Given two parties Alice and Bob, a state $\rho_{AB}$ is said to be separable if it can be written as $\rho_{AB}=\sum_i p_i\rho^i_A\otimes\rho^i_B$, with $p_i$ being probabilities and $\rho^i_A,\rho^i_B$ being density matrices on Alice and Bob respectively. State is entangled iff it is not separable.

Define transpose operation $T(|i\rangle\langle j|) = |j\rangle\langle i|$. A state $\rho_{AB}$ satisfies Positive Partial Transpose (PPT) criteria if $T_A\otimes I_B(\rho_{AB})$ is also a density matrix (has positive eigenvalues, and trace is 1). Such state is called a PPT state. Else the state is a NPT state (its partial transpose has negative eigenvalues).

I have following questions:

  1. Is there an operational difference between separable and entangled states? To be more precise: is there a shared task, in which Alice and Bob can perform upto a limit 'L' if they share a separable state, but can always perform better than 'L' if they have an entangled state? 'L' may be understood as the amount of some resource (shared classical bits/quantum entanglement etc etc) extracted in above task.

  2. Is there an operational difference between PPT and NPT states as well?

  • $\begingroup$ 1) Depends on your definition of "task". If you mean the transmission of classical information, then no, if you mean "quantum state", then yes. $\endgroup$
    – CuriousOne
    Commented May 21, 2015 at 2:42
  • $\begingroup$ Yes, a 'task' can include transmission of quantum state. Then what is known about the ability/capacity of entangled state to transmit a quantum state? $\endgroup$ Commented May 21, 2015 at 4:51
  • $\begingroup$ Do quantum teleportation experiments count? It seems to me that one should be able to completely teleport the quantum state of a sufficiently isolated and nearly interaction-free systems with long enough coherence time (e.g. the state of an atoms or even clusters of ultra-cold atoms) to another one of the same kind. That seems to be the extreme example of this kind to me, but maybe I am wrong. I don't quite see how this could be accomplished without entangled states, but I would love to be corrected, if they are actually not required. $\endgroup$
    – CuriousOne
    Commented May 21, 2015 at 6:08
  • $\begingroup$ In case of teleportation experiment, the correct question to ask is what is the fidelity with which one can send an arbitrary state. It is always possible to 'teleport' a quantum state using a separable state, but that is no different from measuring the quantum state, and sending the outcome to Bob. Details can be found in "arxiv.org/abs/quant-ph/9807091". But the problem here is that bound entangled states behave the same way as separable states, when it comes to teleportation (again you can refer to above paper). Thus teleportation is not best way to distinguish in question 1) above $\endgroup$ Commented May 21, 2015 at 8:14
  • $\begingroup$ That separable states don't exceed the potential of ordinary measurements seems intuitive enough. I am a little surprised that one shouldn't be able to get more out of entangled states, that sounds counterintuitive. I will look at the paper to get an idea where my current ideas about teleportation diverge from reality. $\endgroup$
    – CuriousOne
    Commented May 21, 2015 at 8:33

1 Answer 1


This is a very complex question you are asking and very much depends on the notion of "operational" you consider. This paper, for instance, shows that any entangled state can help to do some task better than without it.

Regarding PPT vs. NPT, e.g. PPT states cannot be distilled, while most (?) NPT states can be distilled. (Whether all NPT states can be distilled is indeed one of the big open problems in entanglement theory.)

  • $\begingroup$ Thanks @NorbertSchuch. I think the link of paper is broken. Can you repost it? Is it true that when local dimensions with Alice and Bob are 2*2 or 2*3, then any non-separable state is distillable? I am guessing this because in 2*2 and 2*3 dimensions, PPT is same as separable. $\endgroup$ Commented May 22, 2015 at 3:44
  • $\begingroup$ @anuraganshu I have fixed the link. That any non-separable 2x2 system can be distilled is shown in arxiv.org/abs/quant-ph/9607009. For 2x3, I don't know. Maybe you should ask a separate question. $\endgroup$ Commented May 22, 2015 at 12:27

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