# Treatment of singlets in the Hubbard model

• A
I'm using the general Hubbard model ($H = U \sum n_{i,\uparrow} n_{i,\downarrow} - t \sum (c^{\dagger}_{i,\sigma} c_{i+1,\sigma} + c^{\dagger}_{i+1,\sigma} c_{i,\sigma})$) to solve for eigenstates of simple quantum dot configurations.
For the case of a double dot with two electrons in singlet configuration, I can solve it in two ways:

(1). Using all possible combination of states:
$| \psi \rangle = a_1 | \uparrow\downarrow , 0 \rangle + a_2 | \uparrow , \downarrow \rangle + a_3 | \downarrow , \uparrow \rangle + a_4 | 0 , \uparrow\downarrow \rangle$
with a matrix Hamiltonian $H = \left[ \begin{array}{cccc} U & -t & -t & 0 \\ -t & 0 & 0 & -t \\ -t & 0 & 0 & -t \\ 0 & -t & -t & U \end{array} \right]$

(2). Grouping the single occupation states into one singlet state:
$| \psi \rangle = a_1 | \uparrow\downarrow , 0 \rangle + a_2 / \sqrt{2} ( | \uparrow , \downarrow \rangle + | \downarrow , \uparrow \rangle ) + a_3 | 0 , \uparrow\downarrow \rangle$
with $H = \left[ \begin{array}{cccc} U & -\sqrt{2}t & 0 \\ -\sqrt{2}t & 0 & -\sqrt{2}t \\ 0 & -\sqrt{2}t & U \end{array} \right]$

The two solutions return the same eigenenergies and states, except (1). has an additional state {0,-1,1,0} with eigenvalue {0}.

For this example it's not so important, but if I do the same on a triple quantum dot, I get more additional eigenstates for method (1). The eigenvalues of these additional states correspond to the eigenvalues I get if I model the system with parallel spins.

For larger systems, this starts to matter with regards to which eigenstate is the lowest energy state, and to replicate ground states in literature I would have to use (2). So what am I doing wrong in (1).?

## Answers and Replies

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

radium
Science Advisor
Education Advisor
The state you have written for 2 is not a spin singlet, it is a spin triplet. The triplet is higher in energy, the ground state should be a spin singlet. If you carry out 1. The triplet will have a zero eigenvalue which is correct.

However If you are considering the large U limit the two doubly occupied states are much higher in energy than the others, so you would want to use the effective Hamiltonian method to project onto the subspace of the lower energy states.