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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?

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  • $\begingroup$ append doesn't return anything, it adds into the circuit. Your circuit is in the circuit variable already; no need to add those things together. $\endgroup$
    – Craig Gidney
    Oct 21, 2020 at 21:18
  • $\begingroup$ oh thank you very much and what about multiplying q or r or p. I can not write that: rho_12 = circuit.append([cirq.H(alice), cirq.CNOT(alice, bob)]) * q @CraigGidney $\endgroup$
    – quest
    Oct 21, 2020 at 21:26
  • $\begingroup$ You can multiply circuits like multiplied_circuit = cirq.Circuit(op1, op2) * 3. $\endgroup$
    – Craig Gidney
    Oct 21, 2020 at 21:30
  • $\begingroup$ Hello, Thanks for the amswer but I want to multiply float number, we can think it like noise so with float number I cannot multiply $\endgroup$
    – quest
    Oct 21, 2020 at 22:06

1 Answer 1

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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
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  • $\begingroup$ Many thanks :) A small question: I wrote the same thing for ghz state too but I have a syntax error and I could not find. Maybe you can see can you have a look? ghz = cirq.Circuit( cirq.H(qubits[0]), for i in range (n-1): cirq.CNOT ( qubits [i] , qubits [i+1]), cirq.measure (* qubits , key ='x'), cirq.final_density_matrix(circuit), ) $\endgroup$
    – quest
    Oct 21, 2020 at 21:49

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