Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

1 files changed, 28 insertions, 28 deletions

diff --git a/source/know/concept/interaction-picture/index.md b/source/know/concept/interaction-picture/index.md
index 05f3ad0..de469fa 100644
--- a/source/know/concept/interaction-picture/index.md
+++ b/source/know/concept/interaction-picture/index.md
@@ -13,25 +13,25 @@ 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 $\hat{L}_H(t)$.
+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,
-both the states $\Ket{\psi_I(t)}$ and the operators $\hat{L}_I(t)$
-evolve in $t$.
+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" $\hat{H}_{1,S}$
-to a time-independent Hamiltonian $\hat{H}_{0,S}$:
+a time-dependent "perturbation" $$\hat{H}_{1,S}$$
+to a time-independent Hamiltonian $$\hat{H}_{0,S}$$:
 
 $$\begin{aligned}
     \hat{H}_S(t)
     = \hat{H}_{0,S} + \hat{H}_{1,S}(t)
 \end{aligned}$$
 
-With $\hat{H}_S(t)$ the full Schrödinger Hamiltonian.
+With $$\hat{H}_S(t)$$ the full Schrödinger Hamiltonian.
 We define the unitary conversion operator:
 
 $$\begin{aligned}
@@ -41,7 +41,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The interaction-picture states $\Ket{\psi_I(t)}$ and operators $\hat{L}_I(t)$
+The interaction-picture states $$\Ket{\psi_I(t)}$$ and operators $$\hat{L}_I(t)$$
 are then defined to be:
 
 $$\begin{aligned}
@@ -57,8 +57,8 @@ $$\begin{aligned}
 
 ## Equations of motion
 
-To find the equation of motion for $\Ket{\psi_I(t)}$,
-we differentiate it and multiply by $i \hbar$:
+To find the equation of motion for $$\Ket{\psi_I(t)}$$,
+we differentiate it and multiply by $$i \hbar$$:
 
 $$\begin{aligned}
     i \hbar \dv{}{t}\Ket{\psi_I}
@@ -68,7 +68,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We insert the Schrödinger equation into the second term,
-and use $\comm{\hat{U}}{\hat{H}_{0,S}} = 0$:
+and use $$\comm{\hat{U}}{\hat{H}_{0,S}} = 0$$:
 
 $$\begin{aligned}
     i \hbar \dv{}{t}\Ket{\psi_I}
@@ -80,7 +80,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Which leads to an analogue of the Schrödinger equation,
-with $\hat{H}_{1,I} = \hat{U} \hat{H}_{1,S} \hat{U}{}^\dagger$:
+with $$\hat{H}_{1,I} = \hat{U} \hat{H}_{1,S} \hat{U}{}^\dagger$$:
 
 $$\begin{aligned}
     \boxed{
@@ -89,7 +89,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Next, we do the same with an operator $\hat{L}_I$
+Next, we do the same with an operator $$\hat{L}_I$$
 to find a description of its evolution in time:
 
 $$\begin{aligned}
@@ -120,8 +120,8 @@ $$\begin{aligned}
 
 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 $t$:
+**generator of translations in time** $$K_S(t, t_0)$$
+brings $$\Ket{\psi_S}$$ from time $$t_0$$ to $$t$$:
 
 $$\begin{aligned}
     \Ket{\psi_S(t)}
@@ -138,8 +138,8 @@ $$\begin{aligned}
     \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 $\hat{K}_I(t_0, t_0) = 1$:
+Inserting this definition into the equation of motion for $$\Ket{\psi_I}$$ yields
+an equation for $$\hat{K}_I$$, with the logical boundary condition $$\hat{K}_I(t_0, t_0) = 1$$:
 
 $$\begin{aligned}
     i \hbar \dv{}{t}\Big( \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)} \Big)
@@ -150,7 +150,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We turn this into an integral equation
-by integrating both sides from $t_0$ to $t$:
+by integrating both sides from $$t_0$$ to $$t$$:
 
 $$\begin{aligned}
     i \hbar \int_{t_0}^t \dv{}{t'}K_I(t', t_0) \dd{t'}
@@ -158,15 +158,15 @@ $$\begin{aligned}
 \end{aligned}$$
 
 After evaluating the left integral,
-we see an expression for $\hat{K}_I$ as a function of $\hat{K}_I$ itself:
+we see an expression for $$\hat{K}_I$$ as a function of $$\hat{K}_I$$ itself:
 
 $$\begin{aligned}
      K_I(t, t_0)
     = 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{K}_I(t', t_0) \dd{t'}
 \end{aligned}$$
 
-By recursively inserting $\hat{K}_I$ once, we get a longer expression,
-still with $\hat{K}_I$ on both sides:
+By recursively inserting $$\hat{K}_I$$ once, we get a longer expression,
+still with $$\hat{K}_I$$ on both sides:
 
 $$\begin{aligned}
      K_I(t, t_0)
@@ -175,9 +175,9 @@ $$\begin{aligned}
 \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 $\hat{H}_{1,I}$:
+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 $$\hat{H}_{1,I}$$:
 
 $$\begin{aligned}
     \hat{K}_I(t, t_0)
@@ -193,8 +193,8 @@ $$\begin{aligned}
 \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 $\hat{K}_I$:
+We recognize the well-known Taylor expansion of $$\exp(x)$$,
+leading us to a final expression for $$\hat{K}_I$$:
 
 $$\begin{aligned}
     \boxed{