Lecture 2: Techniques for quantum algorithms
Macquarie University

1996
Grover’s search algorithm
 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 = 2 𝑛
.

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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 𝑁 times.
Lov Grover
 This is optimal .

Optimality of Grover’s algorithm
1997
 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.

Optimality of Grover’s algorithm
1997
Method of solution:
 Denote operations between oracle calls by 𝑉 ℓ
. The general state produced by the sequence of 𝑘 oracle calls is then
|𝜓 𝑘 𝜔
⟩ = 𝑉 𝑘
𝑈 𝑓
𝑉 𝑘−1
𝑈 𝑓
… 𝑉
2
𝑈 𝑓
𝑉
1
𝑈 𝑓 𝜓


If we did not call the oracle, then the state would be
|𝜓 𝑘 ⟩ = 𝑉 𝑘
𝑉 𝑘−1
… 𝑉
2
𝑉
1
|𝜓⟩
When we sum over all 𝜔 we need large 𝑘 to make the following quantity large:
𝑁 𝑟 𝑘
= 𝜔=1
|𝜓 𝑘 𝜔
⟩ − |𝜓 𝑘 ⟩
2
Bennett, Bernstein,
Brassard, Vazirani
 We then show that, unless 𝑟 𝑘 is large, we cannot obtain the solution with high probability.

1997
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 .

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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 = 𝜓 𝑥
|𝑥⟩ 𝑥=1
Let the 𝑚 success states be 𝜔 𝑛
.
If we were just to measure, the probability of obtaining success would be 𝑝 = 𝜓 𝑥
2 𝑥∈{𝜔 𝑛
}
Classically repeating would require 𝑂(1/𝑝) repetitions.
Amplitude amplification only needs 𝑂(1/ 𝑝) .
Brassard & Høyer

1997
Amplitude amplification
 Method: Alternate reflections about the prepared state |𝑠⟩ and the solution states |𝜔 𝑛
⟩ .
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
As for Grover’s algorithm, the reflection about the superposition state is the oracle 𝑚
𝑈 𝑓
= 𝕀 − 2𝑃 𝜔
, 𝑃 𝜔
= 𝜔 𝑛
⟨𝜔 𝑛
| 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.
𝑈 𝑠
= 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:
𝑈 𝑠
= 𝑈 2 0 0 −𝕀 𝑈
†
Brassard & Høyer

1997
Amplitude amplification



𝑈 𝑠
𝑈 𝑓
= 𝕀 − 2𝑃 𝜔
= 2 𝑠 ⟨𝑠| − 𝕀
The initial state can be written as
𝑁 𝑠 = 𝑥=1 𝜓 𝑥
|𝑥⟩ = 𝜓 𝑥
|𝑥⟩ + 𝑥∉{𝜔 𝑛
}
= cos 𝜙 𝑠 ′ 𝑥∈{𝜔 𝑛
}
+ sin 𝜙 |𝜔⟩ 𝜓 𝑥
|𝑥⟩ where sin
2 𝑠 ′ = 𝜙 = 𝑝 ,
1
1 − 𝑝 𝑥∉{𝜔 𝑛
} 𝜓 𝑥
|𝑥⟩ , 𝜔 =
1 𝑝 𝑥∈{𝜔 𝑛
} 𝜓 𝑥
|𝑥⟩
The action of 𝑈 𝑠 on 𝜔 is
𝑈 𝑠 𝜔 = (2 𝑠 ⟨𝑠| − 𝕀) 𝜔
= 2 𝑠 𝜔 𝑠 − 𝜔
= 2 sin 𝜙 𝑠 − 𝜔
This gives
𝑈 𝑠 𝜔 = 2 sin 𝜙 cos 𝜙 𝑠′ + sin 𝜙 |𝜔⟩ − 𝜔
= 2 sin 𝜙 cos 𝜙 𝑠 ′ − 1 − 2 sin 2 𝜙 𝜔
= sin 2𝜙 𝑠
′
− cos 2𝜙 |𝜔⟩
Brassard & Høyer

