Download Lecture 1: Elementary quantum algorithms

Lecture 1: Elementary
quantum algorithms
Dominic Berry
Macquarie University
Quantum simulator

All classical computers work in fundamentally the same
way.

Quantum systems are fundamentally different. The size
of the problem goes up exponentially in the size of the
system, making them hard to simulate.
1982
Richard Feynman

In 1982 this led Feynman to propose quantum
computers. A quantum system should be able to
efficiently simulate another quantum system.

Efficiently means that the resources
used (time and size of the system)
should scale at most polynomially in
the size of the system to be simulated.

Polynomially means as 𝑥, 𝑥 2 , 𝑥 3 , etc.
Universal quantum computer

A classical computer uses bits, e.g.
001100010010011110100001101101110011

Quantum computers replace this with a quantum state
001100010010011110100001101101110011

Each digit is a two-level quantum system – a qubit.
1985
David Deutsch

Quantum computers can be in superpositions of these states
𝑁
|𝜓〉 =
𝜓𝑥 |𝑥〉
𝑥=1

Represent the state as a vector:
𝜓1
𝜓2
𝜓 ≡ 𝜓3
⋮
𝜓𝑁
Universal quantum computer

Operations on the state can be represented as matrices:
𝑈11 ⋯ 𝑈1𝑁
⋱
⋮
𝑈≡ ⋮
𝑈𝑁1 ⋯ 𝑈𝑁𝑁

The allowed operations on this state are unitary matrices. These satisfy
𝑈†𝑈 = 𝕀

The † indicates the Hermitian conjugate – the transpose and complex
conjugate of the matrix.

Any allowed operation is reversible. Using 𝑈 † on 𝑈|𝜓〉 will restore the
state |𝜓〉.

A special case is permutation matrices – a permutation matrix only
changes one computational basis state to another.
Examples of permutations
1.
The NOT operation, also called an 𝑋, acts on a single qubit
0 1
𝑋≡
1 0
It takes |0〉 to |1〉, and vice versa.
2.
The controlled NOT acts on two qubits. It performs an 𝑋 on the target
qubit if the control qubit is in the state |1〉. Otherwise it does nothing.
1
0
0
0
3.
0
1
0
0
0
0
0
1
0
0
1
0
The Toffoli gate is a NOT controlled by two qubits. It performs an 𝑋
only if both controls are in the state |1〉.
1 0 0 0 0 0 0 0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
1
0
Turning classical into quantum

The NAND gate is universal for classical circuits and acts as below.
𝐴
𝐵


𝑄
𝑨
𝑩
𝑸
0
0
1
0
1
1
1
0
1
1
1
0
We can perform the same operation using a Toffoli gate.
|𝐴〉
|𝐴〉
|𝐵〉
|𝐵〉
|1〉
|𝑄〉
We can convert any classical algorithm into a quantum algorithm,
replacing the NAND gates with Toffolis, and keeping the extra qubits.
Turning classical into quantum

Any function we could calculate on a classical computer we could also
compute on a quantum computer, with the addition of some extra
“garbage” registers.
𝑥 0 ↦ 𝑓 𝑥 |extra data(𝑥)〉

This can be turned into a computation that does this:
𝑈𝑓 𝑥 0 = 𝑥 |𝑓 𝑥 〉

Method:
Add an ancilla, and perform the program as above to give
𝑓 𝑥 |0〉 extra data(𝑥) .
Copy the output of the function into the extra ancilla, to give
𝑓 𝑥 |𝑓(𝑥)〉 extra data(𝑥) .
Invert the program above, giving 𝑥 |𝑓 𝑥 〉 0 .
1.
2.
3.

For ancillas that are not initially zeroed, we can just perform the XOR.
𝑈𝑓 𝑥 𝑏 = 𝑥 |𝑏 ⊕ 𝑓 𝑥 〉
Part of the story: Superposition

Now what if we start the state in a
superposition?
1
2𝑛
2𝑛
𝑥 |0〉 ↦
𝑥=1
1
2𝑛
2𝑛
𝑥 |𝑓(𝑥)〉
𝑥=1

For 𝑛 qubits, we have evaluated an
exponential number of different values of
the function.

But if we measure it, we only ever find one value of the function!
The other part: Interference


We can’t just calculate many values
of a function, we need some way to
interfere them to obtain some global
property of the function.

Just measuring Schrödinger’s cat
would never show it in a
superposition.

