When constructing Hamiltonian matrix for many spins, what is the significance of the order of factors in the outer product?

Question

I'm trying to learning the Density Matrix Renomalization Group (DMRG) method from the book "Strongly Correlated Systems: Numerical Methods". For a two-spin system they build a Hamiltonian from the outer product of the operators for each spin. Starting from the operators, $$ \hat{H} = \hat{S}^{z}_1 \hat{S}^{z}_2 + \frac{1}{2}[\hat{S}^{+}_1\hat{S}^{-}_2 + \hat{S}^{-}_1\hat{S}^{+}_2] $$ one gets the matrix for the Hamiltonian: $$ H_{12} = S^z \otimes S^z + \frac{1}{2}[S^+ \otimes S^- + S^- \otimes S^+] $$ and so far, so good. But then, when they add a third spin, I understand the Hilbert space now consists of 8 basis states so the matrices will be $8 \times 8$, and the book has this formula: $$ H_{3} = H_2 \otimes I_2 + \tilde{S}^z_2 \otimes S^z + \frac{1}{2}[\tilde{S}^+_2 \otimes S^- + \tilde{S}^-_2 \otimes S^+] $$ where the tilde matrices are defined as $\tilde{S}^z_2 = I_2 \otimes S^z$.

So my question is why is the first term of $H_3$ written as $H_2 \otimes I_2$ in that order (which I think I understand, as we are making the Hilbert space bigger) while the z-spin matrix defined in the reverse order, $I_2 \otimes S^z$? What is the meaning or significance in the order of these factors?

EDIT I forgot to mention that the book paints the picture of a pair of spins, and in the first Hamiltonian written above, the #1 operators act on the "left" spin while the #2 operators on the "right" spin. Then, the third spin is added "to the right" of the previous pair. I am not sure if that is relevant to the order of factors.

Answer 1

I believe $I_2$ refers to a $2\times 2$ unit matrix. Thus, in $H_3$, a term like $H_2\otimes I_2$ is (likely) the part of the 3-particle Hamiltonian $H_3$ which acts on the first two particles only.

As to $\tilde S_z^z\otimes S^z$ we have by definition $$ \tilde S_z^z\otimes S^z=I_2\otimes S^z\otimes S^z\, . $$ This is just the operators $S^z$ for particle $2$ multiplied by $S^z$ for particle $3$ etc.

The definitions are in part justified by the recursive way in which the Hilbert space is constructed. The $n$ particle is always added as the last factor of the Hilbert space, so the authors wants to take the various matrix quantities already defined for $n-1$ particles and reuse them for the $n$-particle problem. This requires the introduction of the tilde variables, which are given in (2.12) as $$ \tilde S^z_{i-1}=I_{2^{i-2}}\otimes S^z $$ Thus, for instance, $\tilde S^z_3=I_{4}\otimes S^z$. In this way, matrix for $\tilde S^2_3$ actually acts correctly on the third factor in the product state $\psi_1\otimes \phi_2\otimes \chi_3$. It gets absorbed in to the $3$-particle Hamiltonian $H_3$, which will be multiplied by $I_2$ on the right $H_3\otimes I_2$ as a part of the 4-particle Hamiltonian, etc.