+1 vote
asked by (130 points)

Hi, I want to calculate the ground state of a system made of several fermions.

After getting the coefficient matrix of every two fermions' interaction, I want to put it into the Hubbard sites model.

Should I do antisymmetric for the matrix in advance by myself?

Thanks.

1 Answer

0 votes
answered by (27.6k points)

Hi Zhang,
So I assume you are asking about the coefficients of terms that are summed to make a Hamiltonian? In that case, the key thing is that they are chosen so as to result in a Hermitian Hamiltonian. There is no requirement that the matrix of coefficients has to be antisymmetric per se, but it depends on the operators that this matrix of coefficients is weighting (in terms of whether it would result in a Hermitian Hamiltonian) so you would have to provide more information about the terms for me to say for sure.

Basically you define a Hamiltonian in ITensor just as you would for any other algorithm or in a theoretical / analytical study.

If it helps, please take a look at the sample code called "exthubbard.cc" in the sample/ folder under the main ITensor folder.

Best regards,
Miles

commented by (130 points)
Hi miles,

Actually, I have calculated the coefficients of second quantization for interaction of two electrons. After that, by using operators such as Cdagup and Cdagdn, I may get the MPO I want.

So my queastion is if I should use the coefficients from the initial  tensor without doing antisymmetric or I should use them after doing antisymmetric for the tensor by myself?

Since I can get a seemingly correct answer without doing antisymmetric but a wrong one by doing it, I want to know the reason.

Thanks.
commented by (27.6k points)
Hi Zhang,
I'm not sure if I can answer your question other than to say "the wrong way of calculating the coefficients will give the wrong answer, whereas the right way will give the right answer". You stated that you calculated the coefficients of the Hamiltonian in second quantized form. These are the coefficients that AutoMPO expects to be used to define the Hamiltonian, since it is also working in the second quantized formalism. Any other coefficients will, by definition, give a different Hamiltonian, include other coefficients obtained by antisymmetrization.

So it's really more up to you to tell me why you think that antisymmetrization is needed for Hamiltonian coefficients? It is not a requirement for a fermion Hamiltonian. Rather the requirement is that the Hamiltonian is Hermitian.

Hope that helps!

Miles
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