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I am studying the following hamiltonian, consisting of a double quantum dot system

\begin{align} \hat{H}_{2QD} = \sum_{i=1,2}v_{i}\hat{n}_{i}+\sum_{i=1,2}U_{i}\hat{n}_{i\uparrow}\hat{n}_{i\downarrow} + U_{12}\hat{n}_{i}\hat{n}_{j} \end{align} where $\hat{n}_{i\sigma}=d_{i\sigma}^{\dagger}d_{i\sigma}$ is the number operator, $\hat{n}_i=\hat{n}_{i\uparrow}+\hat{n}_{i\downarrow}$ and $d_{i\sigma}$ ($d_{i\sigma}^\dagger$) are the annihilation (creation) operators for orbital $i$ and spin $\sigma$. Also, $v_{i}$ and $U_{i}$ are the on-site energy (also referred as gate level) and the inter-Coulomb repulsion of the level $i$ and $U_{12}$ is the intra-Coulomb repulsion between the levels 1 and 2.

This hamiltonian is known to be $SU(4)$ invariant, but I'm wondering which transformation explicitly shows this invariance.

Bhagwan rajneesh
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2 Answers2

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If we ignore the orbital index first and consider how the Hubbard interaction $H_h = Un_{\uparrow}n_{\downarrow}$ is SU(2) symmetric. First, we know $n = n_{\uparrow}+n_{\downarrow}$ is SU(2) symmetric because it's the identity matrix in spin representation. And $H_h \sim n^2 = 2n_{\uparrow}n_{\downarrow}+n$, where we used $n_{\sigma}^2 = n_{\sigma}$. We can also explicitly use a Pauli matrix identity $\sum_a\sigma^a_{\alpha\beta}\sigma^a_{\gamma\delta} = 2\delta_{\alpha\delta}\delta_{\beta\gamma}-\delta_{\alpha\beta}\delta_{\gamma\delta}$ to realize ${\bf S}^2 = \sum_a{S^a}{S^a} = \frac{1}{4}(3n-6n_{\uparrow}n_{\downarrow})$.

Introducing back the orbitals, we define $n_t = n_i +n_j$. Notice that $n_t$ is the identity matrix for SU(4) transformations, and $H_{2QD} \sim n_t^2 = n_i^2+n_j^2+2n_in_j$. So if $U_i=U_j = U_{12}$, $H_{2QD}$ is SU(4) symmetric. I don't know a convenient identity to explicitly show it tho.

QBarista
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Probably this is not as smart as QBarista suggested, but you can explicitly check that the Hamiltonian commutes with the 15 generators of the $su(4)$ algebra (or any linear combination of them). Now 6 of them are just spin operators for the $i$-th level, namely $$ \vec{S}_i = \frac{1}{2} \sum_{a,b=\uparrow\downarrow} c^{\dagger}_{ia} \vec{\sigma}_{ab} c_{ib}, $$ where $\vec{\sigma}$ is the vector of Pauli matrices. Another 8 generators are similar to spin $x,y$ operators, but they mix up levels and spins: $$ \frac{1}{2} \left( c^{\dagger}_{1\uparrow} c_{2\uparrow} + \text{hc} \right), \,\,\,\,\,\,\,\,\,\,\, \frac{i}{2} \left( c^{\dagger}_{1\uparrow} c_{2\uparrow} - \text{hc} \right) $$ $$ \frac{1}{2} \left( c^{\dagger}_{1\downarrow} c_{2\downarrow} + \text{hc} \right), \,\,\,\,\,\,\,\,\,\,\, \frac{i}{2} \left( c^{\dagger}_{1\downarrow} c_{2\downarrow} - \text{hc} \right) $$ $$ \frac{1}{2} \left( c^{\dagger}_{1\uparrow} c_{2\downarrow} + \text{hc} \right), \,\,\,\,\,\,\,\,\,\,\, \frac{i}{2} \left( c^{\dagger}_{1\uparrow} c_{2\downarrow} - \text{hc} \right) $$ $$ \frac{1}{2} \left( c^{\dagger}_{1\downarrow} c_{2\uparrow} + \text{hc} \right), \,\,\,\,\,\,\,\,\,\,\, \frac{i}{2} \left( c^{\dagger}_{1\downarrow} c_{2\uparrow} - \text{hc} \right). $$ The last one is a sort of orbital magnetization: $$ n_{1\uparrow} - n_{2\uparrow} $$

Matteo
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