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

1 files changed, 12 insertions, 12 deletions

diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md
index 5f333a2..f454264 100644
--- a/source/know/concept/bell-state/index.md
+++ b/source/know/concept/bell-state/index.md
@@ -24,14 +24,14 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-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/).
 
 More importantly, however,
 is that the Bell states are maximally entangled,
-which we prove here for $\ket{\Phi^{+}}$.
+which we prove here for $$\ket{\Phi^{+}}$$.
 Consider the following pure [density operator](/know/concept/density-operator/):
 
 $$\begin{aligned}
@@ -40,7 +40,7 @@ $$\begin{aligned}
     &= \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:
+The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated as follows:
 
 $$\begin{aligned}
     \hat{\rho}_A
@@ -54,14 +54,14 @@ $$\begin{aligned}
     = \frac{1}{2} \hat{I}
 \end{aligned}$$
 
-This result is maximally mixed, therefore $\ket{\Phi^{+}}$ is maximally entangled.
+This result is maximally mixed, therefore $$\ket{\Phi^{+}}$$ is maximally entangled.
 The same holds for the other three Bell states,
-and is equally true for qubit $B$.
+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 $\Ket{1}$.
+This means that a measurement of qubit $$A$$
+has a 50-50 chance to yield $$\Ket{0}$$ or $$\Ket{1}$$.
 However, due to the entanglement,
-measuring $A$ also has consequences for qubit $B$:
+measuring $$A$$ also has consequences for qubit $$B$$:
 
 $$\begin{aligned}
     \big| \Bra{0}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
@@ -81,10 +81,10 @@ $$\begin{aligned}
     = \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 $\ket{\Phi^{+}}$,
+This was a specific example for $$\ket{\Phi^{+}}$$,
 but analogous results can be found for the other Bell states.