Understanding the Quantum Circuit for Bell State Creation with Dynamic Circuit using Qiskit

Question

qr = QuantumRegister(2)
cr = ClassicalRegister(2)

# Creating a Bell circuit
qc_if_else = QuantumCircuit(qr, cr)

qc_if_else.h(0)
qc_if_else.measure(0, 1)

with qc_if_else.if_test((cr[0], True)) as else_:
    qc_if_else.x(0)
with else_:
    qc_if_else.x(1)

qc_if_else.measure(0, 1)

qc_if_else.draw(output="mpl", idle_wires=False)

I am following this Tutorial on dynamic circuits, what I understood from the above piece of code is that we measure the qubit zero and store the value in the register with index 1 (that is the second register) , and then we test if cr[0] is True which will be False as its zero because we didnt measure anything and store it in cr[0] so it will go to the else block which applies X gate on the qubit with index 1 (the second qubit). and then measure the qubit zero and again store it in the classical register with index[1] (the second register).

Is my understanding of the above circuit correct? I dont understand how the above circuit creates a bell circuit? It would be very helpful if someone can break the circuit and explain.

Answer 1

Hadamard gate creates equal superposition on qubit 0. Qubit 1 remains in state $|0\rangle$. Then qubit 0 is measured. If the result is $|1\rangle$, then this qubit is inverted and it is finally in state $|0\rangle$. State of both qubits is therefore $|00\rangle$. If result of measurement is $|0\rangle$, then qubit 1 is inverted, hence it is in state $|1\rangle$. State of both qubits is therefore $|01\rangle$. This is consistent with the tutorial (up to order of qubits). Since probability of qubit 0 being in either state $|0\rangle$ or $|1\rangle$ is 50%, effectively two qubit state is $|00\rangle + |01\rangle$ (up to normalization constant).

However, this state has nothing to do with Bell state because the state is not entangled. It is clearly separable: $|0\rangle \otimes (|0\rangle + |1\rangle)$.