Improve knowledge base

4 files changed, 339 insertions, 143 deletions

diff --git a/source/know/concept/heisenberg-picture/index.md b/source/know/concept/heisenberg-picture/index.md
index 359ecfe..54bf397 100644
--- a/source/know/concept/heisenberg-picture/index.md
+++ b/source/know/concept/heisenberg-picture/index.md
@@ -8,99 +8,117 @@ categories:
 layout: "concept"
 ---
 
-The **Heisenberg picture** is an alternative formulation of quantum
-mechanics, and is equivalent to the traditionally-taught Schrödinger equation.
+The **Heisenberg picture** is an alternative formulation of quantum mechanics,
+and is equivalent to the traditional Schrödinger equation.
 
-In the Schrödinger picture, the operators (observables) are fixed
-(as long as they do not depend on time), while the state
-$$\Ket{\psi_S(t)}$$ changes according to the Schrödinger equation,
-which can be written using the generator of translations $$\hat{U}(t)$$ like so,
-for a time-independent $$\hat{H}_S$$:
+In the Schrödinger picture,
+time-independent operators are constant by definition,
+and the state $$\Ket{\psi_S(t)}$$ varies as follows, where $$\hat{U}(t)$$
+is the [time evolution operator](/know/concept/time-evolution-operator/):
 
 $$\begin{aligned}
-    \Ket{\psi_S(t)} = \hat{U}(t) \Ket{\psi_S(0)}
-    \qquad \quad
-    \boxed{
-        \hat{U}(t) \equiv \exp\!\bigg(\!-\! i \frac{\hat{H}_S t}{\hbar} \bigg)
-    }
+    \Ket{\psi_S(t)}
+    = \hat{U}(t) \Ket{\psi_S(0)}
 \end{aligned}$$
 
-In contrast, the Heisenberg picture reverses the roles:
-the states $$\Ket{\psi_H}$$ are invariant,
-and instead the operators vary with time.
-An advantage of this is that the basis states remain the same.
+In the Heisenberg picture, the roles are reversed:
+the states $$\Ket{\psi_H}$$ are constants,
+and instead the operators vary in time.
+In some situations this approach can be more convenient,
+and since we usually care about the evolution of observable quantities,
+studying the corresponding operators directly
+may make more sense than finding abstract quantum states.
+Another advantage is that basis states remain fixed,
+which can simplify calculations.
 
-Given a Schrödinger-picture state $$\Ket{\psi_S(t)}$$, and operator
-$$\hat{L}_S(t)$$ which may or may not depend on time, they can be
-converted to the Heisenberg picture by the following change of basis:
+Given a Schrödinger-picture state $$\Ket{\psi_S(t)}$$
+and an operator $$\hat{L}_S(t)$$ that may or may not depend on time,
+they can be converted to the Heisenberg picture by the following transformation:
 
 $$\begin{aligned}
     \boxed{
-        \Ket{\psi_H} \equiv \Ket{\psi_S(0)}
-        \qquad
-        \hat{L}_H(t) \equiv \hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)
+        \Ket{\psi_H}
+        \equiv \Ket{\psi_S(0)}
+    }
+    \qquad\qquad
+    \boxed{
+        \hat{L}_H(t)
+        \equiv \hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)
     }
 \end{aligned}$$
 
-Since $$\hat{U}(t)$$ is unitary, the expectation value of a given operator is unchanged:
+Note that if $$\hat{H}_S$$ is time-independent,
+then it commutes with $$\hat{U}(t)$$,
+meaning $$\hat{H}_H = \hat{H}_S$$,
+so it can simply be labelled $$\hat{H}$$.
+This is not true for time-dependent Hamiltonians.
+
+Since $$\hat{U}(t)$$ is unitary,
+the expectation value of a given operator is unchanged:
 
 $$\begin{aligned}
     \expval{\hat{L}_H}
     &= \matrixel{\psi_H}{\hat{L}_H(t)}{\psi_H}
-    = \matrixel{\psi_S(0)}{\hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)}{\psi_S(0)}
+    \\
+    &= \matrixel{\psi_S(0)}{\hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)}{\psi_S(0)}
     \\
     &= \matrixel{\hat{U}(t) \psi_S(0)}{\hat{L}_S(t)}{\hat{U}(t) \psi_S(0)}
-    = \matrixel{\psi_S(t)}{\hat{L}_S}{\psi_S(t)}
-    = \expval{\hat{L}_S}
+    \\
+    &= \matrixel{\psi_S(t)}{\hat{L}_S}{\psi_S(t)}
+    \\
+    &= \expval{\hat{L}_S}
 \end{aligned}$$
 
 The Schrödinger and Heisenberg pictures therefore respectively
 correspond to active and passive transformations by $$\hat{U}(t)$$
 in [Hilbert space](/know/concept/hilbert-space/).
-The two formulations are thus entirely equivalent,
+The two formulations are entirely equivalent,
 and can be derived from one another,
-as will be shown shortly.
+as we will show shortly.
 
