Exercise 2.9#
Suppose \(V\) is a vector space with basis vector \(|v_1\rangle = |0\rangle\) and \(|v_2\rangle = |1\rangle \), and Pauli operators are linear operator \(V\to V\). From eq. (2.25) in the book, we could conclude that the outer product representation for a linear operator \(A:V\to V\) is given by
(1)#\[
A = \sum_{i,j}\langle v_j |A|v_i\rangle |v_j\rangle\langle v_i|
\]
where \(|v_i\rangle\) is the \(i-\)th basis for vector space \(V\) and \(\langle v_j|A|v_i\rangle\) is the matrix element \(A_{ji}\) of \(A\). According to eq. (1), for Pauli matrices \(\sigma_0,\sigma_1,\sigma_2,\sigma_3\) shown below,
(2)#\[\begin{split}
\sigma_0 = I = \begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix}, \sigma_1 = X = \begin{pmatrix}
0 & 1 \\ 1 & 0
\end{pmatrix}, \sigma_2 = Y = \begin{pmatrix}
0 & -i \\ i & 0
\end{pmatrix}, \sigma_3 = Z = \begin{pmatrix}
1 & 0 \\ 0 & -1
\end{pmatrix}
\end{split}\]
we have their corresponding outer product representation as
(3)#\[\begin{split}
\begin{align}
\sigma_0 = I &= I_{11} |v_1\rangle\langle v_1| + I_{12} |v_1\rangle\langle v_2|+I_{21} |v_2\rangle\langle v_1|+I_{22} |v_2\rangle\langle v_2| \\
&= |v_1\rangle\langle v_1| + |v_2\rangle\langle v_2| \\
&= |0\rangle\langle 0| + |1\rangle\langle 1| \\
\end{align}
\end{split}\]
(4)#\[\begin{split}
\begin{align}
\sigma_1 = X &= X_{11} |v_1\rangle\langle v_1| + X_{12} |v_1\rangle\langle v_2|+X_{21} |v_2\rangle\langle v_1|+X_{22} |v_2\rangle\langle v_2| \\
&= |v_1\rangle\langle v_2| + |v_2\rangle\langle v_1| \\
&= |0\rangle\langle 1| + |1\rangle\langle 0| \\
\end{align}
\end{split}\]
(5)#\[\begin{split}
\begin{align}
\sigma_2 = Y &= Y_{11} |v_1\rangle\langle v_1| + Y_{12} |v_1\rangle\langle v_2|+Y_{21} |v_2\rangle\langle v_1|+Y_{22} |v_2\rangle\langle v_2| \\
&= -i|v_1\rangle\langle v_2| + i|v_2\rangle\langle v_1| \\
&= -i|0\rangle\langle 1| + i|1\rangle\langle 0| \\
\end{align}
\end{split}\]
(6)#\[\begin{split}
\begin{align}
\sigma_3 = Z &= Z_{11} |v_1\rangle\langle v_1| + Z_{12} |v_1\rangle\langle v_2|+Z_{21} |v_2\rangle\langle v_1|+Z_{22} |v_2\rangle\langle v_2| \\
&= |v_1\rangle\langle v_1| - |v_2\rangle\langle v_2| \\
&= |0\rangle\langle 0| - |1\rangle\langle 1| \\
\end{align}
\end{split}\]