# Hubbard model diagonalization in $k$-space

I am trying to diagonalize Hubbard model in real and k-space for spinless fermions. Hubbard model in real space is given as: $$H=-t\sum_{}(c_i^\dagger c_j+h.c.)+U\sum (n_i n_j).$$ 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:

$$H=\begin{bmatrix}0 & -1\\ -1 & 0\end{bmatrix}$$

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 Brillouin 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 https://physics.stackexchange.com/q/352833/140141

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.

• The second term is usual Coulomb interaction, in a general form, not specific to the Hubbard model. Many QFT for condensed matter books treat it - Fetter&Walecka is particularly accessible. Jun 28 at 7:23

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!

• 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. Mar 20, 2018 at 20:22

You have used the spinless fermions. Even the regular Hubbard Hamiltonian with onsite repulsion, the k space representation looks similar. When only the zero center of mass momentum is retained it closely resembles the effective BCS Hamiltonian.

• Please do not use all capital letters in a post.It is normally treated/interpreted as yelling or shouting. Jun 29 at 16:42