# How to generate Hubbard Hamiltonian?

I am a beginner with Hubbard Hamiltonian and my question is very basic: how can I generate the matrix form of the Hubbard Hamiltonian? I know the theory but I don't know how to put it numerically.

$$\hat{H} = -\sum_{\langle ij \rangle\sigma} (\hat{c}_{i\sigma}^{\dagger}\hat{c}_{j\sigma} + H.c.) + U\sum_i \hat{n}_{i\uparrow}\hat{n}_{i\downarrow} + \sum_{i\sigma} V_i\hat{n}_{i\sigma}$$

I mean, assuming that $$\hat{H} = \hat{T} + \hat{U} + \hat{V}$$, how can I generate each Hamiltonian?? Thanks in advance.

• Quick hint: For example just choose occupation number basis (in local single-particle basis) and act with the Hamiltonian and see what happens. – Sunyam Feb 17 '20 at 15:27

## 2 Answers

To be able to write this into a matrix form, we need a priori to escape from the many-body picture. To do so we are going to suppose that for a reason due to the physics of the problem, the spin-down fermions have no dynamics (the spin is represented by $$\sigma$$, we are facing a Fermi-Hubbard model). This means hat we can replace $$\hat n_{\downarrow}$$ by its mean value $$n_{\downarrow}$$.

From this we see that the total Hamiltonian $$\hat H$$ can be decomposed in the following way:

$$T^{(\uparrow)} = -\sum_{} \hat c_{i\uparrow} \hat c^\dagger_{j\uparrow} + H.c.$$

$$V^{(\uparrow)} = \sum_i \hat n_{i\uparrow} V_i.$$

Their spin-down counterpart are simply (sorry for the redundancy):

$$T^{(\downarrow)} = -\sum_{} \hat c_{i\downarrow} \hat c^\dagger_{j\downarrow} + H.c.$$

$$V^{(\downarrow)} = \sum_i \hat n_{i\downarrow} V_i$$

$$U^{(\uparrow)} = U \sum_i \hat n_{i\uparrow} n_{i\downarrow}.$$

This allows us to separate the spin-up and spin-down Hamiltonians:

$$\hat H^{(\uparrow)} = T^{(\uparrow)} + V^{(\uparrow)} + U^{(\uparrow)}$$ $$\hat H^{(\downarrow)} = T^{(\downarrow)} + V^{(\downarrow)}.$$

Note that $$\hat n_{i\downarrow}$$ has lost its hat. By writing this we assume that there is no back-action from the spin-up fermions on the spin-down fermions. Here we just used the assumption of no dynamics for the spin-down fermions.

From this we can first diagonalize $$\hat H^{(\downarrow)}$$ and pick our favorite eigenstate or linear combination of eigensates, take the absolute squared value (which is the density $$n_{\downarrow}$$) and plug it into $$\hat H^{(\uparrow)}$$. We then diagonalize $$\hat H^{(\uparrow)}$$ if we want to know its spectrum. Or we use $$\hat H^{(\uparrow)}$$ to compute the time-evolution of the spin-up fermions in presence of spin-down fermions as a background.

As the occupation number of each the Fermi states is 0 or 1, when you have an $$N$$-site Hubbard model then the Hilbert space has dimension $$2^{N}$$ and the Hamiltonian matrix will be $$2^N$$-by-$$2^N$$. So, for a rather small model with say a 10-by-10 lattice--- i.e. 100 sites--- the matrix is $$2^{100}$$-by-$$2^{100}$$. As $$2^{10}\approx 10^3$$ this is a $$10^{30}$$-by-$$10^{30}$$ matrix. Do you plan to diagonalize this matrix numerically?