The Algorithm

Grover’s quantum search algorithm is defined via the two following unitary operations \[ U\ =\ 1-2|x_s\rangle\langle x_s|\,,\quad V\ =\ 1-2|\psi\rangle\langle\psi|\,. \] Here \[ |\psi\rangle\ =\ \frac{1}{\sqrt{N}}\sum_x |x\rangle\,, \] with states \(|x\rangle\) in the computational basis and \(N=2^n\) with \(n\) the number of qubits. \(x_s\) is the index of the element sougth for.

The unitary operator \(U\) is implemented via an oracle function \(f\) performing the following action \[ |x\rangle|q\rangle\ \to\ |x\rangle|q\oplus f(x)\rangle \] with \[ f(x)\ =\ \begin{cases} 1 & x=x_s\,,\\ 0 & \mathrm{ortherwise}\,.\\ \end{cases} \] Thus, the qubit \(q\) is flipped, if \(f(x)=1\).

The quantum circuit for \(U\) looks as follows

For \(V\) it looks like

Example case \(N=4\)

The case \(n=2\) and \(N=2^2=4\) can be implemented as follows: assume \(x_s=2\), thus we need a function \(f(x) = 1\) for \(x=2\) and \(f(x) = 0\) otherwise. This is achieved as follows:

## oracle for n=2 and x_s=2
oracle <- function(x) {
  x <- X(1) * (CCNOT(c(1,2,3)) *(X(1) * x))
  return(x)
}

The following test should return 1 only for \(x=x_s\)

## case |00>=0
x <- oracle(qstate(3))
measure(x, 3)$value

[1] 0

## case |01>=1
x <- oracle(X(1)*qstate(3))
measure(x, 3)$value

[1] 0

## case |10>=2
x <- oracle(X(2)*qstate(3))
measure(x, 3)$value

[1] 1

## case |11>=3
x <- oracle(X(2)*(X(1)*qstate(3)))
measure(x, 3)$value

[1] 0

The unitaries \(U\) and \(V\) for the \(n=2\) can then be implemented as follows

U <- function(x) {
  x <- oracle(x)
  x <- Z(3) * x
  x <- oracle(x)
  return(x)
}
V <- function(x) {
  for(i in c(1:2)) {
    x <- H(i) * x
  }
  x <- X(1) * (X(2) * x)
  x <- CCNOT(c(1,2,3)) * x
  x <- Z(3) * x
  x <- CCNOT(c(1,2,3)) * x
  x <- X(1) * (X(2) * x)
  for(i in c(1:2)) {
    x <- H(i) * x
  }
  return(x)
}

One application of \(V\cdot U\) looks as follows in the quantum circuit picture

\(N=4\) is the special case where the algorithms returns the correct result with certainty after only a single application of \(V\cdot U\). This is demonstrated in the following example

## prepare psi
psi <- H(1) * ( H(2) * qstate(3))
## apply VU
x <- U(psi)
x <- V(x)
x

   ( -1 )   * |010> 

As expected, the first two qubits (the two rightmost ones) of x are equal to \(x_s\) in binary representation. (The phase is not observable.)