Add DM5.typ

DM4.typ

@@ -1,31 +1,16 @@
+#import "qcs.typ": *
 #set page(
   paper: "a4",
   header: align(center)[
     QCS - DM4 - Augustin LUCAS
   ],
 )
-#import "@preview/showybox:2.0.1": showybox
-#import "@preview/physica:0.9.3": bra, ket
 
-#showybox(
+#question("Assignment Q1",
-  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 Q1",
   ([
     + Let $U=mat(u_(00),u_(01);u_(10),u_(11))$. Write a matrix representation of $U[1]$ and $U[2]$ for $n=2$.
     + For $n=3$, write a matrix representation of CNOT$[3,1]$.
-  ]),
+  ])
 )
 
 + $U[1] &= U times.circle I
@@ -64,21 +49,7 @@
   )$
 
 
-#showybox(
+#question("Assignment Q2",
-  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 Q2",
   ([
     Let $A=1/sqrt(2) mat(-i,-1;1,i)$ and $B=mat(0,1;-1,0)$.
     Which 2-qubit gate can you apply on the first qubits at the end of the circuit

DM5.typ

@@ -0,0 +1,105 @@
+#import "qcs.typ": *
+#set page(
+  paper: "a4",
+  header: align(center)[
+    QCS - DM5 : Phase estimation - Augustin LUCAS
+  ],
+)
+
+
+#question("Assignment Q1", [
+  Show that after applying the phase estimation circuit described in class,
+  we obtain (in the first register) the state
+
+  $ 1/2^m sum_(j=0)^(2^m-1)(sum_(k=0)^(2^m-1)e^(2 pi i k (theta-j 2^(-m)))) $
+])
+
+The phase estimation circuit described in class is:
+
+#{
+  import "@preview/quill:0.4.0": *
+
+  quantum-circuit(
+    lstick($ket(0)$), $H$, slice(label: $ket(Phi_0)$), ctrl(3, show-dot: false),
+    slice(label: $ket(Phi_1)$), mqgate($Q F T_(2^m)^dagger$, n:3), slice(label: $ket(Phi_2)$, n:3), [\ ],
+    setwire(0), midstick($dots.v$), midstick($dots.v$), 2, midstick($dots.v$), [\ ],
+    lstick($ket(0)$), $H$, 1, [\ ],
+    lstick($ket(psi)$), 1, $U$, 3, rstick($$)
+  )
+}
+
+First, $ ket(Phi_0) = ket(+)^(times.circle m) times.circle ket(phi) $
+Then, $ ket(Phi_1) &= gamma_m (U) ket(Phi_0) \
+        &= gamma_m (U) (ket(+)^(times.circle m) times.circle ket(phi)) \
+        &= gamma_m (U) (((ket(0)+ket(1))/sqrt(2))^(times.circle m) times.circle ket(phi)) \
+        &= gamma_m (U) 1/sqrt(2^m) sum_(k=0)^(2^m-1) ket(k) times.circle ket(phi) \
+        &= 1/sqrt(2^m) sum_(k=0)^(2^m-1) gamma_m (U) (ket(k) times.circle ket(phi)) \
+        &= 1/sqrt(2^m) sum_(k=0)^(2^m-1) ket(k) times.circle U^k ket(phi) \
+        &= 1/sqrt(2^m) sum_(k=0)^(2^m-1) e^(2 pi i theta k) ket(k) times.circle ket(phi) $
+
+
+
+With $Q F T_(2^m)^dagger = 1/sqrt(2^m) mat(
+  1, 1, ..., 1;
+  1, omega, ..., omega^(2^m-1);
+  dots.v, dots.v, dots.down, dots.v;
+  1, omega^(2^m-1), ..., (omega^(2^m-1))^(2^m-1);
+)$, where $omega =  e^((2 i pi)/2^m)$
+
+we have $ ket(Phi_2) &=  Q F T_(2^m)^dagger ket(Phi_1) \
