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:

  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.

  • $\begingroup$ 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. $\endgroup$
    – Roger V.
    Jun 28, 2022 at 7:23
  • $\begingroup$ @RogerV. Not at all, in general, Coulomb interaction depends at least on the momentum transfer $q$, and in the equation for $H(k)$ it is momentum independent. $\endgroup$
    – yarchik
    Feb 22 at 14:38
  • $\begingroup$ @yarchik Coulomb interaction depends at least on the momentum transfer $q$ - this is true for distributed states, whereas here we are in the localized representation (i.e., momentum is not a good quantum number.) $\endgroup$
    – Roger V.
    Feb 22 at 14:40
  • $\begingroup$ @RogerV You claim that there is i) Coulomb interaction pertinent (specific) to the Hubbard model, ii)usual Coulomb interaction and iii) distributed Coulomb interaction. Never heard of such classification. Can you define each? $\endgroup$
    – yarchik
    Feb 22 at 14:55
  • $\begingroup$ @RogerV. By the way, why do you claim that momentum is not a good quantum number? $\endgroup$
    – yarchik
    Feb 22 at 14:58

2 Answers 2


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!

  • 2
    $\begingroup$ 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. $\endgroup$ 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.

  • $\begingroup$ Please do not use all capital letters in a post.It is normally treated/interpreted as yelling or shouting. $\endgroup$
    – user320397
    Jun 29, 2022 at 16:42

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