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:
$$H=-t\sum_{<i,j>}(c_i^\dagger c_j+h.c.)+U\sum (n_i n_j).$$
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. $<i,j>$ 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:

*

*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'?

*If you think any article can help me with this problem then please tell me about that.

Thanks a lot.
 A: 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!
A: 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.
