A controlled not gate matrix looks like this:$$\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & 1 & 0 \\\end{bmatrix}$$That works great when you only have two qubits and you want to use the first qubit as the control, and the second qubit as the input to be flipped (or not).Is there a method to convert this matrix to use for instance if you had 3 qubits, and wanted to use qubit 1 as the control and qubit 3 as the input that is possibly flipped?Thinking about it logically, I can come up with this:$$\begin{bmatrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\end{bmatrix}$$Is there a better, more formal/generalized way to convert multi qubit gates to use any specified qubits, rather than the first $N$ in an $N$ qubit circuit?
Applying a multi qubit quantum gate to specific qubits
quantum computing
Expand-n-SwapIf you want to apply a gate to a subset of the qubits:Expand operations up to $2^n$x$2^n$ with the tensor product / Kronecker product.Use swap gates to make the target qubits consecutive.Swap to a simple order, apply, swap back.It can be tedious to do all those large matrix multiplications, but swap matrices are sparse and the idea is very straightforward.Controls are EasierIn the case of adding controls to an operation (which applies to the specific example you gave), there's an easier trick. Just expand the operation like normal, but then replace any parts of the matrix that correspond to control-failed inputs with what the identity matrix would be.This is a bit easier to keep track of if you introduce a fake control value $c$ that overrides the tensor product's usual behavior, so that instead of $\begin{bmatrix} c \end{bmatrix} \otimes U = c \cdot U$, you have $\begin{bmatrix} c \end{bmatrix} \otimes U = c \cdot I$ (in other words: when an entry is $c$ you don't tile $U$ over the entry and scale $U$ by the entry; you use $I$ instead of $U$).Define the operation of a qubit-must-be-ON control to be $C = \begin{bmatrix} c&0 \\ 0&1 \end{bmatrix}$. A no-op is $I$, and a NOT is $X$. Then the operation from your question could be computed like this (assuming big-endian order):$CNOT_{3 \rightarrow 1} = C \otimes I \otimes X$$= \begin{bmatrix} c&0 \\ 0&1 \end{bmatrix} \otimes \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} \otimes \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$$= \begin{bmatrix} c\otimes\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}&0\otimes\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} \\ 0\otimes\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}&1\otimes\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} \end{bmatrix} \otimes \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$$= \begin{bmatrix} \begin{bmatrix} c&0 \\ 0&c \end{bmatrix} & \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} \\ \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}&\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} \end{bmatrix} \otimes \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$$= \begin{bmatrix} c&0&0&0 \\ 0&c&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \otimes \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$(note: going to use a single zero to ambiguously mean a 2x2 zero matrix where convenient)$= \begin{bmatrix} c\otimes \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} &0&0&0 \\ 0&c \otimes \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}&0&0 \\ 0&0&\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}&0 \\ 0&0&0&\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} \end{bmatrix}$$= \begin{bmatrix} \begin{bmatrix} c&0 \\ 0&c \end{bmatrix} &0&0&0 \\ 0&\begin{bmatrix} c&0 \\ 0&c \end{bmatrix}&0&0 \\ 0&0&\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}&0 \\ 0&0&0&\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} \end{bmatrix}$$= \begin{bmatrix}c&0&0&0&0&0&0&0 \\0&c&0&0&0&0&0&0 \\0&0&c&0&0&0&0&0 \\0&0&0&c&0&0&0&0 \\0&0&0&0&0&1&0&0 \\0&0&0&0&1&0&0&0 \\0&0&0&0&0&0&0&1 \\0&0&0&0&0&0&1&0 \end{bmatrix}$(now we're done with the tensor products and don't need the $c$'s anymore)$\rightarrow \begin{bmatrix}1&0&0&0&0&0&0&0 \\0&1&0&0&0&0&0&0 \\0&0&1&0&0&0&0&0 \\0&0&0&1&0&0&0&0 \\0&0&0&0&0&1&0&0 \\0&0&0&0&1&0&0&0 \\0&0&0&0&0&0&0&1 \\0&0&0&0&0&0&1&0 \end{bmatrix}$Make sense?