# Purity of mixed states as a function of radial distance from origin of Bloch ball

@AHusain mentions here that the purity of a qubit state can be expressed as a function of the radius from the center of a Bloch sphere. The state corresponding to the origin is maximally mixed whereas the states corresponding to the boundary points are pure. So far, I haven't seen this fact written anywhere explicitly. What would be the mathematical proof for this claim?

A density matrix $$\rho$$ has the properties of being Hermitian, non-negative and has trace 1.
Any $$2\times 2$$ matrix can be written in the form $$\rho=\frac{n_0\mathbb{I}+\vec{n}\cdot\vec{\sigma}}{2}.$$ The trace being 1 fixes that $$n_0=1$$, while the Hermitian property imposes that $$\vec{n}\in\mathbb{R}^3$$, where $$\vec{\sigma}$$ is the vector of the 3 Pauli matrices. The eigenvalues are $$\frac12(1\pm\sqrt{\vec{n}\cdot\vec{n}}),$$ so it's non-negative iff $$|\vec{n}|\leq 1$$. In the case where the length is 1, the eigenvalues are 0,1 so $$\rho$$ is a projector onto a pure state.
Now let's consider the purity, $$\text{Tr}(\rho^2)$$. In this way of writing, we have $$\text{Tr}(\rho^2)=\frac12(1+\vec{n}\cdot\vec{n})$$ Clearly, the largest value of the purity is 1, for pure states, and the smallest value corresponds to $$\vec{n}\cdot\vec{n}=0$$, i.e. the maximally mixed state.
Now, how do we depict a state on the Bloch Sphere? We simply plot $$\vec{n}$$, so the distance from the centre of the sphere is given by $$|\vec{n}|=\sqrt{2\text{Tr}(\rho^2)-1}$$.
Let's say we have a density matrix of the form $$\rho=\sum_ip_i|\psi_i\rangle\langle\psi_i|,$$ where $$\sum_ip_i=1$$. Trivially, $$\rho$$ has trace 1 and is Hermitian. Non-negative matrices satisfy the property $$\langle\phi|\rho|\phi\rangle\geq 0\qquad\forall|\phi\rangle.$$ So, consider a state $$|\phi\rangle$$: $$\langle\phi|\rho|\phi\rangle=\sum_ip_i|\langle\phi|\psi_i\rangle|^2.$$ Every term on the right-hand side must be non-negative, and hence $$\rho$$ is non-negative.