# Diagonalization of Hubbard model for spinless fermions in 1D $k$-space

In real space we write basis vector for spinless fermions in binary notation for example if there are $$M=4$$ sites in system and $$N=2$$ fermions then basis vectors will be: $$0011, 0101, 0110, 1001, 1010, 1100$$. Hamiltonian in numerical form ($$H=-t\sum_{}(c_j^\dagger c_{j+1}+h.c.)+U\sum_{}n_jn_{j+1}$$) can be written simply using bitwise operations of C/C++, Fortran or MATLAB. One can see hopping part of $$H$$ is off-diagonal and interaction part is diagonal in real space.

When we work in Fourier space Hamiltonain become $$\tilde{H}=\sum_k\epsilon_k\tilde{c_k^\dagger}\tilde{c_k} + \sum_k\tilde{U_k}\tilde{n_k}\tilde{n_{-k}}$$ with $$\epsilon_k=-2t\cos{k}$$ and $$\tilde{U}_k=\frac{1}{L}\sum_j U(j) e^{-ik.j}$$ as explained in this pdf.

## What I can't understand is that how do we define our basis vector in fourier space?

What I have understood from this so for is that let we have a 1D line from $$-\pi$$ to $$+\pi$$ (first brillion zone) on which $$k$$ points are discreetly define. If we have M=4 and N=2 then set of $$k$$-points is $$-\pi$$, $$-\frac{\pi}{2}$$,$$+\frac{\pi}{2}$$, $$+\pi$$
Now considering these 4 points as sites on which fermions can reside our basis vectors can be again given as they were given in real space i.e. $$0011, 0101, 0110, 1001, 1010, 1100$$.
For simplicity I take limit $$U=0$$ and calculate Hamiltonian for both real and fourier space case.
REAL SPACE:
$$H_{R}=-t\begin{bmatrix} 0 & 1 & 0 & 0 & -1 & 0 \\ 1 & 0& 1& 1& 0& -1\\ 0 & 1& 0& 0& 1& 0\\ 0 & 1& 0& 0& 1& 0\\ -1 & 0& 1& 1& 0& 1\\ 0 & -1& 0& 0& 1& 0\\ \end{bmatrix}$$ Let t=1 then Eigenvalues=[-2, -2, -4.4e-16, 0, 2, 2] (using MATLAB function eig())

FOURIER SPACE:

$$\tilde{c_k^{\dagger}}\tilde{c_k}=\tilde{n_k}=$$ number operator in k-space. So our hamiltonian for U=0 should be diagonal with values $$H_{F}= -2t*diagonal[\cos{(\pi/3)}+\cos{\pi}, \cos{(-\pi/3)}+\cos{\pi}, \cos{(-\pi/3)}+\cos{(\pi/3)}, \cos{(-\pi)}+\cos{\pi}, \cos{(-\pi)}+\cos{(\pi/3)}, \cos{(-\pi)}+\cos{(-\pi/3)}]$$

$$=-t\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & -2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}$$

for t=1 eigenvalues=[-2, 1, 1, 1, 1, 4].

results are not matching, I consider there is any fault in my method of defining basis vectors in $$k$$-space. So, please guide be how to properly build basis vectors in $$k$$-space.

• Where did the $\cos(\pi/3)$ come from? $\pi/3$ is not a valid k-vector... Aug 6, 2017 at 20:53

First, the allowed k-vectors are not $-\pi,-\frac{\pi}{2},\frac{\pi}{2},\pi$. The allowed k-vectors are $-\frac{\pi}{2},0,\frac{\pi}{2}, \pi$. In the Brilloin zone, $k=\pi$ and $k=-\pi$ are the same state, so you double counted this state while neglecting $k=0$.
Second, for some reason when you computed $H_F$, you wrote terms like $\cos(\frac{\pi}{3})$ on the diagonal. This is clearly an error, since $\frac{\pi}{3}$ is not an allowed k-value. If you write out $H_F$ more carefully, with the correct k-values, you should get the energies to match like you want.
(Note there could also be an error in your $H_R$, I didn't check it too closely. But fix the k-error and see!)
• Thank you so much. $H_R$ is correct I wast taking k-points wrong. Aug 7, 2017 at 11:22