1997
Amplitude amplification

𝑈 𝑠
𝑈 𝑓
= 𝕀 − 2𝑃 𝜔
𝑈 𝑠
= 2 𝑠 ⟨𝑠| − 𝕀 𝑠 = cos 𝜙 𝑠 ′ + sin 𝜙 |𝜔⟩ sin
2 𝜙 = 𝑝 𝜔 = sin 2𝜙 𝑠 ′ − cos 2𝜙 |𝜔⟩
The action of 𝑈 𝑠 on 𝑠′ is
𝑈 𝑠 𝑠′ = (2 𝑠 ⟨𝑠| − 𝕀) 𝑠′
= 2 𝑠 𝑠′ 𝑠 − 𝑠 ′
= 2 cos 𝜙 𝑠 − |𝑠
′
⟩
 This gives
𝑈 𝑠 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 𝑈 𝑓 is 𝑈 𝑓 𝜓 = cos 𝜃 𝑠′ − sin 𝜃 |𝜔⟩
 Then applying 𝑈 𝑠
= cos 𝜃 cos 2𝜙 𝑠 ′ gives 𝑈 𝑠
𝑈 𝑓 𝜓
+ 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

1997
Amplitude amplification


𝑈 𝑠
𝑈 𝑠
𝑈 𝑓
= 𝕀 − 2𝑃 𝜔
𝑈 𝑠
= 2 𝑠 ⟨𝑠| − 𝕀 𝑠 = cos 𝜙 𝑠 ′ + sin 𝜙 |𝜔⟩ sin
2 𝜙 = 𝑝 𝜔 = sin 2𝜙 𝑠 ′ − cos 2𝜙 |𝜔⟩ 𝑠′ = cos 2𝜙 𝑠
′
+ sin 2𝜙 |𝜔⟩
With the state 𝜓 = cos 𝜃 𝑠′ + sin 𝜃 |𝜔⟩
𝑈 𝑠
𝑈 𝑓 𝜓 = cos(𝜃 + 2𝜙) 𝑠 ′ + sin(𝜃 + 2𝜙) 𝜔
This means that the algorithm gives 𝑘
𝑈 𝑠
𝑈 𝑓 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
Brassard & Høyer

2000
State preparation
 Consider an alternative formulation of the amplitude amplification problem.
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Our algorithm gives us the state
𝑁 𝑠 = 𝑠 𝑥
|𝑥⟩ sin 𝜃 𝑥
0 + cos 𝜃 𝑥
|1⟩ 𝑥=1
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:
𝑈 𝑓
= 𝕀 ⊗ (𝕀 − 2|0⟩⟨0|)
𝑈 𝑠
= 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
−1/2
~ 𝜋
4 𝑝
= 𝜋
4
𝑁 𝑠
2 𝑥 sin
2 𝜃 𝑥 𝑥=1
Lov Grover

2000
State preparation


1.
2.
3.
We want to prepare the state
𝑁 𝜓 = 𝜓 𝑥
|𝑥⟩ 𝑥=1
Consider an oracle that yields 𝑈 𝜓 procedure is as follows: 𝑥 |0⟩ = |𝑥⟩|𝜓 𝑥
⟩ . The
Start with an equal superposition state with two ancillas
𝑁
1
|𝑥⟩ 0 |0⟩
𝑁 𝑥=1
Apply oracle to give
𝑁 𝑥 𝜓 𝑥 𝑥=1
|0⟩
Apply a rotation on the second ancilla controlled by the state of the first ancilla to give
𝑁 𝜓 𝑥
0 + 1 − 𝜓 𝑥
2 |1⟩ 𝑥=1 𝑥 𝜓 𝑥
Lov Grover

