From 9578811c148b429289512d56b0a81c462eb944b9 Mon Sep 17 00:00:00 2001
From: augustin64 <me.git@augustin64.fr>
Date: Fri, 27 Sep 2024 09:26:24 +0200
Subject: [PATCH] Add DM2 & DM3.typ

---
 DM2.typ  | 77 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 DM3.typ  | 63 ++++++++++++++++++++++++++++++++++++++++++++++
 Makefile |  2 ++
 3 files changed, 142 insertions(+)
 create mode 100644 DM2.typ
 create mode 100644 DM3.typ
 create mode 100644 Makefile

diff --git a/DM2.typ b/DM2.typ
new file mode 100644
index 0000000..13e5954
--- /dev/null
+++ b/DM2.typ
@@ -0,0 +1,77 @@
+#set page(
+  paper: "a4",
+  header: align(center)[
+    QCS - DM2 - Augustin LUCAS
+  ],
+)
+
+== Q1
+
+We want to *simplify* the following circuit:
+
+#{
+  import "@preview/quill:0.4.0": *
+
+  quantum-circuit(
+    lstick($|v_0〉$), $H$, ctrl(1), $H$, [\ ],
+    lstick($|v_1〉$), $H$, targ(), $H$
+  )
+}
+
+The gates 
+
+#{
+  import "@preview/quill:0.4.0": *
+
+  quantum-circuit(
+    lstick($$), $H$, rstick($$), [\ ],
+    lstick($$), $H$, rstick($$)
+  )
+}
+
+correspond to the following matrice: $1/2 mat(H, H; H, -H;)$
+
+And CNOT = $mat(1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 0, 1; 0, 0, 1, 0;)$
+
+The circuit then corresponds to the following operation:
+
+$A&=1/2 mat(1, 1, 1, 1; 1, -1, 1, -1; 1, 1, -1, -1; 1, -1, -1, 1)
+        mat(1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 0, 1; 0, 0, 1, 0;)
+    1/2 mat(1, 1, 1, 1; 1, -1, 1, -1; 1, 1, -1, -1; 1, -1, -1, 1) \
+
+ &=1/4 mat(4, 0, 0, 0; 0, 0, 0, 4; 0, 0, 4, 0; 0, 4, 0, 0) \
+ &=mat(1, 0, 0, 0; 0, 0, 0, 1; 0, 0, 1, 0; 0, 1, 0, 0)$
+
+Which gives $cases(
+  |00〉 => |00〉,
+  |01〉 => |11〉,
+  |10〉 => |10〉,
+  |11〉 => |01〉
+)$ that corresponds to an *inversed CNOT* that we can denote:
+
+#{
+  import "@preview/quill:0.4.0": *
+
+  quantum-circuit(
+    lstick($|v_0〉$), targ(), rstick($$), [\ ],
+    lstick($|v_1〉$), ctrl(-1), rstick($$)
+  )
+}
+
+
+== Q2
+
+In the *classical case*, we need $2^(n-1)+1$ queries to determine if $f$ is constant.
+Using only $2^n$ queries, all queries could have the same value with $f$ balanced.
+
+In the *quantum version*, we may use the following circuit:
+#{
+  import "@preview/quill:0.4.0": *
+
+  quantum-circuit(
+    lstick($|0〉$), $H$, mqgate($U_f$, n:4), $H$, meter(), setwire(2), rstick($$), [\ ],
+    lstick($|0〉$), $H$, targ(), $H$, meter(), setwire(2), rstick($$), [\ ],
+    lstick($...$), [\ ],
+    lstick($|1〉$), $H$, targ()
+  )
+}
diff --git a/DM3.typ b/DM3.typ
new file mode 100644
index 0000000..270b490
--- /dev/null
+++ b/DM3.typ
@@ -0,0 +1,63 @@
+#set page(
+  paper: "a4",
+  header: align(center)[
+    QCS - DM3 - Augustin LUCAS
+  ],
+)
+#import "@preview/showybox:2.0.1": showybox
+#import "@preview/physica:0.9.3": bra, ket
+
+#showybox(
+  frame: (
+    border-color: blue.darken(50%),
+    title-color: blue.lighten(60%),
+    body-color: blue.lighten(80%)
+  ),
+  title-style: (
+    color: black,
+    weight: "regular",
+    align: center
+  ),
+  shadow: (
+    offset: 3pt,
+  ),
+  title: "Assignment",
+  ([
+    + Starting from state $ket(Phi)= 1/sqrt(2)(ket(00)+ket(11))$, is it true that for any basis ($ket(v_0)$, ket(v_1)), Alice and Bob will get the same outcome ?
+    + Show that, for the state $1/sqrt(2)(ket(01)-ket(10))$, the outcome of Alice and Bob are opposite in any basis ($ket(v_0), ket(v_1)$)
+  ]),
+)
+
++ Let $ket(v_0),ket(v_1)$ be a basis. $ket(v_0)=mat(alpha; beta)$, $ket(v_1)=mat(gamma;delta)$
+
+  The probability of getting the state $ket(v_0v_1)$ is given by:
+
+  $|(bra(v_0) times.circle bra(v_1))ket(Phi)|^2
+  &=1/2|bra(v_0v_1)ket(00)+bra(v_0v_1)ket(11)|^2
+  &=1/2|alpha gamma + beta delta|^2$
+
+  Then, the probability of getting this state is non-null iff $alpha gamma + beta delta eq.not 0$
+
+  $(ket(v_0), ket(v_1))$ being a base, we have $bra(v_0)ket(v_1) = overline(alpha) gamma + overline(beta) delta = 0$
+
+  Any solution of the following system then shows the possibility of a different outcome:
+  $cases(
+    alpha gamma = -beta delta,
+    overline(alpha) gamma eq.not -overline(beta) delta
+  )$
+
+  $ket(v_0)=1/sqrt(2)mat(1;i)$, $ket(v_1)=1/sqrt(2)mat(1;-i)$ is a solution to this system.
+  Then, it is not true that Alice and Bob will get the same outcome in any basis.
+
++ Let $(ket(v_0), ket(v_1))$ be a basis. $ket(v_0)=mat(alpha; beta)$, $ket(v_1)=mat(gamma;delta)$.
+  Let $ket(Phi)=1/sqrt(2)(ket(01)-ket(10))$
+
+  Let's compute the probability of Alice and Bob getting the same outcome $ket(v_0)$ from state $ket(Phi)$.
+
+  $|bra(v_0v_0)ket(Phi)|^2 &= |alpha beta - beta alpha|^2 = 0$
+
+  And similarly, $|bra(v_1v_1)ket(Phi)|^2 &= |gamma delta - delta gamma|^2 = 0$
+
+  Then, the outcome of Alice and Bob's measurements are opposite in any basis.
+
+
diff --git a/Makefile b/Makefile
new file mode 100644
index 0000000..578ea82
--- /dev/null
+++ b/Makefile
@@ -0,0 +1,2 @@
+clean:
+	rm -f *.aux *.log *.pdf