If we could do a Hadamard on
Schrödinger’s cat, we would turn
We need non-permutation gates:
The Hadamard gate has a matrix
representation
1 1 1
𝐻≡
2 1 −1
0 ↔ (|0〉 + |1〉)/ 2
1 ↔ (|0〉 − |1〉)/ 2
1.
( live + |dead〉)/ 2 to live ,
and
2.
( live − |dead〉)/ 2 to dead
showing we have superposition.
The other part: Interference

We can’t just calculate many values of a function, we need some way to
interfere them to obtain some global property of the function.

Universal sets of gates:
1.
The Hadamard gate
1 1
𝐻≡
2 1
1
−1
together with the Toffoli gate is universal for real unitaries.
2.
The 𝜋/8 or 𝑇 gate
1
0
0 𝑒 𝑖𝜋/4
together with the Hadamard and CNOT is universal.
𝑇≡
The Deutsch algorithm

Problem: Given a function 𝑓(𝑥) mapping bits to bits,
determine if it is one-to-one [i.e. 𝑓 0 ≠ 𝑓(1)].

Classically we need to evaluate both 𝑓(0) and 𝑓(1).

Quantum algorithm:
1.
Start with
1985
David Deutsch
0 |1〉
2.
Perform a Hadamard on both qubits to give
1
0 { 0 − 1 } + 1 { 0 − |1〉}
2
3.
Apply the gate mapping 𝑥 |𝑏〉 ↦ 𝑥 |𝑏 ⊕ 𝑓 𝑥 〉 to give
1
0 { 𝑓 0 − 1 ⊕ 𝑓 0 } + 1 { 𝑓 1 − |1 ⊕ 𝑓(1)〉}
2
4.
Apply Hadamard on qubit 2 to give
1
0 { −1 𝑓 0 } + 1 { −1
2
𝑓 1
} |1〉
The Deutsch algorithm
1.
1985
Start with
0 |1〉
2.
Perform a Hadamard on both qubits to give
1
0 { 0 − 1 } + 1 { 0 − |1〉}
2
3.
Apply the gate mapping 𝑥 |𝑏〉 ↦ 𝑥 |𝑏 ⊕ 𝑓 𝑥 〉 to give
David Deutsch
1
0 { 𝑓 0 − 1 ⊕ 𝑓 0 } + 1 { 𝑓 1 − |1 ⊕ 𝑓 (1)〉}
2
4.
Apply Hadamard on qubit 2 to give
1
0 { −1 𝑓 0 } + 1 { −1If𝑓𝑓1 0} |1〉
If 𝑓(0) = 𝑓(1) then we
≠ 𝑓(1) then we
2
only get this part 
only get this part 
5.
Apply
Hadamard
on
qubit
1
to
give
qubit 1 is in state 0 .
qubit 1 is in state 1 .
1
0 + 1
−1 𝑓 0 + { 0 − |1〉} −1 𝑓 1 1
2
6.
Rearranging gives
1
0 −1 𝑓 0 + −1 𝑓 1 + 1 −1 𝑓 0 − −1 𝑓 1 1
2
The Deutsch algorithm

Problem: Given a function 𝑓(𝑥) mapping 1 bit to 1 bit
𝑓 𝑥 : {0,1} → {0,1}, determine if it is either constant or
balanced.

Classically we need to evaluate both 𝑓(0) and 𝑓(1).
1985
David Deutsch

Deutsch quantum algorithm gives result with one function
evaluation.

Quantum algorithm has two crucial steps:
Calculate the function on both input values simultaneously.
Interfere the result.
1.
2.
The Deutsch algorithm

Problem: Given a function 𝑓(𝑥) mapping 1 bit to 1 bit
𝑓 𝑥 : {0,1} → {0,1}, determine if it is either constant or
balanced.

Classically we need to evaluate both 𝑓(0) and 𝑓(1).
1985
David Deutsch

Deutsch quantum algorithm gives result with one function
evaluation.
Calculate the function
on both input values
simultaneously.
|0〉
Interfere the result.
𝐻
𝐻
𝑈𝑓
|1〉
𝐻
𝐻
measurement
The Deutsch-Jozsa algorithm

Problem: Given a function 𝑓(𝑥) mapping 𝑛 bits to 1 bit
𝑓 𝑥 : 0,1 𝑛 → {0,1}, determine if it is either constant or
balanced. There is a promise that it is either.

