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Pengundur-belakang pengoperasi (OBP) untuk penganggaran nilai jangkaan

Anggaran penggunaan: 16 minit pada pemproses Eagle r3 (NOTA: Ini hanyalah anggaran. Masa larian anda mungkin berbeza.)

# Added by doQumentation — required packages for this notebook
!pip install -q matplotlib numpy qiskit qiskit-addon-obp qiskit-addon-utils qiskit-ibm-runtime rustworkx
# This cell is hidden from users;
# it disables linting rules.
# ruff: noqa

Latar Belakang​

Pengundur-belakang pengoperasi ialah teknik yang melibatkan penyerapan operasi dari hujung Circuit kuantum ke dalam observable yang diukur, yang secara amnya mengurangkan kedalaman Circuit dengan kos tambahan terma dalam observable. Matlamatnya ialah untuk mengundurkan sebanyak mungkin Circuit tanpa membiarkan observable membesar terlalu besar. Pelaksanaan berasaskan Qiskit tersedia dalam addon OBP Qiskit, butiran lanjut boleh didapati dalam dokumentasi yang berkaitan dengan contoh mudah untuk memulakan.

Pertimbangkan contoh Circuit di mana observable O=∑PcPPO = \sum_P c_P P hendak diukur, di mana PP ialah Pauli dan cPc_P ialah pekali. Mari kita tandakan Circuit tersebut sebagai satu kesatuan UU yang boleh dibahagikan secara logik kepada U=UCUQU = U_C U_Q seperti yang ditunjukkan dalam rajah di bawah.

Circuit diagram showing Uq followed by Uc

Pengundur-belakang pengoperasi menyerap kesatuan UCU_C ke dalam observable dengan mengevolusinya sebagai O′=UC†OUC=∑PcPUC†PUCO' = U_C^{\dagger}OU_C = \sum_P c_P U_C^{\dagger}PU_C. Dengan kata lain, sebahagian pengiraan dilakukan secara klasik melalui evolusi observable daripada OO kepada O′O'. Masalah asal kini boleh diformulasi semula sebagai pengukuran observable O′O' untuk Circuit kedalaman rendah baru yang kesatuannya ialah UQU_Q.

Kesatuan UCU_C diwakili sebagai beberapa hirisan UC=USUS−1...U2U1U_C = U_S U_{S-1}...U_2U_1. Terdapat pelbagai cara untuk mentakrifkan hirisan. Sebagai contoh, dalam Circuit contoh di atas, setiap lapisan RzzR_{zz} dan setiap lapisan gate RxR_x boleh dianggap sebagai hirisan individu. Pengundur-belakang melibatkan pengiraan O′=Πs=1S∑PcPUs†PUsO' = \Pi_{s=1}^S \sum_P c_P U_s^{\dagger} P U_s secara klasik. Setiap hirisan UsU_s boleh diwakili sebagai Us=exp(−iθsPs2)U_s = exp(\frac{-i\theta_s P_s}{2}), di mana PsP_s ialah Pauli nn-Qubit dan θs\theta_s ialah skalar. Mudah untuk disahkan bahawa

Us†PUs=Pif [P,Ps]=0,U_s^{\dagger} P U_s = P \qquad \text{if} ~[P,P_s] = 0, Us†PUs=cos(θs)P+isin(θs)PsPif {P,Ps}=0U_s^{\dagger} P U_s = \qquad cos(\theta_s)P + i sin(\theta_s)P_sP \qquad \text{if} ~\{P,P_s\} = 0

Dalam contoh di atas, jika {P,Ps}=0\{P,P_s\} = 0, maka kita perlu melaksanakan dua Circuit kuantum, dan bukannya satu, untuk mengira nilai jangkaan. Oleh itu, pengundur-belakang mungkin meningkatkan bilangan terma dalam observable, yang membawa kepada bilangan pelaksanaan Circuit yang lebih tinggi. Satu cara untuk membenarkan pengundur-belakang yang lebih dalam ke dalam Circuit, sambil mencegah pengoperasi daripada membesar terlalu besar, ialah dengan memangkas terma yang mempunyai pekali kecil, dan bukannya menambahkannya ke dalam pengoperasi. Sebagai contoh, dalam contoh di atas, seseorang mungkin memilih untuk memangkas terma yang melibatkan PsPP_sP dengan syarat θs\theta_s cukup kecil. Pemangkasan terma boleh menghasilkan lebih sedikit Circuit kuantum untuk dilaksanakan, tetapi melakukan demikian mengakibatkan sedikit ralat dalam pengiraan nilai jangkaan akhir yang berkadar dengan magnitud pekali terma yang dipangkas.

