벤치마크 파울리 연산자 투영

계산 기반 상태에서 파울리 현의 작용

계산 기저 상태에 대한 파울리 스트링의 작용은 상당히 단순하며, 그 자체로 하나의 계산 기저 상태입니다. 이는 각 행에 0이 아닌 원소가 단 하나만 있는 파울리 행렬의 구조에서 직접적으로 비롯된 결과입니다. 따라서 큐비트에 대한 이들의 작용은 다음과 같습니다:

\sigma_x |0 \rangle = |1 \rangle

\sigma_x |1 \rangle = |0 \rangle

\sigma_y |0 \rangle = i|1 \rangle

\sigma_y |1 \rangle = -i|0 \rangle

\sigma_z |0 \rangle = |0 \rangle

\sigma_z |1 \rangle = -|1 \rangle

I |0 \rangle = |0 \rangle

I |1 \rangle = |1 \rangle

계산 기저를 나타내는 비트열의 각 비트에는 $x \in \{0, 1 \}$ 가 할당됩니다. 구현을 최대한 가볍게 유지하기 위해, 비트열을 $0\rightarrow \textrm{False}$ 및 $1\rightarrow \textrm{True}$ 변수로bool 표현할 것입니다.

imag계산 기저 상태에서의 각 파울리 연산자의 작용을 나타내기 위해, 이에 세 개의 변수, sign``diag, 를 할당할 것이다.

diag 연산자가 대각 행렬인지 여부를 표시합니다:
- $\textrm{diag}(I) = \textrm{True}$
- $\textrm{diag}(\sigma_x) = \textrm{False}$
- $\textrm{diag}(\sigma_y) = \textrm{False}$
- $\textrm{diag}(\sigma_z) = \textrm{True}$
sign 0 또는 1과 연결된 행렬 원소에서 부호 변화가 있는지 여부를 확인합니다:
- $\textrm{sign}(I) = \textrm{False}$
- $\textrm{sign}(\sigma_x) = \textrm{False}$
- $\textrm{sign}(\sigma_y) = \textrm{True}$
- $\textrm{sign}(\sigma_z) = \textrm{True}$
imag 행렬 원소에 복소수 성분이 있는지 여부를 확인합니다:
- $\textrm{imag}(I) = \textrm{False}$
- $\textrm{imag}(\sigma_x) = \textrm{False}$
- $\textrm{imag}(\sigma_y) = \textrm{True}$
- $\textrm{imag}(\sigma_z) = \textrm{False}$

임의의 파울리 연산자를 $\sigma \in \{ I, \sigma_x, \sigma_y \sigma_z\}$ 로 표기한다. 이때, 계산 기저 상태에 대한 파울리 연산자의 작용은 다음과 같은 논리 연산으로 표현될 수 있다:

\sigma |x \rangle = |x == \textrm{diag}(\sigma) \rangle (-1)^{x\textrm{ and sign}(\sigma)} (i)^{\textrm{imag}(\sigma)}.

이는 임의의 큐비트 수에 대해서도 간단히 일반화될 수 있다.

이것이 제대로 작동하는지 확인해 봅시다:

def connected_element_and_amplitude_bool(
    x: bool, diag: bool, sign: bool, imag: bool
) -> tuple[bool, complex]:
    """
    Finds the connected element to computational basis state |x> under
    the action of the Pauli operator represented by (diag, sign, imag).

    Args:
        x: Value of the bit, either True or False.
        diag: Whether the Pauli operator is diagonal (I, Z)
        sigma: Whether the Pauli operator's rows differ in sign (Y, Z)
        imag: Whether the Pauli operator is purely imaginary (Y)

    Returns:
        A length-2 tuple:
            - The connected element to x, either False or True
            - The matrix element
    """
    return x == diag, (-1) ** (x and sign) * (1j) ** (imag)


sigma_indices = [0, 1, 2, 3]
sigma_string = ["I", "SX", "SZ", "SY"]
sigma_diag = [True, False, True, False]
sigma_sign = [False, False, True, True]
sigma_imag = [False, False, False, True]
qubit_values = [False, True]

for xi in sigma_indices:
    print("-------------------")
    print(sigma_string[xi])
    for x in qubit_values:
        x_p, matrix_element = connected_element_and_amplitude_bool(
            x, sigma_diag[xi], sigma_sign[xi], sigma_imag[xi]
        )
        print(
            "|"
            + str(x)
            + "> -->  |"
            + str(x_p)
            + ">    ME:"
            + str(matrix_element)
        )

Output:

-------------------
I
|False> -->  |False>    ME:(1+0j)
|True> -->  |True>    ME:(1+0j)
-------------------
SX
|False> -->  |True>    ME:(1+0j)
|True> -->  |False>    ME:(1+0j)
-------------------
SZ
|False> -->  |False>    ME:(1+0j)
|True> -->  |True>    ME:(-1+0j)
-------------------
SY
|False> -->  |True>    ME:1j
|True> -->  |False>    ME:(-0-1j)

우리는 40 큐비트 시스템을 위해 방대한 수의 비트열(50 M)을 생성합니다:

import numpy as np
from qiskit_addon_sqd.qubit import sort_and_remove_duplicates

rand_seed = 22
np.random.seed(rand_seed)

# Generate some random bitstrings for testing


def random_bitstrings(n_samples, n_qubits):
    return (
        np.round(np.random.rand(n_samples, n_qubits))
        .astype("int")
        .astype("bool")
    )


n_qubits = 40
bts_matrix = random_bitstrings(50_000_000, n_qubits)

# We need to sort the bitstrings and only keep the unique ones
# NOTE: It is essential for the projection code to have the bitstrings sorted!
bts_matrix = sort_and_remove_duplicates(bts_matrix).astype("bool")

# Final subspace dimension after getting rid of duplicated bitstrings
d = bts_matrix.shape[0]

print("Total number of unique bitstrings: " + str(d))

Output:

Total number of unique bitstrings: 49998839

SQD 파울리 투영 함수의 벤치마크

고려 대상인 파울리 현은 $\sigma_z \otimes ... \otimes \sigma_z$ 입니다.

