64 lines
2.1 KiB
Plaintext
64 lines
2.1 KiB
Plaintext
#set page(
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paper: "a4",
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header: align(center)[
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QCS - DM3 - Augustin LUCAS
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],
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)
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#import "@preview/showybox:2.0.1": showybox
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#import "@preview/physica:0.9.3": bra, ket
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#showybox(
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frame: (
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border-color: blue.darken(50%),
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title-color: blue.lighten(60%),
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body-color: blue.lighten(80%)
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),
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title-style: (
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color: black,
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weight: "regular",
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align: center
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),
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shadow: (
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offset: 3pt,
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),
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title: "Assignment",
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([
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+ 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 ?
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+ Show that, for the state $1/sqrt(2)(ket(01)-ket(10))$, the outcome of Alice and Bob are opposite in any basis ($ket(v_0), ket(v_1)$)
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]),
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)
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+ Let $ket(v_0),ket(v_1)$ be a basis. $ket(v_0)=mat(alpha; beta)$, $ket(v_1)=mat(gamma;delta)$
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The probability of getting the state $ket(v_0v_1)$ is given by:
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$|(bra(v_0) times.circle bra(v_1))ket(Phi)|^2
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&=1/2|bra(v_0v_1)ket(00)+bra(v_0v_1)ket(11)|^2
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&=1/2|alpha gamma + beta delta|^2$
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Then, the probability of getting this state is non-null iff $alpha gamma + beta delta eq.not 0$
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$(ket(v_0), ket(v_1))$ being a base, we have $bra(v_0)ket(v_1) = overline(alpha) gamma + overline(beta) delta = 0$
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Any solution of the following system then shows the possibility of a different outcome:
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$cases(
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alpha gamma = -beta delta,
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overline(alpha) gamma eq.not -overline(beta) delta
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)$
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$ket(v_0)=1/sqrt(2)mat(1;i)$, $ket(v_1)=1/sqrt(2)mat(1;-i)$ is a solution to this system.
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Then, it is not true that Alice and Bob will get the same outcome in any basis.
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+ Let $(ket(v_0), ket(v_1))$ be a basis. $ket(v_0)=mat(alpha; beta)$, $ket(v_1)=mat(gamma;delta)$.
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Let $ket(Phi)=1/sqrt(2)(ket(01)-ket(10))$
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Let's compute the probability of Alice and Bob getting the same outcome $ket(v_0)$ from state $ket(Phi)$.
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$|bra(v_0v_0)ket(Phi)|^2 &= |alpha beta - beta alpha|^2 = 0$
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And similarly, $|bra(v_1v_1)ket(Phi)|^2 &= |gamma delta - delta gamma|^2 = 0$
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Then, the outcome of Alice and Bob's measurements are opposite in any basis.
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