Quantum phase estimation plays an important role as a subroutine in many quantum programs and lays the foundation for the famous Shor's algorithm. The task is usually formulated as:
Given a unitary operator
$U$ and its eigenvector$|u\rangle$ , find the associated eigenvalue such that$U|u\rangle = e^{2\pi i\varphi} |u\rangle$ ( phase$\varphi$ unknown ). Since$U$ is unitary, we can say that$0\leq\varphi<1$ .
To illustrate how phase estimation works, consider an example. Suppose
$$
U =
\begin{bmatrix}
1 &0 \
0 & e^{2i\pi/5}
\end{bmatrix}
$$
with eigenvector
-
Pre-processing
The phase estimation algorithm requires two registers. The first part is called counting register and it contains an
$n$ -qubit state all initialized to$|0\rangle$ . They will estimate the unknown phase$\varphi$ in binary form. The number of$n$ depends on the accuracy we want to achieve. The second part contains$m$ qubits and we need to initialize it to the eigenstate$|u\rangle$ . In our case,$m=1$ and choose$n=3$ to begin with. We would expect bad precision with only 3 counting qubits. Thus, the overall initial state after preprocessing is:$$|\psi_0\rangle = |0\rangle^{\otimes n} \otimes |u\rangle =|000\rangle \otimes |1\rangle$$ We can easily do this in our simulator:qubit_num = 4 shots = 1000 phase = 2 * np.pi / 5 def main(): """ main """ # Create environment env = QuantumEnvironment() # Choose backend machine env.backend(BackendName.LocalBaiduSim2) # Initialize qubits q = [env.Q[i] for i in range(qubit_num)] # Prepare eigenstate |1> = X|0> on the ancilla qubit X(q[3])
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Superposition
$H$ Next, we apply a layer of Hadamard gates to all zero-initialized counting qubits to prepare a uniform superposition state. The resulting superposition state
$|\psi_1 \rangle$ is $$ |\psi_1\rangle = (H|0\rangle)^{\otimes n} \otimes |u\rangle = \frac{1}{\sqrt{2^n}}(|0\rangle + |1\rangle)^{\otimes n} \otimes |u\rangle,. $$ In our case,$n=3$ and$|u\rangle = |1\rangle$ . Hence, $$ |\psi_1\rangle = \frac{1}{\sqrt{8}} (|0\rangle + |1\rangle)\otimes(|0\rangle + |1\rangle)\otimes(|0\rangle + |1\rangle) \otimes |1\rangle,. $$ In our simulator, call the Hadamard gate with# Superposition H(q[0]) H(q[1]) H(q[2])
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Oracle
$\hat{O}$ The oracle used in phase estimation algorithm is a bunch of controlled-unitaries by setting control on
$j$ -th counting qubit and target$U^{2^j}$ on the ancilla qubits in eigenstate$|u\rangle$ . Let's see what happens if we apply$U^{2^j}$ to eigenstate$|u\rangle$ . Set$j = 2$ , we get $$ U^{4}|u\rangle = U^{3}U|u\rangle = e^{2\pi i\varphi} U^{3}|u\rangle = e^{2\pi i(2^1\varphi)}U^{2}|u\rangle = \cdots = e^{2\pi i (2^2\varphi)} |u\rangle $$ Then, we can see a clear recursion relation where $$ U^{2^j}|u\rangle = U^{2^j-1}U|u\rangle = U^{2^j-1}e^{2\pi i\varphi} |u\rangle = \cdots = e^{2 \pi i (2^j \varphi)} |u\rangle, \quad j\in [0, n-1] $$ After that, we need to figure out what controlled-unitaries do. Consider a general quantum state vector for counting qubits, $$ |\phi\rangle = \alpha|0\rangle + \beta|1\rangle $$ Apply the controlled-U gate and we get $$ CU|\phi\rangle \otimes |u\rangle = \alpha|0\rangle \otimes |u\rangle + \beta \cdot U|1\rangle \otimes |u\rangle = \alpha|0\rangle \otimes |u\rangle + \beta \cdot e^{2 \pi i\varphi}|1\rangle \otimes |u\rangle $$ This trick is usually called the phase kickback. Note that the overall effect is adding the unknown phase$\varphi$ to the control qubit by applying a unitary to target ancilla qubits. This is opposite to common sense that the control bit should remain the same. Now, we apply the oracle$\hat{O}$ ( Controlled-$U^{2^j}$) to the whole state$|\psi_1\rangle$ . $$ \hat{O} |\psi_1\rangle = \frac{1}{\sqrt{2^n}}(|0\rangle + e^{2\pi i (2^{n-1}\varphi)}|1\rangle) \otimes\cdots \otimes (|0\rangle + e^{2\pi i (2^{0}\varphi)}|1\rangle) \otimes |u\rangle $$ In our 3-qubit case, $$ |\psi_2\rangle = \hat{O} |\psi_1\rangle =\frac{1}{\sqrt{8}}(|0\rangle + e^{2\pi i (4\varphi)}|1\rangle) \otimes(|0\rangle + e^{2\pi i (2\varphi)}|1\rangle) \otimes (|0\rangle + e^{2\pi i (1\varphi)}|1\rangle) \otimes |1\rangle $$ So far, our quantum circuit looks like:
It may still seem confusing what
$U^{2^j}$ means. It simply means repeating the controlled-unitary$2^j$ times.
