People who code: we want your input. Take the Survey

New answers tagged

2

Is it possible to construct such a black-box gate ?? No, the gate you're describing isn't possible. It's not unitary. You can't condition on amplitude thresholds, you can only condition on orthogonal states.


2

You're basically talking here about a quantum cellular automaton. I found a summary here, which I'm not particularly familiar with but looks good at first glance. The papers that I used in the dim and distant past when thinking about this are here and here. Two caveats: (i) they don't restrict themselves to binary. You could surely make an equivalence if you ...


2

... depending on the real-time error rate, I might want to use different qubits on the hardware. Qiskit provides a transpiler pass that chooses a noise-adaptive layout based on current calibration data for the backend; NoiseAdaptiveLayout. To use it, first transpile your circuit with the parameter layout_method equals "noise_adaptive" ...


3

Why not just use initial_layout method? For instance, from qiskit import IBMQ, QuantumCircuit from qiskit.aqua import QuantumInstance circuit = QuantumCircuit(2,2) circuit.h(0) circuit.x(1) quantum_instance = QuantumInstance(backend= provider.get_backend('ibmq_santiago'), shots = 8192, ...


0

You can use the KAK decomposition method described here. And if you are using Qiskit, this method is implemented in TwoQubitBasisDecomposer class. z = cmath.rect(1, np.pi / 4) # e^(iπ/4) # Your two-qubit entangling gate as numpy array: basis_unitary = 0.5 * np.array([ [1j*z, z, z, -1j*z], [1/z, 1j/z, -1j/z, 1/z], [1/z, -1j/z, 1j/z, 1/z], [-...


5

Welcome to Quantum Computing StackExchange. To see why it would require $2^n$ complex numbers instead of $2n$ to represent a general entangled $n$-qubit state, let's assume we have a 3-qubit system, $(\alpha_1, \beta_1), (\alpha_2, \beta_2), (\alpha_3, \beta_3)$ Where $(\alpha_i, \beta_i)$ denotes the representation of the $i^{th}$ qubit in a classical ...


Top 50 recent answers are included