How do I interpret the CNOT gate control-target structure after a basis-change?

Question

Consider the product of gates acting on two qubits: $$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \end{pmatrix}. $$ This is a controlled Hadamard, then SWAP, then CNOT. On the basis $$ \begin{pmatrix} 00 \\ 01 \\ 10 \\ 11 \end{pmatrix} $$ This leaves us, before the last CNOT, in the basis, $$ \begin{pmatrix} 00 \\ 1 + \\ 01 \\ 1 - \end{pmatrix}. $$ It's hard for me to see the last CNOT as a control on a 1 state anymore. Am I missing something simple, or has the interpretation just been lost since I changed basis? I guess I'm wondering why even make the usual statement that CNOT (or any controlled gate) is used with a ''control'' and ''target'' since in some simple steps I can very rapidly move away from that interpretation. It seems like it should just be interpreted as another SWAP that just moves around the vector elements. Is my understanding of this correct? Thanks.

Answer 1

The normal interpretation of the controlled-not is certainly only relevant to the standard basis. In fact, if you choose a different basis, such as the $|\pm\rangle$ basis, then the role is entirely reversed, due to the circuit identity

So, you might complain that the name is not fair if it only applies to one particular basis. But you have to give it some name to distinguish it from the other gates, and calling it the same as you would call it in classical computation is more than reasonable.

Yes, as you say, it is a permutation matrix (I guess this is what you mean by 'swap') but that hardly distinguishes it from many other permutation matrices.

By the same token, consider the single-qubit $X$ matrix. Also called the bit-flip. In the computational basis, it flips bit values. But in a different basis, such as the Hadamard basis, it doesn't.

I should also mention that I don't think your calculation of your circuit is correct. If you start with the collection $\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}$, the controlled-Hadamard maps these to $\{|00\rangle,|01\rangle,|1+\rangle,|1-\rangle\}$, and the swap then maps them to $\{|00\rangle,|10\rangle,|+1\rangle,|-1\rangle\}$, which is not what you stated. When you apply the controlled-not on these, some of them (the latter two) will be entangled.

Let me run through one specific example: cNOT applied to $|+1\rangle$. The way that I work this out is the expand it as $(|0\rangle+|1\rangle)|1\rangle/\sqrt{2}$. I apply the controlled-not by using linearity and thinking about exactly the description of how it works on basis states: $|01\rangle\mapsto|01\rangle$ and $|11\rangle\mapsto|10\rangle$, so the output is $(|01\rangle+|10\rangle)/\sqrt{2}$.