LOCC vs. separable measurements

Question

I am trying to understand a difference between LOCC and separable measurements.

If I get it right in the paper Quantum Nonlocality without Entanglement it was given a set of pure states, which cannot be distinguished by any LOCC measurement, but by a separable measurement. These states are:

\begin{align*} \{&|1\rangle \otimes |1\rangle, |0\rangle \otimes \frac{|0\rangle+|1\rangle}{\sqrt{2}} , |0\rangle \otimes \frac{|0\rangle-|1\rangle}{\sqrt{2}}, \\ &\frac{|0\rangle+|1\rangle}{\sqrt{2}}\otimes |2\rangle, \frac{|0\rangle-|1\rangle}{\sqrt{2}}\otimes |2\rangle ,|2\rangle \otimes \frac{|1\rangle+|2\rangle}{\sqrt{2}}, \\ &|2\rangle \otimes \frac{|1\rangle-|2\rangle}{\sqrt{2}} , \frac{|1\rangle+|2\rangle}{\sqrt{2}}\otimes |0\rangle, \frac{|1\rangle-|2\rangle}{\sqrt{2}}\otimes |0\rangle \} \end{align*} Am I right with my interpretation:

• the states are distinguished by a separable measurement, as the states are mutually orthogonal and factorised: accordingly one get build an observable with these states as eigenstates (call it $O$) and do a measurement of this observable - as an output one gets in which state the system is. The measurement would be something like $$(|1\rangle\langle1| \otimes |1\rangle \langle1|)\rho(|1\rangle\langle1| \otimes |1\rangle \langle1|) +(|0\rangle\langle0| \otimes \frac{(|0\rangle+|1\rangle)(\langle0|+\langle1|)}{2})\rho(|0\rangle\langle0| \otimes \frac{(|0\rangle+|1\rangle)(\langle0|+\langle1|)}{2}) + .. $$

So the key point is, that all of these states are product states.

• For the LOCC measurement the problem is however, that the states are orthogonal, but only globally and not locally, i.e $O \neq A \otimes B$, with $A,B$ observables. Accordingly, no sequence of measurements first on the Alice side and then on Bob side can differentiate between the states?

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From the comments it seems that the factorisation of the states one wants to differentiate is not so crucial.

Could one give me then a definition of separable measurements and LOCC measurements and explain me (maybe on some example) what is the difference between them?

Answer 1

This question is a couple years old, but I thought I would provide some intuition that I found valuable.

Consider first how to discriminate between the following four states

$$\{|0\rangle\otimes |0\rangle, |0\rangle\otimes |1\rangle, |1\rangle\otimes |0\rangle, |1\rangle\otimes |1\rangle\}$$

where two labs each have only half of this system. One lab can measure the subsystem A, and one lab can measure the subsystem B, both in the standard basis. Then they can correspond with each other via email (classical communication, but after the measurements happened) and find out which of the four it was in. Easy.

Now a slightly harder case is

$$\{|0\rangle\otimes |0\rangle, |0\rangle\otimes |1\rangle, |1\rangle\otimes |+\rangle, |1\rangle\otimes |-\rangle\}$$

where $|+\rangle$ and $|-\rangle$ denote the $X$ basis states,

$$|\pm\rangle = \frac{|0\rangle \pm |1\rangle}{\sqrt 2}.$$

This can still be addressed. The first lab measures their qubit, getting zero or one. They let the second lab know of the result, and then they choose the basis to measure in: $Z$ if it was a zero, and $X$ if it was a one. This requires classical communication between the two measurement steps, but this is still allowed in LOCC. So, this is a LOCC distinguishable operator.

But with the example you gave, there is no way for the two parties to (perfectly) identify which of the nine states they have, using only these LOCC operations! One of them has to go first and perform some measurement on their subsystem, their qutrit. But you can see than any measurement will destroy some of the two states, losing information.

It is a strange state of affairs. The labs have no entanglement, their joint state is completely separate. In fact, this mystery state could have been prepared entirely by non-entangling not-very-quantum means: each lab was independently pranked by an unrelated group and had their qutrit set to some unknown state. (How do the researchers know that their qutrits are in one of these nine states? Well, that's a separate question...) And unlike a problem of finding out whether one's qubit is $|0\rangle, |1\rangle, |+\rangle$ or $|-\rangle$, this information that the labs want to have is entirely observable -- the possibilities are orthogonal. The only issue is that it is not locally observable. To compute an answer, one lab will have to carry a qutrit over to the other lab, preform an entangling two-particle unitary on the pair together, and then measure. (Or, equivalently, send the mystery qutrit over a quantum internet connection, which might as well be carrying it over physically.)

I have this handy and awful graphic to provide intuition:

The blue square in the middle corresponds to the first basis state, $|1\rangle\otimes|1\rangle$. The red corresponds to the mixing of the $|0\rangle\otimes\frac{|0\rangle+|1\rangle}{\sqrt 2}$ and $|0\rangle\otimes\frac{|0\rangle-|1\rangle}{\sqrt 2}$, the second and third given basis state. Then the orange is fourth and fifth, and so on.

In this picture, an entangled state would be something that isn't an axis-aligned box. For example, an entangled Bell state would be $\frac{|0\rangle\otimes|0\rangle+|1\rangle\otimes|1\rangle}{\sqrt 2}$, and so would be mixing the bottom-left box and the middle box. All boxes are axis-aligned, and so are separable states on their own.

But to do a local measurement, someone would need to "cut" their axis first. If person A, say, measures "Is this state spanned by $\{|0\rangle,|1\rangle\}$ or by $\ {|2\rangle\}$?", then they would cut the green box and destroy it, losing information.

Clearly this description isn't exact, since some states that would "mix boxes" could still be entangled if they are 2x2; but I think the intuition helps.

My friend asked me if local Hilbert spaces of dimension at most 2 would mean that your separable measurements are always LOCC. With only two parties, this is the case (two qubits with a separable measurement have a LOCC measurement), but with three parties, it is no longer true. The following picture proves this: