Intuitions about probabilities relating to evolving a two-qubit state through a CNOT gate

Question

If the initial state of $|x_0\rangle = \alpha |0\rangle + \beta |1\rangle$ and $|x_1\rangle =|0\rangle$, and the final state at the barrier is $|10\rangle$ (in the form $|x_1x_0\rangle$), what would the state of this system be in the form $|00\rangle + |01\rangle + |10\rangle + |11\rangle$?

Since $x_0$ is the control qubit, it seems the CNOT gate only does something when $x_0$ is in the $|1\rangle$ state. Does this mean that $x_0$ would effectively be "copied" onto $x_1$? What would the state of this system be?

Answer 1

\begin{align} CNOT |01\rangle &= CNOT \big( |0\rangle \otimes |1\rangle \big) \\ &= \big( |0\rangle \langle 0| \otimes I + |1\rangle \langle 1| \otimes X \big) \big( |0 \rangle \otimes 1\rangle \big) \\ &= \big( |0\rangle \langle 0| \otimes I \big)|0\rangle\otimes|1\rangle + |1\rangle \langle 1| \otimes X \big) |0 \rangle \otimes |1\rangle \\ &= \big(|0\rangle \langle0|\big)|0\rangle \otimes I|1\rangle + \big( |1\rangle \langle 1| \big)|0 \rangle \otimes X|1\rangle\\ &= |0\rangle\otimes|1\rangle + \vec{0} = |01\rangle \end{align}

So when the controlled qubit is in the state $|0\rangle$, the CNOT gate have no effect on the state.

Qiskit uses little endian so $|01\rangle$ here is essentially the same as your $|10\rangle$ (reading it backward). So the CNOT gate doesn't do anything to the state $|10\rangle$ since the controlled qubit is in the state $|0 \rangle$.

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You end with the state $|10\rangle$ in the circuit above at the barrier because your control qubit is in the state $|0\rangle$ and your target qubit is in the state $|1\rangle$ before the CNOT gate. Thus the state stays the same as $|10\rangle$. Also note that the barrier has no physical meaning.