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:
\(M |e_1\rangle = \begin{pmatrix}
1 \\
5 \\
6
\end{pmatrix}\)
Here’s the middle element:
\(\langle e_1 | M | e_1 \rangle = 5\)
And the top-left:
\(\langle e_0 | M | e_0 \rangle = 3\)
To get the \(6\), do:
\(\langle e_2 | M | e_1 \rangle = 6\)