Two-qubit Synthesis

Two-qubit gate synthesis takes in a 4 \times 4 unitary matrix U and synthesizes a circuit with an optimal \text{CNOT} gate count. This pass is implemented as qp.ops.two_qubit_decomposition in PennyLane, which can also be accessed via qp.transforms.unitary_to_rot.

The resulting circuits can take one of the following four forms, depending on how many CNOT gates are necessary.

A, B, C, D \in \text{SU}(2) are general single-qubit gates and, together with the angles \alpha, \beta, \delta \in \mathbb{R}, can be determined algebraically directly from the input matrix U (the exact process is described in the details tab).

In a NISQ setting, where costs for CNOT gates dominate, applying this pass to two-qubit gates may reduce the overall CNOT gate count.

Inputs

• Unitary matrix U \in \mathbb{C}^{4\times 4}

Outputs

• Decomposed circuit with an optimal \text{CNOT} gate count

Example

A tensor product of two operators requires 0 CNOT gates.

import pennylane as qp
import numpy as np

U = qp.matrix(qp.exp(-1j * 0.5 * qp.X(0)), wire_order=range(2))
U @= qp.matrix(qp.exp(-1j * 0.5 * qp.Y(1)), wire_order=range(2))

@qp.transforms.unitary_to_rot
@qp.qnode(qp.device("default.qubit"))
def circuit(U):
    qp.QubitUnitary(U, wires=range(2))
    return qp.expval(qp.Z(0)), qp.expval(qp.Z(1))

>>> print(qp.draw(circuit, level="device")(U))
0: ──RZ(1.57)──RY(1.00)──RZ(11.00)─┤  <Z>
1: ──RZ(0.00)──RY(1.00)──RZ(0.00)──┤  <Z>

Note that a ZYZ decomposition is used to represent the \text{SU}(2) gate on each wire. This leads to a redundancy because the input matrix on wire 1 is already a R_Y rotation.

A general \text{SU}(4) element requires 3 CNOT gates.

>>> SU4_generators = list(qp.pauli.pauli_group(2))
>>> arbitrary_op = qp.dot(np.arange(16), SU4_generators)
>>> U = qp.matrix(qp.exp(-1j * arbitrary_op))
>>> print(qp.draw(circuit, level="device", max_length=80)(U))
0: ──RZ(11.64)──RY(1.86)──RZ(2.97)─╭X──RZ(-0.03)─╭●───────────╭X──RZ(11.57) ···
1: ──RZ(11.68)──RY(2.10)──RZ(5.12)─╰●──RY(1.42)──╰X──RY(1.69)─╰●──RZ(0.04)─ ···

0: ··· ──RY(1.30)──RZ(0.81)─╭GlobalPhase(-0.79)─┤  <Z>
1: ··· ──RY(1.07)──RZ(7.03)─╰GlobalPhase(-0.79)─┤  <Z>

There are intermediate cases with 1 and 2 CNOTs, respectively. Here is an example where 2 CNOT gates are required.

>>> U = qp.matrix(qp.exp(-1j * 0.5 * qp.X(0) @ qp.Y(1)), wire_order=range(2))
>>> print(qp.draw(circuit, level="device")(U))
0: ──RZ(12.57)──RY(1.57)──RZ(11.00)─╭X──RZ(1.00)─╭X──RZ(11.00)──RY(1.57)──RZ(3.14)──┤  <Z>
1: ──RZ(11.00)──RY(1.57)──RZ(1.57)──╰●──RX(0.00)─╰●──RZ(1.57)───RY(1.57)──RZ(11.00)─┤  <Z>

While the CNOT count is optimal, the overall decomposition may not be. This is because this pass always targets a fixed ansatz structure and may introduce some redundant single-qubit gates, as illustrated by this example.

def circ():
    qp.CNOT((0, 1))
    qp.RX(0.5, 0)
    qp.RY(0.5, 1)

>>> print(qp.draw(circ)()) # input circuit
0: ─╭●──RX(0.50)─┤
1: ─╰X──RY(0.50)─┤
>>> U = qp.matrix(circ, wire_order=range(2))()
>>> assert np.allclose(U, qp.matrix(circuit, wire_order=range(2))(U))
>>> print(qp.draw(circuit, level="device", max_length=100)(U)) # synthesized circuit
0: ──RZ(12.14)──RY(0.00)──RZ(11.46)─╭●──RZ(3.10)───RY(0.50)──RZ(11.00)─┤  <Z>
1: ──RZ(1.57)───RY(1.79)──RZ(11.00)─╰X──RZ(11.51)──RY(1.76)──RZ(1.68)──┤  <Z>

Typical usage

This pass is necessary when the input unitary is provided as a matrix rather than a gate decomposition. In a NISQ setting, where costs for CNOT gates dominate, applying this pass to two-qubit gates may reduce the overall CNOT gate count.

References

[1] "Minimal Universal Two-qubit Quantum Circuits", Vivek V. Shende, Igor L. Markov, Stephen S. Bullock, arxiv:quant-ph/0308033, 2003

[2] "Recognizing Small-Circuit Structure in Two-Qubit Operators and Timing Hamiltonians to Compute Controlled-Not Gates", Vivek V. Shende, Stephen S. Bullock, Igor L. Markov, arXiv:quant-ph/0308045, 2003

[3] "Finding Small Two-Qubit Circuits", Vivek V. Shende, Igor L. Markov, and Stephen S. Bullock, preprint, 2004