Applying DMRG to AKLT Hamiltonian.

DMRG and Numerical Methods
1

Hi there to Itensor community, I am a research scholar at IIT BHU Varanasi India, I wanted to apply DMRG to find the ground energy for AKLT model which is similar to Heisenberg Model with a quadratic term, the model is Spin(S=1),please provide some help.

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2

Hi,

You can use the example code in the ITensor documentation here. They have done it for the spin-1 Heisenberg chain with an open boundary. You can add the additional terms to the OpSum() as follows -

os += 1/3,"Sx",j,"Sx",j+1,"Sx",j,"Sx",j+1
os += 1/3,"Sy",j,"Sy",j+1,"Sy",j,"Sy",j+1
os += 1/3,"Sz",j,"Sz",j+1,"Sz",j,"Sz",j+1
os += 1/3,"Sx",j,"Sx",j+1,"Sy",j,"Sy",j+1
os += 1/3,"Sx",j,"Sx",j+1,"Sz",j,"Sz",j+1
os += 1/3,"Sy",j,"Sy",j+1,"Sx",j,"Sx",j+1
os += 1/3,"Sy",j,"Sy",j+1,"Sz",j,"Sz",j+1
os += 1/3,"Sz",j,"Sz",j+1,"Sx",j,"Sx",j+1
os += 1/3,"Sz",j,"Sz",j+1,"Sy",j,"Sy",j+1

Connect the first and last sites if you have a periodic boundary.

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3

Hello sir, Thank you for your response.

This lecture notes (section 3.5) provides exact ground states of AKLT model by projection operator method.
How can I get the same ground state using Itensors and DMRG/MPS technique.

4

Thank you for your response sir, I am a PhD student and working in Quantum Information Theory.
Currently I am trying to solve the AKLT model wherein I have to find its exact ground state energy and eigenstate using DMRG/MPS technique,
it will be great if you have some time to discuss with me regarding the material provided by you on the forum.
Yes my question was answered and was quite helpful,
Thanking you,
Vaibhav Shukla

5

Hi Vaibhav,
If your question is how you can use ITensor DMRG to get the ground state of the AKLT Hamiltonian, then the steps would be:

  1. create the OpSum to then make the MPO of the Hamiltonian using an expression like that given by @tamoghnaRay above.

  2. next, follow steps similar to in our DMRG tutorial to input your Hamiltonian into our dmrg function. That will return the ground state as an MPS.

Note that the returned MPS may not have the same numerical entries that you will see in exact formulas for the AKLT state. This is because MPS have an internal ‘gauge’ freedom where the numbers can be different inside even though the state is the same.

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