4

The Hamiltonian for the two-site Hubbard model is as follows:

$H = - t \sum_{\sigma = \uparrow, \downarrow} (c_{1\sigma}^{\dagger} c_{2\sigma} + \text{h.c.}) + U \sum_{i=1,2}n_{i\uparrow}n_{i\downarrow}$

For homework, I need to diagonalize this Hamiltonian. I don't know how to proceed.

There are 16 possible "natural" basis states, i.e. $\lvert0\rangle$ (empty), $\lvert\uparrow\downarrow\rangle$, $\lvert\uparrow\rangle$, $\lvert\downarrow\rangle$ on each site. I could work out the matrix representation of the Hamiltonian and brute force the diagonalization, but I hope it can be done more elegantly. After searching online, I find that $H$ commutes with total $S_z$ and $N_{\uparrow}$, $N_{\downarrow}$, effectively creating blocks that I can diagonalize separately. How can I use this to solve this problem without resorting to numerical linear algebra?

Kappie001
  • 205
  • This model has another symmetry that could be useful. – TLDR May 02 '16 at 14:44
  • Also, not all of the symmetries that you mentioned before are independent of each other (i.e. one of them may follow from the others). – TLDR May 02 '16 at 15:01
  • Thanks. Turns out we could assume half-filling. Is the symmetry you mean the total amount of spins? Because I ended up solving it by diagonalizing $N = 0, 1, 2, 3, 4$ separately. – Kappie001 May 03 '16 at 07:54
  • Permutation (or translation) symmetry. – TLDR May 03 '16 at 14:26

1 Answers1

2

Assuming half filling (as you indicated in the comments) you can see that, after working out the matrix representation of the Hamiltonian in the occupation basis, you only have to diagonalize a $4$x$4$ matrix corresponding to the states with $S_z = 0$.
There you can make use of the spatial symmetry an introduce a parity operator $M$ which acts in the following way:

Given the state $c_{0\downarrow}^\dagger c_{1\uparrow}^\dagger|0\rangle$

$$M c_{0\downarrow}^\dagger c_{1\uparrow}^\dagger|0\rangle = c_{M(0)\downarrow}^\dagger c_{M(1)\uparrow}^\dagger|0\rangle = c_{1\downarrow}^\dagger c_{0\uparrow}^\dagger|0\rangle$$

Using combinations of the occupation basis states, we can find the eigenstates of $M$, which simplify the Hamiltonian immensely, such that one only has to diagonalize a $2$x$2$ matrix in the end.

The resulting eigenvectors correspond to Triplet and Singlet states.

In this paper arXiv:0807.4878 this is described in more detail.

P-A
  • 410