Lecture 8: Early Quantum Algorithms III

Lecture 8: Early Quantum Algorithms III#

Warning

These lecture notes are a work in progress and are not a replacement for watching the lecture video, it’s intended to be a supplementary reading after watching the lecture.

Focus of Lecture#

After this lecture you should be able to:

  • Understand Grover’s Algorithm.

  • Have an intuition for the difference between quantum and classical random walks.

Grover’s Algorithm#

Grover’s algorithm is an algorithm for searching through an unstructed database. It ofters a quadratic speed-up over classical approaches.

Grover’s Operator#

For a database of size \(N=2^n\) we require \(n\) qubits. We define the oracle, \(O\), which performs the operation

\[ O|x\rangle = (-1)^{f(x)}|x\rangle \]

thus if \(|x\rangle\) is a solution, \(O|x\rangle=-|x\rangle\).

We also define the phase shift which applies the follow transformation:

\[\begin{split} |x\rangle \rightarrow \begin{cases} |0^n\rangle & \text{if } \, x=0^n \\ -|x\rangle & \text{otherwise} \\ \end{cases} \end{split}\]

where here \(|0^n\rangle\) is the state with all \(n\) qubits in \(|0\rangle\), e.g. for \(n=4\), \(|0^4\rangle=|0000\rangle\). The phase shift corresponds to the following unitary:

\[2|0^n\rangle\langle 0^n|-I\]

where \(I\) is the identity matrix. To see this, note that it correctly transforms \(|0^n\rangle\rightarrow |0^n\rangle\) and for \(x\ne 0^n\), \(|x\rangle\rightarrow -|x\rangle\).

Grover’s operator consists in the following steps:

  1. Apply the oracle \(O\)

  2. Apply a Hadamard gate to each qubit

  3. Apply the phase shift

  4. Apply a Hadamard gate to each qubit

Steps 2-4 are called diffusion and can be combined together into unitary

\[ H^{\otimes n}(2|0^n\rangle\langle 0^n|-I)H^{\otimes n}=2|\psi\rangle\langle\psi|-I \]

where we have defined the uniform superposition state:

\[|\psi\rangle=\frac{1}{2^{n/2}}\sum_x|x\rangle\]

e.g. for \(n=2\), \(|\psi\rangle=\frac{1}{2}(|00\rangle+|01\rangle+|10\rangle+|11\rangle)\)

Thus we can define Grover’s operator, G, as

\[ G = (\color{red} 2|\psi\rangle\langle\psi|-I \color{black})\color{blue} O\]

which comprises \(\color{red}\text{Diffusion}\) and an \(\color{blue}\text{Oracle}\).

Grover’s Circuit#

Grover’s algorithm consists in

  1. Apply a Hadamard gate to each qubit (creating a uniform superposition state)

  2. Apply Grover’s operator \(\sim\sqrt{N}\) times

  3. Meaure the qubits

with a probability of at least \(1/2\) a solution will be measured.

../_images/grovers_circuit_full.png

Fig. 61 Grover’s circuit for \(n=3\), here we apply Grover’s operator twice.#

Why does Grover’s algorithm work?#

We define \(M\) to be the number of solutions (i.e. values of \(x\) for which \(f(x)=1\)).

We define an equal superposition of solutions

\[ |A_1\rangle = \frac{1}{\sqrt M} \sum_{x \,|\, f(x)=1}|x\rangle \]

where we some over all bit strings \(x\) which have \(f(x)=1\), and an equal superposition of non-solutions

\[ |A_0\rangle = \frac{1}{\sqrt{N-M}} \sum_{x \,|\, f(x)=0}|x\rangle. \]

Note that we can write the uniform superposition state \(|\psi\rangle\) in terms of these two states

\[\begin{split} \begin{split} |\psi\rangle &= \frac{1}{\sqrt N} \sum_x |x\rangle \\ &= \frac{1}{\sqrt N} \sum_{x \,|\, f(x)=0} |x\rangle + \frac{1}{\sqrt N} \sum_{x \,|\, f(x)=1} |x\rangle \\ &= \sqrt{\frac{N-M}{N}} |A_0\rangle + \sqrt{\frac{M}{N}} |A_1\rangle \\ &= \sqrt{1-\phi} |A_0\rangle + \sqrt{\phi} |A_1\rangle \\ \end{split} \end{split}\]

where we have defined \(\phi=M / N\).

Consider a linear combination of \(|A_0\rangle\) and \(|A_1\rangle\)

\[ |\alpha\rangle = a_0 |A_0\rangle + a_1 |A_1\rangle \]

note that the state the circuit is in before the first application of Grover’s application is \(|\psi\rangle\) and hence is in the form of \(|\alpha\rangle\).