 In the Heisenberg picture, the states are constant,
 so the time-dependent Schrödinger equation is not directly useful.
-Instead, we will use it derive a new equation for $$\hat{L}_H(t)$$.
-The key is that the generator $$\hat{U}(t)$$ is defined from the Schrödinger equation:
+Instead, we use it derive a new equation for $$\hat{L}_H(t)$$,
+with the key being that $$\hat{U}(t)$$ itself
+satisfies the Schrödinger equation by definition:
 
 $$\begin{aligned}
-    \dv{}{t}\hat{U}(t) = - \frac{i}{\hbar} \hat{H}_S(t) \: \hat{U}(t)
+    \dv{}{t} \hat{U}(t)
+    = - \frac{i}{\hbar} \hat{H}_S(t) \: \hat{U}(t)
 \end{aligned}$$
 
 Where $$\hat{H}_S(t)$$ may depend on time. We differentiate the definition of
-$$\hat{L}_H(t)$$ and insert the other side of the Schrödinger equation
-when necessary:
+$$\hat{L}_H(t)$$ and insert the other side of the Schrödinger equation when necessary:
 
 $$\begin{aligned}
-    \dv{}{\hat{L}H}{t}
+    \dv{\hat{L}_H}{t}
     &= \dv{\hat{U}^\dagger}{t} \hat{L}_S \hat{U}
     + \hat{U}^\dagger \hat{L}_S \dv{\hat{U}}{t}
     + \hat{U}^\dagger \dv{\hat{L}_S}{t} \hat{U}
     \\
     &= \frac{i}{\hbar} \hat{U}^\dagger \hat{H}_S (\hat{U} \hat{U}^\dagger) \hat{L}_S \hat{U}
     - \frac{i}{\hbar} \hat{U}^\dagger \hat{L}_S (\hat{U} \hat{U}^\dagger) \hat{H}_S \hat{U}
-    + \Big( \dv{\hat{L}_S}{t} \Big)_H
+    + \bigg( \dv{\hat{L}_S}{t} \bigg)_H
     \\
     &= \frac{i}{\hbar} \hat{H}_H \hat{L}_H
     - \frac{i}{\hbar} \hat{L}_H \hat{H}_H
-    + \Big( \dv{\hat{L}_S}{t} \Big)_H
-    = \frac{i}{\hbar} \comm{\hat{H}_H}{\hat{L}_H} + \Big( \dv{\hat{L}_S}{t} \Big)_H
+    + \bigg( \dv{\hat{L}_S}{t} \bigg)_H
 \end{aligned}$$
 
-We thus get the equation of motion for operators in the Heisenberg picture:
+We thus get the following equation of motion for operators in the Heisenberg picture:
 
 $$\begin{aligned}
     \boxed{
-        \dv{}{t}\hat{L}_H(t) = \frac{i}{\hbar} \comm{\hat{H}_H(t)}{\hat{L}_H(t)} + \Big( \dv{}{t}\hat{L}_S(t) \Big)_H
+        \dv{}{t}\hat{L}_H(t)
+        = \frac{i}{\hbar} \comm{\hat{H}_H(t)}{\hat{L}_H(t)} + \bigg( \dv{}{t}\hat{L}_S(t) \bigg)_H
     }
 \end{aligned}$$
 
-This equation is closer to classical mechanics than the Schrödinger picture:
-inserting the position $$\hat{X}$$ and momentum $$\hat{P} = - i \hbar \: \idv{}{\hat{X}}$$
-gives the following Newton-style equations:
+This result is arguably more intuitive than the Schrödinger picture,
+because it allows us to think about observables (i.e. operators) in a more classical way.
+For example, inserting the position $$\hat{X}$$
+and momentum $$\hat{P} = - i \hbar \: \idv{}{\hat{X}}$$
+gives the following Newton-style relations:
 
 $$\begin{aligned}
     \dv{\hat{X}}{t}
@@ -112,5 +130,7 @@ $$\begin{aligned}
     = - \dv{V(\hat{X})}{\hat{X}}
 \end{aligned}$$
 
