QCS/DM3.typ

#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.