Iterative Quantum Amplitude Estimation#

IQAE(qargs, state_function, eps, alpha, mes_kwargs={})[source]#

Accelerated Quantum Amplitude Estimation (IQAE). This function performs QAE with a fraction of the quantum resources of the well-known QAE algorithm. See Accelerated Quantum Amplitude Estimation without QFT.

The problem of iterative quantum amplitude estimation is described as follows:

Given a unitary operator \(\mathcal{A}\), let \(\ket{\Psi}=\mathcal{A}\ket{0}\ket{\text{False}}\).
Write \(\ket{\Psi}=\sqrt{a}\ket{\Psi_1}\ket{\text{True}}+\sqrt{1-a}\ket{\Psi_0}\ket{\text{False}}\) as a superposition of the orthogonal good and bad components of \(\ket{\Psi}\).
Find an estimate for \(a\), the probability that a measurement of \(\ket{\Psi}\) yields a good state.

Parameters:

qargslist[QuantumVariable]: The list of QuantumVariables which represent the state, the quantum amplitude estimation is performed on. The last variable in the list must be of type QuantumBool.
state_functionfunction: A Python function preparing the state \(\ket{\Psi}\). This function will receive the variables in the list qargs as arguments in the course of this algorithm.
epsfloat: Accuracy \(\epsilon>0\) of the algorithm.
alphafloat: Confidence level \(\alpha\in (0,1)\) of the algorithm.
mes_kwargsdict, optional: The keyword arguments for the measurement function. Default is an empty dictionary.

Returns:

afloat: An estimate \(\hat{a}\) of \(a\) such that

\[\mathbb P\{|\hat{a}-a|<\epsilon\}\geq 1-\alpha\]

Examples

We show the same Numerical integration example which can also be found in the QAE documentation.

We wish to evaluate

\[A=\int_0^1f(x)\mathrm dx.\]

For this, we set up the corresponding state_function acting on the variables in input_list:

from qrisp import QuantumFloat, QuantumBool, control, z, h, ry, IQAE
import numpy as np

n = 6 
inp = QuantumFloat(n,-n)
tar = QuantumBool()
input_list = [inp, tar]

For example, if \(f(x)=\sin^2(x)\), the state_function can be implemented as follows:

def state_function(inp, tar):
    h(inp)

    N = 2**inp.size
    for k in range(inp.size):
        with control(inp[k]):
            ry(2**(k+1)/N,tar)

Finally, we apply IQAE and obtain an estimate \(a\) for the value of the integral \(A=0.27268\).

a = IQAE(input_list, state_function, eps=0.01, alpha=0.01)

>>> a 
0.26782038552705856