Does the CX gate always leave the first qubit unchanged and swap the amplitudes of the second one?

Question

I'm a self-learner of quantum computing and is at the very beginning. I do have some math and coding background though. I'm trying to understand how working with entangled qubits helps performing multiple computations at once, any my first confusion is with the description of the CX gate. It is said, that this is a "controlled not" gate, which swaps the amplitudes of the second qubit only if the first qubit is in the state of $1$. But,

1. this description seems strange to me, as to know in which state the first qubit is I guess we need to measure it,

2. the matrix transformation seems to always leave the first qubit unchanged and swap the amplitudes of the second one no matter what, indeed

$$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{bmatrix} \cdot\begin{bmatrix}\phi_0\\\phi_1\\\psi_0\\\psi_1\end{bmatrix} = \begin{bmatrix}\phi_0\\\phi_1\\\psi_1\\\psi_0\end{bmatrix} $$

Answer 1

1. There is a difference between an agent's knowledge of the state of the system and the state of the system itself in quantum theory. If you would need to know the true state of the system in order to perform the computation then you would need to measure it, and therefore the state would never be in a superposition and only classical computation follows. Nevertheless, quantum theory allows for agents to act in a way that true knowledge of the state is not needed, but where for no matter which case the state would change accordingly. In the CNOT we don't know what is the true state of the system until we measure, but we do know the possibilities or paths that the quantum state might do, and this is something allowed (similarly in a double slit experiment you do not need to know the true path of the particle to know that if the evolution is according to quantum theory one will observe an interference pattern); the CNOT unitary changes those possibilities because it acts on the quantum states. This is very tricky about quantum theory and so it takes a while to get used to this way of reasoning, but importantly is that the dynamics crucially changes if we update $\vert \psi \rangle$ and in quantum theory we are allowed to make unitary changes in $\vert \psi \rangle \to U\vert \psi \rangle$ whithout learning about $\vert \psi \rangle$'s state in a particular moment.
2. If you note, for the state $\vert 00 \rangle$ the vector would read $(1,0,0,0)$ and if the state is $\vert 01 \rangle$ the vector would read $(0,1,0,0)$ therefore even if the matrix is acting, the matrix is changing zero vectors and therefore nothing happens. Your confusion comes from the assumption that those elements in the vector state are always different (or non-zero), but they are linked to the fact that the first qubit has decomposition with terms having $\vert 10 \rangle$, $\vert 11 \rangle$, therefore having a one in the first qubit; when this does not happen, third and fourth entries are equal to zero and the state is left unchanged. Let's see some examples. For the state $\vert 11 \rangle$ the vector would read $(0,0,0,1)$ and therefore one has $(0,0,1,0)$ which is $\vert 10 \rangle$ as wanted. It is also interesting to see the case in which we have a state of the form $\vert+\rangle\vert0\rangle = \frac{1}{\sqrt{2}}(\vert 00 \rangle + \vert 10 \rangle) \stackrel{CNOT}{\to} \frac{1}{\sqrt{2}}(\vert 00 \rangle + \vert 11 \rangle)$. Before the application the state was separable, we have used that matrices are linear operations and this made the state of the system entangled. This is probably one of the reasons why the CNOT gate is so important.