Transcript Lecture 2: Techniques for quantum algorithms

Lecture 2: Techniques
for quantum algorithms
Dominic Berry
Macquarie University
Grover’s search algorithm


1996
Problem: Given a function 𝑓(𝑥) mapping 𝑛 bits to 1 bit
𝑓 𝑥 : 0,1 𝑛 → {0,1}, determine the value of 𝑥 such that 𝑓 𝑥 = 1.
Classically we need to evaluate 𝑂 𝑁 values, 𝑁 = 2𝑛 .


1.
2.
Grover’s algorithm enables a search
with 𝑂 𝑁 queries.
Quantum algorithm has two crucial
steps:
Calculate the function on all values
simultaneously.
Reflect about the equal superposition
state.

These steps are repeated 𝑂
times.

This is optimal.
𝑁
Lov Grover
1997
Optimality of Grover’s algorithm

It is not possible to perform any better.

NP is the class of problems that can be verified in
polynomial time. That is, if given 𝜔, you can verify
that 𝑓 𝜔 = 1 efficiently – but it is hard to find 𝜔.

If it were possible to achieve an exponential
speedup for search, it would mean that a quantum
computer can solve NP problems in polynomial
time.
Bennett, Bernstein,
Brassard, Vazirani

It is not possible to search any faster than 𝑁.

NP problems are still fundamentally hard on
quantum computers.
1997
Optimality of Grover’s algorithm
Method of solution:
 Denote operations between oracle calls by 𝑉ℓ . The
general state produced by the sequence of 𝑘 oracle
calls is then
𝑘〉 = 𝑉 𝑈 𝑉
|𝜓𝜔
𝑘 𝑓 𝑘−1 𝑈𝑓 … 𝑉2 𝑈𝑓 𝑉1 𝑈𝑓 𝜓


If we did not call the oracle, then the state would be
|𝜓 𝑘 〉 = 𝑉𝑘 𝑉𝑘−1 … 𝑉2 𝑉1 |𝜓〉
When we sum over all 𝜔 we need large 𝑘 to make
the following quantity large:
𝑁
𝑘〉
|𝜓𝜔
𝑟𝑘 =
−
Bennett, Bernstein,
Brassard, Vazirani
2
𝑘
|𝜓 〉
𝜔=1

We then show that, unless 𝑟𝑘 is large, we cannot
obtain the solution with high probability.
Amplitude amplification

Problem: Given a function 𝑓(𝑥) mapping 𝑛 bits to 1 bit
𝑓 𝑥 : 0,1 𝑛 → {0,1}, and an algorithm that produces a state that
is a superposition of values of 𝑥, obtain a state that is
superposition only of values of 𝑥 such that 𝑓(𝑥) = 1.

The idea is that 𝑓(𝑥) = 1 corresponds to a success, and we want
to amplify the successful component of the state.

Denote the state prepared by the algorithm
𝑁
𝑠 =
𝜓𝑥 |𝑥〉
𝑥=1

Let the 𝑚 success states be 𝜔𝑛 .

If we were just to measure, the probability of obtaining success
would be
𝑝=
𝜓𝑥
2
𝑥∈{𝜔𝑛 }

Classically repeating would require 𝑂(1/𝑝) repetitions.

Amplitude amplification only needs 𝑂(1/ 𝑝).
1997
Brassard & Høyer
Amplitude amplification

Method: Alternate reflections about the prepared state |𝑠〉 and
the solution states |𝜔𝑛 〉.

As for Grover’s algorithm, the reflection about the superposition
state is the oracle
𝑚
𝑈𝑓 = 𝕀 − 2𝑃𝜔 ,
𝑃𝜔 =
𝜔𝑛 〈𝜔𝑛 |
𝑛=1


In Grover’s algorithm, we can denote the operation that
transforms |0〉 to the equal superposition state by 𝑈. This is just
a tensor product of Hadamards.
The reflection is obtained by performing 𝑈 † , reflecting about |0〉,
then performing 𝑈 again.
𝑈𝑠 = 2 𝑠 〈𝑠| − 𝕀 = 𝑈 2 0 0 −𝕀 𝑈 †

Here we can denote the algorithm by a unitary operation 𝑈 that
transforms |0〉 to |𝑠〉.

