# Can a SLOCC protocol work only in one direction?

Two states $$|\psi\rangle$$ and $$|\phi\rangle$$ are equivalent under SLOCC protocol if $$|\psi\rangle$$ can be converted to $$|\phi\rangle$$ and vice versa via LOCC with a finite probability of success. Thus the protocol can be seen as a two-way street. Can there be a SLOCC protocol which is one-way i.e. $$\psi\rangle$$ can be converted to $$|\phi\rangle$$ with a nonzero probability, but $$\phi\rangle$$ cannot be converted to $$|\psi\rangle$$(or the reverse)?

• It's a bit of a contrived example perhaps, but what about generating a separable state $|\phi\rangle_{AB} = |\phi^{A}\rangle_{A} \otimes |\phi^{B}\rangle_{B}$ from an entangled Bell pair $|\psi\rangle = |00\rangle_{AB} + |11\rangle_{AB}$? Or, more general, any separable measurement?
– JSdJ
Commented May 16 at 9:58
• If this weren't the case, SLOCC equivalence would be a meaningless concept. Commented May 17 at 6:36

For bipartite states, you cannot increase the Schmidt rank under SLOCC (but you can decrease it). So, any pair of bipartite states with different Schmidt ranks serves as an example for a one-way conversion, the simplest being JSdJ's example of a Bell pair to a separable state.

Proof: Vidal tells us that for ordered squared Schmidt coefficients of the initial state ($$\alpha_i: \alpha_i\geq\alpha_{i+1}$$) and target state ($$\beta_i: \beta_i\geq\beta_{i+1}$$), the conversion probability is given by $$\min_l\frac{\sum_{i=l}^n\alpha_i}{\sum_{i=l}^n\beta_i}.$$ Here $$n$$ is the larger of the two Schmidt ranks. So if the initial state has a smaller Schmidt rank, $$\alpha_n=0$$ and the case of $$l=0$$ yields 0. On the other hand, the probability of conversion in the other direction is calculated by $$\min_l\frac{\sum_{i=l}^n\beta_i}{\sum_{i=l}^n\alpha_i}.$$ and the numerator is never 0.

A famous example are the GHZ and the W state, which cannot be converted into each other by SLOCC.