1)I want to use noise model for my state and bit_flip is not defined on cirq.

    rho_13 = cirq.Circuit(
    cirq.CNOT(alice, charlie),

When I wrote this code, I had the error: " bit_flip is not defined" (I used import cirq)

2)For ghz state I wrote this part but I didn't understand I have a syntax error which is very weird maybe I am doing something wrong regarding cirq but I get this error. Can you look at it?

    ghz = cirq.Circuit(  
   for i in range (n-1):
       cirq.CNOT ( qubits [i] , qubits [i+1]),
   cirq.measure (* qubits , key ='x'),. 

The syntax error is just after the "for" and "Invalid syntax"

3)My original state is |GHZ><GHZ|+P*rho. Should I use cirq.final_density_matrix(circuit), in the part of ghz only, or should I use cirq.final_density_matrix(circuit), after defining my all state(I will first define |GHZ><GHZ and then (1-P)rho, and then I will write circuit =|GHZ><GHZ|+(1-P)rho after that cirq.final_density_matrix(circuit)) or should I use cirq.final_density_matrix(circuit) both part for all circuit and |GHZ><GHZ too

  1. I can create entangled state with hadamard and cnot gate but I want to create seperable state. How can I create seperable state with cirq

5)For measurement, Should I measure each part seperately or should I first appending all circuit and then measured together? What I mean is that first I will measure |GHZ><GHZ| and then I will measure P*rho and after that I will add my circuit like circuit =|GHZ><GHZ|+(1-P)rho and simulate or Should I directly add and after that measure the all circuit?

Many thanks from now


1 Answer 1

  1. You are running into basic python issues: trying to call a function with a list [r] instead of a float parameter. Also, the bit_flip channel then needs to be applied on a qubit. A potential fix is here:
import cirq

alice, bob, charlie = cirq.LineQubit.range(3)
rho_13 = cirq.Circuit(
    cirq.CNOT(alice, charlie),
    # you need a probability to have the channel defined
    # then you'll need to apply the channel to a qubit
  1. This is also fraught with Python issues. You are trying to create a circuit - but then passing in a for loop as an argument? cirq.final_density_matrix should not be part of the Circuit construction arguments for sure. Measurement can be part of the circuit but let's deal with that later as you have a question about that too.

  2. Summing states together only makes sense in the density matrix representation. From your three last questions (Producing |GHZ><GHZ| State in Cirq, Multiple Bipartite Entangled State in Cirq and this one) I'm gathering that you would like to put together a state that represents a mixture between the GHZ state and 4 states. So we'll have to

    1. create the density matrix for each of them,
    2. multiply them with the required probabilities
    3. sum them together.
  3. You need to tell us more about what kind of separable state you would like. |000><000| is one of the simplest separable ones - as it is $|0\rangle\langle0| \otimes|0\rangle\langle0|\otimes|0\rangle\langle0|$ - I'm going to assume that that's good enough. But any state that is the result of only local operations (i.e. one qubit operations) should be good enough.

  4. Measurement is not required for your state preparation. If you want to measure your final state, I would add that at the end. Let's cover that as well.

There are two major ways that I can think of to solve this:

  1. using density matrices directly
  2. using mixtures of unitaries

Here is an example for both - at the end the final density matrix is exactly the same.

from typing import Union, Sequence, Tuple, Any

import cirq
import numpy as np

from cirq.type_workarounds import NotImplementedType

# ======== Density matrix based method ============

a, b, c = cirq.LineQubit.range(3)
GHZ_circuit = cirq.Circuit(cirq.H(a),
                           cirq.CNOT(a, b),
                           cirq.CNOT(b, c))

GHZ = cirq.final_density_matrix(GHZ_circuit)

def density_matrix_bipartite_entangled(i, j, qs):
    circuit = biparty_entangle_circuit(i, j, qs)
    return cirq.final_density_matrix(circuit, qubit_order=qs)

def biparty_entangle_circuit(i, j, qs):
    return cirq.Circuit(cirq.H(qs[i]), cirq.CNOT(qs[i], qs[j]))

qs = [a, b, c]

rho01 = density_matrix_bipartite_entangled(0, 1, qs)
rho02 = density_matrix_bipartite_entangled(0, 2, qs)
rho12 = density_matrix_bipartite_entangled(1, 2, qs)

# creates the |+> ⊗ |1> ⊗ |0> state
circuit_separable = cirq.Circuit(cirq.H(a), cirq.X(b))
rho_separable = cirq.final_density_matrix(circuit_separable, qubit_order=qs)

p, q, r, s = 0.5, 0.3, 0.2, 0.1
assert 0 <= q + r + s <= 1
assert 0 <= p <= 1

rho = q * rho01 + r * rho02 + s * rho12 + (1 - q - r - s) * rho_separable
state = p * GHZ + (1 - p) * rho

print(f"final state: \n {state}")

print(cirq.sample_density_matrix(state, indices=[0, 1, 2], repetitions=10))

# ======== Mixture based method ============

class MixtureGate(cirq.Gate):
    def __init__(self, p, q, r, s):
        self.p = p
        self.q = q
        self.r = r
        self.s = s

    def _num_qubits_(self) -> int:
        return 3

    def _mixture_(self) -> Union[Sequence[Tuple[float, Any]],
        p, q, r, s = self.p, self.q, self.r, self.s
        rho01_gate = biparty_entangle_circuit(0, 1, qs).unitary(
        rho02_gate = biparty_entangle_circuit(0, 2, qs).unitary(
        rho12_gate = biparty_entangle_circuit(1, 2, qs).unitary(
        separable_gate = circuit_separable.unitary(

        return [
            (p, GHZ_circuit.unitary()),
            ((1 - p) * q, rho01_gate),
            ((1 - p) * r, rho02_gate),
            ((1 - p) * s, rho12_gate),
            ((1 - p) * (1 - q - r - s), separable_gate),

final_circuit = cirq.Circuit(
    MixtureGate(p, q, r, s)(a, b, c)
circuit_based_state = cirq.final_density_matrix(final_circuit)


# we can do measurements here as well

final_circuit.append(cirq.measure(a, b, c))

r = cirq.DensityMatrixSimulator().run(program=final_circuit, repetitions=10)
print("Measurement results: ", r)

## They are the same, yay!
assert np.allclose(circuit_based_state, state)

  • $\begingroup$ You are just perfect! Thank you very much $\endgroup$
    – quest
    Commented Oct 22, 2020 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.