I was looking at this question but still don't fully understand the distinction between classically correlated mixed states and entangled mixed states.

I understand that a pure state is considered separable if it can be written as $\sum_i p_i \psi_i\otimes \phi_i$ with only one of the $p_i$ being nonzero (and therefore equal to 1). This is quite intuitive to me.

For mixed states, the definition is that the state cannot be written in the form $\sum_i p_i \rho_i\otimes \sigma_i$. The logic that is stated usually is that the definition above is simply classically correlated mixed states i.e. with probability $p_1$, I produce the state $\rho_1\otimes \sigma_1$ and so on.

  1. Is it then correct that the most general way to write the mixed state is $\sum_{i, j} p_{ij}\rho_i\otimes \sigma_j$?
  2. How do I show that the only representation of well know entangled states such as $\rho = \frac{1}{2}(\vert 00\rangle\langle 00\vert + \vert 00\rangle\langle 11\vert + \vert 11\rangle\langle 00\vert + \vert 11\rangle\langle 11\vert)$ cannot be written in the form above? I have to show that no possible $\rho_i$ and $\sigma_i$ exist to put it into the form $\sum_i p_i \rho_i\otimes \sigma_i$ and I'm not sure how to do that.
  • $\begingroup$ Regarding 2., since your example is pure, you can use any argument which applies to pure states, e.g. the Schmidt decomposition. $\endgroup$ Oct 28, 2018 at 22:56
  • $\begingroup$ If I use the Schmidt decomposition for any entangled pure state $\vert \psi\rangle$ and then take $\vert\psi\rangle\langle\psi\vert$, I find an expression indeed. But how does this prove that there does $\textbf{not}$ exist a $\rho_i$ and $\sigma_i$ that allows the form $\sum_i p_i\rho_i\otimes\sigma_i$? $\endgroup$ Oct 29, 2018 at 10:32
  • $\begingroup$ Pure states are extreme points in the set of density operators. They don't allow for non-trivial decompositions. $\endgroup$ Oct 29, 2018 at 12:06


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