-For a proof, see [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/),
-which is closely related to the Heisenberg picture.
+Where the commutators have been treated as known.
+These equations would not be valid in the Schrödinger picture,
+unless we took their expectation value
+to get [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/).
diff --git a/source/know/concept/interaction-picture/index.md b/source/know/concept/interaction-picture/index.md
index de469fa..8428bf3 100644
--- a/source/know/concept/interaction-picture/index.md
+++ b/source/know/concept/interaction-picture/index.md
@@ -13,17 +13,19 @@ is an alternative formulation of quantum mechanics,
 equivalent to both the Schrödinger picture
 and the [Heisenberg picture](/know/concept/heisenberg-picture/).
 
-Recall that Schrödinger lets states $$\Ket{\psi_S(t)}$$ evolve in time,
-but keeps operators $$\hat{L}_S$$ fixed (except for explicit time dependence).
-Meanwhile, Heisenberg keeps states $$\Ket{\psi_H}$$ fixed,
-and puts all time dependence on the operators $$\hat{L}_H(t)$$.
-
-However, in the interaction picture,
+Recall that in the Schrödinger picture,
+the states $$\Ket{\psi_S(t)}$$ evolve in time,
+but time-independent operators $$\hat{L}_S$$ are fixed.
+Meanwhile in the Heisenberg picture,
+the states $$\Ket{\psi_H}$$ are constant,
+and all time dependence is on the operators $$\hat{L}_H(t)$$ instead.
+
+In the interaction picture,
 both the states $$\Ket{\psi_I(t)}$$ and the operators $$\hat{L}_I(t)$$
 evolve in $$t$$.
-This might seem unnecessarily complicated,
-but it turns out be convenient when considering
-a time-dependent "perturbation" $$\hat{H}_{1,S}$$
+This may seem unnecessarily complicated,
+but it turns out to be convenient when considering
+a system with a time-dependent "perturbation" $$\hat{H}_{1,S}$$
 to a time-independent Hamiltonian $$\hat{H}_{0,S}$$:
 
 $$\begin{aligned}
@@ -31,29 +33,43 @@ $$\begin{aligned}
     = \hat{H}_{0,S} + \hat{H}_{1,S}(t)
 \end{aligned}$$
 
-With $$\hat{H}_S(t)$$ the full Schrödinger Hamiltonian.
-We define the unitary conversion operator:
+Despite being called a perturbation,
+$$\hat{H}_{1, S}$$ need not be weak compared to $$\hat{H}_{0, S}$$.
+Basically, any way of splitting $$\hat{H}_S$$ is valid
+as long as $$\hat{H}_{0, S}$$ is time-independent,
+but only a few ways are useful.
+
+We now define the unitary conversion operator $$\hat{U}(t)$$ as shown below.
+Note its similarity to the [time-evolution operator](/know/concept/time-evolution-operator/)
+$$\hat{K}_S(t)$$, but with the opposite sign in the exponent:
 
 $$\begin{aligned}
     \boxed{
         \hat{U}(t)
-        \equiv \exp\!\bigg( i \frac{\hat{H}_{0,S} t}{\hbar} \bigg)
+        \equiv \exp\!\bigg( \frac{i}{\hbar} \hat{H}_{0,S} t \bigg)
     }
 \end{aligned}$$
 
-The interaction-picture states $$\Ket{\psi_I(t)}$$ and operators $$\hat{L}_I(t)$$
-are then defined to be:
+The interaction-picture states $$\Ket{\psi_I(t)}$$
+and operators $$\hat{L}_I(t)$$ are then defined as follows:
 
 $$\begin{aligned}
     \boxed{
         \Ket{\psi_I(t)}
         \equiv \hat{U}(t) \Ket{\psi_S(t)}
-        \qquad
+    }
+    \qquad\qquad
+    \boxed{
         \hat{L}_I(t)
         \equiv \hat{U}(t) \: \hat{L}_S(t) \: \hat{U}{}^\dagger(t)
     }
 \end{aligned}$$
 
+Because $$\hat{H}_{0, S}$$ is time-independent,
+it commutes with $$\hat{U}(t)$$,
+so conveniently $$\hat{H}_{0, I} = \hat{H}_{0, S}$$.
+
+
 
