10
$\begingroup$

Long-range entanglement is characterized by topological order (some kinds of global entanglement properties), and the "modern" definition of topological order is the ground state of the system cannot be prepared by a constant-depth circuit from a product state, instead of ground states dependency and boundary excitations in traditional. Essentially, a quantum state which can be prepared by a constant-depth circuit is called trivial state.

On the other hand, quantum states with long-range entanglement are "robust". One of the most famous corollaries of quantum PCP conjecture which proposed by Matt Hastings is the No Low-energy Trivial States conjecture, and the weaker case proved by Eldar and Harrow two years ago (i.e. NLETS theorem: https://arxiv.org/abs/1510.02082). Intuitively, the probability of a series of the random errors are exactly some log-depth quantum circuit are very small, so it makes sense that the entanglement here is "robust".

It seems that this phenomenon is some kinds of similar to topological quantum computation. Topological quantum computation is robust for any local error since the quantum gate here is implemented by braiding operators which is connected to some global topological properties. However, it needs to point that "robust entanglement" in the NLTS conjecture setting only involved the amount of entanglement, so the quantum state itself maybe changed -- it does not deduce a quantum error-correction code from non-trivial states automatically.

Definitely, long-range entanglement is related to homological quantum error-correction codes, such as the Toric code (it seems that it is related to abelian anyons). However, my question is that are there some connections between long-range entanglement (or "robust entanglement" in the NLTS conjecture setting) and topological quantum computation? Perhaps there exists some conditions regarding when the correspondent Hamiltonian can deduce a quantum error-correction code.

$\endgroup$
  • 1
    $\begingroup$ It is worth mentioning that long-range and robust entanglement are not the same thing. Think of the GHZ state: it requires an $O(N)$ depth circuit to prepare, but certainly is not robust! $\endgroup$ – DaftWullie May 15 '18 at 14:49
  • $\begingroup$ @DaftWullie I clarified the statement. "rubost entanglement" in the NLTS conjecture setting is a confusing term since the state itself may be changed, even the amount of entanglement remains same. $\endgroup$ – Yupan Liu May 22 '18 at 21:37
7
$\begingroup$

There were two simultaneous PRLs published by Kitaev & Preskill and Levin & Wen that I think answer your question.

These use the area law of entanglement seen by states that can be expressed as ground states of a Hamiltonian with only local interactions.

Specifically, suppose you have a 2D system of interacting particles in a pure state. You then single out some region, and calculate the von Neumann entropy of the reduced density matrix for that region. This will essentially be a measure of how entangled the region is with its complement. The area law tells us that this entropy, $S$, should obey

$S = \alpha L - \gamma + \ldots$

Here $L$ is the length of the perimeter of the region. The first term accounts for the fact that correlations in these systems are typically short range, and so the entanglement is mostly composed of correlations between particles on each side of the boundary.

The $\gamma$ term is unaffected by the size or shape of the region, and so represents a contribution of global and topological effects. Whether this is non-zero, and what the value is, tells you about the topologically ordered nature of your entangled system.

The $\ldots$ term just represents contributions that decay as the region increases, and so can be ignored as $L\rightarrow \infty$.

The two papers, and ones based upon them, then find ways to isolate and calculate $\gamma$ for different entangled states. The value is shown to depend on the anyon model for which these entangled states represent the vacuum.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.