The reflection is obtained in exactly the same way:
𝑈𝑠 = 𝑈 2 0 0 −𝕀 𝑈 †
1997
Brassard & Høyer
Amplitude amplification
1997
𝑈𝑓 = 𝕀 − 2𝑃𝜔
𝑈𝑠 = 2 𝑠 〈𝑠| − 𝕀

The initial state can be written as
𝑁
𝑠 =
𝜓𝑥 |𝑥〉 =
𝑥=1
𝜓𝑥 |𝑥〉 +
𝑥∉{𝜔𝑛 }
𝜓𝑥 |𝑥〉
𝑥∈{𝜔𝑛 }
Brassard & Høyer
= cos 𝜙 𝑠 ′ + sin 𝜙 |𝜔〉
where sin2 𝜙 = 𝑝,
1
′
𝑠 =
𝜓𝑥 |𝑥〉 ,
1 − 𝑝 𝑥∉{𝜔 }
𝑛
1
𝜔 =
𝑝
𝜓𝑥 |𝑥〉
𝑥∈{𝜔𝑛 }

The action of 𝑈𝑠 on 𝜔 is
𝑈𝑠 𝜔 = (2 𝑠 〈𝑠| − 𝕀) 𝜔
=2𝑠𝜔 𝑠 − 𝜔
= 2 sin 𝜙 𝑠 − 𝜔

This gives
𝑈𝑠 𝜔 = 2 sin 𝜙 cos 𝜙 𝑠′ + sin 𝜙 |𝜔〉 − 𝜔
= 2 sin 𝜙 cos 𝜙 𝑠 ′ − 1 − 2 sin2 𝜙 𝜔
= sin 2𝜙 𝑠 ′ − cos 2𝜙 |𝜔〉
Amplitude amplification
1997
𝑈𝑓 = 𝕀 − 2𝑃𝜔
𝑈𝑠 = 2 𝑠 〈𝑠| − 𝕀
𝑠 = cos 𝜙 𝑠 ′ + sin 𝜙 |𝜔〉
sin2 𝜙 = 𝑝
𝑈𝑠 𝜔 = sin 2𝜙 𝑠 ′ − cos 2𝜙 |𝜔〉

The action of 𝑈𝑠 on 𝑠′ is
𝑈𝑠 𝑠′ = (2 𝑠 〈𝑠| − 𝕀) 𝑠′
= 2 𝑠 𝑠′ 𝑠 − 𝑠 ′
= 2 cos 𝜙 𝑠 − |𝑠 ′ 〉

This gives
𝑈𝑠 𝑠′ = 2 cos 𝜙 cos 𝜙 𝑠′ + sin 𝜙 |𝜔〉 − 𝑠 ′
= 2 cos 2 𝜙 − 1 s′ + 2 sin 𝜙 cos 𝜙 𝜔
= cos 2𝜙 𝑠 ′ + sin 2𝜙 |𝜔〉

Consider the state
𝜓 = cos 𝜃 𝑠′ + sin 𝜃 |𝜔〉

The action of 𝑈𝑓 is
𝑈𝑓 𝜓 = cos 𝜃 𝑠′ − sin 𝜃 |𝜔〉

Then applying 𝑈𝑠 gives 𝑈𝑠 𝑈𝑓 𝜓
= cos 𝜃 cos 2𝜙 𝑠 ′ + sin 2𝜙 |𝜔〉 − sin 𝜃 sin 2𝜙 𝑠′ − cos 2𝜙 |𝜔〉
= cos 𝜃 cos 2𝜙 − sin 𝜃 sin 2𝜙 𝑠 ′ + cos 𝜃 sin 2𝜙 + sin 𝜃 cos 2𝜙 𝜔
= cos(𝜃 + 2𝜙) 𝑠 ′ + sin(𝜃 + 2𝜙) 𝜔
Brassard & Høyer
Amplitude amplification
𝑈𝑓 = 𝕀 − 2𝑃𝜔
𝑈𝑠 = 2 𝑠 〈𝑠| − 𝕀
𝑠 = cos 𝜙 𝑠 ′ + sin 𝜙 |𝜔〉
sin2 𝜙 = 𝑝
𝑈𝑠 𝜔 = sin 2𝜙 𝑠 ′ − cos 2𝜙 |𝜔〉
𝑈𝑠 𝑠′ = cos 2𝜙 𝑠 ′ + sin 2𝜙 |𝜔〉