Classically we need to evaluate 2𝑛−1 + 1 values of 𝑓(𝑥).
1992
David Deutsch

Deutsch-Jozsa quantum algorithm gives result with one
function evaluation.
0
⊗𝑛
𝐻 ⊗𝑛
𝐻 ⊗𝑛
𝑈𝑓
|1〉
𝐻
measurement
Richard Jozsa
Phase oracles

We have previously assumed that the unitary for the function acts as
𝑈𝑓 𝑥 𝑏 = 𝑥 |𝑏 ⊕ 𝑓 𝑥 〉

In both cases we have needed to turn this into a phase in order to
make the algorithm work, i.e.
1
1
𝑈𝑓 𝑥 (|0〉 − |1〉) =
𝑥 (|𝑓(𝑥)〉 − |1 ⊕ 𝑓(𝑥)〉)
2
2
1
=
−1 𝑓 𝑥 |𝑥〉(|0〉 − |1〉)
2

Instead we can just take the oracle to have this form:
𝑈𝑓 𝑥 = −1 𝑓 𝑥 |𝑥〉
The Deutsch-Jozsa algorithm

Deutsch algorithm: start with
2 𝑥=0
2𝑛
−1
𝑓 𝑥
1
2𝑛
1

1
−1
𝑓 𝑥
𝑥=0
−1
𝑦=0
The probability of measuring 𝑦 is
1
4
−1
𝑓 𝑥
−1
𝑦⋅𝑥
𝑥=0
2𝑛 −1
−1
𝑓 𝑥
𝑥
𝑥=0
2𝑛 −1
2𝑛 −1
−1
𝑓 𝑥
𝑥=0
−1
𝑦⋅𝑥 |𝑦〉
𝑦=0
The probability of measuring 𝑦 is
2
1
𝑥=0

𝑥
Hadamards give
1
2𝑛
𝑦⋅𝑥 |𝑦〉
2𝑛 −1
The function evaluation yields
𝑥
A Hadamard gives
1
2

1
2 𝑥=0

1
𝑥
The function evaluation yields
1

Deutsch-Jozsa algorithm: start with
1
1


1992
1
22𝑛
2
2𝑛 −1
−1
𝑥=0
𝑓 𝑥
−1
𝑦⋅𝑥
1993
Fourier sampling
Bernstein,
Vazirani

Problem: Given a function 𝑓(𝑥) mapping 1 bit to 1 bit 𝑓 𝑥 : {0,1} → {0,1}
such that 𝑓 𝑥 = 𝑠 ⋅ 𝑥, determine 𝑠.

The function evaluation yields
2𝑛 −1
1
2𝑛

𝑓 𝑥
𝑥
𝑥=0
Hadamards give
1
2𝑛

−1
2𝑛 −1
2𝑛 −1
−1
𝑓 𝑥
𝑥=0
−1
𝑦⋅𝑥 |𝑦〉
𝑦=0
The probability of measuring 𝑦 is
1
22𝑛
2
2𝑛 −1
−1
𝑥=0
𝑓 𝑥
−1
𝑦⋅𝑥
1
= 2𝑛
2
2
2𝑛 −1
−1
𝑥=0
(𝑦+𝑠)⋅𝑥
= 𝛿𝑦,𝑠
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. We have a promise that the value is unique.

Classically we need to evaluate 𝑂 𝑁 values, 𝑁 = 2𝑛 .
1996
Lov Grover

Grover’s algorithm enables a search with 𝑂

2.
Quantum algorithm has two crucial steps:
Calculate the function on all values simultaneously.
Reflect about the equal superposition state.

These steps are repeated 𝑂
1.
𝑁 times.
𝑁 queries.
Grover’s search algorithm

We start in the state
𝑠 =

1.
2.
1996
1
𝑁
𝑁
𝑥
𝑥=1
We repeat the following ∼ 𝑁 times:
Apply the function calculation 𝑈𝑓 .
Apply the reflection operation 𝑈𝑠 = 2 𝑠 〈𝑠| − 𝕀.

The system will be in the state |𝜔〉 such that 𝑓 𝜔 = 1.

But how do we perform 𝑈𝑠 ?
Lov Grover
Grover’s search algorithm

But how do we perform 𝑈𝑠 ?

First, invert the operation that prepares |𝑠〉 - just
Hadamards on all qubits.

