Could someone explain, is it possible to make Hubbard Hamiltonian manifestly SU(2) invariant? I know about the interaction term, but how would kinetic (hopping effect) term have to look like?

Here I'm talking about the simplified Fermi-Hubbard Hamiltonian that only has those two terms (hopping effect term which includes the nearest cites and interaction term, where the interaction exists only between the electrons on the same lattice cite).

How could creation and annihilation Fermi operator look like when expressed in terms of spin operators?

I'm looking forward to hearing from someone here.


The Hubbard Model is already manifestly SU(2) spin-rotation invariant because there are no terms that connect the $\uparrow$ and $\downarrow$ sectors (here by connect, I mean change a $\uparrow$ to a $\downarrow$ or vice versa. The full Hamiltonian in terms of fermion annihilation/creation operators is:

$$ H = - t \sum \limits_{t,\sigma} \left( c^\dagger_{i,\sigma} c_{i+1,\sigma} + c^\dagger_{i+1,\sigma} c_{i,\sigma} \right) + U\sum_i n_{i,\uparrow} n_{i,\downarrow}\tag{1}$$

Notice that the hopping term only acts on either the $\uparrow$ or $\downarrow$ sectors, it never changes the **number$$ of $\uparrow$ or $\downarrow$ electrons. The potential ($U$) term does allow interaction between $\uparrow$ and $\downarrow$ electrons, but it is diagonal, so it never changes the total spin of the system.

In terms of spin ladder operators:

The most common way of thinking about the Hubbard model in terms of spin ladder operators is to consider the case of large $U$ at half filling. Half filling means there are $N$ electrons for $N$ sites, and large-$U$ means that the energy cost of double-occupying a site is high.

From this setup, we can do time-independent degenerate perturbation theory, treating the $U$ term as $H_0$ and the hopping as a perturbation. Let's consider ju From this standpoint, we have a set of degenerate ground states with one electron on each site. The energy of this ground state is zero ($E_0 =0$) and all possible combinations of up and down spins are degenerate.

As a rough example, let's consider an isolated pair of Hubbard sites with hamiltonian $H_0$ and hopping perturbation $V$: \begin{eqnarray} H_0 &= & U ( n_{+,1} n_{-,1} + n_{+,2} n_{-,2} ) \tag{2}\\ V &= & -t ( c^\dagger_{\pm,1} c_{\pm,2} + c^\dagger_{\pm,2} c_{\pm,1} )\tag{3} \end{eqnarray} The eigenstates of $H_0$ in the subspace of half filling are \begin{eqnarray} H_0 | 1 \rangle = & H_0 | \pm, \pm \rangle = 0 \tag{4}\\ H_0 |2 \rangle = & H_0 |\pm, \mp \rangle = 0 \tag{5}\\ H_0 |3 \rangle = & H_0 |0,2\rangle = U \tag{6}\\ H_0 |4 \rangle = & H_0 |2,0\rangle= U\tag{7} \end{eqnarray}

This hamiltonian has two degenerate ground states: $|1\rangle,|2\rangle$ with eigenvalues of 0. Here, first order degenerate perturbation theory fails because there the matrix elements $\langle 1|V|2 \rangle = 0$, so we must advance to second-order perturbation theory: $$ E^{(2)}_i = \sum \limits_{i \ne j} \frac{ \langle i | H_0 | j \rangle \langle j | H_0 | i \rangle }{E^{(0)}_i - E^{(0)}_j}\tag{8} $$ In second order, a pair of antiparallel spins can make a virtual transition to a double occupied state and back. This effectively lowers the energy of antiparallel neighbors, while parallel neighbors don't benefit from this virtual transition (also called super exchange). For example, consider state 2:

\begin{align} E^{(2)}_2 &= \frac{ \langle \pm \mp | H_0 | 0 2 \rangle \langle 0 2 | H_0 | \pm \mp \rangle }{0 - U} + \frac{ \langle \pm \mp | H_0 | 2 0 \rangle \langle 2 0 | H_0 | \pm \mp \rangle }{0 - U} \tag{9}\\ \nonumber E^{(2)}_2 &= -\frac{t_0^2 }{U} - \frac{ t_0^2 }{ U} \tag{10}\\ &= - \frac{ 2 t_0^2}{U} \tag{11} \end{align}

This means that $H |2 \rangle = -\frac{2t_0^2}{U} | 2\rangle $ therefore $J = \frac{2t_0^2}{U}$. This is not a rigorous derivation, but it gets the basic idea across.

The equivalent spin-ladder operator Hamiltonian to the large-U half-filling Hubbard model is there for just the Heisenberg model with $J = 2\frac{t^2}{U}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.