pyqpanda_alg.QAOA.dstate¶

Prepare Dicke state D(n,k) in an n-qubit system with k-Hamming-weight. When k=1, this module is equivalent to W state generation. A build-in W state generation supports implementation on a linear architecture. Ref. https://doi.org/10.1109/QCE53715.2022.00027 Ref. https://doi.org/10.1002/qute.201900015

Functions¶

`prepare_dicke_state`(q_list, k[, compress])	Prepare Dicke state.
`linear_w_state`(q_list[, compress])	Prepare W state with a divide-and-conquer algorithm on the linear architecture device.

Module Contents¶

pyqpanda_alg.QAOA.dstate.prepare_dicke_state(q_list, k, compress=True)¶

Prepare Dicke state.

The Dicke state is defined as \(D_{n}^{(k)} = \sum_{hmw(i)=k} |i \rangle\), which is equally superposition state of all states with the same Hamming weight. The method prepare Dicke state with in \(O(k*log(n/k))\) depth in all-to-all connectivity architecture.

Parameters

q_list: List[int], shape (n,)

Qubit addresses. List size is supposed to be the \(n\) of \(D_{n}^{(k)}\).

k : int, k>0

The target Hamming weight of the Dicke state to be prepared, i.e., the \(k\) of \(D_{n}^{(k)}\).

compress : bool, optional

If True, compress the basic gate implementation with simulated control gates otherwise using basic gate implementation; default is True.

Return

circuit : pyqpanda QCircuit

A pyqpanda QCircuit which assumes the input state is all 0.

Raises

ValueError

If the target Hamming weight is larger than the input qubit number (\(k<n\)), or k is invalid (\(k<0\)), or qubit number is 0 (\(n=0\)).

Reference

[1] Bärtschi A, Eidenbenz S. Short-depth circuits for dicke state preparation[C] 2022 IEEE International Conference on Quantum Computing and Engineering (QCE). IEEE, 2022: 87-96. https://doi.org/10.1109/QCE53715.2022.00027

Examples

from pyqpanda3.core import CPUQVM, QProg
from pyqpanda_alg.QAOA import dstate

n = 4
k = 2
machine = CPUQVM()
qubits = list(range(n))
prog = QProg()
prog << dstate.prepare_dicke_state(qubits, k)
print(prog)
machine.run(prog, shots=1)
results = machine.result().get_prob_dict()
for key, prob in results.items():
    key_hmw = key.count('1')
    if key_hmw == k:
        print(key, prob)

The given example illustrates how to prepare the state \(D_4^{(2)}\). The corresponding quantum circuit is:

          ┌─┐     !                               ┌────┐         ! ┌────┐                ┌────┐
q_0:  |0>─┤X├ ────! ────────────── ────────────── ┤CNOT├──── ────! ┤CNOT├ ───────■────── ┤CNOT├
          ├─┤     !                               └──┬┬┴───┐     ! └──┬─┘ ┌──────┴─────┐ └──┬─┘
q_1:  |0>─┤X├ ────! ────────────── ────────────── ───┼┤CNOT├ ────! ───■── ┤RY(1.570796)├ ───■──
          └─┘     ! ┌────────────┐                   │└──┬─┘     ! ┌────┐ └────────────┘ ┌────┐
q_2:  |0>──── ────! ┤RY(2.300524)├ ───────■────── ───┼───■── ────! ┤CNOT├ ───────■────── ┤CNOT├
                  ! └────────────┘ ┌──────┴─────┐    │           ! └──┬─┘ ┌──────┴─────┐ └──┬─┘
q_3:  |0>──── ────! ────────────── ┤RY(0.927295)├ ───■────── ────! ───■── ┤RY(1.570796)├ ───■──
                  !                └────────────┘                !        └────────────┘

And the probability of all possible state are (with possible floating errors):

0.16666666666666663
0.1666666666666667
0.1666666666666667
0.1666666666666667
0.1666666666666667
0.16666666666666663

which include all states with the same Hamming weight \(k=2\).

pyqpanda_alg.QAOA.dstate.linear_w_state(q_list, compress=True)¶

Prepare W state with a divide-and-conquer algorithm on the linear architecture device.

W state is the special Dicke state where \(k=1\). The special case is compatible to Dicke state preparation while it can be formalized on a linear connectivity device with exactly \(n-1\) depth and \(3n-3\) CNOT gates.

Parameters

q_list: List[int], shape (n,)

Qubit addresses. List size is supposed to be the \(n\) of \(D_{n}^{(1)}\).

compress : bool, optional

If True, compress the basic gate implementation with simulated control gates otherwise using basic gate implementation; default is True.

Return

circuit : pyqpanda QCircuit

A pyqpanda QCircuit which assumes the input state is all 0.

Raises

ValueError

If the input qubit number is zero.

Reference

Cruz D, Fournier R, Gremion F, et al. Efficient quantum algorithms for ghz and w states, and implementation on the IBM quantum computer[J]. Advanced Quantum Technologies, 2019, 2(5-6): 1900015.

Example

from pyqpanda3.core import QProg, CPUQVM
from pyqpanda_alg.QAOA import dstate

n = 3
machine = CPUQVM()
qubits = list(range(n))
prog = QProg()
prog << dstate.linear_w_state(qubits, compress=True)
print(prog)
machine.run(prog, shots=1)
results = machine.result().get_prob_dict()
for key, prob in results.items():
    key_hmw = key.count('1')
    if key_hmw == 1:
        print(key, prob)

The example prepare W state on a 3-qubit system which is linearly connected. The corresponding circuit reads as:

          ┌─┐                ┌────┐
q_0:  |0>─┤X├ ───────■────── ┤CNOT├ ────────────── ──────
          └─┘ ┌──────┴─────┐ └──┬─┘                ┌────┐
q_1:  |0>──── ┤RY(1.910633)├ ───■── ───────■────── ┤CNOT├
              └────────────┘        ┌──────┴─────┐ └──┬─┘
q_2:  |0>──── ────────────── ────── ┤RY(1.570796)├ ───■──
                                    └────────────┘

The resulting state should be like (with possible floating errors):

0.3333333333333333
0.3333333333333334
0.3333333333333334

Each of them is one-Hamming-weight.