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: [1][2]

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

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


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

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. [3] What is exactly the physical interpertation of equation 2 in contrast to equation 1?


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

  • $\begingroup$ In the paragraph before Eq. (1) in your Ref. [1] 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. $\endgroup$
    – Anyon
    Aug 9, 2019 at 23:01


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