Q:

Numerical study of Hubbard Model and Spin Charge Separation effect

Hi,
How can i implement the creation operator effect on the ground state(in FORTRAN)?
we calculate the ground state using modified Lanczos method,and we obtain a vector(array) with lots of numbers in it,what does this numbers mean?
As an example for a system with two sites and two electrons,this is the way i build my Hilbert space:

And Here is the Hopping term Matrix:

A:

The first step, probably you have already know, is to get the ground state.
We may write the Hamiltonian on a set of orthogonal basis, e.g., 
$$\left| n_{1,\uparrow} n_{2,\uparrow} \cdots n_{N,\uparrow}, n_{1,\downarrow} n_{2,\downarrow} \cdots n_{N,\downarrow} \right\rangle =\prod_{i} (c_{i,\uparrow}^{\dagger})^{n_{i,\uparrow}} \prod_{i} (c_{i,\downarrow}^{\dagger})^{n_{i,\downarrow}} \left| 0 \right>.$$ Here $\left| 0 \right\rangle$ is the vacuum state. $n_{i,\uparrow}$ and $n_{i,\downarrow}$ can be either 0 or 1. Since the system has $N_e$ electrons, we have $$\sum_{i} (n_{i,\uparrow}+n_{i,\downarrow})=N_{e,\uparrow}+N_{e,\downarrow}=N_e.$$ The dimension of the Hilbert space is given by $D=\sum_{\max(N_{e}-N,0)\leq N_{e,\uparrow} \leq \min(N_e,N)} C_{N}^{N_{e,\uparrow}} C_{N}^{N_e-N_{e,\uparrow}}$. On this basis, the Hamiltonian becomes a $D\times D$ matrix. We can get the ground state by diagonalizing this matrix. Finally the ground state looks like $$\left| \varphi_0 \right\rangle =\sum_{\{n_{1,\uparrow} n_{2,\uparrow} \cdots n_{N,\uparrow}, n_{1,\downarrow} n_{2,\downarrow} \cdots n_{N,\downarrow} \}} f\left(n_{1,\uparrow} n_{2,\uparrow} \cdots n_{N,\uparrow}, n_{1,\downarrow} n_{2,\downarrow} \cdots n_{N,\downarrow}\right)\prod_{i} (c_{i,\uparrow}^{\dagger})^{n_{i,\uparrow}} \prod_{i} (c_{i,\downarrow}^{\dagger})^{n_{i,\downarrow}} \left| 0 \right>. $$ $f\left(n_{1,\uparrow} n_{2,\uparrow} \cdots n_{N,\uparrow}, n_{1,\downarrow} n_{2,\downarrow} \cdots n_{N,\downarrow}\right)$ are the numbers in the vector you get. 
The next step is to add a spin-up electron to the system. Then the total number of electrons becomes $N_e'=N_e+1$. In particular, $N_{e,\uparrow}'=N_{e,\uparrow}+1$ and $N_{e,\downarrow}'=N_{e,\downarrow}$. Now you may set a new set of orthogonal basis with $N_e'$ electrons. The basis can be written similarly as before, except $N_e \rightarrow N_{e}'$. Correspondingly, the dimension of the new Hilbert space becomes $D'$. After that, we express the Hamiltonian and Eq. (2) on the new basis, and do the rest calculations on the new basis. Using second quantization, it is easy to find the relation between the coefficients $f$ in the old basis and the coefficients $f'$ in the new basis. For example, suppose one term in the ground state vector is $$ f c_{1,\uparrow}^{\dagger} c_{3,\downarrow}^{\dagger} \left| 0\right\rangle.$$ Now you want to add a spin-up electron at site 2. The new term becomes $$c_{2,\uparrow}^{\dagger} \left[f c_{1,\uparrow}^{\dagger} c_{3,\downarrow}^{\dagger} \left| 0\right\rangle \right]=-f c_{1,\uparrow}^{\dagger} c_{2,\uparrow}^{\dagger} c_{3,\downarrow}^{\dagger} \left| 0\right\rangle = f' c_{1,\uparrow}^{\dagger} c_{2,\uparrow}^{\dagger} c_{3,\downarrow}^{\dagger} \left| 0\right\rangle.$$ Here I assume one term in your new basis is $c_{1,\uparrow}^{\dagger} c_{2,\uparrow}^{\dagger} c_{3,\downarrow}^{\dagger} \left| 0\right\rangle$. Therefore, in this case $f'=-f$. For each term in the ground state vector, you may do something similar. Finally you will get the new wave function Eq. (2) in the new basis.
In this way, you can implement the creation operator effect on the ground state.

