I'm currently learning on LOCC transformations. In the Dur, 2000 paper, there is a statement that

(...) two pure states $|\psi\rangle$ and $|\phi\rangle$ can be obtained with certainty from each other by means of LOCC if and only if they are related by local unitaries (LU).

From what I've studied, we can transform $|\psi\rangle$ to $|\phi\rangle$ or vice versa by means of LU iff there are unitary matrices relating them. For two-qubit state :

$|\psi\rangle = (U_1\otimes U_2)|\phi\rangle$

So, does LOCC equivalence can be thought as capability of transforming two states by means of LU? If that so, how can the 'Classical Communication'(CC) contribute to the equation? I've looked for an explanation on LOCC, but many reference relating it to majorization, which is pretty abstract and hard to understand for me, a physics student.

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    $\begingroup$ the "with certainty" qualifier is crucial here $\endgroup$
    – glS
    Jan 5, 2023 at 17:45
  • $\begingroup$ So, LOCC transformation is the same as LU transformation if we want to transform one state to another 'with certainty'? $\endgroup$
    – Steve J.
    Jan 5, 2023 at 17:53

1 Answer 1


LOCC is a setting which asks about, for example, transforming $|\psi\rangle$ to $|\phi\rangle$, and what is the maximum probability with which that is achieved.

If you restrict to demanding perfect transformations in both directions, then it turns out that this boils down to local unitary equivalence. You can roughly understand this by realising that if you were to perform a measurement, you remove entanglement from the system, and you can never add that entanglement back in again under LOCC. Thus, including a measurement is not reversible.

So, yes, the CC part is irrelevant in the circumstance where you want a reversible transformation. But once you get to the non-reversible regime (even in the case where you want a transformation with certainty), it's very important because that allows you to make measurements, and send the results to the other person, who can take different actions depending on the outcome. For instance, imagine you have the state $$ |\psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle) $$ and you want to transform it, under LOCC, to $$ \alpha|00\rangle+\beta|11\rangle $$ with certainty. You can do this by Alice performing a measurement comprising two (non-projective) measurement elements $$ M_0=\left(\begin{array}{cc} \alpha & 0 \\ 0 & \beta \end{array}\right),\qquad M_1=\left(\begin{array}{cc} \beta & 0 \\ 0 & \alpha \end{array}\right). $$ Once Alice has measured, Alice and Bob either share the state $$ \alpha|00\rangle+\beta|11\rangle,\qquad \alpha|11\rangle+\beta|00\rangle. $$ Once Alice has used the CC to tell Bob the result, they can either both do nothing or both apply $X$ so that, in either case, the output is the desired state.

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    $\begingroup$ Technicality: I think $M_0$ and $M_1$ are not POVM elements, but (generalized, i.e. not necessarily projective) measurement operators. The completeness relation they satisfy reads $\sum_k M_k^\dagger M_k=I$ whereas POVM elements $E_k$ would satisfy $\sum_k E_k=I$. Also, POVM elements do not allow us to uniquely identify the post-measurement state which is needed for the argument here. $\endgroup$ Jan 6, 2023 at 19:10
  • $\begingroup$ @AdamZalcman Language use in QI is consistently sloppy here (i.e. either measurements are called POVMs, etc.), maybe unless you are a mathematical physicist. $\endgroup$ Jan 8, 2023 at 9:27
  • $\begingroup$ Sorry for late response. Does the local operator performed by Alice (in this case, measurement operator) needs to be normalized and orthogonal? Is there any restriction or some properties for local operator in LOCC needs to meet? $\endgroup$
    – Steve J.
    Jan 8, 2023 at 11:59
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    $\begingroup$ @SteveJ. There's no LOCC restriction the limits what the local operators are that can be performed. Here, I'm just applying a more general measurement formalism which, as Adam Zalcman says, is constrained by $\sum_kM_k^\dagger M_k=I$ (unitaries being the case of a single element). $\endgroup$
    – DaftWullie
    Jan 9, 2023 at 14:29

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