# Can two distinct spatially separated many-body systems in the ground state contain entangled particles?

In particular, I am asking if two distinct many-body systems (e.g. system A and system B) separated at some arbitrary distance will necessarily be found to contain entangled particles (such that particle A1 is entangled with particle B1...etc) simply as a consequence of them both being forced into a ground state? Specifically, can two experimenters go off to separate labs with no preparations made and then force their systems to the ground state and expect to later find through conventional testing of entanglement and sharing of statistical results that in fact they did share entangled particles? Or will any entanglement only be coincidental? The reason I ask is because I'm trying to find a way in which entanglement can be acheived by two separated labs remotely (without having to meet up..though classical comms would be allowed) since entangled states are so fragile. It is impractical, it seems, to meet up to prepare entanglement in one lab and then try to keep it from falling apart as you migrate to separate labs to take advantage of its utility or to test its sensitivity to distance (I'm assuming there is none, of course). Has this been achieved and if so, how is it achieved? Also, I have read a paper that documents that two distinct subsystems of a many-body system are entangled when the many-body system as a whole is in the ground state. What I am asking here is that if that can be possible because the particles making up the many-body system do not have identity, why can't it be true of the particles comprising two separate many-body systems? Why are they not equally indistinguishable, if in fact they are not? Thanks

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No.

Since you say that the two labs are at some arbitrary distance, this means there cannot be any interaction between them (since all known interactions in physics vanish at a long enough distance). This means the total Hamiltonian of the two systems is a sum of the Hamiltonians of each system:

$$H = H_1 + H_2,$$

and as Hamiltonians of different systems, they commute: $[H_1,H_2]=0$.
Each Hamiltonian has its own ground state:

$$H_1 |g_1\rangle = E_{g1} |g_1\rangle \\ H_2 |g_2\rangle = E_{g2} |g_2\rangle,$$

and then the ground state of the total Hamiltonian will just be the tensor product of the two states: $$|g\rangle = |g_1\rangle \otimes |g_2\rangle.$$

This state obviously has no entanglement.

When you talk about two subsystems of the same system, that means you allow an interaction between the systems. This interaction will take the form of an interaction term in the total Hamiltonian:

$$H = H_1 + H_2 + H_{int},$$

And the interaction term needs not commute with either $H_1$ or $H_2$. In this case it is possible that the ground state will have entanglement between the two subsystems.

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Yes in the sense that technically everything is entangled to everything else, if you were to consider a wavefunction representing the entire universe it would have some (probably small) level of entanglement of everything with everything else. However for practical purposes the answer is no.

The bigger issue here is that the entanglement is with all the environment around each of the systems A and B (i.e. the labs, maybe a vacuum chamber?) is much stronger than the entanglement between A and B unless you do something clever in order to link the two together.

Being forced into the ground state means very little, it's just the ground state of the approximate Hamiltonian you consider each system to be influenced by, really it is connected to the other system (and the rest of the world) but by an amount that is usually made to be as small as possible.

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The constraint you are describing defines a class of operations commonly used in quantum information theory called Local Operations Classical Communication, or LOCC. As you outlined, in LOCC you can only operate locally on each of your systems, and can only communicate classically between the two systems.

One of the main reasons this class of operations is interesting is precisely because it can not increase the entanglement between two systems. I've not been able to find a source with a good proof of this, but I think in your scenario it's relatively straightforward:

Consider your system $\rho$ which is a density operator in the space of operators on the joint hilbert space of the two systems.

$$\rho \in \mathbb{L}(\mathbb{H}_A \otimes \mathbb{H}_B)$$

You've stated that your systems start out non-entangled

$$\rho=\rho_A \otimes \rho_B$$ where $\rho_A \in \mathbb{L}(\mathbb{H}_A)$ and $\rho_B \in \mathbb{L}(\mathbb{H}_B)$

Now you only perform an operation $O$ which is local, but may be a function of a classical bit strings $S_A$, $S_B$ sent between parties:

$$O=O_A[S_B] \otimes O_B[S_A]$$ where $O_A[S_B] \in \mathbb{L}(\mathbb{H}_A)$ and $O_B[S_A] \in \mathbb{L}(\mathbb{H}_B)$

$$O \rho =(O_A[S_B] \rho_A) \otimes (O_B[S_A] \rho_B)=\gamma_A[S_B] \otimes \gamma_B[S_A]$$ where $\gamma_A[S_B] \in \mathbb{L}(\mathbb{H}_A)$ and $\gamma_B[S_A] \in \mathbb{L}(\mathbb{H}_B)$

this operation obviously produces a new density operator which is non-entangled, regardless of the bit strings sent, or the operations performed.

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