1
$\begingroup$

I am trying to diagonalize hubbard model in real and K-space for spinless fermions. Hubbard model in real space is given as: $$H=-t\sum_{<i,j>}(c_i^\dagger c_j+h.c.)+U\sum (n_i n_j)$$ I solved this Hamiltonian using MATLAB. It was quite simple. t and U are hopping and interaction potentials. c, $c^\dagger$ and n are annihilation, creation and number operators in real space respectively. The first term is hopping and 2nd is two-body interaction term. is indicating that hopping is possible only to nearest neighbors. To solve this Hamiltonian I break it down as: (for M=# for sites=2 and N=# of particles=1) $$H=-t (c_1^\dagger c_2 + c_2^\dagger c_1)+U n_1 n_2 $$ The basis vectors that can be written in binary notation are: 01, 10 Using t=1, U=1 and above basis the Hamiltonian can be written as:

H=[0 -1

-1 0]

That is correct. I checked with different values of M,N,U and t this MATLAB program give correct results.

In K-space To diagonalize this Hamiltonian in K-space we can perform Fourier transform of operators that will results in: $$H(k)=\sum_k \epsilon_k n_k + U / L \sum_ {k,k,q} c_k^\dagger c_{k-q} c_{k'}^\dagger c_{k'+q}$$ Where $\epsilon_k=-2tcos(k)$. To diagonalize this Hamiltonian I make basis by taking k-points between -pi and +pi (first brillion zone) i.e. for M=2 and N=1 allowed k-points are: [0,pi]

Here first term is simple to solve and I have solved it already but I can't solve the 2nd term as it includes summation over three variables. To get in more details of my attempt you can see Hubbard model diagonalization in 1D K-space for spinless fermions

My question: 1. What is physical significance of 2nd term in H(k) given above? I mean what is it telling about which particles are hopping from where to where? What are limits on q, k and k'?
2. If you think any article can help me with this problem then please tell me about that.

Thanks a lot.

$\endgroup$
2
$\begingroup$

You likely already know this, but for $N=1$ there are no on site interactions because you only have 1 particle, what can it interact with?

To answer your question, in $k$ space the second term in your Hamiltonian corresponds to a scattering event. To clearly see how this works, make a quick change of variables $k\to k+q$ and $k'\to k'-q$. Then you see that your new interaction term is $$\frac{U}{L}\sum_{k,k',q}c^\dagger_{k+q}c^\dagger_{k'-q}c_{k'}c_k$$ which destroys two particles with momenta (wavevectors) $k$ and $k'$ and creates two new particles with momenta (wavevectors) $k+q$ and $k'-q$. Which is exactly what you'd expect from a scattering event between two particles with conservation of momentum. Two initial particles with momenta $k$ and $k'$ interact and leave with momenta $k+q$ and $k'-q$. Like above, when you have only 1 particle this term is trivial - what can this particle scatter with?

The allowed values for $k$, $k'$ and $q$ are just what you'd expect them to be - they can take on any allowed value for momentum in your system $\to \frac{2\pi n}{M}$ with $n\in\{0,\pm 1,\pm2,...,\pm N/2\}$ (remembering that the two endpoints correspond to the same physical mode). Another way to say this, and likely more useful in practice, is that $k$, $k'$ and $q$ run over every site in $k$-space for your system. It's not longer telling you about "which particles are hopping from where to where", you're working in the momentum basis. Anything with a definite momentum is in a superposition of position basis states.

Hope that helps!

$\endgroup$
  • 1
    $\begingroup$ Hi @bRosto3 Thank you so much for your answer. When I posted this question I was quite new to the field. Now I feel like it is a very simple question. I hope it will help someone else as well. Thank you once again. $\endgroup$ – Luqman Saleem Mar 20 '18 at 20:22

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.