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I am creating a 2 qubit entangled state:

    def density_matrix_chsh(i, j, qs):
        circuit = chsh_circuit(i, j, qs)
        return cirq.final_density_matrix(circuit, qubit_order=qs)


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


    qs = [a, b]

    state = density_matrix_chsh(0, 1, qs)

Then I am applying rotation for maximum violation:

th = np.pi/(-2)
th1 = np.pi/(4)
th2 = 0 
random_angles = []
newlist = []
b_big = cirq.kron(cirq.unitary(cirq.ry(th)),cirq.unitary(cirq.ry(th)))
results = []
for choosen in itertools.product([th2, th1], repeat=2): 
        random_angles.append(choosen)
for i in range (1000):
    choosen_combination = random.choice(random_angles)
    a_big = cirq.kron(cirq.unitary(cirq.rz(choosen_combination[0])),cirq.unitary(cirq.rz(choosen_combination[1])))    
    rotated_state = b_big @ a_big @ state @ b_big.conj().T
    measurement_rotated = cirq.measure_density_matrix(rotated_state,indices=[0, 1])
    print("measuring qubits ",measurement_rotated)
    qubit_result = measurement_rotated[0]
    results.append(qubit_result)

For instance here I tried 1000 times and both qubits are either 0 or 1 at the sma time but I could never see one of them 1 and the other one is 0 after measurement. Probably I am doing something wrong and I can't see Thanks for helps

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    $\begingroup$ You'll have a much easier time if you pass a circuit into cirq.Simulator().sample(...) rather than trying to do the matrix math yourself by assembling unitaries ad hoc. E.g. it looks like you applied a_big differently than b_big. $\endgroup$
    – Craig Gidney
    Jan 23, 2021 at 15:17
  • $\begingroup$ well sorry, I did not get your point. I am trying to have chsh measurements so I am rotating my state on bloch sphere with 45° angle and then I am measuring the state and saving the results. But my question is that: I could never see 0 1 or 1 0 for two qubits. I am always seeing 11 or 00. $\endgroup$
    – quest
    Jan 23, 2021 at 18:27
  • $\begingroup$ b_big provides to rotate the state with -90 degree because I have measurement on computational basis $\endgroup$
    – quest
    Jan 23, 2021 at 18:40
  • $\begingroup$ @Balint Pato maybe you have an idea? $\endgroup$
    – quest
    Jan 23, 2021 at 19:40

1 Answer 1

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As Craig said, you applied b_big and a_big differently. The adjoint a_big is missing from your product. Try rotated_state = b_big @ a_big @ state @ a_big.conj().T @ b_big.conj().T. But Craig's right. Using Circuits would be much simpler:

import itertools
import random
import cirq
import numpy as np

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

qs = cirq.NamedQubit("a"), cirq.NamedQubit("b")

th = np.pi/(-2)
th1 = np.pi/(4)
th2 = 0
random_angles = []
newlist = []
b_big = cirq.Circuit(cirq.ry(th)(qs[0]), cirq.ry(th)(qs[1]))
results = []
for choosen in itertools.product([th2, th1], repeat=2):
        random_angles.append(choosen)

print(random_angles)
for i in range (1000):
    choosen_combination = random.choice(random_angles)
    a_big = cirq.Circuit(cirq.rz(choosen_combination[0])(qs[0]),
                         cirq.rz(choosen_combination[1])(qs[1]))
    rotated_state = cirq.Circuit(chsh_circuit(0,1, qs), a_big, b_big, cirq.measure(*qs))
    res = cirq.Simulator().run(program=rotated_state, repetitions=1)
    print("measuring qubits ",res.measurements)
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  • $\begingroup$ if th1 would be np.pi/(2) instead of np.pi/(4), would I need to write rotated_state = b_big @ a_big @ state @ a_big.conj().T @ b_big.conj().T instead of rotated_state = b_big @ a_big @ state @ b_big.conj().T? $\endgroup$
    – quest
    Jan 24, 2021 at 12:36
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    $\begingroup$ No. Because you're working with a density matrix $|\psi\rangle \langle \psi |$ instead of a statevector ($|\psi\rangle$), unitary evolution (gate) $U$ is $U |\psi\rangle \langle \psi | U^\dagger$ instead of $U |\psi\rangle$. So you'll typically need the operator on both "sides". $\endgroup$
    – Balint Pato
    Jan 24, 2021 at 17:05
  • $\begingroup$ so the answer is " yes I need to write for both sides"! and it is not related to angle. I thought mathematically they can cancelled, if they are equal but I saw your point. Thanks! $\endgroup$
    – quest
    Jan 24, 2021 at 18:15

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