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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.

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    $\begingroup$ Try $\rho^2=\rho$ for pure states. If you insist on using $Tr(\rho^2)<1$ for mixed states then you need to use Schwartz. $\endgroup$ Commented Oct 21, 2020 at 21:31
  • $\begingroup$ @ZeroTheHero Thanks for the suggestions, the latter is exactly what I did and I used Cauchy-Schwarz to bound the trace of the product on the RHS which eventually came out to that the trace of the RHS is less than 1 via tedious symbol pushing - my main question is whether there's a more intuitive proof for a statement that seems physically obvious $\endgroup$
    – backstrapp
    Commented Oct 22, 2020 at 14:01
  • $\begingroup$ I'm not sure if this is the intuition you're looking for but consider using the fact that a) the von Neumann entropy is zero iff the state is pure and b) the von Neumann entropy is concave with equality in the concavity iff the convex combination is a trivial one. $\endgroup$
    – rnva
    Commented Oct 23, 2020 at 1:00

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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.

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  • $\begingroup$ Shouldn't there be a $P_k^2$ term when squaring the initial summation? $\endgroup$
    – backstrapp
    Commented Oct 23, 2020 at 16:05
  • $\begingroup$ @backstrapp $P_k$ are pure states, hence projectors, thus $P_k^2=P_k$ $\endgroup$
    – glS
    Commented Oct 23, 2020 at 16:06

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