How to interpret the matrix representation of a quantum gate?

Question

I am trying to understand how the quantum gates work, so I started with the simplest one, the Pauli X gate.

I get that it turns $|0\rangle$ into $|1\rangle$ and $|1\rangle$ to $|0\rangle$.

So my question is how do I read its matrix:

\begin{equation} X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \end{equation}

What maps to it?

Answer 1

Lets understand this first using Bra-Ket notation then we move on to answering : what does a matrix maps by just looking at it.

Lets get some basics out of the way, the vectors $\vert 0 \rangle$ and $\vert 1 \rangle$ are represented by,

\begin{equation} \vert 0 \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \ \ \ \ \ \& \ \ \ \vert 1 \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}. \end{equation}

This implies, \begin{equation} \vert 0 \rangle \langle 1\vert = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \ \ \ \ \ \& \ \ \ \vert 1 \rangle \langle 0\vert = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}. \end{equation}

Now, I can write my $X$ gate as, \begin{equation} X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} = \vert 0 \rangle \langle 1\vert + \vert 1 \rangle \langle 0\vert . \end{equation}

The term $\vert 1 \rangle \langle 0\vert$ maps $\vert 0 \rangle \rightarrow \vert 1 \rangle$, and the term $\vert 0 \rangle \langle 1\vert$ maps $\vert 0 \rangle \rightarrow \vert 1 \rangle$. You can see this by:

\begin{equation} \left(\vert 0 \rangle \langle 1 \vert\right) \vert 1\rangle = \vert 0 \rangle \ \ \ \ \ \& \ \ \ \left(\vert 1 \rangle \langle 0 \vert\right) \vert 0\rangle = \vert 1 \rangle. \end{equation}

Given a matix you can always write it in this bra-ket notation, and easily see how it maps.

As you gain experience working with these it is easy to see that a general matrix pictorially can be represented as, \begin{equation} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a\vert 0 \rangle\langle 0 \vert & b \vert 0 \rangle\langle 1 \vert \\ c\vert 1 \rangle\langle 0 \vert & d\vert 1 \rangle\langle 1 \vert \end{bmatrix} \end{equation} Note the word pictorially, cause the way I wrote it, in standard text it would imply a $4\times4$ matrix. Now given a matrix you can easily work out how each of the non-zero term acts on the basis states $\vert 0 \rangle, \vert 1 \rangle$ ans see where the matrix maps to.

For example, \begin{equation} Z = \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix} \end{equation} has non-zero terms $\vert 0 \rangle\langle0 \vert$ which maps $\vert 0 \rangle \rightarrow \vert 0 \rangle$, and $-\vert 1 \rangle\langle1 \vert$ which maps $\vert 1 \rangle \rightarrow -\vert 1 \rangle$.

Now a homework for you, Understand the mapping of Hadamard gate and then move on to understand $CNOT$ gate.

Note: I did not worry about what $\vert 1 \rangle \langle 0\vert$ maps $\vert 1 \rangle$ to, cause of orthogonality.

EDIT : After writing this I realized maybe you are asking the following

Lets say someone gives you a mapping and ask you to find the matrix that applies the said mapping. Note that $$ \vert x \rangle \langle b \vert $$ maps $\vert b \rangle $ to $\vert x \rangle$. So as an example if one asks you find the matrix that maps $\vert 0 \rangle \rightarrow (\vert 0 \rangle + \vert 1 \rangle)/\sqrt{2}$ and $\vert 1 \rangle \rightarrow (\vert 0 \rangle - \vert 1 \rangle)/\sqrt{2}$ you can easily write the matrix being

$$ \frac{\vert 0 \rangle + \vert 1 \rangle}{\sqrt{2}}\langle 0 \vert + \frac{\vert 0 \rangle - \vert 1 \rangle}{\sqrt{2}}\langle 1 \vert $$

$$ =\frac{1}{\sqrt{2}}(\vert 0 \rangle\langle 0 \vert + \vert 0 \rangle\langle 1 \vert + \vert 1 \rangle\langle 0 \vert - \vert 1 \rangle\langle 1 \vert) = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix} $$