# What do quantum spin hamiltonians describe?

I've learned all particles are either fermions or bosons, obeying their respective operator algebras, and then I've seen Hamiltonians describing models carrying one of these two types of particles. So far it made sense.

But then I started seeing Spin Hamiltonians describing, for example, a chain of spins or something like that... I learned how to do the math by example but didn't really understand what I was doing. Like, how to think about these objects, and what really are these objects? If all there is are either fermions or bosons, what are spins in these Hamiltonians? Also, what are spinless fermions and other variants like that? I'm looking to clarify some concepts in my mind... If you can help with that I'll be glad.

Let's consider the hydrogen molecule $H_2$. In order to form the molecule, we begin with two isolated hydrogen atoms, $H(1)$ and $H(2)$, located at the points $\bf R_1$ and $\bf R_2$, respectively. Their $1s$ orbitals are the only relevant orbitals at low energy and temperature, as the other orbitals have a much larger energy. We take the orbital $\varphi_{1s}({\bf r} -{\bf R}_1)$ of the atom $H(1)$ as the orbital $a$ and the orbital $\varphi_{1s}({\bf r}-{\bf R}_2)$ of the atom $H(2)$ as the orbital $b$. In each $1s$ orbital we can put up to two electrons (one with spin projection $\uparrow$ and the other with $\downarrow$) due to Pauli exclusion principle. If we put one electron in a $1s$ orbital, the energy of this state will be $\varepsilon_{1s}$ ($\sim -13.6 eV$). If we put two electrons in one $1s$ orbital, the energy of this "double occupied" state will be $2\varepsilon_{1s}+U > 0,$ where $U$ is the on-site Coulomb repulsion between both electrons. The Hamiltonian for this system of two isolated hydrogen atoms, in second quantization, will be: $$\hat{H}_{at} = \sum_{\sigma}\varepsilon_{1s}(\hat{c}^\dagger_{a \sigma}\hat{c}_{a\sigma}+ \hat{c}^\dagger_{b \sigma}\hat{c}_{b\sigma})+ U (\hat{n}_{a\uparrow}\hat{n}_{a\downarrow}+\hat{n}_{b\uparrow}\hat{n}_{b\downarrow}),$$ where $\hat{c}$'s are fermionic operators, as it correspond to an electronic system.

Each atom contributes with one electron, so we have two electrons in the system. Due to the Coulomb repulsion $U$, there is just one electron in each atom. So, the ground state is $$|GS_{at}> = \hat{c}^\dagger_{a\sigma}\hat{c}^\dagger_{b\sigma'}|vac>\equiv |a\sigma,b\sigma'>,$$ where $|vac>$ is the vacuum state. The ground state is four-fold degenerate as $\sigma, \sigma'$ can take indistinctly the values $\uparrow, \downarrow,$ and its energy is $E_{gs}=2\varepsilon_{1s}$.

Now, we put the hydrogen atoms closer. Then, there will be a probability that one electron hops from $H(1)$ to $H(2)$ and vice versa, that translates in the Hamiltonian $$\hat{H}=\hat{H}_{at} - t\sum_{\sigma}(\hat{c}^\dagger_{a\sigma}\hat{c}_{b\sigma}+\hat{c}^\dagger_{b\sigma}\hat{c}_{a\sigma}),$$ where $t$ is the probability amplitude for the hopping processes. This Hamiltonian is the two-site version of the well known (fermionic) Hubbard model, that is extensively used in condensed matter physics.

