296 Quantum computing
(d) The operatorUˆZcorresponds to| 0 〉→| 0 〉and
| 1 〉→−| 1 〉, for each qubit. Write down the
resulting state of the system of two qubits.
(e) Show that after anotherUˆCROTthe state is
| 00 〉−| 01 〉−| 10 〉+| 11 〉.
(f) Show that a final Hadamard transformation
yields the required answer.
(g) Repeat the algorithm for another function
with solutionx= 10, to show that it picks
out the required value, i.e.f(x) corresponds
to| 10 〉→−| 10 〉.Comment.This algorithm has a similar ‘one-way’
nature to the factorisation of numbers mentioned
in the text—it is very time consuming to find the
prime factors, but once possible solutions have
been found by the quantum computation it is easy
to verify whether they are indeed factors (or satisfy
the equation) using a classical computer. Some
more complex quantum algorithms do not always
give the ‘correct’ answer but they are still useful
in ‘winnowing the wheat from the chaff’; the al-
gorithms pick out all the required answers (grains
of wheat) along with a few unwanted numbers.
This is not a problem since invalid numbers are
thrown away after checking—the set of numbers
to be checked is much smaller than the initial set
of all possibilities, so the procedure is efficient.(13.5)The quantum Zeno effect
The ancient Greek philosopher Zeno proposed var-
ious arguments against motion, e.g.Achilles and
the tortoisein which the logic leads to the para-
doxical conclusion that the man cannot overtake
the tortoise. In the quantum Zeno effect the evo-
lution of a wavefunction is slowed down by re-
peated quantum measurements on the system, and
this has been described by the colloquial phrase ‘a
watched pot never boils’.
This question is based on the usual treatment of
Rabi oscillations rewritten in terms of the states of
a qubit that starts in| 0 〉at timet= 0. Transitions
between the states are induced by a perturbation
which is characterised by the Rabi frequency Ω:
|ψ〉=cos(
Ωt
2)
| 0 〉−isin(
Ωt
2)
| 1 〉.(a) The state of the qubit is measured after a short
timeτ 1 /Ω. Show that the probability that
the qubit ends up in| 1 〉is (Ωτ/2)^2 1. What
is the probability that the qubit is in| 0 〉after
the measurement?
(b) The state of the qubit is measured at timeτ/2.
What are now the probabilities of| 0 〉and| 1 〉?
(c) After the measurement at timeτ/2 the qubit
will be in either| 0 〉or| 1 〉. The qubit then
evolves for timeτ/2 and its state is measured
again at timeτ. Calculate the probability of
the outcome| 1 〉.
(d) For a sequence ofnmeasurements with a time
interval ofτ/nbetween them, the probability
of having made a transition from| 0 〉to| 1 〉at
timeτis 1/ntimes that in part (a). Verify
this result forn= 3 or, if you can, justify it
for the general case.
(e) Discuss the application of these results to the
measurement of the frequency of a narrow
transition by the quantum jump technique, as
described in Section 12.6. The transition rate
on the weak transition decreases as the mea-
surement periods of excitation on the strong
transition become more frequent. What effect
does this have on the measured line width of
the narrow transition?(13.6)Quantum error correction
Classical computers encode each number with
more than the minimum number of bits neces-
sary; the additional ‘check’ bits allow the system
to detect whether any of the bits has undergone a
random change (caused by noise). In these error
correction codes the binary code for each number
differs from the string of 0s and 1s for any other
possible number by at least two bits, so a random
change of a single bit leads to an invalid binary
code that the computer rejects. In practice, error
correction codes have the binary strings represent-
ing valid entries ‘further apart’ to protect against
errors in several bits. Quantum error correction
(QEC) uses extra qubits to make the computa-
tional states ‘further apart’ (in Hilbert space), so
the system is more robust, i.e. it is harder for the
computational states to become mixed.