In our simulator, the controlled-U gates can be generated easily by setting a general controlled 3-parameter rotation gate, which is defined as $$ R(\theta, \phi, \varphi) = \begin{bmatrix} \cos(\frac{\theta}{2}) & -e^{i\varphi}\sin(\frac{\theta}{2})\ e^{i\phi}\sin(\frac{\theta}{2}) & e^{i(\phi+\varphi)} \cos(\frac{\theta}{2}) \end{bmatrix} $$ It is easy to see that this rotation gate reduces to the given unitary
$U$ when$\varphi' = 2\pi/5$ . $$ R(0, 0, \varphi') = \begin{bmatrix} 1 &0\ 0 & e^{i\varphi'} \end{bmatrix} $$ Here is the code.# Control-U gates CU(0, 0, phase)(q[0], q[3]) CU(0, 0, phase)(q[1], q[3]) CU(0, 0, phase)(q[1], q[3]) CU(0, 0, phase)(q[2], q[3]) CU(0, 0, phase)(q[2], q[3]) CU(0, 0, phase)(q[2], q[3]) CU(0, 0, phase)(q[2], q[3])
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Quantum Fourier Transform (QFT)
We are very close to the answer at this stage but it's in Fourier basis. We need to get back to the familiar computational basis to read out our 3-qubit phase estimation
$\hat{\varphi}$ . We refer the details of QFT to another document. Before we explain what QFT does, let's introduce the notation$\varphi = 0.\varphi_0\cdots \varphi_{n-1}$ to represent the binary fraction$\varphi = \varphi_0/2+\cdots + \varphi_{n-1}/2^n$ and each$\varphi_i \in {0,1}$ . In our 3-qubit case, we choose$n=3$ and the best precision we can reach for$\varphi = 1/5$ is between $$ \varphi_0 =0, \varphi_1 =0, \varphi_2 = 1 \quad \Rightarrow \quad \hat{\varphi}{-} = \frac{1}{2}\varphi_0 + \frac{1}{4}\varphi_1 + \frac{1}{8}\varphi_2 = \frac{1}{8} $$ and $$ \varphi_0 =0, \varphi_1 = 1, \varphi_2 = 0 \quad \Rightarrow \quad \hat{\varphi}{+} = \frac{1}{2}\varphi_0 + \frac{1}{4}\varphi_1 + \frac{1}{8}\varphi_2 = \frac{1}{4} $$ We expect to see the probability amplitude in this range. If we want to increase the accuracy and narrow down this range, we have to increase the counting qubit number. In fact, we can always rewrite the state$|\psi_2 \rangle$ derived from last step as $$ |\psi_2\rangle =\frac{1}{\sqrt{8}}(|0\rangle + e^{2\pi i (0.\varphi_2)}|1\rangle) \otimes(|0\rangle + e^{2\pi i (0.\varphi_1\varphi_2)}|1\rangle) \otimes (|0\rangle + e^{2\pi i (0.\varphi_0\varphi_1\varphi_2)}|1\rangle) \otimes |1\rangle $$ To illustrate this, consider $$ 2^1 \varphi = 2^1* (0.\varphi_0\varphi_1\varphi_2) = 2*( \frac{1}{2}\varphi_0 + \frac{1}{4}\varphi_1 + \frac{1}{8}\varphi_2) = \varphi_0 + \frac{1}{2}\varphi_1 + \frac{1}{4}\varphi_2 = 0.\varphi_1\varphi_2 $$ Similarly, $$ 2^2 \varphi = 2^2* (0.\varphi_0\varphi_1\varphi_2) = 4*( \frac{1}{2}\varphi_0 + \frac{1}{4}\varphi_1 + \frac{1}{8}\varphi_2) = 2\varphi_0 + \varphi_1 + \frac{1}{2}\varphi_2 = 0.\varphi_2 $$ This can be generalized to $$ 2^j \varphi = 2^j* (0.\varphi_0\cdots\varphi_{n-1}) = 2^j*( \frac{1}{2}\varphi_0 + \cdots + \frac{1}{2^n}\varphi_{n-1}) = 2^{j-1}\varphi_0 + \cdots+ \frac{1}{2}\varphi_j + \cdots = 0.\varphi_j\cdots\varphi_{n-1} $$ Now, we can understand what quantum fourier transform (QFT) does! It maps an n-bit binary string in computational basis into a so-called Fourier basis: $$ QFT_n |\varphi_0, \cdots, \varphi_{n-1}\rangle = \frac{1}{\sqrt{2^n}} (|0\rangle + e^{2\pi i (0.\varphi_{n-1})}|1\rangle) \otimes \cdots \otimes (|0\rangle + e^{2\pi i (0.