LagrangeEuler said:
Yes just that. Strange. And how to write ##\sum_{i<j}## if you have vectors ##\vec{i},\vec{j}##?
Yeah, it doesn't make sense if you think of i,j as vectors. The upshot is that you don't have to - the indices are just supposed to label all sites. As long as your system has finite length, width, height (and so on, if you want to) - all you need is one number. It doesn't need to have anything to do with the structure of the lattice or your intuition. See this example for a rectangular grid:
1 2 3 4
5 6 7 8
Every site is characterized by the number i=x+4*(y-1), where x=1,2,3,4 and y=1,2,...
Then, if the actual positions are interesting, we can form the vector \mathbf{r}_i = x \hat{x} + y \hat{y} using modulus operations and the basis vectors of the lattice. Actually, the position vectors are often not that interesting for spin chain systems as the positions are usually fixed, i.e. not dynamical.
LagrangeEuler said:
Notation ##\sum_{\langle i,j \rangle}## looks also confusing for me sometimes. Because I'm never sure
##-\sum_{\langle i,j \rangle}J_{i,j}\vec{S}_i \cdot \vec{S}_j## or ##-\frac{1}{2}\sum_{\langle i,j \rangle}J_{i,j}\vec{S}_i \cdot \vec{S}_j##?
Are you confused about that factor of 1/2? It only changes the definition of J_{ij}, so it's not that important physically. You can just pick either convention at the start of your calculations.
However, it can be important if you want to relate two sums. The factor of 1/2 is then used to avoid double counting. I.e.
\sum_{ij} \mathbf{S}_i \cdot \mathbf{S}_j = 2 \sum_{i<j} \mathbf{S}_i \cdot \mathbf{S}_j
since the first notation contains both \mathbf{S}_1 \cdot \mathbf{S}_2 and \mathbf{S}_2 \cdot \mathbf{S}_1=\mathbf{S}_1 \cdot \mathbf{S}_2 while the second sum only contains one instance of \mathbf{S}_1 \cdot \mathbf{S}_2.
When using the \sum_{\langle i,j\rangle} notation that Jolb described, it generically contains both the ij and the ji terms, so you are in fact counting all symmetric terms twice. In that case using the Hamiltonian
H = -\frac{1}{2} \sum_{\langle i,j \rangle} J_{ij} \mathbf{S}_i \cdot \mathbf{S}_j
makes sense, but again, it's only a scale factor in the end.