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

Question

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 ($H=-t\sum_{<j,j+1>}(c_j^\dagger c_{j+1}+h.c.)+U\sum_{<j,j+1>}n_jn_{j+1}$) 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.

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

My understanding about it:

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].

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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.

Answer 1

I think you've made a couple of mistakes in your allowed k-vectors.

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!)