\varphi_0\cdots\varphi_{n-1})}|1\rangle) $$ This is exactly the form of$|\psi_2\rangle$ we have seen before. Then, we can apply 3-qubit inverse fourier transform to the counting qubits of$|\psi_2 \rangle$ $$ QFT^{-1}3\otimes I |\psi_2 \rangle = |\varphi_0, \varphi_1, \varphi_2\rangle \otimes |u\rangle $$ The matrix representation for this operation is $$ QFT^{-1}{3} = \frac{1}{\sqrt{8}} \begin{bmatrix} 1 &1 &1 &1 &1 &1 &1 &1\ 1 &\sqrt{-i} &-i &-\sqrt{i} &-1 &-\sqrt{-i} &i &\sqrt{i}\ 1 &-i &-1 &i &1 &-i &-1 &i\ 1 &-\sqrt{i} &i &\sqrt{-i} &-1 &i &-i &-\sqrt{-i}\ 1 &-1 &1 &-1 &1 &-1 &1 &-1\ 1 &-\sqrt{-i} &i &\sqrt{i} &-1 &\sqrt{-i} &-i &-\sqrt{i}\ 1 &i &-1 &-i &1 &i &-1 &-i\ 1 &\sqrt{i} &i &-\sqrt{-i} &-1 &-\sqrt{i} &-i &\sqrt{-i} \end{bmatrix} $$ And the circuit model is constructed as
Notice the equality between
$$
R(0, 0, -\pi/4)
= \begin{bmatrix}
1 &0\
0 & e^{-i\frac{\pi}{4}}
\end{bmatrix}
=T^\dagger
$$
and
$$
R(0, 0, -\pi/2)
= \begin{bmatrix}
1 &0\
0 & e^{-i\frac{\pi}{2}}
\end{bmatrix}
= \begin{bmatrix}
1 &0\
0 & -i
\end{bmatrix}
=S^\dagger
$$
In our simulator, this inverse QFT can be implemented by# 3-qubit inverse QFT SWAP(q[0], q[2]) H(q[0]) CU(0, 0, -np.pi / 2)(q[0], q[1]) H(q[1]) CU(0, 0, -np.pi / 4)(q[0], q[2]) CU(0, 0, -np.pi / 2)(q[1], q[2]) H(q[2])
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Read out and measurement
After so many efforts, we finally reach the point where we read out the 3-qubit estimated phase
$\hat{\varphi} = 0.\varphi_0\varphi_1\varphi_2$ in binary fraction.
Combine all the steps above, we get the following measurement results:
Here, we can see that the state with highest probability is $|\varphi_0, \varphi_1, \varphi_2\rangle = |010\rangle$ which corresponds to the estimated phase $$ \hat{\varphi}_{3-qubit} = 0.\varphi_0\varphi_1\varphi_2 = \frac{1}{2}\varphi_0 + \frac{1}{4}\varphi_1 + \frac{1}{8}\varphi_2 = 0.25 $$ which is off by$25%$ from the true value. But this is the best we can get using 3 counting qubits. If we use 5-qubits instead, the result will be much more accurate (error scales as$1/2^n$ ).
Here is the complete code. Please try it, change the parameters and see what happens.
import numpy as np
import sys
sys.path.append('../../..') # "from QCompute import *" requires this
from QCompute import *
matchSdkVersion('Python 3.3.3')
qubit_num = 4
shots = 1000
phase = 2 * np.pi / 5
def main():
"""
main
"""
# Create environment
env = QEnv()
# Choose backend machine
env.backend(BackendName.LocalBaiduSim2)
# Initialize qubits
q = [env.Q[i] for i in range(qubit_num)]
# Prepare eigenstate |1> = X|0> on the ancilla qubit
X(q[3])
# Superposition
H(q[0])
H(q[1])
H(q[2])
# Control-U gates
CU(0, 0, phase)(q[0], q[3])
CU(0, 0, phase)(q[1], q[3])
CU(0, 0, phase)(q[1], q[3])
CU(0, 0, phase)(q[2], q[3])
CU(0, 0, phase)(q[2], q[3])
CU(0, 0, phase)(q[2], q[3])
CU(0, 0, phase)(q[2], q[3])
# 3-qubit inverse QFT
SWAP(q[0], q[2])
H(q[0])
CU(0, 0, -np.pi / 2)(q[0], q[1])
H(q[1])
CU(0, 0, -np.pi / 4)(q[0], q[2])
CU(0, 0, -np.pi / 2)(q[1], q[2])
H(q[2])
# Measurement result
MeasureZ(*env.Q.toListPair())
taskResult = env.commit(shots, fetchMeasure=True)
return taskResult['counts']
if __name__ == '__main__':
main()We can see that the algorithm itself is trying to find the eigenvalue of a given unitary. That's why it is also called eigenvalue estimation. Now, it's your turn to have a try. You may change the phase to