Tutorial ini melaksanakan corak Qiskit untuk mensimulasi dinamik kuantum rantai spin Heisenberg menggunakan qiskit-addon-obp.

Keperluan​

Sebelum memulakan tutorial ini, pastikan anda telah memasang perkara berikut:

  • Qiskit SDK v1.2 atau lebih baru (pip install qiskit)
  • Qiskit Runtime v0.28 atau lebih baru (pip install qiskit-ibm-runtime)
  • OBP Qiskit addon (pip install qiskit-addon-obp)
  • Qiskit addon utils (pip install qiskit-addon-utils)

Persediaan​

import numpy as np
import matplotlib.pyplot as plt

from qiskit.primitives import StatevectorEstimator as Estimator
from qiskit.transpiler.preset_passmanagers import generate_preset_pass_manager
from qiskit.quantum_info import SparsePauliOp
from qiskit.transpiler import CouplingMap
from qiskit.synthesis import LieTrotter

from qiskit_addon_utils.problem_generators import generate_xyz_hamiltonian
from qiskit_addon_utils.problem_generators import (
    generate_time_evolution_circuit,
)
from qiskit_addon_utils.slicing import slice_by_gate_types, combine_slices
from qiskit_addon_obp.utils.simplify import OperatorBudget
from qiskit_addon_obp import backpropagate
from qiskit_addon_obp.utils.truncating import setup_budget

from rustworkx.visualization import graphviz_draw

from qiskit_ibm_runtime import QiskitRuntimeService
from qiskit_ibm_runtime import EstimatorV2, EstimatorOptions

Bahagian I: Rantai spin Heisenberg skala kecil​

Langkah 1: Petakan input klasik kepada masalah kuantum​

Petakan evolusi masa model kuantum Heisenberg kepada eksperimen kuantum.​

Pakej qiskit_addon_utils menyediakan beberapa fungsi yang boleh digunakan semula untuk pelbagai tujuan.

Modul qiskit_addon_utils.problem_generators menyediakan fungsi untuk menjana Hamiltonian seperti Heisenberg pada graf sambungan yang diberikan. Graf ini boleh berupa sama ada rustworkx.PyGraph atau CouplingMap menjadikannya mudah digunakan dalam aliran kerja berpusatkan Qiskit.

Berikut, kita menjana CouplingMap rantai linear 10 Qubit.

num_qubits = 10
layout = [(i - 1, i) for i in range(1, num_qubits)]

# Instantiate a CouplingMap object
coupling_map = CouplingMap(layout)
graphviz_draw(coupling_map.graph, method="circo")

Output of the previous code cell

Seterusnya, kita menjana pengoperasi Pauli yang memodelkan Hamiltonian Heisenberg XYZ.

H^XYZ=∑(j,k)∈E(Jxσjxσkx+Jyσjyσky+Jzσjzσkz)+∑j∈V(hxσjx+hyσjy+hzσjz){\hat{\mathcal{H}}_{XYZ} = \sum_{(j,k)\in E} (J_{x} \sigma_j^{x} \sigma_{k}^{x} + J_{y} \sigma_j^{y} \sigma_{k}^{y} + J_{z} \sigma_j^{z} \sigma_{k}^{z}) + \sum_{j\in V} (h_{x} \sigma_j^{x} + h_{y} \sigma_j^{y} + h_{z} \sigma_j^{z})}

Di mana G(V,E)G(V,E) ialah graf peta gandingan yang diberikan.