 ## Equations of motion
 
@@ -61,17 +77,17 @@ To find the equation of motion for $$\Ket{\psi_I(t)}$$,
 we differentiate it and multiply by $$i \hbar$$:
 
 $$\begin{aligned}
-    i \hbar \dv{}{t}\Ket{\psi_I}
-    &= i \hbar \Big( \dv{\hat{U}}{t} \Ket{\psi_S} + \hat{U} \dv{}{t}\Ket{\psi_S} \Big)
-    \\
-    &= i \hbar \Big( i \frac{\hat{H}_{0,S}}{\hbar} \Big) \hat{U} \Ket{\psi_S} + \hat{U} \Big( i \hbar \dv{}{t}\Ket{\psi_S} \Big)
+    i \hbar \dv{}{t} \Ket{\psi_I}
+    &= i \hbar \dv{\hat{U}}{t} \Ket{\psi_S} + \hat{U} \bigg( i \hbar \dv{}{t}\Ket{\psi_S} \bigg)
 \end{aligned}$$
 
-We insert the Schrödinger equation into the second term,
-and use $$\comm{\hat{U}}{\hat{H}_{0,S}} = 0$$:
+We insert the definition of $$\hat{U}$$ in the first term
+and the Schrödinger equation into the second,
+and use the fact that $$\comm{\hat{H}_{0, S}}{\hat{U}} = 0$$
+thanks to the time-independence of $$\hat{H}_{0, S}$$:
 
 $$\begin{aligned}
-    i \hbar \dv{}{t}\Ket{\psi_I}
+    i \hbar \dv{}{t} \Ket{\psi_I}
     &= - \hat{H}_{0,S} \hat{U} \Ket{\psi_S} + \hat{U} \hat{H}_S \Ket{\psi_S}
     \\
     &= \hat{U} \big( \!-\! \hat{H}_{0,S} + \hat{H}_S \big) \Ket{\psi_S}
@@ -84,129 +100,103 @@ with $$\hat{H}_{1,I} = \hat{U} \hat{H}_{1,S} \hat{U}{}^\dagger$$:
 
 $$\begin{aligned}
     \boxed{
-        i \hbar \dv{}{t}\Ket{\psi_I(t)}
+        i \hbar \dv{}{t} \Ket{\psi_I(t)}
         = \hat{H}_{1,I}(t)  \Ket{\psi_I(t)}
     }
 \end{aligned}$$
 
 Next, we do the same with an operator $$\hat{L}_I$$
-to find a description of its evolution in time:
+in order to describe its evolution in time:
 
 $$\begin{aligned}
-    \dv{}{t}\hat{L}_I
-    &= \dv{\hat{U}}{t} \hat{L}_S \hat{U}{}^\dagger + \hat{U} \hat{L}_S \dv{\hat{U}{}^\dagger}{t} + \hat{U} \dv{\hat{L}_S}{t} \hat{U}{}^\dagger
+    \dv{\hat{L}_I}{t}
+    &= \dv{\hat{U}}{t} \hat{L}_S \hat{U}{}^\dagger + \hat{U} \hat{L}_S \dv{\hat{U}{}^\dagger}{t}
+    + \hat{U} \dv{\hat{L}_S}{t} \hat{U}{}^\dagger
     \\
     &= \frac{i}{\hbar} \hat{U} \hat{H}_{0,S} \big( \hat{U}{}^\dagger \hat{U} \big) \hat{L}_S \hat{U}{}^\dagger
     - \frac{i}{\hbar} \hat{U} \hat{L}_S \big( \hat{U}{}^\dagger \hat{U} \big) \hat{H}_{0,S} \hat{U}{}^\dagger
-    + \Big( \dv{\hat{L}_S}{t} \Big)_I
+    + \bigg( \dv{\hat{L}_S}{t} \bigg)_I
     \\
     &= \frac{i}{\hbar} \hat{H}_{0,I} \hat{L}_I
     - \frac{i}{\hbar} \hat{L}_I \hat{H}_{0,I}
-    + \Big( \dv{\hat{L}_S}{t} \Big)_I
-    = \frac{i}{\hbar} \comm{\hat{H}_{0,I}}{\hat{L}_I} + \Big( \dv{\hat{L}_S}{t} \Big)_I
+    + \bigg( \dv{\hat{L}_S}{t} \bigg)_I
 \end{aligned}$$
 