Then, reflect about |0〉; i.e. apply 2 0 〈0| − 𝕀.

Lastly, repeat the operation that prepares 𝑠 .
|1〉
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
𝐻
1996
Lov Grover
Grover’s search algorithm

What does 𝑈𝑠 do?
𝑁
𝑈𝑠
𝑥=1
𝑁
2
𝜓𝑥 𝑥 =
𝑁
=
𝑁
𝑥
𝑥=1
𝑁
2
𝑁
𝑁
〈𝑦|
𝑦=1
𝑁
𝑥
𝑥=1
𝑁
𝜓𝑧 𝑧 −
𝑧=1
𝑁
𝜓𝑦 −
𝑦=1
=
Lov Grover
𝜓𝑥 𝑥
𝑥=1
(2𝜓 − 𝜓𝑥 ) 𝑥
𝑥=1
Each coefficient is reflected about the mean
1
𝜓=
𝑁
𝑁
𝜓𝑦
𝑦=1
𝜓𝑥 𝑥
𝑥=1
𝑁

1996
Grover’s search algorithm


1996
Example: Take 𝑁 = 4, with solution 𝑥 = 3.
We start in the state
𝑠 = 12 (|1〉 + 2 + 3 + |4〉)

Apply the function calculation 𝑈𝑓 , giving
1
2(|1〉 + 2 − 3 + |4〉)

The mean is
Lov Grover
𝜓 = 14(12 + 12 − 12 + 12) = 14
Grover’s search algorithm


Example: Take 𝑁 = 4, with solution 𝑥 = 3.
We start in the state
𝑠 = 12 (|1〉 + 2 + 3 + |4〉)

Apply the function calculation 𝑈𝑓 , giving
1
2(|1〉 + 2 − 3 + |4〉)

The mean is
Lov Grover
𝜓 = 14(12 + 12 − 12 + 12) = 14

1996
Therefore, when we apply the reflection operation 𝑈𝑠 , we get
1
1
(|1〉
+
2
+
3
+
|4〉)
−
(|1〉 + 2 − 3 + |4〉) = |3〉
2
2
Grover’s search algorithm



1996
Example: Take 𝑁 = 9, with solution 𝑥 = 6.
Reflecting target…
The mean is 7/27.
Lov Grover
Grover’s search algorithm






Example: Take 𝑁 = 9, with solution 𝑥 = 6.
Reflecting target…
The mean is 7/27.
Reflecting about mean…
Reflecting target…
The mean is 17/243.
1996
Lov Grover
Grover’s search algorithm







Example: Take 𝑁 = 9, with solution 𝑥 = 6.
Reflecting target…
The mean is 7/27.
Reflecting about mean…
Reflecting target…
The mean is 17/243.
Reflecting about mean…
Solution with
high probability.
1996
Lov Grover
Grover’s search algorithm

Another way of looking at it: Rotations

Define the ket perpendicular to 𝜔 .
1
′
|𝑠 〉 =
|𝑥〉
𝑁 − 1 𝑥≠𝜔
1996
Lov Grover

Then
𝑠 = cos 𝜙 𝑠′ + sin 𝜙 |𝜔〉
with sin 𝜙 = 1/ 𝑁.

The operation 𝑈𝑓 reflects 𝜔 and leaves all orthogonal
states unchanged.
𝑈𝑓 = 𝕀 − 2 𝜔 〈𝜔|

The action of 𝑈𝑓 on 𝑠′ is
𝑈𝑓 𝑠′ = (𝕀 − 2 𝜔 〈𝜔|) 𝑠′ = 𝑠′
Grover’s search algorithm

Another way of looking at it: Rotations

The action of 𝑈𝑓 on 𝑠′ is
𝑈𝑓 𝑠′ = 𝑠′

The action of 𝑈𝑠 on 𝜔 is
𝑈𝑠 𝜔 = (2 𝑠 〈𝑠| − 𝕀) 𝜔
=2 𝑠𝜔 𝑠 − 𝜔
= 2 sin 𝜙 𝑠 − 𝜔
Now use
𝑠 = cos 𝜙 𝑠′ + sin 𝜙 |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〉


1996
Lov Grover
Grover’s search algorithm

Another way of looking at it: Rotations

The action of 𝑈𝑓 on 𝑠′ is
𝑈𝑓 𝑠′ = 𝑠′

The action of 𝑈𝑠 on 𝜔 is
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 ′ 〉
Now use
𝑠 = cos 𝜙 𝑠′ + sin 𝜙 |𝜔〉
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〉



