How to calculate the bloch vector of a mixed state qubit As I understand there are pure state and mixed state qubits. Pure states can be computed by
$$|\psi\rangle = \cos(\theta/2)|0\rangle + \exp(i \phi) \sin(\theta/2)|1\rangle . $$
Simple enough. You need the angles to find the point on the surface of the sphere.
However, to get the point on the inside you need to use a density vector using identity $I$ and the Pauli matrices?
Why use these matrices and how does it help us find the value?
 A: In general, if $\vert\chi_a\rangle$ and $\vert \chi_b\rangle$ are any two spin states, the density operator
$$
\hat \rho=w_a \vert \chi_a\rangle\langle \chi_a\vert + w_b\vert\chi_b\rangle\langle \chi_b\vert\, ,\qquad w_a+w_b=1\, .
$$
describes a statistical mixture where a fraction $w_a$ of the particles are in the state $\vert\chi_a\rangle$ and a fraction $w_b$ in the state $\vert\chi_b\rangle$.
Note that a Stern-Gerlach magnet oriented along $\vec \langle S_a\rangle =\left(\langle \sigma_{ax}\rangle,\langle\sigma_{ay}\rangle,\langle\sigma_{az}\rangle\right)$ would deflect particles in the state $\vert \chi_a\rangle$ "up", but would split the particles in state $\vert\chi_b\rangle$ into an up and down beam, and conversely a  Stern-Gerlach magnet oriented along $\langle \vec S_b\rangle$ would split the particles in state $\vert\chi_a\rangle$ but only deflect up particles in $\vert\chi_b\rangle$.  This is equivalent to stating that no direction of a Stern-Gerlach magnet would result is an up-only deflection of all particles, or alteratively, that the system described by $\hat \rho$ is not an eigenstate of spin in any direction.
More generally, when $\vert\chi_a\rangle$ and $\vert\chi_b\rangle$ are not orthogonal one can expand on the $\vert \pm\textstyle\frac{1}{2}\rangle$ basis
\begin{align}
\vert{\chi_a}\rangle &=a_+\vert{+\textstyle\frac{1}{2}}\rangle+a_-\vert{-\textstyle\frac{1}{2}}\rangle \\
\vert{\chi_b}\rangle &=b_+\vert{+\textstyle\frac{1}{2}}\rangle+b_-\vert{-\textstyle\frac{1}{2}}\rangle 
\end{align}
so that 
\begin{align}
\vert{\chi_a}\rangle\langle \chi_a\vert = \left( \begin{array}{cc}
a_+a_+^* & a_+a_-^*  \\
a_- a_+^* & a_- a_-^*  \\
 \end{array} \right), 
\end{align}
and similarly from $\vert\chi_b\rangle \langle \chi_b\vert$.  
The density operator becomes 
\begin{align}
\hat \rho&=w_a\left( \begin{array}{cc}
a_+ a_+^* & a_+a_-^*  \\
a_-a_+^* & a_-a_-^*  \\
 \end{array} \right)+w_b\left( \begin{array}{cc}
b_+b_+^* & b_+b_-^*  \\
b_-b_+^* & b_-b_-^*  \\
 \end{array} \right) \\
 &=\left( \begin{array}{cc}
w_a\vert a_+\vert^2+w_b\vert b_+\vert ^2 & w_aa_+a_-^*+w_bb_+b_-^*  \\
w_aa_-a_+^*+w_bb_-b_+^* & w_a\vert  a_-\vert ^2+w_b\vert b_-\vert^2  \\
 \end{array} \right)
\end{align}
You can pick the coefficients $a_\pm$ and $b_\pm$ to your heart's content, making sure that $\vert a_+\vert^2+\vert a_-\vert^2=1$ (likewise for $b_\pm$) and compute using this $\hat\rho$ the vector
$$
\langle \vec S\rangle = \left(\langle \sigma _x\rangle,\langle \sigma _y\rangle,\langle \sigma _z\rangle\right)
$$
with 
$$
\langle \sigma_k\rangle =\hbox{Tr}\left(\hat \rho\sigma_k\right)
$$
and you will find that, unless $w_a=1$ and $w_b=0$ or vice versa, $\rho^2\ne \rho$ and the length of $\langle \vec S\rangle$ is not $1$ so your vector will lie inside the Bloch sphere.  When $\rho^2\ne \rho$, the state is pure and then your vector will be on the surface of the Bloch sphere.
As a simple example, choose $a_+=i, a_-=0$, $b_+=\frac{1}{2}$, $b_-=-\frac{\sqrt{3}}{2}$.  Then
$$
\rho=
\left(\begin{array}{cc}\frac{1}{4}w_b+w_a&-\frac{\sqrt{3}}{4}w_b\\-\frac{\sqrt{3}}{4}w_b&\frac{3}{4}w_b
\end{array}\right)
$$
Note that the trace of this matrix is $w_a+w_b=1$.  It's easy to figure out that
\begin{align}
\langle \vec S_a\rangle &= (0,0,1)\\
\langle \vec S_b\rangle &= \left(-\textstyle\frac{\sqrt{3}}{2},0,-\frac{1}{2}\right)\\
\langle \vec S\rangle &= \left(-\textstyle\frac{\sqrt{3}}{2}w_b,0,(w_a-\frac{w_b}{2})\right)
\end{align}
and that $\langle \vec S\rangle$ has length $\sqrt{w_a^2-w_aw_b+w_b^2}$ so does not lie on the surface of the Bloch sphere unless one of $w_a$ or $w_b$ is $1$ and the other $0$.  In this way, the average $\langle S_{x,y,z}\rangle$ is the statistical average of each $\langle S_{a,xyz}\rangle $ and $\langle S_{b,xyz}\rangle $ with $w_a$ and $w_b$ entering as statistical weights.
