Solving the Two-Site Model

Organizing the Hilbert Space

As a quick reminder, the two-site Hubbard Hamiltonian is:

Since the Hamiltonian conserves both the total number of electrons, $N_e$, and the total spin projection, $S_z$, the full $16\times 16$ Hamiltonian matrix breaks down into smaller, independent blocks for each (Ne, Sz) sector. We can solve each of these small blocks separately—a strategy central to the method of exact diagonalization. We have already solved the most complex block, (Ne=2, Sz=0). Now, let's address the simpler cases.

The Single-Electron Case ($N_e=1$)

Let's start with a single electron in the system. The Hilbert space for a given spin $\sigma$ is two-dimensional, spanned by the states where the electron is on site 1 or site 2:

In this sector, the interaction term $U{\sum }_i{n}_{i\uparrow }{n}_{i\downarrow }$ is always zero, as there is no possibility of double occupancy. The Hamiltonian simplifies to just the hopping term:

Let's construct the Hamiltonian matrix in the basis $\{|1_{\sigma }⟩,|2_{\sigma }⟩\}$. The matrix elements are:

• $⟨1_{\sigma }|H|1_{\sigma }⟩=0$
• $⟨2_{\sigma }|H|2_{\sigma }⟩=0$
• $⟨1_{\sigma }|H|2_{\sigma }⟩=-t$
• $⟨2_{\sigma }|H|1_{\sigma }⟩=-t$

So, the Hamiltonian matrix for a single electron of spin $\sigma$ is:

This is a simple matrix to diagonalize. The eigenvalues are $E=\pm t$. The corresponding eigenstates are the symmetric and antisymmetric linear combinations of the site-basis states:

• Bonding state: $|+{⟩}_{\sigma }=\frac{1}{\sqrt{2}}(|1_{\sigma }⟩+|2_{\sigma }⟩)=\frac{1}{\sqrt{2}}(c_{1\sigma }^{†}+c_{2\sigma }^{†})|0⟩$ with energy $E=-t$.
• Anti-bonding state: $|-{⟩}_{\sigma }=\frac{1}{\sqrt{2}}(|1_{\sigma }⟩-|2_{\sigma }⟩)=\frac{1}{\sqrt{2}}(c_{1\sigma }^{†}-c_{2\sigma }^{†})|0⟩$ with energy $E=+t$.

This result is physically intuitive and should be familiar from elementary quantum chemistry (e.g., the H${}_2^{+}$ ion). The single electron delocalizes across the two sites, forming bonding and anti-bonding "molecular orbitals" with an energy splitting of $2t$. Since this holds for both spin-up and spin-down electrons, we have two states at energy $-t$ and two states at energy $+t$.

The following resource provides a concise derivation of this result.

The Parallel-Spin Case ($N_e=2,S_z=\pm 1$)

Next, let's consider the two remaining states in the $N_e=2$ sector: those with parallel spins.

• For $S_z=+1$, the only possible basis state is $|\uparrow \uparrow ⟩=c_{1\uparrow }^{†}{c}_{2\uparrow }^{†}|0⟩$.
• For $S_z=-1$, the only possible basis state is $|\downarrow \downarrow ⟩=c_{1\downarrow }^{†}{c}_{2\downarrow }^{†}|0⟩$.

Since each of these sectors contains only a single state, these states must be eigenstates of the Hamiltonian. We just need to find their energy. Let's apply the Hamiltonian to $|\uparrow \uparrow ⟩$:

