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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?

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  • $\begingroup$ In which state do you initialize qubit $|x_1 \rangle$? The plain answer to your question is, if $|x_0 \rangle = |0 \rangle$ is your control qubit, the CNOT gate will not change the state of qubit $|x_1 \rangle$. Hence, if you start with state $|x_1 x_0 \rangle = |10 \rangle$, you will end up with the same state at the barrier. $\endgroup$ – Durd3nT Apr 26 at 6:18
  • $\begingroup$ Just for the nomenclature, it is normally an implicit rule to order qubits by ascending index number. This means that, in a circuit, we name the qubits from top to bottom. In you case, we say the state at the barrier is $|01\rangle$ in ordering $|x_0x_1\rangle$ $\endgroup$ – BrockenDuck Apr 26 at 6:52
  • $\begingroup$ So in order to get $|10>$ with my provided ordering, $x_0$ would need to have $\beta = 1$ and $\alpha = 0$, correct? What would the full two qubit state look like at the barrier (e.g. $|00> + |01> + |10> + |11>$)? $\endgroup$ – QuantumLearner Apr 26 at 22:37
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\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 $|0 \rangle$.


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.

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  • $\begingroup$ So how do I end up in state $|10>$ at the barrier? And what would the full state of the system be, e.g. $|00>+|01>+|10>+|11>$? $\endgroup$ – QuantumLearner Apr 26 at 4:10
  • $\begingroup$ I updated my answer... If your state at the barrier is $|10\rangle$ then that means the state before the CNOT gate is $|10\rangle$ since CNOT gate has no effect on this state... because the controlled qubit is in the state $|0\rangle$. This means $|x_0 \rangle = |0\rangle$ and $x_1 = |1\rangle$ and so $|x_1 \rangle \otimes |x_0 \rangle = |1 \rangle \otimes |0\rangle = |10\rangle$ $\endgroup$ – KAJ226 Apr 26 at 4:51
  • $\begingroup$ I just realized I read the image wrong. $x_0$ is the control qubit, not $x_1$. So an initial value of $|0>$ on $x_1$ makes sense. $\endgroup$ – QuantumLearner Apr 26 at 22:34
  • $\begingroup$ I apologize for the confusion, I've updated the question to hopefully be more clear. $\endgroup$ – QuantumLearner Apr 27 at 2:11

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