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This is a really easy question, but my mother language is not English and I get confused quite a lot reading Preskill notes.

What does a bipartite system mean? Is this just that it "lives" in a tensor product of two Hilbert spaces? Does it mean that the system is separable? I just want a clear definition of what bipartite means (without any other special conditions on it), so that I can know for sure. Thanks in advance!

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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, $\mathcal{H}_A\otimes \mathcal{H}_B$. State vectors and density operators on $\mathcal{H}_A\otimes \mathcal{H}_B$ are called bipartite quantum states.

  • Bipartite pure state: Let $\mathcal{H}=\mathcal{H}_A\otimes \mathcal{H}_B$ 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 $|\Psi\rangle_{AB}$ on the composite system $A\cup B$ bipartite, if it is written with respect to the partition $AB$, which means $|\Psi\rangle_{AB} = \sum_{ij}c_{ij} |i\rangle_A|j\rangle_B$, where $\{|i\rangle_A\}$ and $\{|j\rangle_B\}$ are bases in $\mathcal{H}_A$ and $\mathcal{H}_B$ respectively. Note that if $\mathcal{H}_A$ has a basis $\{|i\rangle_A\}$ and $\mathcal{H}_B$ has a basis $\{|j\rangle_B\}$ then then $\{|i\rangle_A\otimes |j\rangle_B\}$ is automatically a basis for $\mathcal{H}_A\otimes \mathcal{H}_B$.

  • Bipartite mixed state: Let $\rho_{AB}$ be a mixed state on a composite system $A\cup B$. Then we say $\rho_{AB}$ is a bipartite mixed state on $\mathcal{H}_A\otimes\mathcal{H}_B$ and write $\rho_{AB}=\sum_{ij}p_{ij}\rho_{A}^i\otimes \rho_{B}^j$ 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 $\{\rho_A^i\otimes \rho_B^j\}$ is automatically a basis for mixed states on $\mathcal{H}_A\otimes \mathcal{H}_B$.

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 $|\Psi\rangle_{AB}$ which cannot be written as $|\Psi_1\rangle_A\otimes |\Psi_2\rangle_B$ for $|\Psi_1\rangle_A\in \mathcal{H}_A$ and $|\Psi_2\rangle_B\in \mathcal{H}_B$. A bipartite state $|\Psi\rangle_{AB}$ 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 $p_k\geq 0$ and $\sum_kp_k=1$) $$\rho_{AB}=\sum_kp_k\rho_A^k\otimes\rho_B^k$$ 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.

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  • $\begingroup$ Thanks a lot, your question cleared up almost everything for me. Just to be sure, for the pure bipartite system you say that: $|\Psi\rangle_{AB} = \sum_{ij}c_{ij} |i\rangle_A|j\rangle_B$, where $\{|i\rangle_A\}$ and $\{|j\rangle_B\}$ are bases in $H_A$ and $H_B$ respectively. So I assume writing it like this does not mean that the bipartite system is seperable (as you mentioned), although you can write the "terms" in $\rho$ like $|i\rangle_A|j\rangle_B$. Being seperable you should be able to write them like $(|i_1\rangle_A + |i_2\rangle_A +....) ( |j_1\rangle_B + |j_2\rangle_B + ...)$, right? $\endgroup$ – CFRedDemon May 4 '19 at 11:44
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    $\begingroup$ @CFRedDemon Almost correct. You can write a separable pure state as $$(a_1|i_1\rangle+a_2|i_2\rangle+\cdots)\otimes (b_1|j_1\rangle+b_2|j_2\rangle+\cdots).$$ An arbitrary $|\Psi\rangle_{AB} = \sum_{ij}c_{ij} |i\rangle_A|j\rangle_B$ cannot be written in that form. For example, you won't be able to write $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ as a tensor product of two single-qubit states. But say $\frac{1}{2}(|00\rangle+|01\rangle+|10\rangle+|11\rangle)$ can be written as $$\left(\frac{|0\rangle+|1\rangle}{\sqrt 2}\right)_A\otimes \left(\frac{|0\rangle+|1\rangle}{\sqrt 2}\right)_B$$. $\endgroup$ – Sanchayan Dutta May 4 '19 at 11:53
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    $\begingroup$ Oh yeah I forgot the constants, but that was just because I typed too fast, I think I get it now. Thanks a lot! I was just getting confused with all the terminology in my notes as and the Internet was not clearing it up. $\endgroup$ – CFRedDemon May 4 '19 at 11:56

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