Which is the ground state of this Hamiltonian for a filling of two electrons ($N_e =2$)? First, let's take a look at the Hilbert space with $N_e=2.$ As the Hamiltonian commutes with $\hat{S}_{tot}^z$ and $\hat{S}^2_{tot}$, their eigenstates can be classified with the total spin and total spin projection quantum numbers. There will be three triplets states $S_{tot}=1$ and three singlets states $S_{tot}=0$. After a little algebra, we get that the ground state is a singlet $$|GS> \propto \left( \frac{|a\uparrow, b\downarrow>-|a\downarrow b\uparrow>}{\sqrt{2}} \right) + \alpha \left(\frac{|a\uparrow, a\downarrow> + |b\uparrow, b\downarrow>}{\sqrt{2}}\right),$$ where $\alpha$ goes to zero as $U/t \to \infty$, and the ground state energy: $$E_{gs} = 2\varepsilon_{1s}+\frac{U}{2}-\sqrt{\left(\frac{U}{2}\right)^2+4t^2} < 2\varepsilon_{1s}.$$ The three triplets $$|a\uparrow,b\uparrow>, |a\downarrow,b\downarrow>, \frac{|a\uparrow,b\downarrow>+|a\downarrow,b\uparrow>}{\sqrt{2}}$$ are eigenstates of the Hamiltonian with energy $E_{tr} = 2\varepsilon_{1s}$. There will be two other singlet eigenvectors: $$|S=0, ex_1> =\frac{1}{\sqrt{2}}\left(|a\uparrow,a\downarrow>-|b\uparrow,b\downarrow>\right), E_{S=0.ex_1} = 2\varepsilon_{1s}+U,$$ and $$|S=0,ex_2> \propto \alpha \left( \frac{|a\uparrow, b\downarrow>-|a\downarrow b\uparrow>}{\sqrt{2}} \right) - \left(\frac{|a\uparrow, a\downarrow> + |b\uparrow, b\downarrow>}{\sqrt{2}}\right), E_{S=0, ex_2} = 2\varepsilon_{1s}+\frac{U}{2}+\sqrt{\left(\frac{U}{2}\right)^2+4t^2} > 2\varepsilon_{1s}+U.$$

If we consider the condition $U \gg t,$ the eigenvectors, according to their energies, will be split in two groups: the low-energy sector, corresponding to the singlet ground state and the three triplet (excited) states, and the high-energy sector, consisting of the other two singlet (excited) eigenstates. Physically, for $U \gg t$ there is one electron localized around each hydrogen atom: charge fluctuations are not favorable due to the Coulomb repulsion between two electrons occupying the same $1s$ orbital. As charge is "frozen" (one electron in each atom), we can forget about electrons hopping from one atom to the other, and the only remaining degree of freedom is the spin. The low energy spectrum calculated before, can be reobtained by means of a pure spin Hamiltonian, the Heisenberg exchange Hamiltonian: $$\hat{H}_{exch} = J\hat{\vec S}_{a} \cdot \hat{\vec S}_{b},$$ where $\hat{\vec S}$ are spin-1/2 angular momentum operators, and the exchange interaction is (for $U \gg t$) $$J = \frac{4t^2}{U}.$$ In this way, the original fermionic problem is mapped in a spin model. Technically, it is said that the charge degree of freedom of the electrons are integrated out.

It is amazing that a spin-dependent interaction (Heisenberg exchange) has its origin in the spin-independent Coulomb repulsion (combined with the particle statistics).

• "In this way, the original fermionic problem is mapped in a spin model." - Thanks, but you kind of showed me how to go from a fermion Hamiltonian to a spin Hamiltonian without really giving me the sense of what a spin model is. Are all spin models about 2 level systems then? I think I'm not being pedantic here, but what I was asking was really about what comes at the end of your answer... Mar 17 '18 at 17:58
• Let's take the electron as an example. It has charge and spin. When, due to energetic considerations (like Mott insulating phase in condensed matter), the charge is "frozen", its spin degree of freedom survives. This is the "spin" that appears in spin models. For example, in a Mott state there is one electron per site, the spin model that describes its low energy physics is the Heisenberg Hamiltonian. Each spin operator in this model take into account the two possible spin projection of the electrons. Spin can be greater than 1/2, so no all spin models are about 2 level systems. Mar 17 '18 at 18:13
• What about integer spin lattices? Mar 17 '18 at 18:19
• There are several examples of transition metal ions that, in condensed matter systems, have spins larger than 1/2. For example, nickel in compounds has a $d^8$ configuration, corresponding to a spin $S=1$. So, if the compound is insulating, its low-energy model will be a Heisenberg-like Hamiltonian with $S=1$. Mar 17 '18 at 18:30
• So, as a tentative answer to my own question can I say spin Hamiltonians can be modelling both interacting fermions and bosons when only the spin degree of freedom is left? Mar 17 '18 at 18:34

Concerning the spinless fermions, it should be considered that Pauli's spin-statistics connection (fermions have half-odd integer spins, bosons have integer spins) applies to Lorentz invariant systems. So, it is possible to have fermions with $S=0$ in non-relativistic systems. Usually, this kind of fermions appears as auxiliary particles in the treatment of many body quantum systems.