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

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?

• 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. Apr 26 at 6:18
• 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 $|x_0x_1\rangle$ Apr 26 at 6:52
• 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. $|00> + |01> + |10> + |11>$)? Apr 26 at 22:37

\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.

• So how do I end up in state $|10>$ at the barrier? And what would the full state of the system be, e.g. $|00>+|01>+|10>+|11>$? Apr 26 at 4:10
• 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 $|x_1 \rangle \otimes |x_0 \rangle = |1 \rangle \otimes |0\rangle = |10\rangle$ Apr 26 at 4:51
• 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. Apr 26 at 22:34
• I apologize for the confusion, I've updated the question to hopefully be more clear. Apr 27 at 2:11