1996
Lov Grover
1996
Grover’s search algorithm

Another way of looking at it: Rotations
𝑈𝑓 𝑠′ = 𝑠′
𝑈𝑠 𝜔 = sin 2𝜙 𝑠′ − cos 2𝜙 |𝜔〉
𝑈𝑠 𝑠′ = cos 2𝜙 𝑠 ′ + sin 2𝜙 |𝜔〉

Consider a state written as
𝜓 = cos 𝜃 𝑠′ + sin 𝜃 |𝜔〉

The action of 𝑈𝑓 on 𝜓 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
Lov Grover
Grover’s search algorithm


Another way of looking at it: Rotations
𝑈𝑓 𝑠′ = 𝑠′
𝑈𝑠 𝜔 = sin 2𝜙 𝑠′ − cos 2𝜙 |𝜔〉
𝑈𝑠 𝑠′ = cos 2𝜙 𝑠 ′ + sin 2𝜙 |𝜔〉
The Grover iteration gives
𝜓 = cos 𝜃 𝑠′ + sin 𝜃 |𝜔〉
′
|𝜔〉 𝑈𝑠 𝑈𝑓 𝜓 = cos(𝜃 + 2𝜙) 𝑠 + sin(𝜃 + 2𝜙) 𝜔
|𝑠〉
𝜙
|𝑠′〉
𝑈𝑓 |𝑠〉
1996
Lov Grover
Grover’s search algorithm


Another way of looking at it: Rotations
𝑈𝑓 𝑠′ = 𝑠′
𝑈𝑠 𝜔 = sin 2𝜙 𝑠′ − cos 2𝜙 |𝜔〉
𝑈𝑠 𝑠′ = cos 2𝜙 𝑠 ′ + sin 2𝜙 |𝜔〉
The Grover iteration gives
𝜓 = cos 𝜃 𝑠′ + sin 𝜃 |𝜔〉
′
|𝜔〉 𝑈𝑠 𝑈𝑓 𝜓 = cos(𝜃 + 2𝜙) 𝑠 + sin(𝜃 + 2𝜙) 𝜔
𝑈𝑠 𝑈𝑓 |𝑠〉
3𝜙
|𝑠〉
𝜙
|𝑠′〉
𝑈𝑓 |𝑠〉
1996
Lov Grover
Grover’s search algorithm


Another way of looking at it: Rotations
𝑈𝑓 𝑠′ = 𝑠′
𝑈𝑠 𝜔 = sin 2𝜙 𝑠′ − cos 2𝜙 |𝜔〉
𝑈𝑠 𝑠′ = cos 2𝜙 𝑠 ′ + sin 2𝜙 |𝜔〉
The Grover iteration gives
𝜓 = cos 𝜃 𝑠′ + sin 𝜃 |𝜔〉
′
|𝜔〉 𝑈𝑠 𝑈𝑓 𝜓 = cos(𝜃 + 2𝜙) 𝑠 + sin(𝜃 + 2𝜙) 𝜔
2
𝑈𝑠 𝑈𝑓 |𝑠〉
𝑈𝑠 𝑈𝑓 |𝑠〉
5𝜙
|𝑠〉
𝜙
|𝑠′〉
𝑈𝑓 |𝑠〉
1996
Lov Grover
Grover’s search algorithm


Another way of looking at it: Rotations
𝑈𝑓 𝑠′ = 𝑠′
𝑈𝑠 𝜔 = sin 2𝜙 𝑠′ − cos 2𝜙 |𝜔〉
𝑈𝑠 𝑠′ = cos 2𝜙 𝑠 ′ + sin 2𝜙 |𝜔〉
The Grover iteration gives
𝜓 = cos 𝜃 𝑠′ + sin 𝜃 |𝜔〉
′
|𝜔〉 𝑈𝑠 𝑈𝑓 𝜓 = cos(𝜃 + 2𝜙) 𝑠 + sin(𝜃 + 2𝜙) 𝜔
3
𝑈𝑠 𝑈𝑓 |𝑠〉
2
𝑈𝑠 𝑈𝑓 |𝑠〉
𝑈𝑠 𝑈𝑓 |𝑠〉
7𝜙
|𝑠〉
𝜙
|𝑠′〉
𝑈𝑓 |𝑠〉
1996
Lov Grover
Grover’s search algorithm


