From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

---
 source/know/concept/no-cloning-theorem/index.md | 12 ++++++------
 1 file changed, 6 insertions(+), 6 deletions(-)

(limited to 'source/know/concept/no-cloning-theorem/index.md')

diff --git a/source/know/concept/no-cloning-theorem/index.md b/source/know/concept/no-cloning-theorem/index.md
index d4ca0d4..a91ae6f 100644
--- a/source/know/concept/no-cloning-theorem/index.md
+++ b/source/know/concept/no-cloning-theorem/index.md
@@ -10,13 +10,13 @@ layout: "concept"
 ---
 
 In quantum mechanics, the **no-cloning theorem** states
-there is no general way to make copies of an arbitrary quantum state $\ket{\psi}$.
+there is no general way to make copies of an arbitrary quantum state $$\ket{\psi}$$.
 This has profound implications for quantum information.
 
 To prove this theorem, let us pretend that a machine exists
 that can do just that: copy arbitrary quantum states.
-Given an input $\ket{\psi}$ and a blank $\ket{?}$,
-this machines turns $\ket{?}$ into $\ket{\psi}$:
+Given an input $$\ket{\psi}$$ and a blank $$\ket{?}$$,
+this machines turns $$\ket{?}$$ into $$\ket{\psi}$$:
 
 $$\begin{aligned}
     \ket{\psi} \ket{?}
@@ -24,7 +24,7 @@ $$\begin{aligned}
     \ket{\psi} \ket{\psi}
 \end{aligned}$$
 
-We can use this device to make copies of the basis vectors $\ket{0}$ and $\ket{1}$:
+We can use this device to make copies of the basis vectors $$\ket{0}$$ and $$\ket{1}$$:
 
 $$\begin{aligned}
     \ket{0} \ket{?}
@@ -36,7 +36,7 @@ $$\begin{aligned}
     \ket{1} \ket{1}
 \end{aligned}$$
 
-If we feed this machine a superposition $\ket{\psi} = \alpha \ket{0} + \beta \ket{1}$,
+If we feed this machine a superposition $$\ket{\psi} = \alpha \ket{0} + \beta \ket{1}$$,
 we *want* the following behaviour:
 
 $$\begin{aligned}
@@ -47,7 +47,7 @@ $$\begin{aligned}
     &= \Big( \alpha^2 \ket{0} \ket{0} + \alpha \beta \ket{0} \ket{1} + \alpha \beta \ket{1} \ket{0} + \beta^2 \ket{1} \ket{1} \Big)
 \end{aligned}$$
 
-Note the appearance of the cross terms with a factor of $\alpha \beta$.
+Note the appearance of the cross terms with a factor of $$\alpha \beta$$.
 The problem is that the fundamental linearity of quantum mechanics
 dictates different behaviour:
 
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