A density operator $\rho$ for the pure or mixed state of a qubit can be written in the following general form:
$$\rho = \begin{pmatrix} a+b & c-id \\ c+id & a-b \end{pmatrix}$$
I know $\rho$ is a Hermitian operator and can be decomposed in terms of Pauli matrices. But how is the generalized form on the right hand side of the equation produced?
The general density matrix $\rho$ for a qubit is a $2 \times 2$ complex Hermitian matrix, therefore it can be expanded over the basis $\{I, \sigma_x, \sigma_y, \sigma_z\}$(*), which is an orthogonal basis of the Hilbert space of $2 \times 2$ complex Hermitian matrices:
$$\rho=aI+c\sigma_x+d\sigma_y+b\sigma_z$$
where $a,b,c,d \in \mathbb R$. We therefore have $$\rho = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}+ \begin{pmatrix} 0 & c \\ c & 0 \end{pmatrix}+ \begin{pmatrix} 0 & -id \\ id & 0 \end{pmatrix}+ \begin{pmatrix} b & 0 \\ 0 & -b \end{pmatrix}= \begin{pmatrix} a+b & c-id \\ c+id & a-b \end{pmatrix} $$
However, the density matrix should also satisfy $Tr(\rho)=1$, i.e.
$$2a=1 \to a=1/2$$
we therefore have the more usual form
$$\rho=\frac 1 2 \begin{pmatrix} 1+z & x-iy \\ x+iy & 1-z \end{pmatrix}$$
where $z=2b$,$x=2c$,$y=2d$.
(*) $\sigma_i$ are the Pauli matrices
The way to work out the decomposition is to set your density matrix $\rho$ to be equal to a linear combination of the Pauli matrices: $$ \rho = \sum_{i=0}^3\alpha_i\sigma_i. $$ where $\sigma_0=I$ and and the 1,2,3 matrices are the $x,y,z$ Pauli matrices and the alphas are real numbers. You then do some algebra to calculate the alphas.
