Fock Space Basis for Half-Filled Two-Site Hubbard Model

The Hilbert Space of the Two-Site Model

Let's consider a system with just two sites, labeled 1 and 2. In the language of second quantization, the fundamental building blocks are the single-particle states. For our system, there are four such states, which we can define by specifying the site and the spin of an electron:

• An electron at site 1 with spin up: $|\uparrow ,-⟩\equiv c_{1\uparrow }^{†}|0⟩$
• An electron at site 1 with spin down: $|\downarrow ,-⟩\equiv c_{1\downarrow }^{†}|0⟩$
• An electron at site 2 with spin up: $|-,\uparrow ⟩\equiv c_{2\uparrow }^{†}|0⟩$
• An electron at site 2 with spin down: $|-,\downarrow ⟩\equiv c_{2\downarrow }^{†}|0⟩$

Here, $|0⟩$ represents the vacuum state, the state with no electrons.

Any possible state of the system, a Fock state, is determined by specifying whether each of these four single-particle states is occupied or not. Since fermions obey the Pauli exclusion principle, each state can be either empty (occupation number 0) or occupied by one electron (occupation number 1). This gives a total of $2^4=16$ possible states in the full Fock space.

As we discussed when covering the model's symmetries, the Hubbard Hamiltonian conserves the total number of particles, $N={\sum }_{i,\sigma }{n}_{i\sigma }$. This is a powerful constraint. It means the full 16x16 Hamiltonian matrix is block-diagonal, with separate blocks for $N=0,1,2,3,$ and $4$. We can analyze each sector independently.

The Half-Filled Sector: N = 2

The most interesting physics often occurs at or near half-filling. For our two-site model, the maximum electron capacity is four. Therefore, half-filling corresponds to a total of $N=2$ electrons. Our task is to find all possible ways to place two indistinguishable fermions into the four available single-particle slots. From combinatorics, the number of ways to choose 2 slots from 4 is given by the binomial coefficient:

So, the Hilbert space for the half-filled two-site model is six-dimensional. Our next step is to write down these six basis states explicitly. A systematic way to do this is to consider two cases: both electrons on the same site, or on different sites.

1. Both electrons on one site (double occupancy): Pauli exclusion forbids two electrons of the same spin on the same site. Thus, they must have opposite spins.

   • Doubly occupied site 1: One spin-up and one spin-down electron are on site 1. We write this state as $c_{1\uparrow }^{†}{c}_{1\downarrow }^{†}|0⟩$.
   • Doubly occupied site 2: One spin-up and one spin-down electron are on site 2. We write this state as $c_{2\uparrow }^{†}{c}_{2\downarrow }^{†}|0⟩$.

2. One electron on each site:

   • Both spins up: An electron with spin-up on site 1 and another with spin-up on site 2. This state is $c_{2\uparrow }^{†}{c}_{1\uparrow }^{†}|0⟩$.
   • Both spins down: An electron with spin-down on site 1 and another with spin-down on site 2. This state is $c_{2\downarrow }^{†}{c}_{1\downarrow }^{†}|0⟩$.
   • Opposite spins: One electron is spin-up and the other is spin-down. There are two possibilities for this:
     • Spin-up on site 1, spin-down on site 2: $c_{2\downarrow }^{†}{c}_{1\uparrow }^{†}|0⟩$.
     • Spin-down on site 1, spin-up on site 2: $c_{2\uparrow }^{†}{c}_{1\downarrow }^{†}|0⟩$.

Note the convention used in writing these states. To ensure a consistent sign for our basis vectors, we adopt a fixed ordering for the creation operators. A common convention is to order by site index first (larger index to the left) and then by spin (up-spin to the left of down-spin), but any consistent rule works.

The following resource provides an excellent and detailed breakdown of these six states.

The table you've just studied is the core of today's lesson. It provides a complete, explicit basis for the Hilbert space we need. Let's briefly summarize the six states, categorizing them by the total spin projection $S_z=\frac{1}{2}{\sum }_i(n_{i\uparrow }-n_{i\downarrow })$.