If we apply Grover’s operator to \(|\alpha\rangle\) we get

\[\begin{split} \begin{split} G|\alpha\rangle &= (2|\psi\rangle\langle \psi|-I)O (a_0 |A_0\rangle + a_1 |A_1\rangle) \\ &= (2|\psi\rangle\langle \psi|-I)(a_0 |A_0\rangle - a_1 |A_1\rangle) \\ &= 2a_0 |\psi\rangle\langle \psi|A_0\rangle - 2 a_1 |\psi\rangle\langle \psi|A_1\rangle - a_0 |A_0\rangle + a_1 |A_1\rangle \\ &= 2a_0 \sqrt{1-\phi} |\psi\rangle - 2 a_1 \sqrt{\phi} |\psi\rangle - a_0 |A_0\rangle + a_1 |A_1\rangle \\ &= \left[a_0(1-2\phi) - 2 a_1 \sqrt{\phi(1-\phi)} \right] |A_0\rangle + \left[a_1(1- 2\phi) + 2a_0 \sqrt{\phi(1-\phi)} \right] |A_1\rangle .\\ \end{split} \end{split}\]

Note that \(0 \le \phi \le 1\) and hence \(-1 \le 1-2\phi \le 1\), this means we can write

\[ 1-2\phi = \cos \theta \]

and hence

\[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (1-2\phi )^2} = 2\sqrt{\phi(1-\phi)}. \]

Thus using this new \(\theta\) variable we get

\[\begin{split} G|\alpha\rangle = \left[a_0\cos \theta - 2 a_1 \sin \theta \right] |A_0\rangle + \left[a_1\cos \theta + 2a_0 \sin \theta \right] |A_1\rangle .\\ \end{split}\]

In terms of the coefficients \(a_0\) and \(a_1\), applying Grover’s operator to \(|\alpha\rangle\) results in the transformation

\[\begin{split} \begin{pmatrix} a_0\\ a_1\\ \end{pmatrix} \rightarrow \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix} \begin{pmatrix} a_0\\ a_1\\ \end{pmatrix}. \end{split}\]

In the 2-D plane with \(|A_0\rangle\) being the x-axis and \(|A_1\rangle\) being the y-axis, this corresponds to an anti-clockwise rotation of angle \(\theta\).

../_images/grover_rotation.drawio.svg

Fig. 62 Applying Grover’s operator to state \(|\alpha\rangle\) results in an anti-clockwise rotation.#

Let’s suppose that the number of solutions \(M\) is small, then \(\phi=M/N\) must also be small. Thus we can approximate \(\phi(1-\phi)\approx \phi\) and hence \(\sin \theta = 2\sqrt{\phi(1-\phi)} \approx 2\sqrt{\phi}\). But if \(\phi\) is small then we can approximate \(\sin \theta\approx \theta\) and hence we end up with

\[ \theta \approx 2\sqrt{\phi} = 2\sqrt{\frac{M}{N}}. \]

Recall that the first step of Grover’s algorithm is to apply a Hadamard gate to each qubit, creating the uniform superposition state \(|\psi\rangle\), if \(M\) is small then \(|\psi\rangle\approx |A_0\rangle\). Thus we start very close to the \(|A_0\rangle\) state, recall that Grover’s operator rotates anticlockwise by angle \(\theta\), if we wish to end up near the \(|A_1\rangle\) we need to rotate by \(90\) degrees or \(\pi/2\) radians. Thus the number of rotations we require, \(k\), is

\[ k = \frac{\pi}{2 \theta} = \frac{\pi}{4} \sqrt{\frac{N}{M}}. \]

Thus we have derived the \(O(\sqrt N)\) scaling of Grover’s algorithm.

Classical Random Walks#

Classical random walks can be used to model:

  • The stock market (Economics)

  • The random motion of molecules (Chemistry)

  • Population dynamics (Biology)

For the simplest random walk we are restricted to move in a 1-dimensional line. At each time step we randomly take a step left or right with equal probability.

If we start at the origin where will we be after \(n\) steps? Let \(p_n(x)\) be the probability of being at position \(x\) after \(n\) steps.

After \(1\) step we either have taken a single step left or right, so the possible paths are L and R and hence:

  • \(p_1(-1)=1/2\)

  • \(p_1(1)=1/2\)

After \(2\) steps the possible paths are LL, LR, RL and RR and hence:

  • \(p_2(-2)=1/4\)

  • \(p_2(0)=1/2\)

  • \(p_2(2)=1/4\)

After \(3\) steps the possible paths are LLL, LLR, LRL, LRR, RLL, RLR, RRL and RRR and hence:

  • \(p_3(-3)=1/8\)

  • \(p_3(-1)=3/8\)

  • \(p_3(1)=3/8\)

  • \(p_3(3)=1/8\)

It can be shown that \(p_n(x)\) has the following properties

  • The mean is zero, \(\mu=\sum_x p_n(x) x=0\)

  • The variance is \(n\), \(\sigma^2=\sum_x p_n(x) (x-\mu)^2=n\)

Additionally when \(n\) is large the distribution tends towards a bell curve.

