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I'm attempting to examine the Classical Fisher Information (CFI) for a 1-qubit system in comparison to a 2-qubit system.(PennyLane)

I anticipated that the CFI for the 2-qubit system would be double(at least) that of the 1-qubit. However, in both cases, the CFI is the same, equal to 1.

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Both are flat as 1(1e-8 + 1 = 1.00000001...)

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Circuit for 1-qubit

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Circuit for 2-qubit

In the code:

  • RY(pi/2)
  • Time Evolution for a duration t using Pauli_Z for the 1-qubit and ZZ interaction for the 2-qubit
  • RY(pi/2)
  • Set the sweeping range from 1e-4 to 3*pi + 1e-4, with a step of 1e-1
  • Calculate and plot the CFI.

Would anyone be able to assist me with this issue? Thank you in advance.

import pennylane as qml
from pennylane import numpy as np

# qnode device
dev1 = qml.device('default.mixed', wires = 1)
dev2 = qml.device('default.mixed', wires = 2)

coeffs = [-0.5]
obs_1 = [qml.PauliZ(0)]
obs_2 = [qml.PauliZ(0) @ qml.PauliZ(1)]

hamiltonian_1 = qml.Hamiltonian(coeffs, obs_1)  # Define Hamiltonian
hamiltonian_2 = qml.Hamiltonian(coeffs, obs_2)  # Define Hamiltonian

# 1-qubit circuit
@qml.qnode(dev1)
def circuit_1(phi):    
    qml.RY(np.pi/2, wires=0)
    
    qml.ApproxTimeEvolution(hamiltonian_1, phi, 1)
    
    qml.RY(np.pi/2, wires=0)
    
    return qml.density_matrix(wires=0) 


# 2-qubit circuit
@qml.qnode(dev2)
def circuit_2(phi):    
    qml.RY(np.pi/2, wires=0)
    qml.RY(np.pi/2, wires=1)
    
    qml.ApproxTimeEvolution(hamiltonian_2, phi, 1)  
    
    qml.RY(np.pi/2, wires=0)
    qml.RY(np.pi/2, wires=1)

    return qml.density_matrix(wires=[0, 1]) 


# Sweep for 1e-4 ~ 3pi + 1e-4
PHI = np.arange(1e-4, 3*np.pi + 1e-4, 1e-1)
Data = np.zeros((3, len(PHI)))

DUMMY = np.zeros(len(PHI))

for i in range(len(PHI)):
    Data[0][i] = qml.qinfo.classical_fisher(circuit_1)(PHI[i])
    Data[1][i] = qml.qinfo.classical_fisher(circuit_2)(PHI[i])

# Plot
plt.plot(PHI, Data[0][:], label = '1-qubit')
plt.plot(PHI, Data[1][:], label = '2-qubit')
plt.xlabel('Time')
plt.ylabel('CFI')
plt.grid()

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  • $\begingroup$ Why would you expect the Fisher information to double? From a quick manual calculation it seems that it should indeed be equal for both circuits. $\endgroup$ Nov 25, 2023 at 19:02

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