How to prove that a state is pure if and only if it cannot be expressed as a convex combination

Question

Define a pure state to be one that can be expressed in the form $ \rho = | \psi \rangle \langle \psi |$. How can we show that this is equivalent to $ \rho \neq p \sigma_0 + (1-p) \sigma_1 $ for any distinct states $ \sigma_0 \neq \sigma_1 $ and $ 0 < p < 1 $.

Intuitively this statement makes sense to me - I've managed to make some progress by using the fact that $ \mathrm{tr} (\rho ^2 ) = 1 \iff \; \rho \; \text{is pure} $ and then applying various bounds on the trace of a product to reduce it to the case where $ \sigma _0 $ and $\sigma_1 $ are themselves pure states. My only problem this hardly seems very enlightening and I was wondering if there was a more elegant proof for what seems like such a natural physical statement.

Answer 1

An arbitrary state $\rho$ can be written as a convex mixture of pure states, $\rho=\sum_k p_k P_k$ with $P_k$ pure states and $p_k\ge0$ with $\sum_k p_k=1$. Taking the eigendecomposition of $\rho$, we can assume that $P_k$ have orthogonal support.

The state $\rho$ is pure iff it has a single non-zero eigenvalue equal to $+1$ (i.e. iff it is a projector). Therefore $$\left(\sum_k p_k P_k\right)^2=\sum_k p_k^2 P_k + \sum_{i\neq j}p_i p_j \underbrace{P_i P_j}_{=0}= \sum_k p_k^2 P_k=\sum_k p_k P_k.$$ It follows that $p_k^2=p_k$. But $\sum_k p_k=1$ implies that $p_k\in[0,1]$. It must then be the case that $p_k\in\{0,1\}$, and therefore there is a single non-zero $p_k$ equal to $+1$, and $\rho$ is pure.