Phase Estimation Algorithm

Rotation Matrix

We use a rotation matrix \[ U\ =\ \begin{pmatrix} c & s \\ -s & c \\ \end{pmatrix} \] with \(c=\cos(\alpha)\), \(s=\sin(\alpha)\) and a real-valued angle \(\alpha\) as an example. \(U\) has eigenvalues \[ \lambda_\pm\ =\ c\pm \mathrm{i} s\ =\ e^{\pm i \alpha}\,. \] Thus, \(\phi=\alpha/(2\pi)\). The corresponding eigenvectors are of the form \[ u_\pm\ =\ \begin{pmatrix} 1 \\ \pm\mathrm{i}\\ \end{pmatrix}\,. \]

Phase Estimation

We use

t=6

in the second register which allows us with probability \(1-\epsilon\) to get the correct phase up to \(t-\left\lceil \log\left(2+\frac{1}{2\epsilon}\right)\right\rceil\) digits. Let us choose

epsilon <- 1/4
## note the log in base-2
digits <- t-ceiling(log(2+1/(2*epsilon))/log(2)) 
digits

[1] 4

and therefore expect an error of less than

2^(-digits)

[1] 0.0625

We start with qubit 1 in state \(u_+\)

x <- S(1) * (H(1) * qstate(t+1, basis=""))

and we define the gate corresponding to \(U\)

alpha <- pi*3/7
s <- sin(alpha)
c <- cos(alpha)
## note that R fills the matrix columns first
M <- array(as.complex(c(c, -s, s, c)), dim=c(2,2)) 
Uf <- sqgate(bit=1, M=M, type=paste0("Uf"))

Now we apply the Hadamard gate to qubits 2,,t+1

for(i in c(2:(t+1))) {
  x <- H(i) * x
}

and the controlled \(U_f\)

for(i in c(2:(t+1))) {
  x <- cqgate(bits=c(i, 1),
              gate=sqgate(bit=1,
                          M=M, type=paste0("Uf", 2^(i-2)))) * x
  M <- M %*% M
}
plot(x)

Next we apply the inverse Fourier transform

x <- qft(x, inverse=TRUE, bits=c(2:(t+1)))
plot(x)

\(x\) is now the state \(|\tilde\varphi\rangle|u\rangle\). \(|\tilde\varphi\rangle\) is not necessarily a pure state. The next step is a projective measurement of \(|\tilde\varphi\rangle\)

xtmp <- measure(x)
cbits <- genStateNumber(which(xtmp$value==1)-1, t+1)
phi <- sum(cbits[1:t]/2^(1:t))

cbits[1:t]

[1] 0 0 1 1 1 0

phi

[1] 0.21875

Note that we can measure the complete state, because \(|u\rangle\) is not entangled to the rest. We find that usually

phi-alpha/(2*pi)

[1] 0.004464286

is indeed smaller than the maximal deviation \(2^{-\mathrm{digits}}=\) 0.0625 we expect. The distribution of probabilities over the states in \(|\tilde\varphi\rangle\) is given as follows (factor 2 from dropping \(|u\rangle\))

plot(2*abs(x@coefs[seq(1,128,2)])^2, type="l",
     ylab="p", xlab="state index")

Starting from a random state

The algorithm also works in case the specific eigenvector cannot be prepared. Starting with a random initial state \(|\psi\rangle = \sum_u c_u |u\rangle\), we may apply the very same algorithm and we will find the approximation to the phase \(\varphi_u\) with probability \(|c_u|^2(1-\epsilon)\).

We prepare the second register in the state \[ \begin{pmatrix} 1\\ 1\\ \end{pmatrix}\ =\ (1-i) u_+ + (1+i) u_-\,. \]

x <- (H(1) * qstate(t+1, basis=""))

This implies that we will find both \(\varphi_u\) with equal probability.

for(i in c(2:(t+1))) {
  x <- H(i) * x
}
M <- array(as.complex(c(c, -s, s, c)), dim=c(2,2)) 
for(i in c(2:(t+1))) {
  x <- cqgate(bits=c(i, 1),
              gate=sqgate(bit=1,
                          M=M, type=paste0("Uf", 2^(i-2)))) * x
  M <- M %*% M
}
x <- qft(x, inverse=TRUE, bits=c(2:(t+1)))

measurephi <- function(x, t) {
  xtmp <- measure(x)
  cbits <- genStateNumber(which(xtmp$value==1)-1, t+1)
  phi <- sum(cbits[1:t]/2^(1:t))
  return(invisible(phi))
}
phi <- measurephi(x, t=t)
2*pi*phi

[1] 1.374447

phi-c(+alpha, 2*pi-alpha)/2/pi

[1]  0.004464286 -0.566964286

We can draw the probability distribution again and observe the two peaks corresponding to the two eigenvalues

plot(abs(x@coefs)^2, type="l",
     ylab="p", xlab="state index")

Let’s measure 1000 times, which is easily possible in our simulator

phi <- c()
for(i in c(1:N)) {
  phi[i] <- measurephi(x, t)
}
hist(phi, breaks=2^t, xlim=c(0,1))
abline(v=c(alpha/2/pi, 1-alpha/2/pi), lwd=2, col="red")

The red vertical lines indicate the true values.