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Consider a quantum spin chain of length $N$. Each site/spin has the local Hilbert space $\mathbb{C}^d$ and so for the whole chain the Hilbert space is $(\mathbb{C}^d)^{\otimes N}$. Now for periodic boundaries, we impose that the Hamiltonian $H$ be (for simplicity let the interaction be only nearest neighbor) $$H = \sum_i^N h_{i,i+1}, \qquad \text{with }\quad h_{N,N+1} = h_{N,1}. $$ Explicitly, each local interaction acts on any state $|\Psi \rangle \in (\mathbb{C}^d)^{\otimes N}$ as $$I_1 \otimes I_2 \otimes \cdots \otimes I_{i-1} \otimes h_{i,i+1} \otimes I_{i+2} \otimes \cdots I_{N-1} \otimes I_N, \qquad \text{for } i=1,2,..,N-1, $$ where $I_j=I$ is just the identity operator/matrix (acting on $\mathbb{C}^d$) and the subscript is just to be explicit on what site it acts.

My question is, how about $h_{N,N+1}=h_{N,1}$? Is it $I_2 \otimes I_3 \otimes \cdots \otimes h_{N,1}$? If so, I am confused on how it acts on the Hilbert space, i.e. how does one differentiate $(I_2 \otimes I_3 \otimes \cdots \otimes h_{N,1})$ from $(I_1 \otimes I_2 \otimes \cdots \otimes h_{N,N+1})$ if one write it in terms of a matrix especially when doing a numerical calculation?

Or is it better to work on $(\mathbb{C}^d)^{\otimes N+1}$ and put a constraint that the 1st spin and $(N+1)$th spin are always on the same state?

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  • $\begingroup$ I think your notation is confusing you a little bit. Maybe it would be better to write something like $I_1 \otimes I_2 \otimes \ldots \otimes \mathcal{O}_i \otimes \mathcal{O}_{i+1} \ldots \otimes I_N$ so that it is transparent that the interaction term between the end and beginning of the chain is $\mathcal{O}_1 \otimes I_2 \otimes \ldots \otimes I_{N-1} \otimes \mathcal{O}_N$ $\endgroup$ – d_b Jun 30 at 12:03
  • $\begingroup$ Correct me if I wrong, but I think not all two-body (two-spin in my case) operators can be written in that way, i.e. $O_{i,i+1}=O_i \otimes O_{i+1}$. That is actually why I am asking this question, specifically for those cases. $\endgroup$ – git-able Jun 30 at 12:24
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    $\begingroup$ @git-able Not all two-body operators can be written as $O_i \otimes O_{i+1}$. However, all two-body can be written as a linear combination of such product operator terms, i.e., $O_{i, i+1}= \sum_a O^{(a)}_i \otimes O^{(a)}_{i+1}$. And then the comment of d_b above (with considering such a linear combination instead of a single product) should explain what we mean by the $h_{N, 1}$ operators. $\endgroup$ – Zoltan Zimboras Jun 30 at 12:55
  • $\begingroup$ @ZoltanZimboras Thanks, this -- $h_{N,1} = \sum_a (O_1^{(a)} \otimes I_2 \otimes \cdots \otimes I_{N-1} \otimes O_N^{(a)})$ -- makes more sense. $\endgroup$ – git-able Jun 30 at 13:06
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So just to extend my comment as an answer: Consider a two-body Hamiltonian $h$ defined on $\mathbb{C}^d \otimes \mathbb{C}^d$, it can always be expanded as $h=\sum_p A^{(p)} \otimes B^{(p)}$, where $A^{(p)}$ and $B^{(p)}$ are operators acting on $\mathbb{C}^d$. Now a translation-invariant spin-chain Hamiltonian defined on $N$ sites with periodic boundary condition, where the local two-body interaction is given by the term $h$, can be written as $$H = \sum_i^N h_{i,i+1}, \qquad \text{with }\quad h_{N,N+1} = h_{N,1}, $$ where for the case $i<N$, we have $$h_{i, i+1}=\sum_{p} \, I_1 \otimes I_2 \otimes \cdots \otimes I_{i-1} \otimes A^{(p)} \otimes B^{(p)} \otimes I_{i+2} \otimes \cdots I_{N-1} \otimes I_N, \qquad \text{for } i=1,2,..,N-1, $$ while for $i=N$ $$h_{N, N+1}=h_{N,1}=\sum_{p} \, B^{(p)} \otimes I_2 \otimes \cdots \otimes I_{N-1} \otimes I_{N-1} \otimes A^{(p)}. $$

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You seem to be confused by the basis-dependent definition of a tensor product.

$$A \otimes B = \left( \begin{array}{cccc} a_{11} b_{11} & a_{11}b_{12} & a_{12}b_{11} & a_{12}b_{12} \\ a_{11} b_{21} & a_{11}b_{22} & a_{12}b_{21} & a_{12}b_{22} \\ a_{21} b_{11} & a_{21}b_{12} & a_{22}b_{11} & a_{22}b_{12} \\ a_{21} b_{21} & a_{21}b_{22} & a_{22}b_{21} & a_{22}b_{22} \end{array}\right). $$

If you have a matrix $A_{13}$ on a Hilbert space ${\cal H_1} \otimes {\cal H_3}$ and a matrix $I_2$ on a Hilbert space ${\cal H}_2$, the above definition of a tensor product does not apply. You cannot write the tensor product the way it is written above. I'm not going to write down the $8 \times 8 $ matrix for $A_{13} \otimes I$, but the first two rows are

$$ \left(\begin{array}{cccccccc} a_{1111} & a_{1112} & 0 & 0 & a_{1211} & a_{1212} & 0 & 0\\ a_{1121} & a_{1122} & 0 & 0 & a_{1221} & a_{1222} & 0 & 0 \end{array}\right). $$

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