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diff --git a/source/know/concept/bells-theorem/index.md b/source/know/concept/bells-theorem/index.md
index 3b71dbf..a01bf9e 100644
--- a/source/know/concept/bells-theorem/index.md
+++ b/source/know/concept/bells-theorem/index.md
@@ -13,7 +13,7 @@ layout: "concept"
 cannot be explained by theories built on
 so-called **local hidden variables** (LHVs).
 
-Suppose that we have two spin-1/2 particles, called $A$ and $B$,
+Suppose that we have two spin-1/2 particles, called $$A$$ and $$B$$,
 in an entangled [Bell state](/know/concept/bell-state/):
 
 $$\begin{aligned}
@@ -22,23 +22,23 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Since they are entangled,
-if we measure the $z$-spin of particle $A$, and find e.g. $\Ket{\uparrow}$,
-then particle $B$ immediately takes the opposite state $\Ket{\downarrow}$.
+if we measure the $$z$$-spin of particle $$A$$, and find e.g. $$\Ket{\uparrow}$$,
+then particle $$B$$ immediately takes the opposite state $$\Ket{\downarrow}$$.
 The point is that this collapse is instant,
-regardless of the distance between $A$ and $B$.
+regardless of the distance between $$A$$ and $$B$$.
 
 Einstein called this effect "action-at-a-distance",
 and used it as evidence that quantum mechanics is an incomplete theory.
-He said that there must be some **hidden variable** $\lambda$
-that determines the outcome of measurements of $A$ and $B$
+He said that there must be some **hidden variable** $$\lambda$$
+that determines the outcome of measurements of $$A$$ and $$B$$
 from the moment the entangled pair is created.
 However, according to Bell's theorem, he was wrong.
 
-To prove this, let us assume that Einstein was right, and some $\lambda$,
+To prove this, let us assume that Einstein was right, and some $$\lambda$$,
 which we cannot understand, let alone calculate or measure, controls the results.
 We want to know the spins of the entangled pair
-along arbitrary directions $\vec{a}$ and $\vec{b}$,
-so the outcomes for particles $A$ and $B$ are:
+along arbitrary directions $$\vec{a}$$ and $$\vec{b}$$,
+so the outcomes for particles $$A$$ and $$B$$ are:
 
 $$\begin{aligned}
     A(\vec{a}, \lambda) = \pm 1
@@ -46,8 +46,8 @@ $$\begin{aligned}
     B(\vec{b}, \lambda) = \pm 1
 \end{aligned}$$
 
-Where $\pm 1$ are the eigenvalues of the Pauli matrices
-in the chosen directions $\vec{a}$ and $\vec{b}$:
+Where $$\pm 1$$ are the eigenvalues of the Pauli matrices
+in the chosen directions $$\vec{a}$$ and $$\vec{b}$$:
 
 $$\begin{aligned}
     \hat{\sigma}_a
@@ -59,8 +59,8 @@ $$\begin{aligned}
     = b_x \hat{\sigma}_x + b_y \hat{\sigma}_y + b_z \hat{\sigma}_z
 \end{aligned}$$
 
-Whether $\lambda$ is a scalar or a vector does not matter;
-we simply demand that it follows an unknown probability distribution $\rho(\lambda)$:
+Whether $$\lambda$$ is a scalar or a vector does not matter;
+we simply demand that it follows an unknown probability distribution $$\rho(\lambda)$$:
 
 $$\begin{aligned}
     \int \rho(\lambda) \dd{\lambda} = 1
@@ -68,8 +68,8 @@ $$\begin{aligned}
     \rho(\lambda) \ge 0
 \end{aligned}$$
 
-The product of the outcomes of $A$ and $B$ then has the following expectation value.
-Note that we only multiply $A$ and $B$ for shared $\lambda$-values:
+The product of the outcomes of $$A$$ and $$B$$ then has the following expectation value.
+Note that we only multiply $$A$$ and $$B$$ for shared $$\lambda$$-values:
 this is what makes it a **local** hidden variable:
 
 $$\begin{aligned}
@@ -83,7 +83,7 @@ which both prove Bell's theorem.
 
