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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.

Solarflare0
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1 Answers1

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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.

DaftWullie
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