# Tunneling elements in the Hubbard model

Consider the tunneling Hamiltonian in the Hubbard model for a 1D lattice of quantum dots.

\begin{align}\hat{H}_t=t\displaystyle\sum_{i,j,\sigma}c_{i,\sigma}^{\dagger}c_{j\sigma}+c^{\dagger}_{j,\sigma}c_{i,\sigma}\hspace{5mm}\text{where } i\neq j\label{one} \tag{1}\end{align}

where $$i,j\in {1,2}$$ (we will only be looking in the case of 2 dots) are the index of the dots and $$\sigma \in {\uparrow,\downarrow}$$ is the spin of the electron and $$t$$ is the tunnel coupling between the dots.

This term is also referred to as the "kinetic" term as it conveys the hopping of an electron from one site $$i$$ to a neighbour site $$j=i+1$$ or $$j=i-1$$. From a physical perspective this term doesn't prefer the hopping of a spin up $$\sigma=\uparrow$$ or a spin down state $$\sigma=\downarrow$$. However when looking into literature of lateral semi-conductor quantum dots, the hamiltonian for the hopping is conveyed as

$$\hat{H}_t=t\displaystyle\sum_{i,j,\sigma}c_{i,\sigma}^{\dagger}c_{j\sigma}-c^{\dagger}_{j,\sigma}c_{i,\sigma}\hspace{5mm}\text{where } i\neq j \tag{2}$$

where there seems to be a distinction between which spin is favourable to hop.

For example the Hamiltonian in the following basis: 

$$\psi_i\in\{|\downarrow, \downarrow\rangle,|\uparrow, \downarrow\rangle,|\downarrow, \uparrow\rangle,|\uparrow, \uparrow\rangle,|\uparrow\downarrow, 0\rangle,|0, \uparrow\downarrow\rangle\}$$

is given by: $$H_t=\langle\psi_i|\hat{H}_t|\psi_j\rangle=\left(\begin{array}{cccccc}{0} & {0} & {0} & {0} & {0} & {0} \\ {0} & {0} & {0} & {0} & {t} & {t} \\ {0} & {0} & {0} & {0} & {-t} & {-t} \\ {0} & {0} & {0} & {0} & {0} & {0} \\ {0} & {t} & {-t} & {0} & {0} & {0} \\ {0} & {t} & {-t} & {0} & {0} & {0}\end{array}\right)$$

where

\begin{align} \langle\psi_2|\hat{H}_t|\psi_5\rangle &=\langle\uparrow, \downarrow|\hat{H}_t|\uparrow\downarrow, 0\rangle =t \\ \langle\psi_2|\hat{H}_t|\psi_6\rangle &=\langle\uparrow, \downarrow|\hat{H}_t|0,\uparrow\downarrow\rangle =t \\ \langle\psi_3|\hat{H}_t|\psi_5\rangle &=\langle\downarrow, \uparrow|\hat{H}_t|\uparrow\downarrow, 0\rangle =-t \\ \langle\psi_3|\hat{H}_t|\psi_6\rangle &=\langle\downarrow, \uparrow|\hat{H}_t|0,\uparrow\downarrow\rangle =-t \end{align}

From these equations it seems that when there are two electrons in a single dot $$\{|\uparrow\downarrow, 0\rangle,|0, \uparrow\downarrow\rangle\}$$, the spin down $$\sigma=\downarrow$$ state seems to be the preferred electron to hop as the matrix element is $$-t$$ and not $$t$$.

My question is how is this explained from a physical perspective? How can you create a preference for what spin to hop? Usually in the Hubbard model all the tunneling elements have the same sign because of equation 1.  What is exactly the physical interpertation of equation 2 in contrast to equation 1?

References:

 https://arxiv.org/abs/1010.0164, equation 2
 https://arxiv.org/abs/1411.5760, equation 1
 https://arxiv.org/abs/0807.4878, equation 22

• In the paragraph before Eq. (1) in your Ref.  they talk about inducing a spin-dependent phase shift to the states with anti-parallel spins that'd give rise to the sign structure you see here. I'm not sure how that is engineered in the quantum dot system though. Aug 9, 2019 at 23:01