../_images/random_walk_paths.png

Fig. 63 Some random walk trajectories after 1000 steps.#

Quantum Random Walks - Introduction#

Firstly we need a way of encoding position into a set of qubits. If we have \(n\) qubits then each basis state can serve as a position by converting the binary string into an integer. E.g. the state \(|0110\rangle\) would be position 6, \(|6\rangle\).

For \(n=2\) the positions would be:

Basis State

Position

00

0

01

1

10

2

11

3

Consider two unitaries; one for a ‘step’ to the left and another for a ‘step’ to the right

\[\begin{split} \begin{split} U_L|j\rangle &= |j-1\rangle \\ U_R|j\rangle &= |j+1\rangle \\ \end{split} \end{split}\]

Note that since \(U_LU_R=I\) the two are hermitian conjugates of each other: \(U_L=U_R^\dagger\), \(U_R=U_L^\dagger\) i.e. \(U_L\) undoes \(U_R\).

Thus for example in \(n=2\) case \(U_R |01\rangle = |10\rangle\), i.e. a step from position 1 to 2.

We have an issue though, what will happen when we apply \(U_L\) on the lowest position (\(0\))? Or \(U_R\) to the highest position (\(2^n-1\))?

I.e. in the \(n=2\) case what is

\[\begin{split} \begin{split} U_L|0\rangle &=? \\ U_R|3\rangle &=? \\ \end{split} \end{split}\]

To ensure the operators are unitary we need the positions to be on a circle, i.e.

\[\begin{split} \begin{split} U_L|0\rangle &= |2^n-1\rangle \\ U_R|2^n-1\rangle &= |0\rangle \\ \end{split} \end{split}\]
../_images/ring-quantum-walk.drawio.svg

Fig. 64 The \(U_L\) and \(U_R\) operators on a circle.#

We want to find a unitary that does a ‘walk’ similar to the classical random walk.

Here’s one way to implement a quantum walk. We need a circuit composed of two registers \(|x,y\rangle\) where

  • The first register comprises a single qubit.

  • The second register encodes position (as before).

Consider the unitary \(U\) where

\[\begin{split} \begin{split} U|0, j\rangle &= (I\otimes U_L)|0, j\rangle = |0,j-1\rangle \\ U|1, j\rangle &= (I\otimes U_R)|1, j\rangle = |1,j+1\rangle \\ \end{split}\end{split}\]

i.e. if the qubit in the first register is 0 (1) we step to the left (right).

By itself, this unitary doesn’t do anything very interesting. For example if we start in the state \(|0,j\rangle\) then applying \(U\) \(k\) times results in state \(|0,j-k\rangle\).

Instead, we combine with unitary with a Hadamard gate on the first register: \(U(H\otimes I)\)

\[ U(H\otimes I)|0, j\rangle = \frac{1}{\sqrt 2} \left(U|0, j\rangle + U|1, j\rangle \right) = \frac{1}{\sqrt 2} \left(|0, j-1\rangle + |1, j+1\rangle \right) \]

i.e. \(50\%\) in position \(j-1\) and \(50\%\) in \(j+1\). This looks like a random walk!

Quantum Random Walks - Circuit#

The Quantum random walk algorithm is

  • Prepare an initial state, e.g. \(|0,j\rangle\)

  • Apply \(U(H\otimes I)\) for a given number of steps

  • Measure the second register

  • Find the position by converting the measurement to an integer

../_images/random_walk_circuit.png

Fig. 65 The circuit for Quantum random walks.#

This looks very similar to Grover’s algorithm!

Quantum Random Walks - Results#

../_images/q_walk_asym.png

Fig. 66 Results for initial state: \(|0,0\rangle\)#

../_images/q_walk_sym.png

Fig. 67 Results for initial state: \(\frac{1}{\sqrt 2}(|0,0\rangle+i|1,0\rangle)\)#

Thus we see important differences between classical and quantum random walks (recall \(n\) is the number of steps):

Classical

Quantum

Distribution Symmetric

Always

Sometimes

Distribution Shape

Gaussian

Non-Gaussian

Variance (\(\sigma^2\))

\(n\)

\(\propto n^2\)

  • The fact that the variance of the Quantum distibution is \(n^2\) means the Quantum walks are ‘faster’ than the classical

  • The \(n^2\) variance can be seen by the fact that there are peaks far from the origin.

  • The differences between the classical and quantum cases are caused by interference.

  • Thus we see that for this simple classical algorithm the quantum version displays very different phenomena.

Use cases of quantum random walks include:

  • Finding marked items on a graph faster than classical search (similar to Grover’s algorithm but adapted to spatial structures like grids).

  • Evaluating whether a set of elements are distinct faster than classical algorithms.