Another way of looking at it: Rotations
𝑈𝑓 𝑠′ = 𝑠′
𝑈𝑠 𝜔 = sin 2𝜙 𝑠′ − cos 2𝜙 |𝜔〉
𝑈𝑠 𝑠′ = cos 2𝜙 𝑠 ′ + sin 2𝜙 |𝜔〉
The Grover iteration gives
𝜓 = cos 𝜃 𝑠′ + sin 𝜃 |𝜔〉
′
|𝜔〉 𝑈𝑠 𝑈𝑓 𝜓 = cos(𝜃 + 2𝜙) 𝑠 + sin(𝜃 + 2𝜙) 𝜔
3
𝑈𝑠 𝑈𝑓 |𝑠〉
4
𝑈𝑠 𝑈𝑓 |𝑠〉
2
𝑈𝑠 𝑈𝑓 |𝑠〉
𝑈𝑠 𝑈𝑓 |𝑠〉
9𝜙
|𝑠〉
𝜙
|𝑠′〉
𝑈𝑓 |𝑠〉
1996
Lov Grover
Grover’s search algorithm


Another way of looking at it: Rotations
𝑈𝑓 𝑠′ = 𝑠′
𝑈𝑠 𝜔 = sin 2𝜙 𝑠′ − cos 2𝜙 |𝜔〉
𝑈𝑠 𝑠′ = cos 2𝜙 𝑠 ′ + sin 2𝜙 |𝜔〉
The Grover iteration gives
𝜓 = cos 𝜃 𝑠′ + sin 𝜃 |𝜔〉
′
|𝜔〉 𝑈𝑠 𝑈𝑓 𝜓 = cos(𝜃 + 2𝜙) 𝑠 + sin(𝜃 + 2𝜙) 𝜔
3
5
𝑈𝑠 𝑈𝑓 |𝑠〉
𝑈𝑠 𝑈𝑓 |𝑠〉
2
𝑈𝑠 𝑈𝑓 |𝑠〉
11𝜙
𝑈𝑠 𝑈𝑓 |𝑠〉
|𝑠〉
𝜙
|𝑠′〉
𝑈𝑓 |𝑠〉
1996
Lov Grover
Grover’s search algorithm


Another way of looking at it: Rotations
In general
𝑈𝑠 𝑈𝑓

1996
𝑘
𝑠 = cos[ 2𝑘 + 1 𝜙] 𝑠 ′ + sin[ 2𝑘 + 1 𝜙] 𝜔
For solution we want
2𝑘 + 1 𝜙 ≈ 𝜋/2
⇒
2𝑘 + 1 arcsin 1/ 𝑁 ≈ 𝜋/2
Lov Grover
|𝜔〉
3
5
𝑈𝑠 𝑈𝑓 |𝑠〉
𝑈𝑠 𝑈𝑓 |𝑠〉
2
𝑈𝑠 𝑈𝑓 |𝑠〉
11𝜙
𝑘opt
𝜋
1
≈
𝑁−
4
2
𝑈𝑠 𝑈𝑓 |𝑠〉
|𝑠〉
𝜙
|𝑠′〉
𝑈𝑓 |𝑠〉
For 𝑁 = 42
𝑘opt ≈ 4.58996 …
1996
Grover’s search algorithm



What about multiple targets?
In general we have 𝑚 values of 𝑥 such that 𝑓(𝑥) = 1.
Let us define
𝑚
1
|𝜔〉 =
|𝜔𝑛 〉
𝑚
𝑛=1

where 𝜔𝑛 are the values of 𝑥 such that 𝑓(𝑥) = 1.
Define a modified |𝑠 ′ 〉
1
′
|𝑠 〉 =
|𝑥〉
𝑁 − 𝑚 𝑥∉{𝜔 }
𝑛

Define projector
𝑚
𝑃𝜔 =
𝜔𝑛 〈𝜔𝑛 |
𝑛=1
Then
𝑠 = cos 𝜙 𝑠′ + sin 𝜙 |𝜔〉

Lov Grover
and reflector
𝑈𝑓 = 𝕀 − 2𝑃𝜔
with sin 𝜙 = 𝑚/𝑁.
The analysis is the same, except with the modified value of 𝜙.
𝑘opt
𝜋
≈
𝑁
4
becomes
𝑘opt
𝜋 𝑁
≈
4 𝑚