Hi,

I am new to the field. I am wondering how we choose the "Lx" parameter in a 2D iDMRG calculation. It seems like in the literature there are not much discussion on it. For example in https://journals.aps.org/prb/pdf/10.110 ... .94.121111 Ly is reported but Lx is not specified. Is this because we always assume translational invariance along the x-axis so Lx=1? If this is the case, how do we compute the NN correlations beyond the x=0 cell such as those shown in Fig. 1 in the above work or even long-range correlations in Tenpy? Also, is it possible to compute the entanglement spectrum along a cut at some finite x value with Lx=1 only?

Many thanks!

Yang-hao

## Cell choice and long range correlations in a 2D iDMRG calculation

### Re: Cell choice and long range correlations in a 2D iDMRG calculation

If you expect translation invariance along the cylinder direction (and you Hamiltonian itself is also translation invariant in x-direction), Lx=1 is certainly a natural choice. Using Lx=1 does not hinder the calculation of correlation lengths - in general, if you have an infinite MPS, it just implies that you have a unit cell that repeats, but you can insert operators to be measured anywhere. The trick is to insert as many unit-cells as needed, and use the properties of the canonical form to collapse the (infinite) contractions on the left/right.

As an example, consider a Lx=1, Ly=3 square lattice, where you would have a three-site MPS unit cell with (right-canonical) tensor B0, B1, B2.

Say you want to measure \( \left\langle X_{x=5,y=1} Y_{x=7,y=0} \right\rangle =_{TI} \left\langle X_{x,y=1} Y_{x+2,y=0} \right\rangle \), then you would get a tensor network diagram looking like this:
Here, the second line used the canonical form properties to collabse contractions of B's from the right into identities, and the left to the squared singular values.

For further details, see also Chapter 4 of arXiv:1805.00055

As an example, consider a Lx=1, Ly=3 square lattice, where you would have a three-site MPS unit cell with (right-canonical) tensor B0, B1, B2.

Say you want to measure \( \left\langle X_{x=5,y=1} Y_{x=7,y=0} \right\rangle =_{TI} \left\langle X_{x,y=1} Y_{x+2,y=0} \right\rangle \), then you would get a tensor network diagram looking like this:

Code: Select all

```
x,y= -1,0 -1,1 -1,2 0,0 0,1 0,2 1,0 1,1 1,2 2,0 2,1 2,2 3,0 3,1 3,2
... ---B0---B1---B2---B0---B1---B2---B0---B1---B2---B0---B1---B2---B0---B1---B2--- ...
| | | | | | | | | | | | | | |
| | | | X | | | | Y | | | | |
| | | | | | | | | | | | | | |
... ---B0*--B1*--B2*--B0*--B1*--B2*--B0*--B1*--B2*--B0*--B1*--B2*--B0*--B1*--B2*-- ...
--S--B1---B2---B0---B1---B2---B0---
(can. form) | | | | | | | |
= | X | | | | Y |
| | | | | | | |
--S--B1*--B2*--B0*--B1*--B2*--B0*--
```

For further details, see also Chapter 4 of arXiv:1805.00055

### Re: Cell choice and long range correlations in a 2D iDMRG calculation

Thank you for the detailed explanation! It is very helpful.