# To increase accuracy of DMRG code

**URL:** <https://itensor.discourse.group/t/to-increase-accuracy-of-dmrg-code/501>\
**Category:** DMRG and Numerical Methods\
**Created:** [November 7, 2022, 10:03am UTC](https://itensor.discourse.group/t/to-increase-accuracy-of-dmrg-code/501 "2022-11-07T10:03:25Z")\
**Posts on this page:** 1\
**Showing post:** 10

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**Author:** ![Vladimir](https://avatars.discourse-cdn.com/v4/letter/v/94ad74/32.png) [@Vladimir](https://itensor.discourse.group/u/Vladimir)\
**Post date:** [November 18, 2022, 8:43am UTC](https://itensor.discourse.group/t/to-increase-accuracy-of-dmrg-code/501/10 "2022-11-18T08:43:45Z")

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> [@miles](#):
>
> I agree with Vladimir. I did some experiments myself and here’s what I think is going on:
> 
> 1. in the infinite L limit, it’s known that for the XXZ model the Neel order only develops for \Delta \> 1 where the Z term is \Delta S^z\_j S^z\_{j+1}.
> 2. however, there is no guarantee for a finite-sized system whether this ordering will happen; it is only a limiting statement for large enough system size L
> 3. therefore your system could be on the other side of the phase transition (or not) and so the only way to tell is to study various system sizes and extrapolate to the thermodynamic limit (alternatively you can try even larger \Delta)

This isn’t exactly what I had in mind. Let me go into more detail:

1. Whenever a Hamiltonian has some symmetry, its eigenstates can always be chosen to also be eigenstates of the symmetry operator. Hence, its non-degenerate eigenstates _must_ be eigenstates of the symmetry, i.e. exhibit no symmetry breaking.

2. The XXZ model has a Z2 symmetry. For infinite \Delta it becomes the classical Ising model, whose ground states are the two symmetry-related Neel states. Since they’re degenerate, one of them is as good a ground state as their symmetric or anti-symmetric superpositions. However, at any finite \Delta the S^+\_jS^-\_{j+1}​ terms introduce mixing between the Neel states, opening a gap and leaving a Z2 symmetric state with \langle S^z\_i \rangle =0 as the unique ground state (and a Z2 anti-symmetric state as the first excited state) for any finite L and \Delta. This can be easily checked using exact diagonalization for L=16. The gap decays exponentially as system size is increased since mixing the Neel states requires an L order term in perturbation theory, thus allowing for spontaneous symmetry breaking in the thermodynamic limit.

3. Specifically for L=16 and \Delta=2, the gap between the ground state and the first excited state is about 0.07, or 4e-3 per site. The energy difference between the ground state and the best symmetry broken state would be of that order of magnitude, i.e. well within the resolution of a high quality DMRG run. When the system size is increased or the run quality decreased, DMRG finds a symmetry broken state instead of the exact ground state.

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