 The result is analogous to the equation of motion in the Heisenberg picture:
 
 $$\begin{aligned}
     \boxed{
-        \dv{}{t}\hat{L}_I(t)
-        = \frac{i}{\hbar} \comm{\hat{H}_{0,I}(t)}{\hat{L}_I(t)} + \Big( \dv{}{t}\hat{L}_S(t) \Big)_I
+        \dv{}{t} \hat{L}_I(t)
+        = \frac{i}{\hbar} \comm{\hat{H}_{0,I}(t)}{\hat{L}_I(t)} + \bigg( \dv{}{t}\hat{L}_S(t) \bigg)_I
     }
 \end{aligned}$$
 
+In other words, in the interaction picture,
+the "simple" time-dependence (from $$\hat{H}_{0, S}$$) is given to the operators,
+and the "complicated" dependence (from $$\hat{H}_{1, S}$$) to the states.
+This means that the difficult part of a problem
+can be solved in isolation in a kind of Schrödinger picture.
 
-## Time evolution operator
 
-Recall that an alternative form of the Schrödinger equation is as follows,
-where a **time evolution operator**  or
-**generator of translations in time** $$K_S(t, t_0)$$
-brings $$\Ket{\psi_S}$$ from time $$t_0$$ to $$t$$:
 
-$$\begin{aligned}
-    \Ket{\psi_S(t)}
-    = \hat{K}_S(t, t_0) \Ket{\psi_S(t_0)}
-    \qquad \quad
-    \hat{K}_S(t, t_0)
-    \equiv \exp\!\Big( \!-\! i \frac{\hat{H}_S (t - t_0)}{\hbar} \Big)
-\end{aligned}$$
+## Time evolution operator
 
-We want to find an analogous operator in the interaction picture, satisfying:
+What about the time evolution operator $$\hat{K}_S(t)$$?
+Its interaction version $$\hat{K}_I(t)$$
+is unsurprisingly obtained by the standard transform
+$$\hat{K}_I = \hat{U} \hat{K}_S \hat{U}^\dagger$$:
 
 $$\begin{aligned}
     \Ket{\psi_I(t)}
-    \equiv \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)}
-\end{aligned}$$
-
-Inserting this definition into the equation of motion for $$\Ket{\psi_I}$$ yields
-an equation for $$\hat{K}_I$$, with the logical boundary condition $$\hat{K}_I(t_0, t_0) = 1$$:
-
-$$\begin{aligned}
-    i \hbar \dv{}{t}\Big( \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)} \Big)
-    &= \hat{H}_{1,I}(t) \Big( \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)} \Big)
+    &= \hat{U}(t) \Ket{\psi_S(t)}
     \\
-    i \hbar \dv{}{t}\hat{K}_I(t, t_0)
-    &= \hat{H}_{1,I}(t) \hat{K}_I(t, t_0)
-\end{aligned}$$
-
-We turn this into an integral equation
-by integrating both sides from $$t_0$$ to $$t$$:
-
-$$\begin{aligned}
-    i \hbar \int_{t_0}^t \dv{}{t'}K_I(t', t_0) \dd{t'}
-    = \int_{t_0}^t \hat{H}_{1,I}(t') \hat{K}_I(t', t_0) \dd{t'}
-\end{aligned}$$
-
-After evaluating the left integral,
-we see an expression for $$\hat{K}_I$$ as a function of $$\hat{K}_I$$ itself:
-
-$$\begin{aligned}
-     K_I(t, t_0)
-    = 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{K}_I(t', t_0) \dd{t'}
+    &= \hat{U}(t) \: \hat{K}_S(t) \: \hat{U}^\dagger(t) \Ket{\psi_I(0)}
+    \\
+    &\equiv \hat{K}_I(t) \Ket{\psi_I(0)}
 \end{aligned}$$
 
-By recursively inserting $$\hat{K}_I$$ once, we get a longer expression,
-still with $$\hat{K}_I$$ on both sides:
+But we can do better. By inserting this definition of $$\hat{K}_I$$
+into the interaction picture's analogue of Schrödinger's equation,
+we get the following relation for $$\hat{K}_I$$:
 
 $$\begin{aligned}
-     K_I(t, t_0)
-    = 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'}
-    + \frac{1}{(i \hbar)^2} \int_{t_0}^t \hat{H}_{1,I}(t') \int_{t_0}^{t'} \hat{H}_{1,I}(t'') \hat{K}_I(t'', t_0) \dd{t''} \dd{t'}
+    i \hbar \dv{}{t} \hat{K}_I(t)
+    &= \hat{H}_{1,I}(t) \: \hat{K}_I(t)
 \end{aligned}$$
 
