Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,873 answers , 20,701 comments
1,470 users with positive rep
502 active unimported users
More ...

How to calculate energy in two-band Hubbard model

+ 3 like - 0 dislike
54 views

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 $\epsilon_{k}$.

-------o-------o---------------o-------o----

    |<-a->|<----b---->|

         bilattice

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?

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user qfzklm
asked Apr 28, 2014 in Theoretical Physics by UnknownToSE (505 points) [ no revision ]
retagged May 4, 2014
It is not clear what you call a bilattice. Do you have a formula (in real space) for it ?

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user Adam

1 Answer

+ 2 like - 0 dislike

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

       A       B               A       B
-------o-------o---------------o-------o----

       |<--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
-------o-------o-------o-------o-------o
      i-1      i      i+1     i+2     i+3
       |<--a-->|<--a-->|<--a-->|

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

       A       B               A       B
-------o-------o---------------o-------o----
       i       i              i+1     i+1
       |<--a-->|<------b------>|

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.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user ChocoPouce
answered Apr 28, 2014 by ChocoPouce (20 points) [ no revision ]
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?

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user qfzklm
@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.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user ChocoPouce
@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 $E(k) = \pm \sqrt{t_s^2 + t_l^2 + 2t_s t_l \cos(k(a+b))}$.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user ChocoPouce

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\varnothing$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...