With the state
𝜓 = cos 𝜃 𝑠′ + sin 𝜃 |𝜔〉
𝑈𝑠 𝑈𝑓 𝜓 = cos(𝜃 + 2𝜙) 𝑠 ′ + sin(𝜃 + 2𝜙) 𝜔

This means that the algorithm gives
𝑈𝑠 𝑈𝑓

𝑘
𝑠 = cos[ 2𝑘 + 1 𝜙] 𝑠 ′ + sin 2𝑘 + 1 𝜙 |𝜔〉
The optimal number of iterations is therefore
𝜋
2𝑘 + 1 ≈
2𝜙
𝜋
𝑘≈
4 𝑝
1997
Brassard & Høyer
State preparation

Consider an alternative formulation of the amplitude amplification
problem.

Our algorithm gives us the state
2000
𝑁
𝑠 =
𝑠𝑥 |𝑥〉 sin 𝜃𝑥 0 + cos 𝜃𝑥 |1〉
𝑥=1
Lov Grover

We want to remove the |1〉 component – i.e., we want the effect
of a measurement on the ancilla with result 0.

We use these two reflections:
𝑈𝑓 = 𝕀 ⊗ (𝕀 − 2|0〉〈0|)
𝑈𝑠 = 2 𝑠 〈𝑠| − 𝕀

This is now identical to standard amplitude amplification. The
oracle is just returning success if the ancilla is in state |0〉.

The number of steps is
~
𝜋
4 𝑝
=
𝜋
4
−1/2
𝑁
𝑠𝑥2 sin2 𝜃𝑥
𝑥=1
State preparation

We want to prepare the state
2000
𝑁
𝜓 =
𝜓𝑥 |𝑥〉
𝑥=1

Consider an oracle that yields 𝑈𝜓 𝑥 |0〉 = |𝑥〉|𝜓𝑥 〉. The
procedure is as follows:
1.
Start with an equal superposition state with two ancillas
𝑁
1
𝑁 𝑥=1
2.
Apply oracle to give
|𝑥〉 0 |0〉
𝑁
𝑥 𝜓𝑥 |0〉
𝑥=1
3.
Apply a rotation on the second ancilla controlled by the state of
the first ancilla to give
𝑁
𝑥 𝜓𝑥
𝑥=1
𝜓𝑥 0 + 1 − 𝜓𝑥 2 |1〉
Lov Grover
2000
State preparation

We want to prepare the state
𝑁
𝜓 =
𝜓𝑥 |𝑥〉
4.
𝑠 =
𝑥=1

Consider an oracle that yields
𝑈𝜓 𝑥 |0〉 = |𝑥〉|𝜓𝑥 〉. The procedure is:
1.
Start with an equal superposition state
with two ancillas
1
𝑁
𝑁 𝑥=1
2.
3.
The number of steps required in the
amplitude amplification is then
𝜋
𝜋 𝑁
~
=
4 𝑝
4

This problem may be used to encode a
search problem; for example, where
𝜓𝜔 = 1 for an unknown 𝜔. That means
that this complexity is optimal for worstcase problems.

For specific problems, the complexity
may be improved upon.
Apply a rotation on the second ancilla
controlled by the state of the first ancilla
1
𝑁
𝑁 𝑥=1
𝑥 𝜓𝑥
𝜓𝑥 0 + 1 − 𝜓𝑥 2 |1〉
𝜓𝑥 0 + 1 − 𝜓𝑥 2 |1〉

𝑁
𝑁 𝑥=1
𝑁 𝑥=1
𝑥 0
The procedure in steps 1 to 4 then
prepares the state |𝑠〉 that can be used
in the amplitude amplification
procedure described earlier.
|𝑥〉 0 |0〉
𝑥 𝜓𝑥 |0〉
1
𝑁
5.
Apply oracle to give
1
Invert the oracle to give
Quantum phase estimation

Often in quantum algorithms, one has a unitary operator 𝑈, and
wants to find an eigenvalue.