 ## Bell inequality
 
-If $\vec{a} = \vec{b}$, then we know that $A$ and $B$ always have opposite spins:
+If $$\vec{a} = \vec{b}$$, then we know that $$A$$ and $$B$$ always have opposite spins:
 
 $$\begin{aligned}
     A(\vec{a}, \lambda)
@@ -98,8 +98,8 @@ $$\begin{aligned}
     = - \int \rho(\lambda) \: A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda}
 \end{aligned}$$
 
-Next, we introduce an arbitrary third direction $\vec{c}$,
-and use the fact that $( A(\vec{b}, \lambda) )^2 = 1$:
+Next, we introduce an arbitrary third direction $$\vec{c}$$,
+and use the fact that $$( A(\vec{b}, \lambda) )^2 = 1$$:
 
 $$\begin{aligned}
     \Expval{A_a B_b} - \Expval{A_a B_c}
@@ -109,7 +109,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Inside the integral, the only factors that can be negative
-are the last two, and their product is $\pm 1$.
+are the last two, and their product is $$\pm 1$$.
 Taking the absolute value of the whole left,
 and of the integrand on the right, we thus get:
 
@@ -121,7 +121,7 @@ $$\begin{aligned}
     &\le \int \rho(\lambda) \dd{\lambda} - \int \rho(\lambda) A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \dd{\lambda}
 \end{aligned}$$
 
-Since $\rho(\lambda)$ is a normalized probability density function,
+Since $$\rho(\lambda)$$ is a normalized probability density function,
 we arrive at the **Bell inequality**:
 
 $$\begin{aligned}
@@ -131,18 +131,18 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Any theory involving an LHV $\lambda$ must obey this inequality.
+Any theory involving an LHV $$\lambda$$ must obey this inequality.
 The problem, however, is that quantum mechanics dictates the expectation values
-for the state $\Ket{\Psi^{-}}$:
+for the state $$\Ket{\Psi^{-}}$$:
 
 $$\begin{aligned}
     \Expval{A_a B_b} = - \vec{a} \cdot \vec{b}
 \end{aligned}$$
 
 Finding directions which violate the Bell inequality is easy:
-for example, if $\vec{a}$ and $\vec{b}$ are orthogonal,
-and $\vec{c}$ is at a $\pi/4$ angle to both of them,
-then the left becomes $0.707$ and the right $0.293$,
+for example, if $$\vec{a}$$ and $$\vec{b}$$ are orthogonal,
+and $$\vec{c}$$ is at a $$\pi/4$$ angle to both of them,
+then the left becomes $$0.707$$ and the right $$0.293$$,
 which clearly disagrees with the inequality,
 meaning that LHVs are impossible.
 
@@ -152,8 +152,8 @@ meaning that LHVs are impossible.
 The **Clauser-Horne-Shimony-Holt** or simply **CHSH inequality**
 takes a slightly different approach, and is more useful in practice.
 
-Consider four spin directions, two for $A$ called $\vec{a}_1$ and $\vec{a}_2$,
-and two for $B$ called $\vec{b}_1$ and $\vec{b}_2$.
+Consider four spin directions, two for $$A$$ called $$\vec{a}_1$$ and $$\vec{a}_2$$,
+and two for $$B$$ called $$\vec{b}_1$$ and $$\vec{b}_2$$.
 Let us introduce the following abbreviations:
 
 $$\begin{aligned}
@@ -196,7 +196,7 @@ $$\begin{aligned}
     + \bigg|\! \int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg|
 \end{aligned}$$
 
-Using the fact that the product of $A$ and $B$ is always either $-1$ or $+1$,
+Using the fact that the product of $$A$$ and $$B$$ is always either $$-1$$ or $$+1$$,
 we can reduce this to:
 
 $$\begin{aligned}
@@ -209,7 +209,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Evaluating these integrals gives us the following inequality,
-which holds for both choices of $\pm$:
+which holds for both choices of $$\pm$$:
 
 $$\begin{aligned}
     \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
@@ -235,8 +235,8 @@ $$\begin{aligned}
     &\ge \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} + \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big|
 \end{aligned}$$
 
