Quantum computing: CNOT gate

Question

A CNOT gate can be written as follow:
$$\begin{pmatrix}1 & 0 & 0 & 0 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & 0 & 1 \\\ 0 & 0 & 1 & 0 \end{pmatrix}$$

So using the CNOT gate on $ \psi = |0\rangle$ and $ \phi = |1\rangle $ can be written as: $$\mathrm{CNOT}(\psi, \phi) =\begin{pmatrix}1 & 0 & 0 & 0 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & 0 & 1 \\\ 0 & 0 & 1 & 0 \end{pmatrix} . \begin{pmatrix} 1 \\\ 0 \\\ 0 \\\ 1 \end{pmatrix}$$

But the heck is that the control qubit ($\psi$) is on state $|0\rangle$, so the target qubit ($\phi$) should not be flipped.

I we do the maths, the result system = $ \begin{pmatrix} 1 \\\ 0 \\\ 1 \\\ 0 \end{pmatrix} $

The target qubit is flipped even if the control qubit is $|0\rangle$, I think that I'm doing something wrong with the maths but I don't know what..

Answer 1

You're doing a direct sum instead of a tensor product of the two vectors; as such, your input state is wrong.

A direct sum operates by concatenating the two vectors: $$ \psi \oplus \phi = \begin{pmatrix}\psi_1\\\psi_2\end{pmatrix} \oplus \begin{pmatrix}\phi_1\\\phi_2\end{pmatrix} = \begin{pmatrix} \begin{pmatrix}\psi_1\\\psi_2\end{pmatrix}\\ \begin{pmatrix}\phi_1\\\phi_2\end{pmatrix} \end{pmatrix} = \begin{pmatrix}\psi_1\\\psi_2\\\phi_1\\\phi_2\end{pmatrix} . $$ In a tensor product, on the other hand, you form a 'vector of vectors', whose entries are products of the components of the initial two vectors: $$ \psi \otimes \phi = \begin{pmatrix}\psi_1\\\psi_2\end{pmatrix} \otimes \begin{pmatrix}\phi_1\\\phi_2\end{pmatrix} = \begin{pmatrix} \psi_1\begin{pmatrix}\phi_1\\\phi_2\end{pmatrix}\\ \psi_2\begin{pmatrix}\phi_1\\\phi_2\end{pmatrix} \end{pmatrix} = \begin{pmatrix}\psi_1\phi_1\\\psi_1\phi_2\\\psi_2\phi_1\\\psi_2\phi_2\end{pmatrix} . $$ You need to be doing the latter, not the former, so that you work in the computational basis, $$ \left\{ |00\rangle = |0\rangle\otimes|0\rangle, |01\rangle = |0\rangle\otimes|1\rangle, |10\rangle = |1\rangle\otimes|0\rangle, |11\rangle = |1\rangle\otimes|1\rangle \right\}. $$ That means, in particular, that the state $|\psi\rangle\otimes |\phi\rangle = |0\rangle \otimes |1\rangle = |01\rangle$ gets represented by the vector $$ \begin{pmatrix} 1\begin{pmatrix}0\\1\end{pmatrix}\\ 0\begin{pmatrix}0\\1\end{pmatrix} \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, $$ which remains invariant under the CNOT matrix you've (correctly) given.

More generally, you seem to be having trouble interpreting vectors in a column format. So, to be clear: the amplitudes in each entry correspond to the probability amplitude that the system is in the corresponding basis state, so that e.g. the state $$ \begin{pmatrix} \alpha \\ \beta \\ \gamma \\ \delta \end{pmatrix} \leftrightarrow \alpha |00\rangle + \beta |01\rangle + \gamma |10\rangle + \delta |11\rangle $$ has a probability $|\beta|^2$ to be in the state $|01\rangle = |0\rangle\otimes|1\rangle$, and so on.

Answer 2

For the CNOT gate matrix as you've written it, the basis for the state space is $\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}$. As such, the input state you started with was actually $|00\rangle + |11\rangle$, and you ended, as you should, with $|00\rangle+|10\rangle$. If you wanted to start with the state $|01\rangle$, you should have input $(0,1,0,0)^T$. Then your output would be $(0,1,0,0)^T$, so the CNOT gate takes $|01\rangle$ to $|01\rangle$, as expected.

Answer 3

In the matrix form of the CNOT gate, the first qubit is assumed to be the control and the second one the target. Therefore, in the vector you provided, $|{C}\rangle = |1\rangle$ and $|T\rangle = |0\rangle$, resulting in $|{C}\rangle = |1\rangle$ and $|T\rangle = |1\rangle$.