# Get a qubit operator describing the Heisenberg XYZ model
hamiltonian = generate_xyz_hamiltonian(
    coupling_map,
    coupling_constants=(np.pi / 8, np.pi / 4, np.pi / 2),
    ext_magnetic_field=(np.pi / 3, np.pi / 6, np.pi / 9),
)
print(hamiltonian)
SparsePauliOp(['IIIIIIIXXI', 'IIIIIIIYYI', 'IIIIIIIZZI', 'IIIIIXXIII', 'IIIIIYYIII', 'IIIIIZZIII', 'IIIXXIIIII', 'IIIYYIIIII', 'IIIZZIIIII', 'IXXIIIIIII', 'IYYIIIIIII', 'IZZIIIIIII', 'IIIIIIIIXX', 'IIIIIIIIYY', 'IIIIIIIIZZ', 'IIIIIIXXII', 'IIIIIIYYII', 'IIIIIIZZII', 'IIIIXXIIII', 'IIIIYYIIII', 'IIIIZZIIII', 'IIXXIIIIII', 'IIYYIIIIII', 'IIZZIIIIII', 'XXIIIIIIII', 'YYIIIIIIII', 'ZZIIIIIIII', 'IIIIIIIIIX', 'IIIIIIIIIY', 'IIIIIIIIIZ', 'IIIIIIIIXI', 'IIIIIIIIYI', 'IIIIIIIIZI', 'IIIIIIIXII', 'IIIIIIIYII', 'IIIIIIIZII', 'IIIIIIXIII', 'IIIIIIYIII', 'IIIIIIZIII', 'IIIIIXIIII', 'IIIIIYIIII', 'IIIIIZIIII', 'IIIIXIIIII', 'IIIIYIIIII', 'IIIIZIIIII', 'IIIXIIIIII', 'IIIYIIIIII', 'IIIZIIIIII', 'IIXIIIIIII', 'IIYIIIIIII', 'IIZIIIIIII', 'IXIIIIIIII', 'IYIIIIIIII', 'IZIIIIIIII', 'XIIIIIIIII', 'YIIIIIIIII', 'ZIIIIIIIII'],
              coeffs=[0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j,
 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j,
 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j,
 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j,
 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j,
 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j,
 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j, 1.04719755+0.j,
 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j,
 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j,
 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j,
 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j,
 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j])

Daripada pengoperasi Qubit, kita boleh menjana Circuit kuantum yang memodelkan evolusi masanya. Sekali lagi, modul qiskit_addon_utils.problem_generators datang membantu dengan fungsi yang berguna untuk melakukan tepat itu:

circuit = generate_time_evolution_circuit(
    hamiltonian,
    time=0.2,
    synthesis=LieTrotter(reps=2),
)
circuit.draw("mpl", style="iqp", scale=0.6)

Output of the previous code cell

Langkah 2: Optimumkan masalah untuk pelaksanaan perkakasan kuantum​

Cipta hirisan Circuit untuk pengundur-belakang​

Ingat, fungsi backpropagate akan mengundurkan keseluruhan hirisan Circuit pada satu masa, jadi pilihan cara menghiris boleh memberi impak terhadap prestasi pengundur-belakang untuk masalah tertentu. Di sini, kita akan mengumpulkan gate yang sama jenis ke dalam hirisan menggunakan fungsi slice_by_gate_types.

Untuk perbincangan yang lebih terperinci tentang penghirisan Circuit, semak panduan cara-cara ini daripada pakej qiskit-addon-utils.

slices = slice_by_gate_types(circuit)
print(f"Separated the circuit into {len(slices)} slices.")
Separated the circuit into 18 slices.

Hadkan saiz maksimum pengoperasi semasa pengundur-belakang​

Semasa pengundur-belakang, bilangan terma dalam pengoperasi umumnya akan mendekati 4N4^N dengan cepat, di mana NN ialah bilangan Qubit. Apabila dua terma dalam pengoperasi tidak bertukar-ganti secara Qubit-wise, kita memerlukan Circuit berasingan untuk mendapatkan nilai jangkaan yang sepadan dengannya. Sebagai contoh, jika kita mempunyai observable 2-Qubit O=0.1XX+0.3IZ−0.5IXO = 0.1 XX + 0.3 IZ - 0.5 IX, maka oleh kerana [XX,IX]=0[XX,IX] = 0, pengukuran dalam satu asas sudah cukup untuk mengira nilai jangkaan bagi dua terma ini. Namun, IZIZ anti-bertukar-ganti dengan dua terma lain. Jadi kita memerlukan pengukuran asas yang berasingan untuk mengira nilai jangkaan IZIZ. Dengan kata lain, kita memerlukan dua, dan bukannya satu, Circuit untuk mengira ⟨O⟩\langle O \rangle. Apabila bilangan terma dalam pengoperasi bertambah, terdapat kemungkinan bilangan pelaksanaan Circuit yang diperlukan juga bertambah.

Saiz pengoperasi boleh dibatasi dengan menentukan argumen kata kunci operator_budget bagi fungsi backpropagate, yang menerima instans OperatorBudget.

