0
$\begingroup$

I have a short question about Time evolving block decimation (TEBD). During a lecture I was told that this method is very efficient in evolving 1D quantum spin systems with only nearest neighbor interactions. Unfortunately no one bother explaining why this algorithm is so efficient in this case and what exactly its limitations are. Wikipedia only has the following sentence in the introduction

The algorithm [...] is highly efficient when the amount of entanglement in the system is limited, a requirement fulfilled by a large class of quantum many-body systems in one dimension.

but without any further explanation why. So, what exactly is the huge advantage of TEBD when working with 1D quantum spin systems compared to other, more well known, approaches to evolve a system in time?

$\endgroup$
1
  • $\begingroup$ An excellent obscure, and obscured, question! $\endgroup$ Apr 24 at 4:38
0
$\begingroup$

TEBD is an approximation method that relies on an ansatz for the wavefunction in terms of matrix-product states. If you consider the formula for MPS in the link, you see that the different degrees of freedom are coupled in a "one-by-one style".

This works good for spin chains as these use short-range interaction potentials, typically nearest neighbor, and because in 1D, theres no ambiguity in which DOF to include next.

For example start from a two-spin system. TEBD is exact for this case, and corresponds to the Schmidt decomposition, which is basically a SVD representation. If one now adds a further particle, it's like coupling two exact subsystems (the spins at sites 1 and 2 and the spins at sites 2 and 3) to a three-spin system This works well if spin 1 and 3 are not too closely entangled, which is the case because of the NN-interaction, and because in 1D there are no other sites whose interaction has to be included further.

$\endgroup$
2
  • $\begingroup$ Thank you for the answer! Questions: 1) "If you consider the formula for MPS in the link, you see that the different degrees of freedom are coupled in a "one-by-one style"." I'm sorry, but I don't know what you mean by "one-by-one" style here. Could you elaborate? 2) "start from a two-spin system. TEBD is exact for this case," What do you mean here? TEBD evolves a state in time, the Schmidt decomposition dividies a matrix into three matrices. $\endgroup$
    – Sito
    Aug 2 '20 at 18:00
  • $\begingroup$ Yes, the Schmidt decomposition uses three matrices to describe the state, and for two particles this representation is exact, i.e. it is as flexible as the original wavefunction. For three particles, there is no similar decomposition which is exact. One thus applies a heuristic ansatz which connects the first and second particle, the second and third particle, and so on. This is what I loosely called "one-by-one style". And this works well for a 1D chain with short-range interaction, as in this case it is quite natural and unambiguous to couple particles in this way. $\endgroup$
    – davidhigh
    Aug 2 '20 at 18:53

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.