It might be a very easy question for you, but I am confused and I need helps.

In the simplest Hubbard model at one-dimensional lattice, I ignore the $U$ term and only remain the hopping term. $$H=-t\sum_{<ij>}{c^{\dagger}_{i} c_{j}}$$ where $c^{\dagger}_i$ and $c_i$ are creation operator and annihilation operator on site $i$, respectively.

And the Fourier transformation is defined as $$c_{k}=\frac{1}{\sqrt{V}}\sum_{i}{c_{i} e^{i k r_{i}}}$$ here, the summing index $i$ is subcript, and it is different from the imaginary unit $i$.

Finally, we get $$H(k)=\sum_{k}{\epsilon_{k} c^{\dagger}_{k} c_{k}}$$ by sum over all site index $i$, and $<ij>$ gives geometric factor which is included in energy $\epsilon_{k}$.




Here, my qusetion is, for bilattice, $\epsilon_{k}$ should give two energy bond -- one $k$ should give two differnet energy. I don't know how to calculate it.

Could you help me?

  • $\begingroup$ It is not clear what you call a bilattice. Do you have a formula (in real space) for it ? $\endgroup$ – Adam Apr 28 '14 at 14:19

When considering a bilattice you need to distinguish two type of sites.

       A       B               A       B


For instance you can denote the two kind of sites with letters $A$ and $B$ as it is shown above. Then you have now two different creation and destruction operators . $c_{iA}^{\dagger}$ and $c_{iA}$ in order to create or annihilate a particle in $A$ sites and $c_{iB}^{\dagger}$ and $c_{iB}$ for $B$ sites.

You have to be careful with indices $i$. With a simple lattice the sites had indices like that:

       A       A       A       A       A
      i-1      i      i+1     i+2     i+3

But know the indices are different since there is two kinds of sites:

       A       B               A       B
       i       i              i+1     i+1

All this changes expressions of Fourier transforms and hamiltonian: $$ c_{kA}=\frac{1}{\sqrt{V/2}}\sum_{i}{c_{iA} e^{i k r_{i}}} \quad \text{where} \quad r_i=i*(a+b) $$ $$ c_{kB}=\frac{1}{\sqrt{V/2}}\sum_{i}{c_{iB} e^{i k r_{i}}} \quad \text{where} \quad r_i=i*(a+b)+a $$ EDIT:The volume of the system has to be divided by two in Fourier transforms since there is now two sites in each primitive cell (there were only one before). END EDIT

The hamiltonian now takes the form: $$ H=-\sum_{i}{t_s (c^{\dagger}_{iA} c_{iB} + c^{\dagger}_{iB} c_{iA})+t_l (c^{\dagger}_{iB} c_{(i+1)A} + c^{\dagger}_{(i+1)A} c_{iB})} $$ where I have considered a one-dimensional problem. Since the distance between sites is not always the same, you might want to consider two different hopping parameters: $t_s$ for short jumps and $t_l$ for long ones.

EDIT: I had forgotten terms in the Hamiltonian, nearest neighbor hopping terms must be present in both ways $(A,i)\rightarrow (B,i)$ and $(B,i) \rightarrow(A,i)$. The long jumps are $(A,i+1)\rightarrow (B,i)$ and $(B,i)\rightarrow (A,i+1)$. END EDIT

If you want to obtain a diagonal form for your Hamiltonian, you can try to find a $2\times2$ matrix $M$ such that: $$ H=-\sum_{k}{(c^{\dagger}_{kA} c^{\dagger}_{kB})M\binom{c_{kA}}{c_{kB}}} $$ $M$ will contain hopping parameters $t_s$ and $t_l$, once you have diagonalized $M$ your problem is solved.

  • $\begingroup$ Thank you so much! I tried your method, and got an answer. I calculate the matrix $M=\left( \begin{array}{} 0 & t_s e^{ika} + t_l e^{-ikb} \\ t_s e^{-ika} + t_l e^{ikb} & 0 \\ \end{array} \right)$ and energy is $E(k)=\pm |t_s e^{ika} + t_l e^{-ikb}|$. Am I right? $\endgroup$ – qfzklm Apr 29 '14 at 8:14
  • $\begingroup$ @qfzklm I had forgotten hermitian conjugate terms in the Hamiltonian, those terms represent the other way for a jump between two sites. I guess you'll have to preform the calculations again... Sorry. Also don't forget to divide the volume/number of particles by 2 in your Fourier transforms. $\endgroup$ – ChocoPouce Apr 29 '14 at 8:54
  • $\begingroup$ @qfzklm Actually you were right, you didn't forgot any terms. I check your $M$ matrix with the edited hamiltonian and it works (don't forget prefactors depending on volume). But I find a different energy $E(k) = \pm \sqrt{t_s^2 + t_l^2 + 2t_s t_l \cos(k(a+b))}$. $\endgroup$ – ChocoPouce Apr 29 '14 at 9:11

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