Problem : Express the matrix $ \begin{pmatrix} 1 & 3 \\ 3 & 4 \end{pmatrix} $ in terms of the Pauli matrices and the identity matrix.
The Pauli matrices are : $$ \sigma_0 = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad \sigma_1 = \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_2 = \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_3 = \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$
We express the given matrix $ A = \begin{pmatrix} 1 & 3 \\ 3 & 4 \end{pmatrix} $ as a linear combination of the Pauli matrices:
$$ A = a_0 I + a_1 \sigma_x + a_2 \sigma_y + a_3 \sigma_z $$
To find the coefficients $ a_0, a_1, a_2, a_3 $, we use the property that the Pauli matrices (including the identity matrix) form a basis for 2x2 Hermitian matrices.
$$ a_0 = \frac{1}{2} \text{Tr}(A) = \frac{1}{2} (1 + 4) = \frac{5}{2} $$
$$ a_1 = \frac{1}{2} \text{Tr}(A \sigma_x) = \frac{1}{2} \text{Tr} \left( \begin{pmatrix} 1 & 3 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right) = \frac{1}{2} \text{Tr} \left( \begin{pmatrix} 3 & 1 \\ 4 & 3 \end{pmatrix} \right) = \frac{1}{2} (3 + 3) = 3 $$
$$ a_2 = \frac{1}{2} \text{Tr}(A \sigma_y) = \frac{1}{2} \text{Tr} \left( \begin{pmatrix} 1 & 3 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \right) = \frac{1}{2} \text{Tr} \left( \begin{pmatrix} 3i & -i \\ 4i & -3i \end{pmatrix} \right) = \frac{1}{2} (3i - 3i) = 0 $$
$$ a_3 = \frac{1}{2} \text{Tr}(A \sigma_z) = \frac{1}{2} \text{Tr} \left( \begin{pmatrix} 1 & 3 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \right) = \frac{1}{2} \text{Tr} \left( \begin{pmatrix} 1 & -3 \\ 3 & -4 \end{pmatrix} \right) = \frac{1}{2} (1 - 4) = -\frac{3}{2} $$
So the matrix can be written as:
$$ A = \frac{5}{2} I + 3 \sigma_x - \frac{3}{2} \sigma_z $$