Untuk mengawal jumlah sumber tambahan (masa) yang diperuntukkan, kita hadkan bilangan maksimum kumpulan Pauli bertukar-ganti secara Qubit-wise yang dibenarkan untuk dimiliki oleh observable yang telah diundurkan. Di sini kita tentukan bahawa pengundur-belakang harus berhenti apabila bilangan kumpulan Pauli bertukar-ganti secara Qubit-wise dalam pengoperasi melebihi 8.

op_budget = OperatorBudget(max_qwc_groups=8)

Undurkan hirisan dari Circuit​

Pertama kita tentukan observable sebagai MZ=1N∑i=1N⟨Zi⟩M_Z = \frac{1}{N} \sum_{i=1}^N \langle Z_i \rangle, dengan NN ialah bilangan Qubit. Kita akan mengundurkan hirisan dari Circuit evolusi masa sehingga terma dalam observable tidak lagi boleh digabungkan ke dalam lapan atau lebih sedikit kumpulan Pauli bertukar-ganti secara Qubit-wise.

observable = SparsePauliOp.from_sparse_list(
    [("Z", [i], 1 / num_qubits) for i in range(num_qubits)],
    num_qubits=num_qubits,
)
observable
SparsePauliOp(['IIIIIIIIIZ', 'IIIIIIIIZI', 'IIIIIIIZII', 'IIIIIIZIII', 'IIIIIZIIII', 'IIIIZIIIII', 'IIIZIIIIII', 'IIZIIIIIII', 'IZIIIIIIII', 'ZIIIIIIIII'],
              coeffs=[0.1+0.j, 0.1+0.j, 0.1+0.j, 0.1+0.j, 0.1+0.j, 0.1+0.j, 0.1+0.j, 0.1+0.j,
 0.1+0.j, 0.1+0.j])

Di bawah anda akan melihat bahawa kita telah mengundurkan enam hirisan, dan terma-terma digabungkan ke dalam enam dan bukan lapan kumpulan. Ini menunjukkan bahawa mengundurkan satu lagi hirisan akan menyebabkan bilangan kumpulan Pauli melebihi lapan. Kita boleh mengesahkan bahawa ini memang benar dengan memeriksa metadata yang dikembalikan. Juga perhatikan bahawa dalam bahagian ini transformasi Circuit adalah tepat. Iaitu, tiada terma observable baru O′O' yang dipangkas. Circuit yang telah diundurkan dan pengoperasi yang telah diundurkan memberikan hasil yang sama persis dengan Circuit asal dan pengoperasi asal.

# Backpropagate slices onto the observable
bp_obs, remaining_slices, metadata = backpropagate(
    observable, slices, operator_budget=op_budget
)
# Recombine the slices remaining after backpropagation
bp_circuit = combine_slices(remaining_slices)

print(f"Backpropagated {metadata.num_backpropagated_slices} slices.")
print(
    f"New observable has {len(bp_obs.paulis)} terms, which can be combined into {len(bp_obs.group_commuting(qubit_wise=True))} groups."
)
print(
    f"Note that backpropagating one more slice would result in {metadata.backpropagation_history[-1].num_paulis[0]} terms "
    f"across {metadata.backpropagation_history[-1].num_qwc_groups} groups."
)
print("The remaining circuit after backpropagation looks as follows:")
bp_circuit.draw("mpl", fold=-1, scale=0.6)
Backpropagated 6 slices.
New observable has 60 terms, which can be combined into 6 groups.
Note that backpropagating one more slice would result in 114 terms across 12 groups.
The remaining circuit after backpropagation looks as follows:

Output of the previous code cell

Seterusnya, kita akan menentukan masalah yang sama dengan kekangan yang sama terhadap saiz observable output. Namun kali ini, kita memperuntukkan bajet ralat untuk setiap hirisan menggunakan fungsi setup_budget. Terma Pauli dengan pekali kecil akan dipangkas dari setiap hirisan sehingga bajet ralat dipenuhi, dan lebihan bajet akan ditambahkan ke bajet hirisan berikutnya. Perhatikan bahawa dalam kes ini, transformasi akibat pengundur-belakang adalah anggaran kerana sebahagian terma dalam pengoperasi dipangkas.

