Diagonalizing the Half-Filled Hamiltonian

Welcome back to our study of the two-site Hubbard model. In our last lesson, we took the essential first step of mapping out the Hilbert space for this system at half-filling. We identified the six key basis states and saw how they can be grouped according to conserved quantum numbers like total particle number $N$ and total spin projection $S_{z}$ .

This lesson will build directly on that foundation. Our goal is to bridge the gap from the abstract Hubbard Hamiltonian operator to a concrete numerical problem we can solve. We will construct the Hamiltonian matrix using the basis we developed. As you know from your quantum mechanics courses, once we have this matrix, finding the system's energy eigenvalues and eigenstates becomes a standard linear algebra problem: matrix diagonalization. This process will allow us to see, for the first time, the exact energy spectrum of an interacting electron system.

From Operator to Matrix

Let's begin by recalling the Hubbard Hamiltonian for our two-site system (sites labeled 1 and 2):

H = - t σ \sum (c_{2 σ}^{†} c_{1 σ} + c_{1 σ}^{†} c_{2 σ}) + U i = 1, 2 \sum n_{i ↑} n_{i ↓}

The fundamental task of "solving" a quantum system is to solve the time-independent Schrödinger equation, $H ∣Ψ ⟩ = E ∣Ψ ⟩$ . In the "exact diagonalization" approach, this is achieved by representing the Hamiltonian $H$ as a matrix in a chosen basis and then finding its eigenvalues $E$ and eigenvectors $∣Ψ ⟩$ .

The elements of the Hamiltonian matrix, $H_{ij}$ , are given by the inner product $H_{ij} = ⟨ ψ_{i} ∣ H ∣ ψ_{j} ⟩$ , where $∣ ψ_{i} ⟩$ and $∣ ψ_{j} ⟩$ are the basis states we enumerated in the previous lesson.

Exact Diagonalization for spin systems

Before we start computing matrix elements, it's helpful to review the general principle of representing a quantum Hamiltonian as a matrix. This video from "Dr Mitchell's physics channel" provides a clear overview. While the example uses spin systems, the concepts are perfectly general.

Watch the segments that explain what it means to solve the Schrödinger equation by matrix diagonalization and how to define and construct the elements of the Hamiltonian matrix in a given basis. Notice the emphasis on the matrix being Hermitian, which is a property we will rely on.

Exploiting Symmetries: Block Diagonalization

In the previous lesson, we found that the 6 basis states for the half-filled ( $N = 2$ ) system can be divided into three groups based on their total spin-z projection, $S_{z}$ :

$S_{z} = + 1$ : $∣ ↑, ↑ ⟩$ (1 state)
$S_{z} = - 1$ : $∣ ↓, ↓ ⟩$ (1 state)
$S_{z} = 0$ : A subspace of 4 states: $∣ ↑↓, - ⟩$ , $∣ -, ↑↓ ⟩$ , $∣ ↑, ↓ ⟩$ , $∣ ↓, ↑ ⟩$ .

Since the Hubbard Hamiltonian commutes with the total $S_{z}$ operator, $S_{z}^{tot}$ , there can be no matrix elements connecting states with different $S_{z}$ values. This means our 6x6 Hamiltonian matrix is block-diagonal, composed of smaller, independent matrices for each $S_{z}$ sector.

This image is a schematic of a sparse Hamiltonian matrix, similar to what we will construct. The colored blocks along the diagonal represent non-zero matrix elements within subspaces defined by conserved quantities (like $S_z$), while the white areas are all zeros.

This block-diagonal structure is a massive simplification. Instead of diagonalizing a large 6x6 matrix, we can diagonalize each block separately.

Exact Diagonalization for spin systems

To reinforce this crucial concept, the same video explains how conserved quantities lead to this block-diagonal form.

Focus on the part that discusses how conserved quantum numbers simplify the problem by breaking the full Hamiltonian into smaller blocks. Then, watch the concrete example of a two-spin system, where conservation of total spin-z leads to a 4x4 matrix breaking into a 1x1, a 2x2, and another 1x1 block. This is perfectly analogous to our current problem.

Constructing the Matrix Blocks

Let's now compute the matrix elements for each block.

The $S_{z} = \pm 1$ Sectors

These two sectors are the simplest, as they each contain only one state.

