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I want to find ground state energy (as well as wavefunction) for spinless $tV$ model using Real-Space Renormalization Group (RSRG) approximation. The $tV$ model is defined as $$H=H_t+H_{int}=-t\sum_{i=1}^N (c_i^\dagger c_{i+1}+c_{i+1}^\dagger c_i) + V\sum_{i=1}^N n_i n_{i+1}$$ where $n_i$ is number operator.

The RSRG works on hypothesis that the groundstate of a system is composed of low-energy states of the system's bipartitions. And an algorithm of RSRG is:

  1. Construct Hamiltonian $H_N$ for exactly diagonalizable $N$ sites
  2. Diagonalize $H_N=\sum E_i|E_i\rangle\langle E_i|$ where $E_i$ are eigenvalues in incrasing order
  3. Apply a projector $P$ on $H_N$ to find space spanned by lowest $m$ eigenstates, $P=\sum_{i=1}^m |E_i\rangle\langle E_i|$
  4. Calculate projected Hamiltonian $\tilde{H}_N=P^\dagger H_N P$
  5. Construct Hamiltonian of size $2N$ by $$H_{2N}=\tilde{H}_N\otimes I + I\otimes \tilde{H}_N + \tilde{H}_{int}$$ where $\tilde{H}_{int}=\tilde{A}_N\otimes \tilde{B}_N$ and $\tilde{A}_N(\tilde{B}_N)$ are the projected operator acting on each bipartition $\tilde{A}=P^\dagger A P(\tilde{B}=P^\dagger B P)$
  6. repeat step 2-5 until desired system size is acheived

My attempt

For simplicity, I considered a system at half-filling and used the binary basis to write $H$ for 4 sites, $H_4$. Using $t=1, V=1$, $H_4$ is

Step:1

$$H_4= \begin{bmatrix} 1 & -1 & 0 & 0 & 1 & 0 \\ -1 & 0 & -1 & -1 & 0 & 1 \\ 0 & -1 & 1 & 0 & -1 & 0 \\ 0 & -1 & 0 & 1 & -1 & 0 \\ 1 & 0 & -1 & -1 & 0 & -1 \\ 0 & 1 & 0 & 0 & -1 & 1 \\ \end{bmatrix} $$

Step:2-3

It is easy to diagonalize $H_4$ and calculate $P$ for $m=3$ $$P= \begin{bmatrix} 0.69 & 0.24 & 0 & 0 & -0.24 & 0.31\\ 0.24 & 0.62 & 0.24 & 0.24 & 0 & -0.24\\ 0 & 0.24 & 0.19 & 0.19 & 0.24 & 0\\ 0 & 0.24 & 0.19 & 0.19 & 0.24 & 0\\ -0.24 & 0 & 0.24 & 0.24 & 0.62 & 0.24\\ 0.31 & -0.24 & 0 & 0 & 0.24 & 0.69 \end{bmatrix} $$

Step:4 is also straightforward.

Step:5

I am having trouble in calculating the $\tilde{H}_{int}$. What are $A$ and $B$ operators here? Are they just equal to $H_{int}=diagonal[1,0,1,1,0,1]$ i.e. $\tilde{H}_{int}=[P^\dagger H_{int} P] \otimes [P^\dagger H_{int} P]$

Another little confusion: After projecting out $m$ (which is 3 in my attempt) low-energy eigenstates, the size of Hamiltonian is still $6\times 6$ (it is not changed). If the size of Hamiltonian is not changed then how is RSRG efficient in saving the computational memory?

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