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For this question, fix three qubits $q_1, q_2, q_3$. I'll use the notation $U_{123} \in SU(8)$ to denote an arbitrary quantum circuit/unitary on the three qubits, and $U_{12}, U_{23} \in SU(4)$ to denote arbitrary circuits/unitaries on nearest-neighbour qubits $q_1,q_2$ and $q_2, q_3$ respectively. Let $|\psi\rangle = U_{23}U_{12}|000\rangle$.

In $|\psi\rangle$, it is possible for qubits $q_1$ and $q_3$ to be entangled with each other (for example, the GHZ state $1/\sqrt{2}(|000\rangle + |111\rangle)$ can be constructed using a circuit of the form $U_{23}U_{12}$ acting on $|000\rangle$). However, there are no entangling gates that act directly on qubits $q_1$ and $q_3$ in the circuit that defines $|\psi\rangle$. In other words, entanglement between $q_1$ and $q_3$ can only be created by "passing the entanglement through $q_2$". My question is: does this create some restriction on the entanglement between $q_1$ and $q_3$ that can be characterized mathematically?


Something that might be relevant is, I think it is true is that $SU(8)$ unitaries can be approximated by nearest-neighbour $SU(4)$ unitaries, if I allow for arbitrarily many of them: $$ U_{123} = \prod_{i=1}^{K}U_{23}^i U_{12}^i.$$ (the $i$'s are indices, not exponents) This is a consequence of the fact that $\text{CNOT}$ gates plus rotations are computationally universal, together with the fact that the $\text{SWAP}$ circuits are in $SU(4)$, meaning that I can swap qubits $1,2,3$ around and create arbitary circuits using nearest-neighbour circuits with swapped qubits. But what is unique about the $K=1$ case is that you lose the ability to swap qubits $1,2$ or $2,3$, because the restriction in the definition of the circuit implies that you cannot swap them back.


There is a related question on whether entanglement is transitive, but my question is different because here, I am entangling qubits $q_2$ and $q_3$ after qubits $q_1$ and $q_2$ are already entangled. In that question the only assumption that is made is that $q_1, q_2$ and $q_2, q_3$ are entangled, and there they are asking the opposite question of whether it is necessary that $q_1$ and $q_3$ are entangled.

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  • $\begingroup$ +1 from the parts that I understand, but $q_1,q_2,q_3$ are just arbitrary qubits, not necessarily nearest neighbors, right? You have three qubits and the GHZ circuit is pretty small, so it’s reasonable that you don’t need a gate for each pair of qubits... $\endgroup$
    – Mark S
    Jun 2 at 2:21
  • $\begingroup$ I'm using the phrase "nearest neighbours" in the abstract sense, to refer to qubits that are adjacent in my indexing, so that $q_1$ and $q_3$ are not nearest neighbours. The GHZ circuit is just an example to motivate my question; define $U_{12} = CNOT(1,2)H(1)$ and $U_{23} = CNOT(2,3)$, then $|GHZ\rangle = U_{23}U_{12}|000\rangle$. Here, the entanglement between $q_1$ and $q_3$ is "indirect" since no subset of gates acts only on $q_1$ and $q_3$. My question is whether circuits like this can entangle $q_1$ and $q_3$ in a general way, or only in a restricted form. $\endgroup$ Jun 2 at 3:12
  • $\begingroup$ What is the distinction between a general way and a restricted form? Many quantum computers will only have a NN topology once built, relying on SWAPs as you kind of suggest to do much of the work. Are you kind of asking about the amount of entanglement that could be done without SWAPs? $\endgroup$
    – Mark S
    Jun 2 at 3:26
  • $\begingroup$ Essentially yes. The circuit I am considering is an arbitrary circuit on $q_{1}, q_{2}$, followed by an arbitrary circuit on $q_{2}, q_{3}$, and that's the end of the circuit. In principle, one could include swaps, but the utility of applying swaps is lost here, since for the swap to be useful, you must be able to swap back once you're done. However, if I swap 1 and 2, after applying gates to 3, I can no longer swap back in the circuit that I defined. $\endgroup$ Jun 2 at 3:37
  • $\begingroup$ I see, I think. You are asking if you can explore the entirety of the Hilbert space on three qubits with one and two qubit gates but wherein you initially act on two of the three qubits, then act on one of the two and the other not involved in the first set of gates. With such a small Hilbert space I suspect you might never have to have a gate acting on q1 and q3 together but that’s just a guess. $\endgroup$
    – Mark S
    Jun 2 at 4:07
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I'm going to label my 3 parties by A, B, C (just to be awkward). If we consider a bipartition of A|BC, we know that a general state can be written in the form $$ U_A\otimes I_{BC}(\alpha|0\rangle_A|\psi_0\rangle_{BC}+\beta|1\rangle_A|\psi_1\rangle_{BC}) $$ where $|\psi_i\rangle$ are mutually orthogonal and may be entangled. This is just the Schmidt decomposition.

Let's start by creating a state $$ U_A\otimes I_{BC}(\alpha|00\rangle+\beta|11\rangle)_{AB}|0\rangle_C. $$ We can certainly do this with a unitary $U_{AB}$. Now we just need to define a unitary $U_{BC}$ such that $$ U_{BC}|00\rangle=|\psi_0\rangle,\qquad U_{BC}|10\rangle=|\psi_1\rangle. $$ Such unitaries certainly exist - these relations define two columns of the $4\times 4$ matrix, and we just have to complete the columns using any orthonormal basis.

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