Untuk membolehkan pemangkasan ini, kita perlu menetapkan bajet ralat kita seperti berikut:

truncation_error_budget = setup_budget(max_error_per_slice=0.005)

Perhatikan bahawa dengan memperuntukkan ralat 5e-3 setiap hirisan untuk pemangkasan, kita dapat mengeluarkan 1 hirisan lagi dari Circuit, sambil kekal dalam bajet asal lapan kumpulan Pauli bertukar-ganti dalam observable. Secara lalai, backpropagate menggunakan norma L1 pekali yang dipangkas untuk membatasi jumlah ralat yang ditimbulkan daripada pemangkasan. Untuk pilihan lain rujuk panduan cara-cara tentang menentukan p_norm.

Dalam contoh tertentu ini di mana kita telah mengundurkan tujuh hirisan, jumlah ralat pemangkasan tidak sepatutnya melebihi (5e-3 ralat/hirisan) * (7 hirisan) = 3.5e-2. Untuk perbincangan lanjut tentang pengagihan bajet ralat merentasi hirisan anda, semak panduan cara-cara ini.

# Run the same experiment but truncate observable terms with small coefficients
bp_obs_trunc, remaining_slices_trunc, metadata = backpropagate(
    observable,
    slices,
    operator_budget=op_budget,
    truncation_error_budget=truncation_error_budget,
)

# Recombine the slices remaining after backpropagation
bp_circuit_trunc = combine_slices(
    remaining_slices_trunc, include_barriers=False
)

print(f"Backpropagated {metadata.num_backpropagated_slices} slices.")
print(
    f"New observable has {len(bp_obs_trunc.paulis)} terms, which can be combined into {len(bp_obs_trunc.group_commuting(qubit_wise=True))} groups.\n"
    f"After truncation, the error in our observable is bounded by {metadata.accumulated_error(0):.3e}"
)
print(
    f"Note that backpropagating one more slice would result in {metadata.backpropagation_history[-1].num_paulis[0]} terms "
    f"across {metadata.backpropagation_history[-1].num_qwc_groups} groups."
)
print("The remaining circuit after backpropagation looks as follows:")
bp_circuit_trunc.draw("mpl", scale=0.6)
Backpropagated 7 slices.
New observable has 82 terms, which can be combined into 8 groups.
After truncation, the error in our observable is bounded by 3.266e-02
Note that backpropagating one more slice would result in 114 terms across 12 groups.
The remaining circuit after backpropagation looks as follows:

Output of the previous code cell

Kita perhatikan bahawa pemangkasan membolehkan kita mengundurkan lebih jauh tanpa peningkatan bilangan kumpulan bertukar-ganti dalam observable.

Kini setelah kita mempunyai ansatz yang dikurangkan dan observable yang dikembangkan, kita boleh mentranspile eksperimen kita ke Backend.​

Di sini kita akan menggunakan komputer kuantum IBM® 127-Qubit untuk menunjukkan cara mentranspile ke Backend QPU.

service = QiskitRuntimeService()
backend = service.least_busy(
    operational=True, simulator=False, min_num_qubits=127
)
pm = generate_preset_pass_manager(backend=backend, optimization_level=1)

# Transpile original experiment
circuit_isa = pm.run(circuit)
observable_isa = observable.apply_layout(circuit_isa.layout)

# Transpile backpropagated experiment
bp_circuit_isa = pm.run(bp_circuit)
bp_obs_isa = bp_obs.apply_layout(bp_circuit_isa.layout)

# Transpile the backpropagated experiment with truncated observable terms
bp_circuit_trunc_isa = pm.run(bp_circuit_trunc)
bp_obs_trunc_isa = bp_obs_trunc.apply_layout(bp_circuit_trunc_isa.layout)

Kita cipta Primitive Unified Bloc (PUB) untuk setiap tiga kes.

pub = (circuit_isa, observable_isa)
bp_pub = (bp_circuit_isa, bp_obs_isa)
bp_trunc_pub = (bp_circuit_trunc_isa, bp_obs_trunc_isa)

Langkah 3: Laksanakan menggunakan primitif Qiskit​

Kira nilai jangkaan​

Akhirnya, kita boleh menjalankan eksperimen yang telah diundurkan dan membandingkannya dengan eksperimen penuh menggunakan StatevectorEstimator tanpa hingar.

ideal_estimator = Estimator()

# Run the experiments using Estimator primitive to obtain the exact outcome
result_exact = (
    ideal_estimator.run([(circuit, observable)]).result()[0].data.evs.item()
)
print(f"Exact expectation value: {result_exact}")
Exact expectation value: 0.8871244838989416

