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?
Best Answer
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