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.)