# 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?

## 1 Answer

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.

• Yes. It is really nice. Thanks, @ZeroTheHero. Could you suggest some reference providing the explanation along similar lines? – W. Voltera Aug 31 '18 at 15:30
• @user149973 the canonical reference is by Karl Blum, Density Matrix Theory and Applications, springer.com/us/book/9783642205606 – ZeroTheHero Sep 1 '18 at 0:24
• @ZeroTheHero Nice answer! How to calculate $w_a$ and $w_b$? – Nigel1 Nov 18 '20 at 21:59
• @Nigel1 they would be given as initial data of the problem. – ZeroTheHero Nov 18 '20 at 22:16