2000
State preparation


1.
2.
3.
We want to prepare the state
𝑁 𝜓 = 𝜓 𝑥
|𝑥⟩ 𝑥=1
Consider an oracle that yields
𝑈 𝜓 𝑥 |0⟩ = |𝑥⟩|𝜓 𝑥
⟩ . The procedure is:
Start with an equal superposition state with two ancillas
𝑁
1
𝑁 𝑥=1
|𝑥⟩ 0 |0⟩
Apply oracle to give
𝑁
1 𝑥 𝜓 𝑥
𝑁 𝑥=1
|0⟩
Apply a rotation on the second ancilla controlled by the state of the first ancilla
𝑁
1 𝑥 𝜓 𝑥 𝜓 𝑥
0 + 1 − 𝜓 𝑥
2 |1⟩
𝑁 𝑥=1
4.
Invert the oracle to give
𝑁
1 𝑠 = 𝑥 0 𝜓 𝑥
0 + 1 − 𝜓 𝑥
𝑁 𝑥=1
2 |1⟩
5.
The procedure in steps 1 to 4 then prepares the state |𝑠⟩ that can be used in the amplitude amplification procedure described earlier.
 The number of steps required in the amplitude amplification is then 𝜋
~
4 𝑝
= 𝜋 𝑁
4

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This problem may be used to encode a search problem; for example, where 𝜓 𝜔
= 1 for an unknown 𝜔 . That means that this complexity is optimal for worstcase problems.
For specific problems, the complexity may be improved upon.

1995
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 = 𝜓 𝑘
|𝜆 𝑘
⟩ 𝑘=1 where e 𝑖𝜆 𝑘 eigenvector, one wants to obtain an estimate,
𝜆 𝑘 probability is an eigenvalue of 𝑈 and |𝜆 𝑘 𝜓 𝑘
2
.
⟩ is the corresponding
, of 𝜆 𝑘 with
 Alternatively, one wants to place the estimate in an ancillary register in a coherent way to obtain (approximately) the state
𝑁 𝜓 𝑘
|𝜆 𝑘
⟩| 𝜆 𝑘
⟩ 𝑘=1
Alexei Kitaev

bits of
𝜆
1995
Quantum phase estimation

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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
2 𝑛
1
−1 𝑒
2 𝑛 𝑥=0 𝑖𝑥𝜆 |𝑥⟩
|0⟩
|0⟩
|0⟩
|0⟩
|0⟩
|0⟩
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
Inverse
QFT
|𝜆⟩ 𝑈 𝑈 2 𝑈 4 𝑈 8 𝑈 16 𝑈 32
Alexei Kitaev

1995
Quantum phase estimation
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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
2 𝑛
−1
1 𝑒 𝑖𝑥𝜆 |𝑥⟩
Alexei Kitaev
It therefore gives
2 𝑛 𝑥=0
The inverse QFT is the operation
2 𝑛
1
−1
2 𝑛 𝑥,𝑦=0 𝑒 −2𝜋𝑖𝑥𝑦/2 𝑛 𝑦 ⟨𝑥|
2 𝑛
−1
1
2 𝑛 𝑥,𝑦=0 𝑒
−2𝜋𝑖𝑥𝑦/2 𝑛 𝑒 𝑖𝑥𝜆 𝑦
Net result is state 𝑘
𝑁 𝜓 = becomes
𝑁 𝑘=1 𝜓 𝑘=1 𝜆 𝑘 𝜓 𝑘
| 𝜆
|𝜆 𝑘
⟩ 𝑘
⟩
For 𝜆 = 2𝜋𝑚/2 𝑛
, the sum over 𝑥 is nonzero only for 𝑦 = 𝑚 , giving the state |𝑚⟩ .
That is, for 𝜆 = 2𝜋 × 0. 𝜆
1 𝜆
2 𝜆
3
… 𝜆 𝑛 𝜆
1
, the output state is 𝜆
2
… |𝜆 𝑛
⟩



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 = 𝜓 𝑘
|𝜆 𝑘
⟩ ↦ 𝜓 𝑘 𝑘=1 𝑘=1 𝜆 𝑘
| 𝜆 𝑘
⟩
The eigenvalue 𝑓(𝜆) can then be imposed, giving (ignoring error)
𝑁 𝜓 𝑘 𝑓(𝜆 𝑘
) 𝜆 𝑘 𝑘=1
| 𝜆 𝑘
⟩
Inverting the phase estimation gives
𝑁 𝜓 𝑘 𝑓(𝜆 𝑘
) 𝜆 𝑘 𝑘=1
= 𝑂|𝜓⟩

