[eigenket]

Circuits "without entanglement" are divided up into fixed size blocks. I.e. we have n qubits divided up into blocks of size d. Within each block we can do whatever you like, but in circuits without entanglement you will have no gates which cross between blocks (i.e. for every multiple-qubit gate all of its qubits are within the same block).

[dabeddabed]

Ah I see, searching a bit I think you're referring to multipartite entanglement (https://en.wikipedia.org/wiki/Multipartite_entanglement) don't understand much from what is written there but I believe I got the gist from from your comment, thanks a lot

[eigenket]

Not necessarily multipartite entanglement. Bipartite entanglement is covered in this scenario as well.

They key point is that if a circuit can be divided up into pieces which aren't entangled with each other then you can understand the individual pieces separately, which is very nice. On the other hand if the system can't be divided up in this way you don't really have any choice except deal with the whole thing as one big blob, which is difficult.

[dabeddabed]

Thanks for clarifying, reading your first comment again you say that you can cook up quite easily some other examples of islands like the 3 ones in the article, for the sake of curiosity I would be interested if you could give another comment expanding on that too.

[eigenket]

The obvious dumb classical algorithm for simulating a quantum computer with n qubits runs in O(2^n) time and space. Since 2^log(n) = n is polynomial it follows that "log-space" quantum circuits, I.e. quantum circuits with a logarithmic number of qubits are efficiently classically simulable. These log-space circuits are an obvious "island" in the metaphor of the article and adding more qubits is the "quantum resource which let's you swim in the sea".

Another obvious example is what you get if there is no superpositions at any time. I.e. at every step in the computation the state is a computational basis state. Since we label computational basis states with binary strings, with a little checking you can see that circuits with no superpositions are exactly the same as reversible classical Boolean/logic circuits. These can obviously be efficiently classical simulated so they form another island. Here the relevant quantum resource that frees you from the island is what is called coherence.

There are probably a bunch more examples. I think you can invent one where all your quantum logic gates have to have at least a certain amount of noise in them, but I haven't checked the details on that.

[dabeddabed]

Would be incorrect to consider the example where there is no superposition like a special case of the island with no entanglement? Because you should be able to classically simulate superposed states as long as these are not entangled so states with no superposition should be a particular subset.

[eigenket]

I don't think this argument really works, because you can also classically simulate superposed states as long as they are stabilizer states, or fermionic gaussian states, or log-space states or whatever else. Similarly you can also classically simulate states with arbitrarily large amounts of stabilzer magic if they happen to be fermionic gaussian states or whatever.

I think you can sort-make the argument you want to make, because all computational basis states are also product states, just like all computational basis states are also stabilizer states and also (if they have the right parity) fermionic gaussian states. Where it goes wrong is when you start thinking about gates/circuits instead of states, because (e.g.) CX or CCC..CX gates with arbitrary numbers of controls are (in general) highly entangling gates, but they map computational basis states to other computational basis states.

Does this make sense?