#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.