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

1 files changed, 13 insertions, 13 deletions

diff --git a/source/know/concept/heisenberg-picture/index.md b/source/know/concept/heisenberg-picture/index.md
index 59ed2da..359ecfe 100644
--- a/source/know/concept/heisenberg-picture/index.md
+++ b/source/know/concept/heisenberg-picture/index.md
@@ -13,9 +13,9 @@ mechanics, and is equivalent to the traditionally-taught 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 $\hat{H}_S$:
+$$\Ket{\psi_S(t)}$$ changes according to the Schrödinger equation,
+which can be written using the generator of translations $$\hat{U}(t)$$ like so,
+for a time-independent $$\hat{H}_S$$:
 
 $$\begin{aligned}
     \Ket{\psi_S(t)} = \hat{U}(t) \Ket{\psi_S(0)}
@@ -26,12 +26,12 @@ $$\begin{aligned}
 \end{aligned}$$
 
 In contrast, the Heisenberg picture reverses the roles:
-the states $\Ket{\psi_H}$ are invariant,
+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.
 
-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
+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:
 
 $$\begin{aligned}
@@ -42,7 +42,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Since $\hat{U}(t)$ is unitary, the expectation value of a given operator is unchanged:
+Since $$\hat{U}(t)$$ is unitary, the expectation value of a given operator is unchanged:
 
 $$\begin{aligned}
     \expval{\hat{L}_H}
@@ -55,7 +55,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The Schrödinger and Heisenberg pictures therefore respectively
-correspond to active and passive transformations by $\hat{U}(t)$
+correspond to active and passive transformations by $$\hat{U}(t)$$
 in [Hilbert space](/know/concept/hilbert-space/).
 The two formulations are thus entirely equivalent,
 and can be derived from one another,
@@ -63,15 +63,15 @@ as will be shown 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 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:
 
 $$\begin{aligned}
     \dv{}{t}\hat{U}(t) = - \frac{i}{\hbar} \hat{H}_S(t) \: \hat{U}(t)
 \end{aligned}$$
 
-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
+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:
 
 $$\begin{aligned}
@@ -99,7 +99,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 This equation is closer to classical mechanics than the Schrödinger picture:
-inserting the position $\hat{X}$ and momentum $\hat{P} = - i \hbar \: \idv{}{\hat{X}}$
+inserting the position $$\hat{X}$$ and momentum $$\hat{P} = - i \hbar \: \idv{}{\hat{X}}$$
 gives the following Newton-style equations:
 
 $$\begin{aligned}