1996
Simulation of Hamiltonians



In general, the problem of simulating a physical system is that of simulating Hamiltonian evolution: 𝑑 𝑑𝑡 𝜓 𝑡
= −𝑖𝐻(𝑡)|𝜓⟩
For a constant Hamiltonian, we would have 𝜓 𝑡
= 𝑒 −𝑖𝐻𝑡 |𝜓
0
⟩
A typical physical system consists of many subsystems:
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 = 𝐻 𝑘 𝑘

1996
Simulation of Hamiltonians


We want to simulate the evolution 𝜓 𝑡
= 𝑒 −𝑖𝐻𝑡 |𝜓
0
⟩
With a Hamiltonian a sum of terms:

𝐻 = 𝐻 𝑘 𝑘
Each individual 𝐻 𝑘 has low dimension, so could be efficiently simulated on its own.
 That is, we can perform 𝑒 −𝑖𝐻 𝑘 𝑡
Seth Lloyd
 For short times we can use the approximation that 𝑒 −𝑖𝐻
1 𝛿𝑡 𝑒 −𝑖𝐻
2 𝛿𝑡 … 𝑒 −𝑖𝐻
𝐾−1 𝛿𝑡 𝑒 −𝑖𝐻
𝐾 𝛿𝑡 ≈ 𝑒 −𝑖𝐻𝛿𝑡

1996
Simulation of Hamiltonians
 For short times we can use the approximation that 𝑒 −𝑖𝐻
1 𝛿𝑡 𝑒 −𝑖𝐻
2 𝛿𝑡 … 𝑒 −𝑖𝐻
𝐾−1 𝛿𝑡 𝑒 −𝑖𝐻
𝐾 𝛿𝑡 ≈ 𝑒 −𝑖𝐻𝛿𝑡




This approximation is because 𝑒 −𝑖𝐻
1 𝛿𝑡 𝑒 −𝑖𝐻
2 𝛿𝑡 … 𝑒 −𝑖𝐻
𝐾−1 𝛿𝑡 𝑒 −𝑖𝐻
𝐾 𝛿𝑡
= 𝕀 − 𝑖𝐻
1 𝛿𝑡 + 𝑂 𝛿𝑡 2 𝕀 − 𝑖𝐻
2 𝛿𝑡 + 𝑂 𝛿𝑡 2
… 𝕀 − 𝑖𝐻
𝐾 𝛿𝑡 + 𝑂 𝛿𝑡 2
= 𝕀 − 𝑖𝐻
1 𝛿𝑡 − 𝑖𝐻
2
= 𝕀 − 𝑖𝐻𝛿𝑡 + 𝑂 𝛿𝑡 𝛿𝑡 … − 𝑖𝐻
𝐾
2 𝛿𝑡 + 𝑂 𝛿𝑡 2
= 𝑒 −𝑖𝐻𝛿𝑡 + 𝑂(𝛿𝑡 2 )
…
If we divide long time 𝑡 into 𝑟 intervals, then 𝑒 −𝑖𝐻𝑡 = 𝑒 −𝑖𝐻𝑡/𝑟 𝑟 = 𝑒 −𝑖𝐻
1 𝑡/𝑟 𝑒 −𝑖𝐻
2 𝑡/𝑟 … 𝑒 −𝑖𝐻
𝐾 𝑡/𝑟 + 𝑂 𝑡/𝑟
= 𝑒 −𝑖𝐻
1 𝑡/𝑟 𝑒 −𝑖𝐻
2 𝑡/𝑟 … 𝑒 −𝑖𝐻
𝐾 𝑡/𝑟 𝑟 + 𝑂 𝑡 2 /𝑟
2
Seth Lloyd 𝑟
Typically, we want to simulate a system with some maximum allowable error 𝜀 .
Then we need 𝑟 ∝ 𝑡 2 /𝜀 .