-The quantity on the right-hand side is sometimes called the **CHSH quantity** $S$,
-and measures the correlation between the spins of $A$ and $B$:
+The quantity on the right-hand side is sometimes called the **CHSH quantity** $$S$$,
+and measures the correlation between the spins of $$A$$ and $$B$$:
 
 $$\begin{aligned}
     \boxed{
@@ -244,7 +244,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The CHSH inequality places an upper bound on the magnitude of $S$
+The CHSH inequality places an upper bound on the magnitude of $$S$$
 for LHV-based theories:
 
 $$\begin{aligned}
@@ -258,15 +258,15 @@ $$\begin{aligned}
 
 Quantum physics can violate the CHSH inequality, but by how much?
 Consider the following two-particle operator,
-whose expectation value is the CHSH quantity, i.e. $S = \expval{\hat{S}}$:
+whose expectation value is the CHSH quantity, i.e. $$S = \expval{\hat{S}}$$:
 
 $$\begin{aligned}
     \hat{S}
     = \hat{A}_2 \otimes \hat{B}_1 + \hat{A}_2 \otimes \hat{B}_2 + \hat{A}_1 \otimes \hat{B}_1 - \hat{A}_1 \otimes \hat{B}_2
 \end{aligned}$$
 
-Where $\otimes$ is the tensor product,
-and e.g. $\hat{A}_1$ is the Pauli matrix for the $\vec{a}_1$-direction.
+Where $$\otimes$$ is the tensor product,
+and e.g. $$\hat{A}_1$$ is the Pauli matrix for the $$\vec{a}_1$$-direction.
 The square of this operator is then given by:
 
 $$\begin{aligned}
@@ -292,15 +292,15 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Spin operators are unitary, so their square is the identity,
-e.g. $\hat{A}_1^2 = \hat{I}$. Therefore $\hat{S}^2$ reduces to:
+e.g. $$\hat{A}_1^2 = \hat{I}$$. Therefore $$\hat{S}^2$$ reduces to:
 
 $$\begin{aligned}
     \hat{S}^2
     &= 4 \: (\hat{I} \otimes \hat{I}) + \comm{\hat{A}_1}{\hat{A}_2} \otimes \comm{\hat{B}_1}{\hat{B}_2}
 \end{aligned}$$
 
-The *norm* $\norm{\hat{S}^2}$ of this operator
-is the largest possible expectation value $\expval{\hat{S}^2}$,
+The *norm* $$\norm{\hat{S}^2}$$ of this operator
+is the largest possible expectation value $$\expval{\hat{S}^2}$$,
 which is the same as its largest eigenvalue.
 It is given by:
 
@@ -321,7 +321,7 @@ $$\begin{aligned}
     \le 2
 \end{aligned}$$
 
-And $\norm{\comm{\hat{B}_1}{\hat{B}_2}} \le 2$ for the same reason.
+And $$\norm{\comm{\hat{B}_1}{\hat{B}_2}} \le 2$$ for the same reason.
 The norm is the largest eigenvalue, therefore:
 
 $$\begin{aligned}
@@ -336,7 +336,7 @@ $$\begin{aligned}
 
 We thus arrive at **Tsirelson's bound**,
 which states that quantum mechanics can violate
-the CHSH inequality by a factor of $\sqrt{2}$:
+the CHSH inequality by a factor of $$\sqrt{2}$$:
 
 $$\begin{aligned}
     \boxed{
@@ -359,8 +359,8 @@ $$\begin{aligned}
     \hat{B}_2 = \frac{\hat{\sigma}_z - \hat{\sigma}_x}{\sqrt{2}}
 \end{aligned}$$
 
-Using the fact that $\Expval{A_a B_b} = - \vec{a} \cdot \vec{b}$,
-it can then be shown that $S = 2 \sqrt{2}$ in this case.
+Using the fact that $$\Expval{A_a B_b} = - \vec{a} \cdot \vec{b}$$,
+it can then be shown that $$S = 2 \sqrt{2}$$ in this case.