# Can separable, but non-valid "states", be used as a resource to generate entanglement?

Suppose two parties, Alice and Bob, who are separated share the "state" $$\sigma_{AB}=|0\rangle\langle1|_A\otimes |0\rangle\langle1|_B$$. Of course, this not a valid quantum state as it does not satisfy trace 1 and is not Hermitian.

Suppose Alice and Bob have additional qubits 1 and 2 respectively, which are not initially entangled: they are in some separable state $$\rho_1\otimes \rho_2$$. Can they use LOCC operations as well as access to $$|0\rangle\langle1|_A\otimes |0\rangle\langle1|_B$$ to entangle qubits 1 and 2?

Naturally if $$\sigma_{AB}$$ were a valid separable state, then this would not be possible. However, here we relaxing the assumption $$\sigma_{AB}$$ is a valid state.

It's not totally clear if this question is well defined, as the output state of whatever process you perform will not be a valid quantum state. But if we insist on "ignoring" the state of qubits A and B, maybe one can say something about the separability of the states when only considering qubits 1 and 2?

I am also somewhat aware that one definition of entangled states is the set of states which are not able to be generated from separable states with LOCC. However, it is not clear to me that this definition works in this case.

• if it's not a state, what do you mean that they have "access" to it?
– glS
Nov 6, 2022 at 13:35
• @glS As I comment below this can be phrased cleanly mathematically. Of course, physically it does not make much sense. Nov 7, 2022 at 0:30

Since $$\tfrac14\lvert 00\rangle\langle 00\rvert+ \tfrac14\lvert 11\rangle\langle 11\rvert+ \tfrac14\lvert 11\rangle\langle 00\rvert+ \tfrac14\lvert 00\rangle\langle 11\rvert$$ is the maximally entangled state (up to a normalization factor), Alice and Bob can "create it by LOCC" if they have at their disposal the "unphysical states" $$\lvert 11\rangle\langle 00\rvert$$ and $$\lvert 00\rangle\langle 11\rvert$$.