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I encounter a problem in understanding the mapping between spinful and spineless fermions. This method is discussed on TenPy. Suppose our system is described by 1D spinful fermion Hubbard Hamiltonian. We can use the following method to map a 1D spin chain as a spineless fermions on a ladder(the attached picture). Each rung (blue box) forms a spin half site which is composed of two fermion sites (the circles in the picture) for spin-up and spin-down. Originally we write the 1D spin chain Hubbard Hamiltonian as follow: \begin{equation} H = -t \sum_{i} \big( c^{\dagger}_{i \uparrow} c_{i+1 \uparrow} + c^{\dagger}_{i \downarrow} c_{i+1 \downarrow} + h.c. \big) + U\sum_{i} n_{i \uparrow} n_{i \downarrow} \end{equation} When we introduce the ordering on the ladder( suppose the first spin-up site is site 0). We can rewrite the Hamiltonian in this new ordering: \begin{equation} H = -t \sum_{i\in even} \big( c^{\dagger}_{i \uparrow} c_{i+2 \uparrow} + c^{\dagger}_{i+1 \downarrow} c_{i+3 \downarrow} +h.c. \big) + U\sum_{i\in even} n_{i \uparrow} n_{i+1 \downarrow} \end{equation}

My problem is that even we introduce a new ordering of the spin chain to ladder, the Hamiltonian is still consisted of spinful operators like $c_{i \uparrow} ,c^{\dagger}_{i \downarrow} $. Therefore, I want to know in what sense this method can map a spinful fermions to spineless fermions. I would appreciate any response of this question.

[Graphical representation of spinful fermions on a ladder2

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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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