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