/quantum information – What does “bipartite” mean? (via Qpute.com)
quantum information - What does "bipartite" mean?

quantum information – What does “bipartite” mean? (via Qpute.com)


What does a bipartite system mean?

I’ll summarize the main definitions below (adapted from Quantiki).

  • Bipartite system and states: If Alice’s subsystem is described by the Hilbert space $mathcal{H}_A$ and Bob’s is described by $mathcal{H}_B$, the compound bipartite system is described by the tensor product of the two spaces, $mathcal{H}_Aotimes mathcal{H}_B$. State vectors and density operators on $mathcal{H}_Aotimes mathcal{H}_B$ are called bipartite quantum states.

  • Bipartite pure state: Let $mathcal{H}=mathcal{H}_Aotimes mathcal{H}_B$ be a Hilbert space defined as a tensor product of two Hilbert spaces $mathcal{H}_A$ and $mathcal{H}_B$. We call some pure state $|Psirangle_{AB}$ on the composite system $Acup B$ bipartite, if it is written with respect to the partition $AB$, which means $|Psirangle_{AB} = sum_{ij}c_{ij} |irangle_A|jrangle_B$, where ${|irangle_A}$ and ${|jrangle_B}$ are bases in $mathcal{H}_A$ and $mathcal{H}_B$ respectively. Note that if $mathcal{H}_A$ has a basis ${|irangle_A}$ and $mathcal{H}_B$ has a basis ${|jrangle_B}$ then then ${|irangle_Aotimes |jrangle_B}$ is automatically a basis for $mathcal{H}_Aotimes mathcal{H}_B$.

  • Bipartite mixed state: Let $rho_{AB}$ be a mixed state on a composite system $Acup B$. Then we say $rho_{AB}$ is a bipartite mixed state on $mathcal{H}_Aotimesmathcal{H}_B$ and write $rho_{AB}=sum_{ij}p_{ij}rho_{A}^iotimes rho_{B}^j$ where ${rho_A^i}$ and ${rho_B^j}$ are bases of density operators in $A$ and $B$ respectively. Note that if $mathcal{H}_A$ has a basis of density operators ${rho_A^i}$ and $mathcal{H}_B$ has a basis of density operators ${rho_B^j}$ then then ${rho_A^iotimes rho_B^j}$ is automatically a basis for mixed states on $mathcal{H}_Aotimes mathcal{H}_B$.

Is this just that it “lives” in a tensor product of two Hilbert spaces?

Almost, yes.

Does it mean that the system is separable?

No, not necessarily. You might want to read the Wikipedia page on Separable state.

There exist bipartite pure states $|Psirangle_{AB}$ which cannot be written as $|Psi_1rangle_Aotimes |Psi_2rangle_B$ for $|Psi_1rangle_Ain mathcal{H}_A$ and $|Psi_2rangle_Bin mathcal{H}_B$. A bipartite state $|Psirangle_{AB}$ will separable if and only if it lies in the image of the Segre embedding $$iota: mathbb{C}P^1 times mathbb{C}P^1 to mathcal{P}(mathcal{H}_A otimes mathcal{H}_B) cong mathbb{C}P^3,$$ $$((x_1 : y_1),(x_2 : y_2)) mapsto (x_1x_2:x_1y_2:y_1x_2:y_1y_2)$$ as discussed here.

Similarly, not every mixed state $rho_{AB}$ can be written as a convex combination of product states (i.e. given $p_kgeq 0$ and $sum_kp_k=1$) $$rho_{AB}=sum_kp_krho_A^kotimesrho_B^k$$ where ${rho_A^k}$ and ${rho_B^k}$ are bases of mixed states on subsystems $A$ and $B$ respectively.

Interestingly, unlike bipartite pure states, deciding whether an arbitrary density matrix $rho_{AB}$ is a NP-hard problem cf. Gurvits, 2003 and this TCS SE discussion. So in general there are no nice set of rules to distinguish between entangled and separable bipartite mixed states. This is also an open research problem.

For a detailed treatment of quantum entanglement go through arXiv:quant-ph/0702225.


This is a syndicated post. Read the original post at Source link .