_{na}$,
+and we assume that the driving frequency is close to resonance $\omega \approx \omega_{ma}$,
+such that the second term dominates the first, which can then be neglected.
+We thus get:
+
+$$\begin{aligned}
+    c_m^{(1)}(t)
+    &= i \frac{V_{ma}}{2 \hbar} \frac{\exp\!\big(i t (\omega_{ma} \!-\! \omega) \big) - 1}{\omega_{ma} - \omega}
+    \\
+    &= i \frac{V_{ma}}{2 \hbar}
+    \frac{\exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big) - \exp\!\big(\!-\! i t (\omega_{ma} \!-\! \omega) / 2 \big)}{\omega_{ma} - \omega}
+    \: \exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big)
+    \\
+    &= - \frac{V_{ma}}{\hbar}
+    \frac{\sin\!\big( t (\omega_{ma} \!-\! \omega) / 2 \big)}{\omega_{ma} - \omega}
+    \: \exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big)
+\end{aligned}$$
+
+Taking the norm squared yields the **transition probability**:
+the probability that a particle that started in state $\ket{a}$
+will be found in $\ket{m}$ at time $t$:
+
+$$\begin{aligned}
+    \boxed{
+        P_{a \to m}
+        = |c_m^{(1)}(t)|^2
+        = \frac{|V_{ma}|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_{ma} - \omega) t / 2 \big)}{(\omega_{ma} - \omega)^2}
+    }
+\end{aligned}$$
+
+The result would be the same if $\hat{H}_1 \equiv V \cos\!(\omega t)$.
+However, if instead $\hat{H}_1 \equiv V \exp\!(- i \omega t)$,
+the result is larger by a factor of $4$,
+which can cause confusion when comparing literature.
+
+In any case, the probability oscillates as a function of $t$
+with period $T = 2 \pi / (\omega_{ma} \!-\! \omega)$,
+so after one period the particle is certain to be back in $\ket{a}$.
+
+However, when regarded as a function of $\omega$,
+the probability takes the form of
+a sinc-function centred around $(\omega_{ma} \!-\! \omega)$,
+so it is highest for transitions with energy $\hbar \omega = E_m \!-\! E_a$.
+
+Also note that the sinc-distribution becomes narrower over time,
+which roughly means that it takes some time
+for the system to "notice" that
+it is being driven periodically.
+In other words, there is some "inertia" to it.
+
+
+
 ## References
 1.  D.J. Griffiths, D.F. Schroeter,
     *Introduction to quantum mechanics*, 3rd edition,
     Cambridge.
+2.  R. Shankar,
+    *Principles of quantum mechanics*, 2nd edition,
+    Springer.
diff --git a/content/know/concept/time-independent-perturbation-theory/index.pdc b/content/know/concept/time-independent-perturbation-theory/index.pdc
index 3be3cd5..d2a879f 100644
--- a/content/know/concept/time-independent-perturbation-theory/index.pdc
+++ b/content/know/concept/time-independent-perturbation-theory/index.pdc
@@ -14,7 +14,7 @@ markup: pandoc
 
 # Time-independent perturbation theory
 
-**Time-independent perturbation theory**, sometimes also called
+**Time-independent perturbation theory**, also known as
 **stationary state perturbation theory**, is a specific application of
 perturbation theory to the time-independent Schrödinger
 equation in quantum physics, for