Multiple Bipartite Entangled State in Cirq

Question

I am trying to create this state: rho = = q . rho_{1,2} + r . rho_{2,3} + s . rho{1,3} + (1-q-r-s) . rho_separable

And I wrote this code:

   import random
import numpy as np
import cirq
circuit, circuit2, circuit3   = cirq.Circuit()
p = 0.2
q = 0.1
r = 0.3
alice, bob, charlie = cirq.LineQubit.range(1, 4)
rho_12 = circuit.append([cirq.H(alice), cirq.CNOT(alice, bob)]) 
#circuit.append([cirq.H(alice), cirq.CNOT(alice, bob)]) 
rho_23 = circuit.append([cirq.H(bob), cirq.CNOT(bob, charlie)]) 
rho_13 = circuit.append([cirq.H(alice), cirq.CNOT(alice, charlie)]) 
circuit = rho_12 + rho_23 + rho_13
print(circuit)

In here I have 2 problem:

1)This line is not working: circuit = rho_12 + rho_23 + rho_13

2)I cannot multiply the state with p or q or r. What I mean is that I can't write this line:

rho_12 = circuit.append([cirq.H(alice), cirq.CNOT(alice, bob)]) * q

Could you please show me how I can write this state?

score 0 · Accepted Answer · answered Oct 21 '20 at 21:28

You seem to think append is returning a circuit, instead of modifying the circuit you called it on. circuit.append(op) doesn't return anything, it adds an operation to circuit.

alice, bob, charlie = cirq.LineQubit.range(1, 4)
circuit = cirq.Circuit()
circuit.append([cirq.H(alice), cirq.CNOT(alice, bob)])
circuit.append([cirq.H(bob), cirq.CNOT(bob, charlie)]) 
...

Alternatively, you can make a new circuit for each of the pieces and then add them together:

rho_12 = cirq.Circuit(
    cirq.H(alice),
    cirq.CNOT(alice, bob),
)
...
circuit = rho_12 + rho_23 + rho_13

Multiple Bipartite Entangled State in Cirq

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