# Hamiltonian density of a spin Hamiltonian

A many-body Hamiltonian in second quantization is written as $$H = \int d\vec r \Psi^\dagger_{\vec r} H_1 (r) \Psi_{\vec r}$$ where $$H_1(r)$$ is one-body Hamiltonian. For example, for a non-interacting electronic system $$H_1(r) = -\frac{\hbar^2\nabla^2}{2m}$$. So, we get $$H = -\frac{\hbar^2}{2m} \int d\vec r \Psi^+_{\vec r} \nabla^2 \Psi_{\vec r} =\int d\vec r \mathcal{E}$$

here Hamiltonian (or energy) density $$\mathcal{E}$$ is taken as $$\mathcal{E} = -\frac{\hbar^2}{2m}\Psi^\dagger_{\vec r} \nabla^2\Psi_{\vec r} \ .$$ Now, let we have a many-spin model, written in tight-binding notation as $$H = J\sum_{ij}\textbf{S}_i\cdot \textbf{S}_j \ .$$ In this notation how do we write Hamiltonian density?

To obtain the Hamiltonian density you can write something like $$H = \sum_i H_i,$$ where $$H_i$$ is a Hamiltonian centered around site $$i$$. For the Hamiltonian you mention it'd simply be $$H_i = J\mathbf{S}_i \cdot \sum_j \mathbf{S}_j.$$ Note that this looks just like an effective field acting on $$\mathbf{S}_i$$, but there's no escaping that the field is due to other spins. That is, $$H_i$$ cannot be written as a one-body Hamiltonian unless you make a mean-field approximation. Now, regardless of whether you make that approximation or not, interpreting $$H_i$$ as an energy density makes most sense when the sum is restricted to some set of local terms, e.g. if $$j$$ and $$i$$ are nearest neighbors. But I wrote the sum unrestricted here, like in your post. Finally, if you want to, you can take a continuum limit (lattice spacing $$a\rightarrow 0$$) so that $$H_i \rightarrow H(\vec{r})$$.
• Thank you for the answer. I am considering the nearest-neighbor interactions only. Can you please give some comments on how to take continuum limit? Will it be $H(\vec r) = \int d\vec r' J \vec S (r) \cdot \vec S(r')$ Commented Sep 18, 2021 at 17:53