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Consider universal resources for measurement based quantum computation, as defined here:

We are now ready to formulate the following definition. A family $\Psi$ is called a universal resource for MQC if for each state $|\phi\rangle$ on $n$ qubits there exists a state $|\psi\rangle \in \Psi$ on m qubits, with $m \geq n$, such that the transformation $$|\psi\rangle 􏰆\rightarrow |\phi\rangle|+\rangle ^{m−n}$$ is possible deterministically (with probability 1) by LOCC.

However, note that $|\phi\rangle|+\rangle ^{m−n}$ is a product state. Its Schmidt coefficient is $1$. Thus, no matter what state $|\psi\rangle$ is, its Schmidt coefficients will always be majorized by $1$. Hence, by Nielsen's theorem, the transformation

$$|\psi\rangle 􏰆\rightarrow |\phi\rangle|+\rangle ^{m−n}$$

will always be possible by LOCC (it may be an inefficient LOCC protocol, but the authors explicitly remark they do not care about efficiency and just care about universality.)

Doesn't it make this definition trivial?

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That only works if you consider the bipartition of the bit that's going to be $|\phi\rangle$, versus the other $m-n$ qubits. Thus, to make that work, you have to be able to do untaries over the entirety of those two systems, which are almost certain to be entangling unitaries, and therefore often cannot be implemented by LOCC (except by consuming some additional entanglement resource).

So no, it's not trivial.

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  • $\begingroup$ I’m not sure I follow. Once we bipartition $|\psi\rangle$, why can’t we implement unitaries over each of the two partitions using LOCC? Each such unitary would be a local operation. $\endgroup$
    – BlackHat18
    Nov 30, 2021 at 8:01
  • $\begingroup$ Yes, that's the argument for Nielsen's theorem. But the baseline question is whether you can achieve the transformation using LOCC on a finer-grained bipartition (i.e. each individual qubit). There are plenty of transformations that I can realise if I artificially lump two qubits together and allow myself transformations as compared to having to perform LOCC transformations on the two qubits individually. $\endgroup$
    – DaftWullie
    Nov 30, 2021 at 8:09
  • $\begingroup$ So, when we bipartition $|\psi\rangle$, and look at the two halves (let them be A and B), the local operations we are allowed, either on half A or on half B, are only adaptive single qubit unitaries? Where do they state that in the paper? $\endgroup$
    – BlackHat18
    Nov 30, 2021 at 8:53
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    $\begingroup$ First line of the abstract. $\endgroup$
    – DaftWullie
    Nov 30, 2021 at 10:09

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