$S_{z} = + 1$ : The basis is $∣ ψ_{1} ⟩ = ∣ ↑, ↑ ⟩ = c_{2 ↑}^{†} c_{1 ↑}^{†} ∣0 ⟩$ .
To find the 1x1 matrix (which is just a single number, the eigenvalue), we calculate $⟨ ψ_{1} ∣ H ∣ ψ_{1} ⟩$ .
- Hopping term: The hopping operator $- t (c_{2 σ}^{†} c_{1 σ} + c_{1 σ}^{†} c_{2 σ})$ attempts to move an electron. For example, $c_{2 ↑}^{†} c_{1 ↑}$ tries to move the spin-up electron from site 1 to site 2. But site 2 is already occupied by a spin-up electron. Due to the Pauli exclusion principle, this action results in zero. The same applies to all other hopping possibilities. So, the kinetic energy is 0.
- Interaction term: The operator $U \sum_{i} n_{i ↑} n_{i ↓}$ gives a contribution of $U$ for each doubly occupied site. In the state $∣ ↑, ↑ ⟩$ , there are no doubly occupied sites. So, the interaction energy is 0.
  Therefore, $H ∣ ↑, ↑ ⟩ = 0$ . This state is an eigenstate with energy $E = 0$ .
$S_{z} = - 1$ : The basis is $∣ ψ_{2} ⟩ = ∣ ↓, ↓ ⟩ = c_{2 ↓}^{†} c_{1 ↓}^{†} ∣0 ⟩$ .
By identical reasoning, $H ∣ ↓, ↓ ⟩ = 0$ . This is also an eigenstate with energy $E = 0$ .

These two states form two of the three components of a spin triplet, which we will analyze further in a later lesson.

The $S_{z} = 0$ Sector

This is the most interesting sector, with a 4-dimensional basis. As discussed, a powerful technique is to use a basis that respects the symmetries of the problem. The Hamiltonian is symmetric under the exchange of sites $1 \leftrightarrow 2$ . We can therefore construct basis states with definite parity (even or odd) under this exchange. This will further block-diagonalize our 4x4 matrix into two 2x2 blocks.

The following document provides an elegant demonstration of this procedure.

6 Introduction to the Hubbard Model - Robert Eder ...

This lecture note by Robert Eder provides a concise but thorough solution of the two-site Hubbard model, or "Hubbard dimer". We will follow its construction of the Hamiltonian matrix.

Please read section 3, "The Hubbard dimer: solution by exact diagonalization". Focus on the first half of the section, up to the point where the eigenstates and energies for both parity sectors are stated. Pay close attention to how the even (P=+1) and odd (P=-1) basis states are defined in equation (14) and how the 2x2 matrix for the even sector is constructed and solved.

Let's re-trace the key steps from the reading and fill in some details.

1. Symmetrized Basis States ( $S_{z} = 0$ )
We recombine our original four $S_{z} = 0$ states into combinations with definite parity:

Even Parity ( $P = + 1$ ) states:
- $∣1 + ⟩ = \frac{1}{2} (c_{1 ↑}^{†} c_{2 ↓}^{†} + c_{2 ↑}^{†} c_{1 ↓}^{†}) ∣0 ⟩$ , representing delocalized, opposite-spin electrons.
- $∣2 + ⟩ = \frac{1}{2} (c_{1 ↑}^{†} c_{1 ↓}^{†} + c_{2 ↑}^{†} c_{2 ↓}^{†}) ∣0 ⟩$ , representing a superposition of doubly occupied sites.
Odd Parity ( $P = - 1$ ) states:
- $∣1 - ⟩ = \frac{1}{2} (c_{1 ↑}^{†} c_{2 ↓}^{†} - c_{2 ↑}^{†} c_{1 ↓}^{†}) ∣0 ⟩$
- $∣2 - ⟩ = \frac{1}{2} (c_{1 ↑}^{†} c_{1 ↓}^{†} - c_{2 ↑}^{†} c_{2 ↓}^{†}) ∣0 ⟩$

2. The Even-Parity 2x2 Matrix
The Hamiltonian for the even-parity sector is a 2x2 matrix with basis ${∣1 + ⟩, ∣2 + ⟩}$ . The matrix elements are:

$⟨ 1 + ∣ H ∣1 + ⟩ = 0$ (No double occupancy, so $U = 0$ . The hopping term moves the system to state $∣2 + ⟩$ , so this diagonal element is zero).
$⟨ 2 + ∣ H ∣2 + ⟩ = U$ (The state is a superposition of double occupancies, so the interaction energy is $U$ . The hopping term moves the system to state $∣1 + ⟩$ ).
$⟨ 1 + ∣ H ∣2 + ⟩ = ⟨ 2 + ∣ H ∣1 + ⟩ = - 2 t$ (The hopping term connects these two states).