Given a state in the form
𝑁
𝜓 =
𝜓𝑘 |𝜆𝑘 〉
𝑘=1
where e𝑖𝜆𝑘 is an eigenvalue of 𝑈 and |𝜆𝑘 〉 is the corresponding
eigenvector, one wants to obtain an estimate, 𝜆𝑘 , of 𝜆𝑘 with
probability 𝜓𝑘 2 .

1995
Alternatively, one wants to place the estimate in an ancillary
register in a coherent way to obtain (approximately) the state
𝑁
𝜓𝑘 |𝜆𝑘 〉|𝜆𝑘 〉
𝑘=1
Alexei Kitaev
Quantum phase estimation

Consider an initial eigenstate 𝜓 = |𝜆〉.

The Hadamard gives ( 0 + |1〉)/ 2.

The controlled 𝑈 ℓ operation gives ( 0 + 𝑒 𝑖ℓ𝜆 |1〉)/ 2.

The state of the set of 𝑛 qubits is
1
2𝑛
|0〉
|0〉
|0〉
|0〉
|0〉
|0〉
|𝜆〉
2𝑛 −1
𝑒 𝑖𝑥𝜆 |𝑥〉
Alexei Kitaev
𝑥=0
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
Inverse
QFT
𝑈
1995
𝑈 2 𝑈 4 𝑈 8 𝑈16 𝑈 32
bits
of 𝜆
1995
Quantum phase estimation

Consider an initial eigenstate 𝜓 = |𝜆〉.

The Hadamard gives ( 0 + |1〉)/ 2.

The controlled 𝑈 ℓ operation gives ( 0 + 𝑒 𝑖ℓ𝜆 |1〉)/ 2.

The state of the set of 𝑛 qubits is
2𝑛 −1
1
2𝑛

Net result is state
𝑁
𝑛
𝑒 −2𝜋𝑖𝑥𝑦/2 𝑦 〈𝑥|
𝜓 =
𝜓𝑘 |𝜆𝑘 〉
𝑘=1
becomes
𝑁
It therefore gives
1
2𝑛

𝑥=0
2𝑛 −1
2𝑛 𝑥,𝑦=0

Alexei Kitaev
The inverse QFT is the operation
1

𝑒 𝑖𝑥𝜆 |𝑥〉
2𝑛 −1
𝑛
𝑒 −2𝜋𝑖𝑥𝑦/2 𝑒 𝑖𝑥𝜆
𝑦
𝑥,𝑦=0
𝜓𝑘 𝜆𝑘 |𝜆𝑘 〉
𝑘=1
For 𝜆 = 2𝜋𝑚/2𝑛 , the sum over 𝑥 is nonzero only for 𝑦 = 𝑚, giving the state |𝑚〉.
That is, for 𝜆 = 2𝜋 × 0. 𝜆1 𝜆2 𝜆3 … 𝜆𝑛 , the output state is
𝜆1 𝜆2 … |𝜆𝑛 〉
Operation conversion

We can implement one unitary 𝑉, but we want to implement a different
operation 𝑂.

The operations share eigenstates, but the eigenvalues are related by a
function.
𝑉 𝜆 = 𝑒 𝑖𝜆 𝜆
𝑂 𝜆 = 𝑓 𝜆 |𝜆〉

Implement the unitary V many times and perform phase estimation to give
𝑁
𝑁
𝜓 =
𝜓𝑘 |𝜆𝑘 〉
𝑘=1

↦
𝜓 𝑘 𝜆𝑘 | 𝜆𝑘 〉
𝑘=1
The eigenvalue 𝑓(𝜆) can then be imposed, giving (ignoring error)
𝑁
𝜓𝑘 𝑓(𝜆𝑘 ) 𝜆𝑘 |𝜆𝑘 〉
𝑘=1

