 # Tom Hodson

Physicist, Programmer, Maker and Baker

# Particle-Hole Symmetry

The Hubbard and FK models on a bipartite lattice have particle-hole (PH) symmetry $$\mathcal{P}^\dagger H \mathcal{P} = - H$$, accordingly they have symmetric energy spectra. The associated symmetry operator $$\mathcal{P}$$ exchanges creation and annihilation operators along with a sign change between the two sublattices. In the language of the Hubbard model of electrons $$c_{\alpha,i}$$ with spin $$\alpha$$ at site $$i$$ the particle hole operator corresponds to the substitution of new fermion operators $$d^\dagger_{\alpha,i}$$ and number operators $$m_{\alpha,i}$$ where

$d^\dagger_{\alpha,i} = \epsilon_i c_{\alpha,i}$ $m_{\alpha,i} = d^\dagger_{\alpha,i}d_{\alpha,i},$

the lattices must be bipartite because to make this work we set $$\epsilon_i = +1$$ for the A sublattice and $$-1$$ for the even sublattice  .

The entirely filled state $$\ket{\Omega} = \sum_{\alpha,i} c^\dagger_{\alpha,i} \ket{0}$$ becomes the new vacuum state $d_{i\sigma} \ket{\Omega} = (-1)^i c^\dagger_{i\sigma} \sum_{j\rho} c^\dagger_{j\rho} \ket{0} = 0.$

The number operator $$m_{\alpha,i} = 0,1$$ now counts holes rather than electrons $m_{\alpha,i} = c^{\phantom{\dagger}}_{\alpha,i} c^\dagger_{\alpha,i} = 1 - c^\dagger_{\alpha,i} c^{\phantom{\dagger}}_{\alpha,i}.$

In the case of nearest neighbour hopping on a bipartite lattice this transformation also leaves the hopping term unchanged because $$\epsilon_i \epsilon_j = -1$$ when $$i$$ and $$j$$ are on different sublattices: $d^\dagger_{\alpha,i} d_{\alpha,j} = \epsilon_i \epsilon_j c^{\phantom{\dagger}}_{\alpha,i} c^\dagger_{\alpha,j} = c^\dagger_{\alpha,i} c^{\phantom{\dagger}}_{\alpha,j}.$

Defining the particle density $$\rho$$ as the number of fermions per site: $\rho = \frac{1}{N} \sum_i \left( n_{i \uparrow} + n_{i \downarrow} \right).$

The PH symmetry maps the Hamiltonian to itself with the sign of the chemical potential reversed and the density inverted about half-filling: $\text{PH} : H(t, U, \mu) \rightarrow H(t, U, -\mu)$ $\rho \rightarrow 2 - \rho.$

The Hamiltonian is symmetric under PH at $$\mu = 0$$ and so must all the observables, hence half-filling $$\rho = 1$$ occurs here. This symmetry and known observable acts as a useful test for the numerical calculations.

Next Section: Markov Chain Monte Carlo

# Bibliography


C. Gruber and D. Ueltschi, The Falicov-Kimball model, in Encyclopedia of mathematical physics. 1, 1, (Elsevier, Amsterdam, 2006).