This gives the 2x2 matrix shown in the reading:

H_{P = + 1} = (0 - 2 t - 2 t U)

3. The Odd-Parity 2x2 Matrix
For the odd-parity sector, with basis ${∣1 - ⟩, ∣2 - ⟩}$ , a similar calculation shows that the off-diagonal hopping element is zero: $⟨ 1 - ∣ H ∣2 - ⟩ = 0$ . This means the matrix is already diagonal!

$⟨ 1 - ∣ H ∣1 - ⟩ = 0$
$⟨ 2 - ∣ H ∣2 - ⟩ = U$
The states $∣1 - ⟩$ and $∣2 - ⟩$ are already eigenstates with energies $E = 0$ and $E = U$ , respectively.

Diagonalizing the Matrix and Finding Eigenstates

Now we solve the eigenvalue problem for the only non-trivial block, the even-parity matrix. We need to find the eigenvalues $E$ and eigenvectors $(u, v)$ that satisfy:

(0 - 2 t - 2 t U) (u v) = E (u v)

The characteristic equation is $\det(H - E\mathbb{I}}) = 0$ , which gives:

(- E) (U - E) - (- 2 t) (- 2 t) = 0 ⟹ E^{2} - U E - 4 t^{2} = 0

Solving this quadratic equation for $E$ gives the two eigenvalues:

E_{\pm} = \frac{U \pm U ^{2} + 16 t ^{2}}{2} = \frac{U}{2} \pm (\frac{U}{2})^{2} + 4 t^{2}

These are the energies of the remaining two eigenstates of the system. The corresponding eigenvectors $∣ Ψ_{\pm} ⟩ = u_{\pm} ∣1 + ⟩ + v_{\pm} ∣2 + ⟩$ can be found by substituting the eigenvalues back into the matrix equation. The ground state of the entire system corresponds to the lower energy, $E_{-}$ .

6 Introduction to the Hubbard Model - Robert Eder ...

The document by Eder also gives the explicit result for the ground state energy and the coefficients of the eigenvector.

Now, look at the equations labeled (15) in the same section. This gives you the expression for the ground state energy E_0 (which is our E_-) and the coefficients u and v that define the ground state wave function in terms of the even-parity basis states.

Summary of Results

We have successfully diagonalized the Hamiltonian for the two-site, half-filled Hubbard model. Let's collect all our results. The system has 6 energy eigenstates:

Energy Eigenvalue	Eigenstate(s)	Degeneracy	$S_{z}$	Parity
$E_{0} = \frac{U}{2} - (\frac{U}{2})^{2} + 4 t^{2}$	$	\Psi_0\rangle = u	1+\rangle + v	2+\rangle$ (ground state)
$E_{1} = 0$	$	\uparrow, \uparrow\rangle $,$	\downarrow, \downarrow\rangle $,$	1-\rangle$ (triplet excited state)
$E_{2} = U$	$	2-\rangle$	1	0
$E_{3} = \frac{U}{2} + (\frac{U}{2})^{2} + 4 t^{2}$	$	\Psi_3\rangle = v	1+\rangle - u	2+\rangle$ (highest energy state)

This table contains the complete solution to the problem. We have found the exact energy spectrum.

Conclusion

In this lesson, we transitioned from an abstract operator to a concrete matrix representation of the Hubbard Hamiltonian for the two-site model at half-filling. By exploiting the conservation of total spin projection $S_{z}$ and parity, we simplified a 6x6 matrix problem into several much smaller, manageable blocks.

Key Takeaways:

Solving the Schrödinger equation via exact diagonalization means constructing and diagonalizing the Hamiltonian matrix in a chosen basis.
Conserved quantities (symmetries) make the Hamiltonian matrix block-diagonal, dramatically simplifying the problem.
For the two-site Hubbard model at half-filling, the 6x6 Hamiltonian breaks down into two 1x1 blocks ( $S_{z} = \pm 1$ ) and one 4x4 block ( $S_{z} = 0$ ).
The 4x4 $S_{z} = 0$ block can be further block-diagonalized into two 2x2 matrices using a basis with definite parity.
Solving the resulting eigenvalue problems yields the complete set of 6 energy eigenvalues and their corresponding eigenstates.

We now have the exact solution in hand. In the next lesson, we will begin to analyze these results. We will examine the ground state wave function and its energy as a function of the crucial ratio $U / t$ , which will give us our first quantitative glimpse into the competition between electron delocalization and Coulomb repulsion.

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