Inverting the phase estimation gives
𝑁
𝜓𝑘 𝑓(𝜆𝑘 ) 𝜆𝑘 = 𝑂|𝜓〉
𝑘=1
Simulation of Hamiltonians

In general, the problem of simulating a physical system is
that of simulating Hamiltonian evolution:
𝑑
𝜓 = −𝑖𝐻(𝑡)|𝜓〉
𝑑𝑡 𝑡

For a constant Hamiltonian, we would have
𝜓𝑡 = 𝑒 −𝑖𝐻𝑡 |𝜓0 〉

A typical physical system consists of many subsystems:
1996
Seth Lloyd

Each subsystem is of limited
dimension 𝑑, but with 𝑛 subsystems
the overall dimension is 𝑑 𝑛 .

A typical Hamiltonian is a sum of local
terms on individual subsystems, and
interaction terms on 2 subsystems.
𝐻=
𝐻𝑘
𝑘
1996
Simulation of Hamiltonians

We want to simulate the evolution
𝜓𝑡 = 𝑒 −𝑖𝐻𝑡 |𝜓0 〉

With a Hamiltonian a sum of terms:
𝐻=
𝐻𝑘
𝑘

Each individual 𝐻𝑘 has low dimension, so could be
efficiently simulated on its own.

Seth Lloyd
That is, we can perform
𝑒 −𝑖𝐻𝑘𝑡

For short times we can use the
approximation that
𝑒 −𝑖𝐻1𝛿𝑡 𝑒 −𝑖𝐻2 𝛿𝑡 … 𝑒 −𝑖𝐻𝐾−1 𝛿𝑡 𝑒 −𝑖𝐻𝐾 𝛿𝑡 ≈ 𝑒 −𝑖𝐻𝛿𝑡
1996
Simulation of Hamiltonians

For short times we can use the approximation that
𝑒 −𝑖𝐻1 𝛿𝑡 𝑒 −𝑖𝐻2𝛿𝑡 … 𝑒 −𝑖𝐻𝐾−1𝛿𝑡 𝑒 −𝑖𝐻𝐾 𝛿𝑡 ≈ 𝑒 −𝑖𝐻𝛿𝑡




This approximation is because
𝑒 −𝑖𝐻1𝛿𝑡 𝑒 −𝑖𝐻2 𝛿𝑡 … 𝑒 −𝑖𝐻𝐾−1 𝛿𝑡 𝑒 −𝑖𝐻𝐾 𝛿𝑡
= 𝕀 − 𝑖𝐻1 𝛿𝑡 + 𝑂 𝛿𝑡 2 𝕀 − 𝑖𝐻2 𝛿𝑡 + 𝑂 𝛿𝑡 2 …
… 𝕀 − 𝑖𝐻𝐾 𝛿𝑡 + 𝑂 𝛿𝑡 2
= 𝕀 − 𝑖𝐻1 𝛿𝑡 − 𝑖𝐻2 𝛿𝑡 … − 𝑖𝐻𝐾 𝛿𝑡 + 𝑂 𝛿𝑡 2
= 𝕀 − 𝑖𝐻𝛿𝑡 + 𝑂 𝛿𝑡 2
= 𝑒 −𝑖𝐻𝛿𝑡 + 𝑂(𝛿𝑡 2 )
If we divide long time 𝑡 into 𝑟 intervals, then
𝑟
𝑒 −𝑖𝐻𝑡 = 𝑒 −𝑖𝐻𝑡/𝑟 = 𝑒 −𝑖𝐻1 𝑡/𝑟 𝑒 −𝑖𝐻2𝑡/𝑟 … 𝑒 −𝑖𝐻𝐾 𝑡/𝑟 + 𝑂 𝑡/𝑟 2
𝑟
−𝑖𝐻
𝑡/𝑟
−𝑖𝐻
𝑡/𝑟
−𝑖𝐻
𝑡/𝑟
1
2
𝐾
= 𝑒
𝑒
…𝑒
+ 𝑂 𝑡 2 /𝑟
Typically, we want to simulate a system with some maximum
allowable error 𝜀.
Then we need
𝑟 ∝ 𝑡 2 /𝜀.
Seth Lloyd
𝑟