# Simple examples of bras and kets

Quantum computing involves lots of matrix multiplication. After seeing definitions, sometimes you just need a few simple examples.

Here’s a matrix:

$$\begin{pmatrix} 3 & 1 & 4\\ 1 & 5 & 9\\ 2 & 6 & 5 \end{pmatrix}$$

We can pull out the second row by multiplying as follows:

$$\begin{pmatrix} 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 3 & 1 & 4\\ 1 & 5 & 9\\ 2 & 6 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 5 & 9 \end{pmatrix}$$

And the third row:

$$\begin{pmatrix} 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & 1 & 4\\ 1 & 5 & 9\\ 2 & 6 & 5 \end{pmatrix} = \begin{pmatrix} 2 & 6 & 5 \end{pmatrix}$$

Or the second column:

$$\begin{pmatrix} 3 & 1 & 4\\ 1 & 5 & 9\\ 2 & 6 & 5 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \\ 6 \end{pmatrix}$$

Here’s the middle element:

$$\begin{pmatrix} 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 3 & 1 & 4\\ 1 & 5 & 9\\ 2 & 6 & 5 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = 5$$

And the top-left element:

$$\begin{pmatrix} 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} 3 & 1 & 4\\ 1 & 5 & 9\\ 2 & 6 & 5 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = 3$$

### Bra-kets

Again, this time using Dirac notation.

Let $$M= \begin{pmatrix} 3 & 1 & 4\\ 1 & 5 & 9\\ 2 & 6 & 5 \end{pmatrix}$$

And define the following kets:

$$|e_o\rangle = \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}$$

$$|e_1\rangle = \begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}$$

$$|e_2\rangle = \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}$$

This means we get the following bras (yes, they are really called that):

$$\langle e_o | = \begin{pmatrix} 1 & 0 & 0 \end{pmatrix}$$

$$\langle e_1 | = \begin{pmatrix} 0 & 1 & 0 \end{pmatrix}$$

$$\langle e_2 | = \begin{pmatrix} 0 & 0 & 1 \end{pmatrix}$$

We can pull out the second row of $$M$$ like so:

$$\langle e_1 | M = \begin{pmatrix}1 & 5 & 9\end{pmatrix}$$

And the third row:

$$\langle e_2 | M = \begin{pmatrix}2 & 6 & 5\end{pmatrix}$$

Or the second column:

$$M |e_1\rangleΒ = \begin{pmatrix} 1 \\ 5 \\ 6 \end{pmatrix}$$

Here’s the middle element:

$$\langle e_1 | M | e_1 \rangleΒ = 5$$

And the top-left:

$$\langle e_0 | M | e_0 \rangleΒ = 3$$

To get the $$6$$, do:

$$\langle e_2 | M | e_1 \rangle = 6$$