From aeacfca5aea5df7c107cf0c12e72ab5d496c96e1 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Tue, 3 Jan 2023 19:48:17 +0100
Subject: More improvements to knowledge base

---
 source/know/concept/bell-state/index.md | 41 +++++++++++++++++----------------
 1 file changed, 21 insertions(+), 20 deletions(-)

(limited to 'source/know/concept/bell-state')

diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md
index f454264..fa289de 100644
--- a/source/know/concept/bell-state/index.md
+++ b/source/know/concept/bell-state/index.md
@@ -16,16 +16,16 @@ $$\begin{aligned}
     \boxed{
         \begin{aligned}
             \ket{\Phi^{\pm}}
-            &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{0}_B \pm \Ket{1}_A \Ket{1}_B \Big)
+            &= \frac{1}{\sqrt{2}} \Big( \ket{0}_A \ket{0}_B \pm \ket{1}_A \ket{1}_B \Big)
             \\
             \ket{\Psi^{\pm}}
-            &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{1}_B \pm \Ket{1}_A \Ket{0}_B \Big)
+            &= \frac{1}{\sqrt{2}} \Big( \ket{0}_A \ket{1}_B \pm \ket{1}_A \ket{0}_B \Big)
         \end{aligned}
     }
 \end{aligned}$$
 
-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/).
 
@@ -37,7 +37,7 @@ Consider the following pure [density operator](/know/concept/density-operator/):
 $$\begin{aligned}
     \hat{\rho}
     = \ket{\Phi^{+}} \bra{\Phi^{+}}
-    &= \frac{1}{2} \Big( \Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big) \Big( \Bra{0}_A \Bra{0}_B + \Bra{1}_A \Bra{1}_B \Big)
+    &= \frac{1}{2} \Big( \ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big) \Big( \bra{0}_A \bra{0}_B + \bra{1}_A \bra{1}_B \Big)
 \end{aligned}$$
 
 The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated as follows:
@@ -45,12 +45,12 @@ The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated
 $$\begin{aligned}
     \hat{\rho}_A
     &= \Tr_B(\hat{\rho})
-    = \sum_{b = 0, 1} \Bra{b}_B \Big( \ket{\Phi^{+}} \bra{\Phi^{+}} \Big) \Ket{b}_B
+    = \sum_{b = 0, 1} \bra{b}_B \Big( \ket{\Phi^{+}} \bra{\Phi^{+}} \Big) \ket{b}_B
     \\
-    &= \sum_{b = 0, 1} \Big( \Ket{0}_A \Inprod{b}{0}_B + \Ket{1}_A \Inprod{b}{1}_B \Big)
-    \Big( \Bra{0}_A \Inprod{0}{b}_B + \Bra{1}_A \Inprod{1}{b}_B \Big)
+    &= \sum_{b = 0, 1} \Big( \ket{0}_A \inprod{b}{0}_B + \ket{1}_A \inprod{b}{1}_B \Big)
+    \Big( \bra{0}_A \inprod{0}{b}_B + \bra{1}_A \inprod{1}{b}_B \Big)
     \\
-    &= \frac{1}{2} \Big( \Ket{0}_A \Bra{0}_A + \Ket{1}_A \Bra{1}_A \Big)
+    &= \frac{1}{2} \Big( \ket{0}_A \bra{0}_A + \ket{1}_A \bra{1}_A \Big)
     = \frac{1}{2} \hat{I}
 \end{aligned}$$
 
@@ -59,35 +59,36 @@ The same holds for the other three Bell states,
 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 $$\Ket{1}$$.
+has a 50-50 chance to yield $$\ket{0}$$ or $$\ket{1}$$.
 However, due to the entanglement,
 measuring $$A$$ also has consequences for qubit $$B$$:
 
 $$\begin{aligned}
-    \big| \Bra{0}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
-    &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{0}{0}_B + \Inprod{0}{1}_A \Inprod{0}{1}_B \Big)^2
+    \big| \bra{0}_A \! \bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \inprod{0}{0}_A \inprod{0}{0}_B + \inprod{0}{1}_A \inprod{0}{1}_B \Big)^2
     = \frac{1}{2}
     \\
-    \big| \Bra{0}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
-    &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{1}{0}_B + \Inprod{0}{1}_A \Inprod{1}{1}_B \Big)^2
+    \big| \bra{0}_A \! \bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \inprod{0}{0}_A \inprod{1}{0}_B + \inprod{0}{1}_A \inprod{1}{1}_B \Big)^2
     = 0
     \\
-    \big| \Bra{1}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
-    &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{0}{0}_B + \Inprod{1}{1}_A \Inprod{0}{1}_B \Big)^2
+    \big| \bra{1}_A \! \bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \inprod{1}{0}_A \inprod{0}{0}_B + \inprod{1}{1}_A \inprod{0}{1}_B \Big)^2
     = 0
     \\
-    \big| \Bra{1}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
-    &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{1}{0}_B + \Inprod{1}{1}_A \Inprod{1}{1}_B \Big)^2
+    \big| \bra{1}_A \! \bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
+    &= \frac{1}{2} \Big( \inprod{1}{0}_A \inprod{1}{0}_B + \inprod{1}{1}_A \inprod{1}{1}_B \Big)^2
     = \frac{1}{2}
 \end{aligned}$$
 
-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 $$\ket{\Phi^{+}}$$,
 but analogous results can be found for the other Bell states.
 
 
+
 ## References
 1.  J.B. Brask,
     *Quantum information: lecture notes*,
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