-And so on. Note the ordering of the integrals and integrands:
-upon closer inspection, we see that the $$n$$th term is
-a [time-ordered product](/know/concept/time-ordered-product/) $$\mathcal{T}$$
-of $$n$$ factors $$\hat{H}_{1,I}$$:
+In other words, $$\hat{K}_I$$ can be said to also obey
+the standard equation of motion for states, despite being an operator.
+We integrate both sides and use $$\hat{K}_I(0) = 1$$:
 
 $$\begin{aligned}
-    \hat{K}_I(t, t_0)
-    &= 1 + \int_{t_0}^t \hat{H}_{1,I}(t_1) \dd{t_1}
-    + \frac{1}{2} \int_{t_0}^{t} \int_{t_0}^{t_1} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \hat{H}_{1,I}(t_2) \Big\} \dd{t_1} \dd{t_2}
-    + \: ...
-    \\
-    &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n}
-    \int_{t_0}^{t} \cdots \int_{t_0}^{t_n} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \cdots \hat{H}_{1,I}(t_n) \Big\} \dd{t_1} \cdots \dd{t_n}
-    \\
-    &= \sum_{n = 0}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n}
-    \mathcal{T} \bigg\{ \bigg( \int_{t_0}^{t} \hat{H}_{1,I}(t') \dd{t'} \bigg)^n \bigg\}
+     K_I(t)
+    = 1 + \frac{1}{i \hbar} \int_0^t \hat{H}_{1,I}(\tau) \: \hat{K}_I(\tau) \dd{\tau}
 \end{aligned}$$
 
-This construction is occasionally called the **Dyson series**.
-We recognize the well-known Taylor expansion of $$\exp(x)$$,
-leading us to a final expression for $$\hat{K}_I$$:
+This equation can be recursively inserted into itself forever.
+We recognize the resulting so called *Dyson series*
+from the derivation of $$\hat{K}_S(t)$$
+for time-dependent Hamiltonians in the Schrödinger picture
+([given here](/know/concept/time-evolution-operator/)),
+so we know that the result is given by:
 
 $$\begin{aligned}
     \boxed{
-        \hat{K}_I(t, t_0)
-        = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \bigg\}
+        \hat{K}_I(t)
+        = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_0^t \hat{H}_{1,I}(\tau) \dd{\tau} \bigg) \bigg\}
     }
 \end{aligned}$$
 
+Where $$\mathcal{T}$$ is the
+[time-ordering meta-operator](/know/concept/time-ordered-product/),
+which is conventionally written in this way
+to say that it applies to the terms of a Taylor expansion of $$\exp(x)$$.
+This means that the evolution of a quantum state in the interaction picture
+is determined by the perturbation $$\hat{H}_{1, I}$$.
+
 
 
 ## References
 1.  H. Bruus, K. Flensberg,
     *Many-body quantum theory in condensed matter physics*,
     2016, Oxford.
-
diff --git a/source/know/concept/self-steepening/index.md b/source/know/concept/self-steepening/index.md
index 409f6c9..015aa40 100644
--- a/source/know/concept/self-steepening/index.md
+++ b/source/know/concept/self-steepening/index.md
@@ -97,7 +97,9 @@ $$E \equiv \int_{-\infty}^\infty |A|^2 \dd{t}$$ anymore,
 which is often used to quantify simulation errors.
 Fortunately, another value can then be used instead:
 it can be shown that the "photon number" $$N$$
-is still conserved, defined as:
+is still conserved, defined like so,
+where $$\omega$$ is the absolute frequency
+(as opposed to the relative frequency $$\Omega$$):
 
 $$\begin{aligned}
     \boxed{
@@ -193,9 +195,9 @@ pulls the pulse apart before a shock can occur.
 The early steepening is observable though.
 
