Exercises for Chapter 13 297The above figure shows a particular QEC scheme.
On the left-hand side, Qubit 1 starts in|x〉and
Qubits 2 and 3 both start in| 0 〉. The letters (a) to
(d) indicate the part of this question correspond-
ing to each stage.(a) To represent the qubit|x〉=a| 0 〉+b| 1 〉this
quantum error correction code uses Ψin =
a| 000 〉+b| 111 〉. Show that two CNOT gates
acting on Qubits 2 and 3, with Qubit 1 as the
control qubit in both cases, as indicated in the
figure, encode|x〉in this way.
(b) Suppose that a perturbation of Qubit 1 causes
a bit-change error so that the state of the
three qubits becomes α(a| 000 〉+b| 111 〉)+
β(a| 100 〉+b| 011 〉). Hereβis the amplitude
of the unwanted state mixed into the origi-
nal state (and normalisation determinesα).
Show that after another two CNOT gates we
get (a| 0 〉+b| 1 〉)α| 00 〉+(a| 1 〉+b| 0 〉)β| 11 〉.(c) The measurement of the states of Qubits 2
and 3 has two possible results. One possibility
is to find| 11 〉, which means that Qubit 1 has
changed and in this case the error is corrected
by applying a (conditional) NOT operation to
Qubit 1, i.e.| 0 〉↔| 1 〉if and only if the mea-
surement of the other qubits gives| 11 〉.What
is the other possible state of Qubits 2 and 3
resulting from the measurement? Verify that
Qubit 1 ends up in the original state|x〉after
this stage.
(d) Qubits 2 and 3 are both reset to| 0 〉.Twofur-
ther CNOT gates recreate Ψinfrom|x〉,just
as in stage (a).This scheme corrects ‘bit flip’ errors in any of the
qubits. Show this by writing out what happens in
stages (b), (c) and (d) for the stateα(a| 000 〉+b| 111 〉)+β(a| 100 〉+b| 011 〉)
+γ(a| 010 〉+b| 101 〉)+δ(a| 001 〉+b| 110 〉).Web site:
http://www.physics.ox.ac.uk/users/foot
This site has answers to some of the exercises, corrections and other supplementary information.