Kita akan menggunakan resilience_level = 2 untuk contoh ini.

options = EstimatorOptions()
options.default_precision = 0.011
options.resilience_level = 2

estimator = EstimatorV2(mode=backend, options=options)
job = estimator.run([pub, bp_pub, bp_trunc_pub])

Langkah 4: Proses pasca dan kembalikan hasil ke format klasik yang dikehendaki​

result_no_bp = job.result()[0].data.evs.item()
result_bp = job.result()[1].data.evs.item()
result_bp_trunc = job.result()[2].data.evs.item()

std_no_bp = job.result()[0].data.stds.item()
std_bp = job.result()[1].data.stds.item()
std_bp_trunc = job.result()[2].data.stds.item()
print(
    f"Expectation value without backpropagation: {result_no_bp} ± {std_no_bp}"
)
print(f"Backpropagated expectation value: {result_bp} ± {std_bp}")
print(
    f"Backpropagated expectation value with truncation: {result_bp_trunc} ± {std_bp_trunc}"
)
Expectation value without backpropagation: 0.8033194665993642
Backpropagated expectation value: 0.8599808781259016
Backpropagated expectation value with truncation: 0.8868736004169483
methods = [
    "No backpropagation",
    "Backpropagation",
    "Backpropagation w/ truncation",
]
values = [result_no_bp, result_bp, result_bp_trunc]
stds = [std_no_bp, std_bp, std_bp_trunc]

ax = plt.gca()
plt.bar(methods, values, color="#a56eff", width=0.4, edgecolor="#8a3ffc")
plt.axhline(result_exact)
ax.set_ylim([0.6, 0.92])
plt.text(0.2, 0.895, "Exact result")
ax.set_ylabel(r"$M_Z$", fontsize=12)
Text(0, 0.5, '$M_Z$')

Output of the previous code cell

Bahagian B: Naik taraf skala!​

Mari kita guna Operator Backpropagation untuk mengkaji dinamik Hamiltonian bagi Rantai Spin Heisenberg 50-Qubit.

Langkah 1: Petakan input klasik kepada masalah kuantum​

Kita pertimbangkan Hamiltonian 50-Qubit H^XYZ\hat{\mathcal{H}}_{XYZ} untuk masalah berskala besar ini dengan nilai yang sama untuk pekali JJ dan hh seperti dalam contoh skala kecil. Observable MZ=1N∑i=1N⟨Zi⟩M_Z = \frac{1}{N} \sum_{i=1}^N \langle Z_i \rangle juga sama seperti sebelum ini. Masalah ini berada di luar kemampuan simulasi brute-force klasik.

num_qubits = 50
layout = [(i - 1, i) for i in range(1, num_qubits)]

# Instantiate a CouplingMap object
coupling_map = CouplingMap(layout)
graphviz_draw(coupling_map.graph, method="circo")

Output of the previous code cell

hamiltonian = generate_xyz_hamiltonian(
    coupling_map,
    coupling_constants=(np.pi / 8, np.pi / 4, np.pi / 2),
    ext_magnetic_field=(np.pi / 3, np.pi / 6, np.pi / 9),
)
print(hamiltonian)
SparsePauliOp(['IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIXXI', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIYYI', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZZI', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIXXIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIYYIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZZIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIXXIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIYYIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZZIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIXXIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIYYIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZZIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIXXIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIYYIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZZIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIXXIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIYYIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZZIIIIIIIIIII', 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 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j,
 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j,
 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j,
 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j,
 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j,
 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j,
 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j,
 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j,
 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j,
 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j,
 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j,
 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j,
 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j,
 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j,
 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j,
 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j,
 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j,
 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j,
 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j,
 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j,
 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j,
 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j,
 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j,
 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j,
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 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j,
 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j,
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 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j,
 1.57079633+0.j, 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j,
 0.39269908+0.j, 0.78539816+0.j, 1.57079633+0.j, 0.39269908+0.j,
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 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j,
 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j,
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 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
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 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j,
 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j,
 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j,
 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j,
 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j,
 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j,
 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j,
 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j,
 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j,
 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j,
 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j,
 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j,
 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j,
 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j,
 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j,
 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j,
 1.04719755+0.j, 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j,
 0.52359878+0.j, 0.34906585+0.j, 1.04719755+0.j, 0.52359878+0.j,
 0.34906585+0.j])
observable = SparsePauliOp.from_sparse_list(
    [("Z", [i], 1 / num_qubits) for i in range(num_qubits)],
    num_qubits,
)
observable
SparsePauliOp(['IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZ', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZI', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IIZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'IZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII', 'ZIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII'],
              coeffs=[0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j,
 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j,
 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j,
 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j,
 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j,
 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j,
 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j, 0.02+0.j,
 0.02+0.j])