 A simulation of self-steepening without dispersion is illustrated below
-for the following initial power distribution,
+for the following Gaussian power distribution,
 with $$T_0 = 25\:\mathrm{fs}$$, $$P_0 = 3\:\mathrm{kW}$$,
-$$\beta_2 = 0$$, $$\gamma = 0.1/\mathrm{W}/\mathrm{m}$$,
+$$\beta_2 = 0$$, $$\gamma_0 = 0.1/\mathrm{W}/\mathrm{m}$$,
 and a vacuum carrier wavelength $$\lambda_0 \approx 73\:\mathrm{nm}$$
 (the latter determined by the simulation's resolution settings):
 
diff --git a/source/know/concept/time-evolution-operator/index.md b/source/know/concept/time-evolution-operator/index.md
new file mode 100644
index 0000000..f489ac6
--- /dev/null
+++ b/source/know/concept/time-evolution-operator/index.md
@@ -0,0 +1,184 @@
+---
+title: "Time evolution operator"
+sort_title: "Time evolution operator"
+date: 2024-10-15
+categories:
+- Quantum mechanics
+- Physics
+layout: "concept"
+---
+
+In general, given a system whose governing equation is known,
+the **time evolution operator** $$\hat{U}(t, t_0)$$
+transforms the state at time $$t_0$$ to the one at time $$t$$.
+Although not specific to it,
+this is most often used in quantum mechanics,
+as governed by the Schrödinger equation:
+
+$$\begin{aligned}
+    i \hbar \dv{}{t} \ket{\psi(t)}
+    = \hat{H}(t) \ket{\psi(t)}
+\end{aligned}$$
+
+Such that the definition of $$\hat{U}(t)$$ is as follows,
+where we have set $$t_0 = 0$$:
+
+$$\begin{aligned}
+    \ket{\psi(t)}
+    = \hat{U}(t) \ket{\psi(0)}
+\end{aligned}$$
+
+Clearly, $$\hat{U}(t)$$ must be unitary.
+The goal is to find an expression that satisfies this relation.
+
+
+
+## Time-independent Hamiltonian
+
+We start by inserting the definition of $$\hat{U}(t)$$
+into the Schrödinger equation:
+
+$$\begin{aligned}
+    \dv{}{t} \hat{U}(t) \ket{\psi(0)}
+    = - \frac{i}{\hbar} \hat{H} \: \hat{U}(t) \ket{\psi(0)}
+\end{aligned}$$
+
+If we hide the state $$\ket{\psi(0)}$$,
+then $$\hat{U}(t)$$ can be said to satisfy the equation in its own right:
+
+$$\begin{aligned}
+    \dv{}{t} \hat{U}(t)
+    = - \frac{i}{\hbar} \hat{H} \: \hat{U}(t)
+\end{aligned}$$
+
+If the Hamiltonian $$\hat{H}$$ is time-independent,
+this is straightforward to integrate, yielding:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{U}(t)
+        = \exp\!\bigg( \!-\! \frac{i}{\hbar} t \hat{H} \bigg)
+    }
+\end{aligned}$$
+
+And the generalization to $$t_0 \neq 0$$ is trivial,
+since we can just shift the time axis:
+
+$$\begin{aligned}
+    \hat{U}(t, t_0)
+    = \exp\!\bigg( \!-\! \frac{i}{\hbar} (t - t_0) \hat{H} \bigg)
+\end{aligned}$$
+
+
+
+## Time-dependent Hamiltonian
+
+Even when $$\hat{H}$$ is time-dependent,
+$$\hat{U}(t)$$ can be said to satisfy the Schrödinger equation:
+
+$$\begin{aligned}
+    \dv{}{t} \hat{U}(t)
+    = - \frac{i}{\hbar} \hat{H}(t) \: \hat{U}(t)
+\end{aligned}$$
+
+Integrating from $$0$$ to $$t$$,
+and using $$\hat{U}(0) = 1$$ (which should be clear from its definition):
+
+$$\begin{aligned}
+    \hat{U}(t)
+    = 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \: \hat{U}(\tau_1) \dd{\tau_1}
+\end{aligned}$$
+
+This is a self-consistent equation for $$\hat{U}(t)$$.
+We can recursively insert it into itself, yielding:
+
+$$\begin{aligned}
+    \hat{U}(t)
+    &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1)
+    \bigg( 1 + \frac{1}{i \hbar} \int_0^{\tau_1} \hat{H}(\tau_2) \: \hat{U}(\tau_2) \dd{\tau_2} \bigg) \dd{\tau_1}
+    \\
+    &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \dd{\tau_1}
+    + \frac{1}{(i \hbar)^2} \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \: \hat{U}(\tau_2) \dd{\tau_2} \dd{\tau_1}
