After thinking this problem a few day, I realise that how to map the spinful fermionic system to spinless fermions system. The main idea is to separate the local site Hilbert space to spin up and spin down:
\begin{equation}
\begin{split}
\mathcal{H}_{i} &= \text{span} \{ |0\rangle, \uparrow \rangle , |\downarrow \rangle, |\uparrow \downarrow\rangle \} \\
&= V_{i \uparrow} \otimes V_{i \downarrow} \\
&= \text{span} \{ |0 \rangle , |\uparrow\rangle \} \otimes \text{span} \{ |0 \rangle , |\downarrow\rangle \}
\end{split}
\end{equation}
Firstly, we remind ourselves that we are now working on a spinful chain. Therefore, the corresponding Hamiltonian is the usual Hubbard Hamiltonian:
\begin{equation}
H = -t \sum_{i } ( c^{\dagger}_{i \uparrow } c_{i+1 \uparrow } + c_{i \downarrow }c_{i+1 \downarrow } + h.c.) + U\sum_{i} n_{i \uparrow} n_{i\downarrow} ~~,~~ i \text{ is the ordering on chain}
\end{equation}
Then, we move to our spin ladder case. We want to map our spinful fermionic system. The advantage is that we can convert a spinful system to spinless system. But in what sense is it spinless? This is the question that I want to address below. Recall that for a single site, we can decompose our Hilbert space into two parts: spin up and spin down. In the spin ladder case, we re-order our labelling from chain to this ladder. For instance, we look at the 1st blue box above: We denote the first spin-up site is site 0 and the first spin-down site is spin-down. Having this ordering, we realise our local site Hilbert space on a chain is spilt in to spin-up and spin-down sites:
\begin{equation}
\mathcal{H}_{i} \rightarrow \mathcal{H}_{i',i'+1} = V_{i' \uparrow} \otimes V_{i'+1 \downarrow}
\end{equation}
where $i, i'$ are the ordering on chain and ladder respectively. Recall that we want to know in what sense the ladder maps a spinful system to spinless. If we look at the Hilbert space of the ladder case, we immediately know why since the ordering inherently contain the information of spin-up and spin-down. Therefore, if we add a $c^{\dagger}_{i}$ on even sites on the ladder, it is automatically equivalent to create a spin-up fermion on this ladder. For the spin-ladder, we can rewrite our Hamiltonian like above:
\begin{equation}
H = -t \sum_{i' } ( c^{\dagger}_{i' \uparrow } c_{i'+2 \uparrow } + c_{i' \downarrow }c_{i'+2 \downarrow } + h.c.) + U\sum_{i' \in even } n_{i; \uparrow} n_{'i+1\downarrow} ~~,~~ i' \text{ is the ordering on ladder}
\end{equation}
Suppose we have two sites on the chain, it automatically translates to 4 sites on the ladder.
\begin{equation}
\mathcal{H}_{1} \otimes \mathcal{H}_{2} \rightarrow ( V_{0' \uparrow} \otimes V_{1' \downarrow}) \otimes ( V_{2' \uparrow} \otimes V_{3' \downarrow})
\end{equation}
Therefore, for the spin-up hopping term:
\begin{equation}
c^{\dagger}_{1 \uparrow}c_{2 \uparrow} \rightarrow c^{\dagger}_{0'} \otimes I_{1'} \otimes c_{2'} \otimes I_{3'}
\end{equation}
Therefore, this mapping provides us a way to map a spinful fermion chain to spinless fermion ladder. where $c, c^{\dagger}$ are the spinless fermionic operator.
Regarding the computation efficiency, the spin ladder may be better than chain since local site dimension sis 2 rather than 4. However, in iDMRG implementation, the hopping term of spin ladder is not nearest-neighbour interaction. It may not be trivial if we use the usual iDMRG method to compute. May be MPO is a better choice.
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