+        &= 1/sqrt(2^m) sum_(k=0)^(2^m-1) e^(2 pi i theta k) Q F T_(2^m)^dagger ket(k) \
+        &= 1/sqrt(2^m) sum_(k=0)^(2^m-1) e^(2 pi i theta k )
+        (1/sqrt(2^m) sum_(j=0)^(2^m-1) e^(-2 pi i k j 2^(-m)) ket(j)) \
+        &= 1/2^m sum_(j=0)^(2^m-1)
+        (sum_(k=0)^(2^m-1) e^(2 pi i k (theta-2^(-m)j))) ket(j) $
+
+
+#question("Assignment Q2", [
+  We now measure this state in the standard basis.
+  Compute the probability $p_j$ of obtaining outcome $j$.
+])
+
+$ p_j &= abs(braket(Phi_2, j))^2 \
+  &= abs(braket(1/2^m sum_(l=0)^(2^m-1)
+    (sum_(k=0)^(2^m-1) e^(2 pi i k (theta-2^(-m)l))) l, j))^2 \
+  &= abs( 1/2^m sum_(l=0)^(2^m-1) sum_(k=0)^(2^m-1)
+      e^(2 pi i k (theta-2^(-m)l)) braket(l, j) )^2 \
+  &= abs( 1/2^m sum_(k=0)^(2^m-1)
+      e^(2 pi i k (theta-2^(-m)j)) braket(j, j) )^2 \
+  &= abs( 1/2^m sum_(k=0)^(2^m-1)
+      (e^(2 pi i (theta-2^(-m)j)))^k)^2 \
+  &= 1/2^(2m) abs(
+      ( 1-e^(2 pi i (2^m theta - j)) )
+      /(1-e^(2 pi i (theta - 2^(-m)j)))
+    )^2 $
+
+#question("Assignment Q3", [
+  Suppose we are aiming for a precision of $t<m$ bits, i.e.,
+  we would like the outcome to be some $j$ such that $abs(theta - j 2^(-m)) <= 2^(-t-1)$.
+  Compute a lower bound on the probability of obtaining such an outcome.
+  The bound should be such that if $t$ is fixed and $m$ grows, the probability of success goes to $1$.
+
+  For this you may use the following inequality without proof :
+  for $gamma in [-pi, pi], abs(1-e^(i gamma)) >= 2/pi abs(gamma)$.
+])
+
+/*
+$ bb(P)(abs(theta - j 2^(-m)) <= 2^(-t-1)) &= bb(P)(theta in ]j 2^(-m)-2^(-t-1), j 2^(-m)+2^(-t-1)[) \
+  &= bb(P)(union_(k=j 2^(-m)-2^(-t-1))^(j 2^(-m)+2^(-t-1))(theta=k)) \
+  &= sum_(k=j 2^(-m)-2^(-t-1))^(j 2^(-m)+2^(-t-1))bb(P)(theta=k) \
+  &= sum_(k=0)^(2^(-t))bb(P)(theta=k+j 2^(-m)-2^(-t-1)) \
+  &= sum_(k=0)^(2^(-t))p_(k+j 2^(-m)-2^(-t-1)) \
+  &= 1/2^(2m)sum_(k=0)^(2^(-t)) 
+      abs( 1-e^(2 pi i (2^m theta - j)) )^2
+      /abs(1-e^(2 pi i (theta - 2^(-m)j)))^2 \
+  &<= 1/2^(2m)sum_(k=0)^(2^(-t)) 
+      abs( 1-e^(2 pi i (2^m theta - j)) )^2
+      /(2/pi abs(2 pi (theta - 2^(-m)j)))^2 \
+  &= 1/2^(2m)sum_(k=0)^(2^(-t)) 
+      abs( e^(pi i (2^m theta - j))(e^(pi i (2^m theta - j)) - e^(-pi i (2^m theta - j))) )^2
+      /(2/pi abs(2 pi (theta - 2^(-m)j)))^2 \
+  &= 1/2^(2m)sum_(k=0)^(2^(-t)) 
+      abs(2 i sin(pi i (2^m theta -j)) e^(pi i (2^m theta - j)) )^2
+      /(2/pi abs(2 pi (theta - 2^(-m)j)))^2
+
+  $*/

qcs.typ

@@ -0,0 +1,22 @@
+#import "@preview/showybox:2.0.1": showybox
+#import "@preview/physica:0.9.3": *
+
+#let question(title, content) = {
+  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: title,
+    (content)
+  )
+}