+    \\
+    &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \dd{\tau_1}
+    + \frac{1}{(i \hbar)^2} \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1}
+    + \frac{1}{(i \hbar)^3} \int_0^t \cdots \: \dd{\tau_1}
+\end{aligned}$$
+
+And so on.
+Let us take a closer look at the third (i.e. second-order) term in this series,
+noting that the integrals are ordered such that $$\tau_2 < \tau_1$$ always.
+We can exploit this fact to introduce several
+[Heaviside step functions](/know/concept/heaviside-step-function/) $$\Theta(t)$$:
+
+$$\begin{aligned}
+    &\quad \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1}
+    \\
+    &= \frac{1}{2} \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1}
+    + \frac{1}{2} \int_0^t \hat{H}(\tau_2) \int_0^{\tau_2} \hat{H}(\tau_1) \dd{\tau_1} \dd{\tau_2}
+    \\
+    &= \frac{1}{2} \int_0^t \! \hat{H}(\tau_1)
+    \int_0^{\tau_1} \! \Theta(\tau_1 \!-\! \tau_2) \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1}
+    + \frac{1}{2} \int_0^t \! \hat{H}(\tau_2)
+    \int_0^{\tau_2} \! \Theta(\tau_2 \!-\! \tau_1) \hat{H}(\tau_1) \dd{\tau_1} \dd{\tau_2}
+    \\
+    &= \frac{1}{2} \int_0^t \int_0^t
+    \bigg( \Theta(\tau_1 \!-\! \tau_2) \: \hat{H}(\tau_1) \: \hat{H}(\tau_2)
+    + \Theta(\tau_2 \!-\! \tau_1) \: \hat{H}(\tau_1) \: \hat{H}(\tau_2) \bigg) \dd{\tau_1} \dd{\tau_2}
+    \\
+    &= \frac{1}{2} \int_0^t \int_0^t \mathcal{T} \Big\{ \hat{H}(\tau_1) \: \hat{H}(\tau_2) \Big\} \dd{\tau_1} \dd{\tau_2}
+\end{aligned}$$
+
+Where we have recognized the
+[time-ordering meta-operator](/know/concept/time-ordered-product/) $$\mathcal{T}$$.
+The above procedure is easy to generalize to the higher-order terms,
+so we arrive at the following expression for $$\hat{U}(t)$$:
+
+$$\begin{aligned}
+    \hat{U}(t)
+    &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \dd{\tau_1}
+    + \frac{1}{2} \frac{1}{(i \hbar)^2} \iint_0^t \mathcal{T} \Big\{ \hat{H}(\tau_1) \: \hat{H}(\tau_2) \Big\} \dd{\tau_2} \dd{\tau_1}
+    \\
+    &\qquad+ \frac{1}{6} \frac{1}{(i \hbar)^3} \iiint_0^t
+    \mathcal{T} \Big\{ \hat{H}(\tau_1) \: \hat{H}(\tau_2) \: \hat{H}(\tau_3) \Big\} \dd{\tau_3} \dd{\tau_2} \dd{\tau_1}
+    + \: ...
+    \\
+    &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n} \int_0^t \!\cdots\! \int_0^t
+    \mathcal{T} \Big\{ \hat{H}(\tau_1) \cdots \hat{H}(\tau_n) \Big\} \dd{\tau_n} \cdots \dd{\tau_1}
+\end{aligned}$$
+
+This result is sometimes called a **Dyson series**.
+Convention allows us to write it as follows,
+despite such a use of $$\mathcal{T}$$ looking a bit strange:
+
+$$\begin{aligned}
+    \hat{U}(t)
+    &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n}
+    \mathcal{T} \bigg\{ \bigg( \int_0^t \hat{H}(\tau) \dd{\tau} \bigg)^n \bigg\}
+\end{aligned}$$
+
+Here, we recognize the Taylor expansion of $$\exp(x)$$,
+leading us to the desired result:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{U}(t)
+        = \mathcal{T} \bigg\{ \exp\!\bigg( \!-\! \frac{i}{\hbar} \int_0^t \hat{H}(\tau) \dd{\tau} \bigg) \bigg\}
+    }
+\end{aligned}$$
+
+Where once again $$\mathcal{T}$$ is being used according to convention.
+Finally, the time axis can be shifted arbitrarily,
+so many authors write the evolution operator from $$t_0$$ to $$t$$ as $$\hat{U}(t, t_0)$$:
+
+$$\begin{aligned}
+    \hat{U}(t, t_0)
+    = \mathcal{T} \bigg\{ \exp\!\bigg( \!-\! \frac{i}{\hbar} \int_{t_0}^t \hat{H}(\tau) \dd{\tau} \bigg) \bigg\}
+\end{aligned}$$
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.