1
$\begingroup$

I'm attempting to model a 1D system that consists of 4 lattice sites. When only considering the nearest-neighbor interactions, the Hamiltonian is simply

$$ H_1 = \sum\limits_{i=1}^4 E(i,i+1) $$ where $E(i,j)$ describes the interaction between sites $i,j$ and with periodic boundary conditions.

I now want to consider next nearest-neighbor interactions, so my Hamiltonian becomes

$$ H_2 = H_1 + \sum\limits_{i=1}^4E(i,i+2). $$

Here comes the question: If I consider the case where I only have species at sites 1 and 3, am I double counting the interaction if I say the Hamiltonian under next nearest-neighbor interactions is

$$ H_2 = 2E(1,3) $$ where here I'm assuming $E(1,3)=E(3,1)$.

On one hand, if I iterate through the $i=1,2,3,4$ I encounter both $E(1,3)$ and $E(3,1)$. However, this also feels like trying to count the interactions between many particles and having to divide by 2 due to double counting.

$\endgroup$
1
$\begingroup$

When considering these problems, the energy of interaction between two particles must be accounted only once. While in your first Hamiltonian $H_1$ the sum doesn't double count, in your second Hamiltonian the sum is wrong and you have to adjust it to consider only one term per pair of interacting particles. The correct Hamiltonian would be: $$ H_2 = H_1 +\sum_{i=2}^3E(i,i+2) $$ Then in your specific case with species at sites 1 and 3 the Hamiltonian would read $H_2=E(3,1)$.

$\endgroup$
  • 1
    $\begingroup$ I was thinking along these lines, but this paper had me second guessing: arxiv.org/pdf/quant-ph/0403026.pdf In it, the interaction Hamiltonian is summed from 1,...,L for both terms. Does this mean they are double counting? $\endgroup$ – Grant Cates May 17 at 6:10
  • $\begingroup$ I can't read it now (I'll do when at home!) but may be they are assuming that $L>4$ so there is no double counting. $\endgroup$ – Lith May 17 at 7:34
  • $\begingroup$ they don’t appear to make such an assumption, but the term for next nearest-neighbor interactions has a prefactor that characterizes the interaction strength, so potentially that is what is used to eliminate double counting. $\endgroup$ – Grant Cates May 17 at 9:24
  • $\begingroup$ It could be, but note also that they define periodic conditions as $\sigma_1=\sigma_{L+1}$, so strictly speaking $\sigma_{L+2}\neq \sigma_2$. That's another way to get rid of double counting. $\endgroup$ – Lith May 17 at 14:59
  • $\begingroup$ Interesting. What would be the second component of the second term $\sigma_{j+2}$ for $j=L$, if only the “first” and “last” lattice site experience PBC. I always thought this type of boundary condition was essentially like a ring. $\endgroup$ – Grant Cates May 18 at 7:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.