비트스트링 행렬을 슬라이싱하여 다양한 부분공간 차원을 고려합니다. 다양한 부분공간 크기에 대해 부분공간 투영에 소요되는 시간을 측정합니다.

import time

from qiskit.quantum_info import Pauli
from qiskit_addon_sqd.qubit import matrix_elements_from_pauli

pauli = Pauli("Z" * n_qubits)

# Different subspace sizes to test
d_list = np.linspace(d / 1000, d, 20).astype("int")

# To store the walltime
time_array = np.zeros(20)

for i in range(20):
    int_bts_matrix = bts_matrix[: d_list[i], :]
    time_1 = time.time()
    _ = matrix_elements_from_pauli(int_bts_matrix, pauli)
    time_array[i] = time.time() - time_1
    print(f"Iteration {i} took {round(time_array[i], 6)}s")

Output:

Iteration 0 took 0.201246s
Iteration 1 took 0.348222s
Iteration 2 took 0.576333s
Iteration 3 took 0.78356s
Iteration 4 took 1.016162s
Iteration 5 took 1.305325s
Iteration 6 took 1.392751s
Iteration 7 took 1.632433s
Iteration 8 took 1.826521s
Iteration 9 took 2.02903s
Iteration 10 took 2.297458s
Iteration 11 took 2.588042s
Iteration 12 took 2.738746s
Iteration 13 took 2.906144s
Iteration 14 took 3.148833s
Iteration 15 took 3.323253s
Iteration 16 took 3.664171s
Iteration 17 took 3.680663s
Iteration 18 took 4.008313s
Iteration 19 took 4.173532s

import matplotlib.pyplot as plt

# Data for energies plot
x1 = d_list
y1 = time_array

# Plot energies
plt.title("Runtime vs subspace dimension 40 qubits")
plt.xlabel("Subspace dimension (millions)")
plt.ylabel("Wall time [s]")
plt.xticks([1e7, 2e7, 3e7, 4e7, 5e7], [str(i) for i in [10, 20, 30, 40, 50]])
plt.plot(x1, y1, marker=".", markersize=20)
plt.tight_layout()
plt.show()

Output:

이제 60 큐비트에 대해서도 같은 과정을 수행합니다:

n_qubits = 60
bts_matrix = random_bitstrings(50_000_000, n_qubits)

# We need to sort the bitstrings and just keep the unique ones
bts_matrix = sort_and_remove_duplicates(bts_matrix).astype("bool")

# Final subspace dimension after getting rid of duplicated bitstrings
d = bts_matrix.shape[0]

print("Total number of unique bitstrings: " + str(d))

Output:

Total number of unique bitstrings: 50000000

pauli = Pauli("Z" * n_qubits)

# Different subspace sizes to test
d_list = np.linspace(d / 1000, d, 20).astype("int")

# It is better to do this once
row_array = np.arange(d)

# To store the walltime
time_array = np.zeros(20)

for i in range(20):
    int_bts_matrix = bts_matrix[: d_list[i], :]
    int_row_array = row_array[: d_list[i]]
    time_1 = time.time()
    _ = matrix_elements_from_pauli(int_bts_matrix, pauli)
    time_array[i] = time.time() - time_1
    print(f"Iteration {i} took {round(time_array[i], 6)}s")

Output:

Iteration 0 took 0.236567s
Iteration 1 took 0.424116s
Iteration 2 took 0.673399s
Iteration 3 took 0.905164s
Iteration 4 took 1.168936s
Iteration 5 took 1.454204s
Iteration 6 took 1.74778s
Iteration 7 took 1.920795s
Iteration 8 took 2.259994s
Iteration 9 took 2.550674s
Iteration 10 took 2.681287s
Iteration 11 took 3.04411s
Iteration 12 took 3.293262s
Iteration 13 took 3.471247s
Iteration 14 took 3.726639s
Iteration 15 took 4.072854s
Iteration 16 took 4.221037s
Iteration 17 took 4.498535s
Iteration 18 took 4.741108s
Iteration 19 took 5.159038s

# Data for energies plot
x1 = d_list
y1 = time_array

fig, axs = plt.subplots(1, 1, figsize=(6, 6))

# Plot energies
axs.plot(x1, y1, marker=".", markersize=20)
axs.set_title("Runtime vs subspace dimension 60 qubits")
axs.set_xlabel("Subspace dimension (millions)")
plt.xticks([1e7, 2e7, 3e7, 4e7, 5e7], [str(i) for i in [10, 20, 30, 40, 50]])
axs.set_ylabel("Wall time [s]")

plt.tight_layout()
plt.show()

Output:

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