- #1
supernano
- 6
- 0
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[ <br /> \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[ <br /> \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).?
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[ <br /> \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[ <br /> \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).?