1. Interaction term: $U{\sum }_i{n}_{i\uparrow }{n}_{i\downarrow }|\uparrow \uparrow ⟩$. Since there is no site with both a spin-up and a spin-down electron, the number operator $n_{i\uparrow }{n}_{i\downarrow }$ gives zero for both $i=1$ and $i=2$. The energy from the $U$ term is zero.
2. Hopping term: $-t(c_{1\uparrow }^{†}{c}_{2\uparrow }+c_{2\uparrow }^{†}{c}_{1\uparrow })|\uparrow \uparrow ⟩=-t(c_{1\uparrow }^{†}{c}_{2\uparrow }{c}_{1\uparrow }^{†}{c}_{2\uparrow }^{†}+c_{2\uparrow }^{†}{c}_{1\uparrow }{c}_{1\uparrow }^{†}{c}_{2\uparrow }^{†})|0⟩$. Let's look at the first term: $c_{1\uparrow }^{†}{c}_{2\uparrow }{c}_{1\uparrow }^{†}{c}_{2\uparrow }^{†}|0⟩=-c_{1\uparrow }^{†}{c}_{1\uparrow }^{†}{c}_{2\uparrow }{c}_{2\uparrow }^{†}|0⟩=0$, because $(c_{1\uparrow }^{†}{)}^2=0$ due to the Pauli exclusion principle. The same applies to all other hopping terms.

Physically, the Pauli principle forbids an electron from hopping to a site that is already occupied by an electron of the same spin. Thus, the hopping term gives zero.

Since both the interaction and hopping terms yield zero when acting on these states, their energy is simply $E=0$.

This completes our analysis. In the previous lesson, we found a triplet of states with degenerate energy. These two states, $|\uparrow \uparrow ⟩$ and $|\downarrow \downarrow ⟩$, are the $S_z=+1$ and $S_z=-1$ components of that triplet. The $S_z=0$ component, $|1-⟩=\frac{1}{\sqrt{2}}(c_{1\uparrow }^{†}{c}_{2\downarrow }^{†}-c_{2\uparrow }^{†}{c}_{1\downarrow }^{†})|0⟩$, also has energy $E=0$, confirming the three-fold degeneracy of the triplet state for any value of $U$ and $t$.

The following resource confirms this finding.

The Complete Energy Spectrum

We have now diagonalized the Hamiltonian in every sector of the Hilbert space. We can assemble our findings into a complete energy spectrum. The table below summarizes all 16 energy eigenvalues.

Note the energies for $N_e=3$. You can derive these by solving the problem for a single hole instead of a single electron, which reveals the model's particle-hole symmetry. A hole hopping from site 1 to 2 is like an electron hopping from 2 to 1, giving the same $\mp t$ energies, but now every state has one doubly occupied site, adding a constant energy $U$ to the total.

This full solution is one of the classic results in condensed matter theory. The ability to find every single eigenvalue and eigenstate, even for just two sites, provides a powerful and complete benchmark for the much more complex approximate and computational methods we will need for larger systems.

Conclusion

In this lesson, we completed our analytical solution of the two-site Hubbard model by addressing the sectors outside of half-filling.

Key Takeaways:

• The Hamiltonian's conservation of particle number $N_e$ and spin projection $S_z$ allows the Hilbert space to be broken down into small, manageable blocks.
• In the single-electron ($N_e=1$) sector, the electron delocalizes to form bonding and anti-bonding states with energies $\mp t$, respectively. The interaction $U$ plays no role.
• In the two-electron, parallel-spin ($N_e=2,S_z=\pm 1$) sector, the Pauli exclusion principle prevents hopping. With no double occupancy, the interaction term is also zero. Therefore, these states are trivial eigenstates with energy $E=0$.
• By combining the results from all sectors, we obtained the complete 16-state energy spectrum of the Hubbard dimer.

The two-site model has served its purpose. We have extracted a remarkable amount of physics from it, including charge localization, the Mott gap, and the emergence of superexchange. However, to tackle systems with more atoms—where fascinating collective phenomena like magnetism and superconductivity can truly emerge—we can no longer solve things exactly with pen and paper.

In the next module, we will transition from analytical solutions to computational methods. We will learn how to implement Exact Diagonalization on a computer, a technique that leverages the same block-diagonal structure we used here to numerically solve for the ground state of larger Hubbard clusters than is possible by hand.