In urbach/qsimulatR: A Quantum Computer Simulator

library(knitr)
library(qsimulatR)
knitr::opts_chunk$set(fig.align='center',
                      comment='')

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

and therefore expect an error of less than

2^(-digits)

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,\dots,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]
phi

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)

is indeed smaller than the maximal deviation $2^{-\mathrm{digits}}=$ r 2^(-digits) 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
phi-c(+alpha, 2*pi-alpha)/2/pi

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 r N=1000 r N 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")