@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 $|\vec{n}|=\sqrt{2\text{Tr}(\rho^2)-1}$.

Proof of the basic properties

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.


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