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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:

[itex]| \psi \rangle = a_1 | \uparrow\downarrow , 0 \rangle + a_2 | \uparrow , \downarrow \rangle + a_3 | \downarrow , \uparrow \rangle + a_4 | 0 , \uparrow\downarrow \rangle[/itex]

with a matrix Hamiltonian [itex]H = \left[

\begin{array}{cccc} U & -t & -t & 0 \\ -t & 0 & 0 & -t \\ -t & 0 & 0 & -t \\ 0 & -t & -t & U \end{array} \right][/itex]

(2). Grouping the single occupation states into one singlet state:

[itex]| \psi \rangle = a_1 | \uparrow\downarrow , 0 \rangle + a_2 / \sqrt{2} ( | \uparrow , \downarrow \rangle + | \downarrow , \uparrow \rangle ) + a_3 | 0 , \uparrow\downarrow \rangle[/itex]

with [itex]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][/itex]

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).?