Untuk masalah berskala besar ini, kita dah tetapkan masa evolusi sebagai 0.20.2 dengan 44 langkah trotter. Masalah ini dipilih supaya ia melebihi kemampuan simulasi brute-force klasik, tapi boleh disimulasikan menggunakan kaedah rangkaian tensor. Ini membolehkan kita mengesahkan keputusan yang diperoleh melalui backpropagation pada komputer kuantum dengan keputusan ideal.

Nilai jangkaan ideal untuk masalah ini, seperti yang diperoleh melalui simulasi rangkaian tensor, ialah ≃0.89\simeq 0.89.

circuit = generate_time_evolution_circuit(
    hamiltonian,
    time=0.2,
    synthesis=LieTrotter(reps=4),
)
circuit.draw("mpl", style="iqp", fold=-1, scale=0.6)

Output of the previous code cell

Langkah 2: Optimumkan masalah untuk pelaksanaan perkakasan kuantum​

slices = slice_by_gate_types(circuit)
print(f"Separated the circuit into {len(slices)} slices.")
Separated the circuit into 36 slices.

Kita tetapkan max_error_per_slice kepada 0.005 seperti sebelumnya. Walau bagaimanapun, memandangkan bilangan hirisan untuk masalah berskala besar ini jauh lebih tinggi berbanding masalah berskala kecil, membenarkan ralat 0.005 per hirisan mungkin akan menghasilkan ralat backpropagation keseluruhan yang besar. Kita boleh hadkan ini dengan menentukan max_error_total yang mengikat jumlah ralat backpropagation, dan kita tetapkan nilainya kepada 0.03 (lebih kurang sama seperti dalam contoh berskala kecil).

Untuk contoh berskala besar ini, kita benarkan nilai yang lebih tinggi untuk bilangan kumpulan komuting, dan tetapkannya kepada 15.

op_budget = OperatorBudget(max_qwc_groups=15)
truncation_error_budget = setup_budget(
    max_error_total=0.03, max_error_per_slice=0.005
)

Jom kita dapatkan dulu litar backpropagasi dan observable tanpa sebarang pemangkasan.

bp_obs, remaining_slices, metadata = backpropagate(
    observable, slices, operator_budget=op_budget
)
bp_circuit = combine_slices(remaining_slices)

print(f"Backpropagated {metadata.num_backpropagated_slices} slices.")
print(
    f"New observable has {len(bp_obs.paulis)} terms, which can be combined into {len(bp_obs.group_commuting(qubit_wise=True))} groups."
)
print(
    f"Note that backpropagating one more slice would result in {metadata.backpropagation_history[-1].num_paulis[0]} terms "
    f"across {metadata.backpropagation_history[-1].num_qwc_groups} groups."
)
print("The remaining circuit after backpropagation looks as follows:")
bp_circuit.draw("mpl", fold=-1, scale=0.6)
Backpropagated 7 slices.
New observable has 634 terms, which can be combined into 12 groups.
Note that backpropagating one more slice would result in 1246 terms across 27 groups.
The remaining circuit after backpropagation looks as follows:

Output of the previous code cell

Sekarang dengan membenarkan pemangkasan, kita peroleh:

bp_obs_trunc, remaining_slices_trunc, metadata = backpropagate(
    observable,
    slices,
    operator_budget=op_budget,
    truncation_error_budget=truncation_error_budget,
)

# Recombine the slices remaining after backpropagation
bp_circuit_trunc = combine_slices(
    remaining_slices_trunc, include_barriers=False
)

print(f"Backpropagated {metadata.num_backpropagated_slices} slices.")
print(
    f"New observable has {len(bp_obs_trunc.paulis)} terms, which can be combined into {len(bp_obs_trunc.group_commuting(qubit_wise=True))} groups.\n"
    f"After truncation, the error in our observable is bounded by {metadata.accumulated_error(0):.3e}"
)
print(
    f"Note that backpropagating one more slice would result in {metadata.backpropagation_history[-1].num_paulis[0]} terms "
    f"across {metadata.backpropagation_history[-1].num_qwc_groups} groups."
)
print("The remaining circuit after backpropagation looks as follows:")
bp_circuit_trunc.draw("mpl", fold=-1, scale=0.6)
Backpropagated 10 slices.
New observable has 646 terms, which can be combined into 14 groups.
After truncation, the error in our observable is bounded by 2.998e-02
Note that backpropagating one more slice would result in 1226 terms across 29 groups.
The remaining circuit after backpropagation looks as follows:

Output of the previous code cell

Kita dapati bahawa membenarkan pemangkasan membawa kepada backpropagasi tiga hirisan tambahan. Kita boleh mengesahkan kedalaman 2-Qubit bagi litar asal, litar backpropagasi, dan litar backpropagasi dengan pemangkasan selepas transpilasi.

# Transpile original experiment
circuit_isa = pm.run(circuit)
observable_isa = observable.apply_layout(circuit_isa.layout)

# Transpile the backpropagated experiment
bp_circuit_isa = pm.run(bp_circuit)
bp_obs_isa = bp_obs_trunc.apply_layout(bp_circuit_isa.layout)

# Transpile the backpropagated experiment with truncated observable terms
bp_circuit_trunc_isa = pm.run(bp_circuit_trunc)
bp_obs_trunc_isa = bp_obs_trunc.apply_layout(bp_circuit_trunc_isa.layout)
print(
    f"2-qubit depth of original circuit: {circuit_isa.depth(lambda x:x.operation.num_qubits==2)}"
)
print(
    f"2-qubit depth of backpropagated circuit: {bp_circuit_isa.depth(lambda x:x.operation.num_qubits==2)}"
)
print(
    f"2-qubit depth of backpropagated circuit with truncation: {bp_circuit_trunc_isa.depth(lambda x:x.operation.num_qubits==2)}"
)
2-qubit depth of original circuit: 48
2-qubit depth of backpropagated circuit: 40
2-qubit depth of backpropagated circuit with truncation: 36

Langkah 3: Jalankan menggunakan primitif Qiskit​

pubs = [
    (circuit_isa, observable_isa),
    (bp_circuit_isa, bp_obs_isa),
    (bp_circuit_trunc_isa, bp_obs_trunc_isa),
]
options = EstimatorOptions()
options.default_precision = 0.01
options.resilience_level = 2
options.resilience.zne.noise_factors = [1, 1.2, 1.4]
options.resilience.zne.extrapolator = ["linear"]

estimator = EstimatorV2(mode=backend, options=options)
job = estimator.run(pubs)

Langkah 4: Proses pasca dan kembalikan keputusan ke format klasik yang dikehendaki​

result_no_bp = job.result()[0].data.evs.item()
result_bp = job.result()[1].data.evs.item()
result_bp_trunc = job.result()[2].data.evs.item()
print(f"Expectation value without backpropagation: {result_no_bp}")
print(f"Backpropagated expectation value: {result_bp}")
print(f"Backpropagated expectation value with truncation: {result_bp_trunc}")
Expectation value without backpropagation: 0.7887194658035515
Backpropagated expectation value: 0.9532818300978584
Backpropagated expectation value with truncation: 0.8913400398926913
methods = [
    "No backpropagation",
    "Backpropagation",
    "Backpropagation w/ truncation",
]
values = [result_no_bp, result_bp, result_bp_trunc]

ax = plt.gca()
plt.bar(methods, values, color="#a56eff", width=0.4, edgecolor="#8a3ffc")
plt.axhline(0.89)
ax.set_ylim([0.6, 0.98])
plt.text(0.2, 0.895, "Exact result")
ax.set_ylabel(r"$M_Z$", fontsize=12)
Text(0, 0.5, '$M_Z$')

Output of the previous code cell

Tinjauan tutorial​

Sila ambil tinjauan ringkas ini untuk memberikan maklum balas tentang tutorial ini. Pandangan anda akan membantu kami memperbaiki kandungan dan pengalaman pengguna kami.

Pautan ke tinjauan

Note: This survey is provided by IBM Quantum and relates to the original English content. To give feedback on doQumentation's website, translations, or code execution, please open a GitHub issue.

Source: IBM Quantum docs — updated 27 Apr 2026
English version on doQumentation — updated 7 Mei 2026
This translation based on the English version of 9 Apr 2026