78 lines
1.8 KiB
Plaintext
78 lines
1.8 KiB
Plaintext
#set page(
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paper: "a4",
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header: align(center)[
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QCS - DM2 - Augustin LUCAS
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],
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)
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== Q1
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We want to *simplify* the following circuit:
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#{
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import "@preview/quill:0.5.0": *
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quantum-circuit(
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lstick($|v_0〉$), $H$, ctrl(1), $H$, [\ ],
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lstick($|v_1〉$), $H$, targ(), $H$
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)
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}
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The gates
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#{
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import "@preview/quill:0.5.0": *
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quantum-circuit(
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lstick($$), $H$, rstick($$), [\ ],
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lstick($$), $H$, rstick($$)
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)
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}
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correspond to the following matrice: $1/2 mat(H, H; H, -H;)$
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And CNOT = $mat(1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 0, 1; 0, 0, 1, 0;)$
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The circuit then corresponds to the following operation:
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$A&=1/2 mat(1, 1, 1, 1; 1, -1, 1, -1; 1, 1, -1, -1; 1, -1, -1, 1)
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mat(1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 0, 1; 0, 0, 1, 0;)
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1/2 mat(1, 1, 1, 1; 1, -1, 1, -1; 1, 1, -1, -1; 1, -1, -1, 1) \
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&=1/4 mat(4, 0, 0, 0; 0, 0, 0, 4; 0, 0, 4, 0; 0, 4, 0, 0) \
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&=mat(1, 0, 0, 0; 0, 0, 0, 1; 0, 0, 1, 0; 0, 1, 0, 0)$
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Which gives $cases(
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|00〉 => |00〉,
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|01〉 => |11〉,
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|10〉 => |10〉,
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|11〉 => |01〉
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)$ that corresponds to an *inversed CNOT* that we can denote:
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#{
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import "@preview/quill:0.5.0": *
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quantum-circuit(
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lstick($|v_0〉$), targ(), rstick($$), [\ ],
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lstick($|v_1〉$), ctrl(-1), rstick($$)
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)
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}
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== Q2
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In the *classical case*, we need $2^(n-1)+1$ queries to determine if $f$ is constant.
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Using only $2^n$ queries, all queries could have the same value with $f$ balanced.
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In the *quantum version*, we may use the following circuit:
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#{
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import "@preview/quill:0.5.0": *
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quantum-circuit(
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lstick($|0〉$), $H$, mqgate($U_f$, n:4), $H$, meter(), setwire(2), rstick($$), [\ ],
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lstick($|0〉$), $H$, targ(), $H$, meter(), setwire(2), rstick($$), [\ ],
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lstick($...$), [\ ],
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lstick($|1〉$), $H$, targ()
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)
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}
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