• $S_z=+1$ sector (dimension 1):
  1. $|\uparrow ,\uparrow ⟩\equiv c_{2\uparrow }^{†}{c}_{1\uparrow }^{†}|0⟩$
• $S_z=-1$ sector (dimension 1):
  2. $|\downarrow ,\downarrow ⟩\equiv c_{2\downarrow }^{†}{c}_{1\downarrow }^{†}|0⟩$
• $S_z=0$ sector (dimension 4):
  3. $|\uparrow \downarrow ,-⟩\equiv c_{1\uparrow }^{†}{c}_{1\downarrow }^{†}|0⟩$ (Doubly occupied site 1)
  4. $|-,\uparrow \downarrow ⟩\equiv c_{2\uparrow }^{†}{c}_{2\downarrow }^{†}|0⟩$ (Doubly occupied site 2)
  5. $|\uparrow ,\downarrow ⟩\equiv c_{2\downarrow }^{†}{c}_{1\uparrow }^{†}|0⟩$
  6. $|\downarrow ,\uparrow ⟩\equiv c_{2\uparrow }^{†}{c}_{1\downarrow }^{†}|0⟩$

This grouping by $S_z$ is not just for organizational purposes. Since the Hubbard Hamiltonian conserves total spin (and thus also its $z$-component), it will have no matrix elements connecting states with different $S_z$. This means the 6x6 Hamiltonian matrix for the $N=2$ sector further breaks down into smaller, independent blocks for each $S_z$ value, simplifying the diagonalization process significantly.

From Fock States to Bit-Strings

Your interest in computational physics makes the "computer" representation in the reading from Avella and Mancini particularly relevant. This method of mapping quantum states to integers is the foundation of the exact diagonalization (ED) technique we will explore in a later module. Let's see how it works.

First, we establish a fixed ordering for our single-particle basis. A natural choice is:
[site 1/spin ↓, site 1/spin ↑, site 2/spin ↓, site 2/spin ↑]

A Fock state can now be represented by a 4-bit binary string, where a '1' in a position means the corresponding single-particle state is occupied, and a '0' means it is empty.

Let's take the state $|\uparrow ,\downarrow ⟩\equiv c_{2\downarrow }^{†}{c}_{1\uparrow }^{†}|0⟩$.

• Is site 1/spin ↓ occupied? No (0).
• Is site 1/spin ↑ occupied? Yes (1).
• Is site 2/spin ↓ occupied? Yes (1).
• Is site 2/spin ↑ occupied? No (0).

Reading from right to left (a common convention in physics for mapping to integers), the bit-string representation is 0110. The corresponding integer is $0\cdot 2^3+1\cdot 2^2+1\cdot 2^1+0\cdot 2^0=6$.

Let's do one more: the doubly occupied state $|\uparrow \downarrow ,-⟩\equiv c_{1\uparrow }^{†}{c}_{1\downarrow }^{†}|0⟩$.

• The bit string is 0011, which corresponds to the integer 3.

This mapping is a one-to-one correspondence between the physical Fock states and a set of integers. This is extremely powerful because it allows us to represent the basis states of our Hilbert space as a simple array of integers in a computer program. All operations, like applying the Hamiltonian to a state, then become manipulations of these bit-strings.

Conclusion

In this lesson, we have taken the first crucial step toward solving the two-site Hubbard model. By systematically analyzing the constraints of particle number and spin, we have constructed a complete, six-dimensional basis for the physically rich half-filled sector.

Key Takeaways:

• The full Hilbert space of the two-site model has $2^4=16$ states.
• By focusing on the half-filled sector ($N=2$), we reduce the problem to a more manageable 6-dimensional Hilbert space.
• The six basis states can be categorized by their configuration (doubly occupied vs. singly occupied sites) and by their total spin projection $S_z$.
• This basis can be represented in multiple ways: using creation operators, pictorially, or as bit-strings, the last of which is fundamental for computational approaches.

With our basis vectors now clearly defined, we are ready to determine how the Hamiltonian acts on them. In the next lesson, we will use this basis to construct the 6x6 Hamiltonian matrix. This will transform the abstract operator equation into a concrete matrix eigenvalue problem, which we will then solve to find the exact energies and eigenstates of the two-site Hubbard model.