ベンチマーク・パウリ演算子の射影

計算上の基底状態におけるパウリ弦の作用

計算基底状態に対するパウリ弦の作用はごく単純であり、それ自体が単一の計算基底状態である。これは、各行に非ゼロの要素が1つしか含まれないパウリ行列の構造から直接導かれる結果である。したがって、クビットに対する彼らの作用は次のようになる：

\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

bool計算基底を表すビット列の各ビットには、 $x \in \{0, 1 \}$ というラベルが付けられます。実装をできるだけ軽量にするため、これらのビット列を $0\rightarrow \textrm{False}$ および $1\rightarrow \textrm{True}$ という変数で表現します。

imag計算基底状態における各パウリ演算子の作用を表すために、これに 3 つの変数、 diag sign、を割り当てます。

diag 演算子が対角であるかどうかを示すラベル：
- $\textrm{diag}(I) = \textrm{True}$
- $\textrm{diag}(\sigma_x) = \textrm{False}$
- $\textrm{diag}(\sigma_y) = \textrm{False}$
- $\textrm{diag}(\sigma_z) = \textrm{True}$
sign 0 または 1 のいずれかに接続された行列要素に符号の変化があるかどうかを判定します：
- $\textrm{sign}(I) = \textrm{False}$
- $\textrm{sign}(\sigma_x) = \textrm{False}$
- $\textrm{sign}(\sigma_y) = \textrm{True}$
- $\textrm{sign}(\sigma_z) = \textrm{True}$
imag 行列要素に複素成分が含まれているかどうかを判定します：
- $\textrm{imag}(I) = \textrm{False}$
- $\textrm{imag}(\sigma_x) = \textrm{False}$
- $\textrm{imag}(\sigma_y) = \textrm{True}$
- $\textrm{imag}(\sigma_z) = \textrm{False}$

任意のパウリ演算子を $\sigma \in \{ I, \sigma_x, \sigma_y \sigma_z\}$ と表記する。計算基底状態に対するパウリ演算子の作用は、以下の論理演算によって表すことができる：

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

これは、任意の数の量子ビットに対しても容易に一般化できる。

これがうまくいくか確認してみましょう：

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)

40キュービットのシステムに対して、膨大な数のビット列（50 M）を生成します：

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パウリ投影関数のベンチマーク

ここで検討しているパウリ弦は、 $\sigma_z \otimes ... \otimes \sigma_z$ である。

ビット文字列の行列をスライスすることで、さまざまな部分空間の次元が検討される。さまざまな部分空間のサイズについて、部分空間の射影にかかる時間を測定する。

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:

次に、60キュービットについても同様の処理を行います：

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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