# How can I converge to a continuum harmonic trap ground state?

+1 vote

I would like to investigate bosons in a 1D harmonic trap, and was hoping that this could be done by using DMRG in the continuum limit of having many more sites than particles.

As a test case, I have simulated one boson in a trap with the Hamiltonian

$\hat H_N = -t \sum_{i=1}^{M-1} (\hat a_i^\dagger \hat a_{i+1} + H.c.) + \sum_{i=1}^M \Big[ \frac{\omega}{2} (x_i - x_c)^2 \hat n_i\Big]$

However, the density, as measured with

$\langle \psi | \hat n_i |\psi \rangle$

does not converge towards the harmonic oscillator ground state for up to 100 sweeps on 1000 sites with small lattice constant in a system large enough that the particle does not feel the walls of the box. All other DMRG parameters were also generously set. It seems that for t >> \omega, the system tends towards the particle in a box ground state (which converges nicely if there is no trap), and for t <= \omega, the system tends towards a much more localized ground state than the harmonic oscillator ground state, see the figure which has \omega = 2t. Note that the DMRG potential is the potential felt at every point as output by the DMRG code.

I am aware that iTensor DMRG is not designed for the continuum limit, but it has been used to investigate that limit in several published papers. I cannot find any DMRG parameters whose adjustment changes any of this, apart from those energies. Can you help me figure out what is going on?

commented by (70.1k points)
Hi, I plan to answer your question soon & thanks for your patience. One question for you though: are you scaling âtâ in some way with the lattice constant âaâ? Or just choosing various t values. If the second one, then it may not be that you are taking the continuum limit toward the Hamiltonian that you are expecting to reach.

Another question is whether the large number of sweeps is being done to reach the density you do end up getting, or because you were hoping it would eventually change to the analytic density?

Thanks - this will just help me to give a better answer.
commented by (280 points)
Hi Miles,

First of all, thank you for your incredible work on iTensor and especially on maintaining this support page through the years! It is immensely helpful.

So, first of all I would just like to see that for N particles on M sites, taking N<<M leads to a density which is a close approximation to the continuum solution density. To converge towards the continuum energy, I would have to take into account the relationship between t and a, but it is my understanding that for just getting the right density, only the length scale of the potential really matters. Thus I am just choosing somewhat arbitrary t for now, and making sure that in each point, the value of the potential corresponds to that of the continuum potential in the same point.

But since you ask, I guess the shape of the density is an interplay between the kinetic and potential energy, and scaling them wrong might lead to the wrong density? In that case, how come I find the right particle in a box density for basically arbitrary t?

Sweeps are very cheap for N<<M, and sometimes a large number is required to converge. For a very weak trap, that is the case, but for a trap like the one pictured, not many sweeps are required. I mostly mentioned it to say that it certainly had enough sweeps to converge on a ground state.

Thank you again for taking the time to answer, and please let me know if you need anything else.
commented by (70.1k points)
Hi, thanks for the kind words!

For your case where there is both a potential and a hopping, the ratio of these two couplings is of crucial importance. The density and essentially every other property will depend on this ratio. So it's not correct that the density will only depend on the potential. Please check this by taking t = 1/(2*a^2) where a is the lattice spacing as is done in this paper for example: https://arxiv.org/abs/1107.2394

I think the reason you got the right density for the particle in the box is that for that case the potential at each point is either zero or infinity. (Let's just think of it as being zero and the system length as being finite.) Then the ratio of the potential to the hopping is always zero. So this ratio will be  the same no matter what you take t to be, and changing t will only change the overall energy just like in the case where the potential v is non-zero and you change t while keeping v/t fixed.

Yes agreed these kinds of simulations can take a huge number of sweeps to converge! My experience is that particles can act sort of diffusively, only spreading out and moving rather gradually with each DMRG sweep. So you have to use tricks like good initial states or lots of sweeps at smaller bond dimensions at the beginning, or more fancy tricks like coarse-to-fine graining multigrid style or RG transformations. See for example: https://arxiv.org/abs/1203.6363

Best,
Miles