Eigenenergy of triangular molecule

[physicsxanime]

Homework Statement

See the screenshot next

Relevant Equations

Eigenenergy E is $\hat{H}\psi = E\psi$
electron is fermion, so it should be antisymmetrized

I am preparing for graduate prelim exam:
My attempt is that, I have three sites for each of the electron to be at, each of them are 1s orbital. Also, electron has a spin 1/2. So, I think the Hilbert space would be quite large, I have state of both electron on each site 1, 2, 3 with singlet spins. Then there are 3 more state of electron occupying a different site. My confusion here is then, since both electrons are in s state, they are in symmetric state, which forces the spin to be in singlet which is anti-symmetric. Then how do I get the triplet state energy? Also the Hilbert space is now size of 6, How am I diagonalizing 6x6 matric?

B is to add a U term in the diagonals of singlet state, which is the electron on same site?This would split the eigenenergy?

 

[DrClaude]

Since A says to neglect Coulomb repulsion, start with finding the eigenstates of a single electron, then you can put the two electrons in these eigenstates while taking care of building anti-symmetric states.

 

[physicsxanime]

Yes, of course. I am just stupid. Let's shift the energy scale so that the atomic energy at 1s is 0.
Then the Hamiltonian is
$$
\begin{pmatrix}
0 & 1 & 1\\
1 & 0 & 1\\
1 & 1 & 0
\end{pmatrix}
$$
The eigenenergies and corresponding eigenvectors are $\lambda_1 = -1$, $\lambda_2 = -1$, $\lambda_3 = 2$, $v_1 = (-1,0,1)^T$, $v_2 = (-1,1,0)^T$, $v_3 = (1,1,1)^T$

Note that $v_1$ and $v_2$ are anti-symmetric while $v_3$ is symmetric.
Therefore, the singlet state is paired up with symmetric state and the lowest eigenvalue is using $v_1$ and $v_2$ or any linear combination of the two on each of the electrons. So this state has eigen value of -2.
Whereas the triplet state is aired up with antisymmetric state and the lowest eigenvalue is using $v_1$ and $v_2$ or any linear combination on one electron and the other one using $v_3$. Hence, the eigenvalue is 1.

B, hmmm, only some of the singlet state is having large splitting. But anyway, one can use degenerate perturbation theory to prove the assertion, the correction must be of order $U$, so we know there is a large splitting without doing the calculation. And the large splitting is from singlet state forcing a symmetric spatial state and hence gaining U by putting them on the same site.

 

[DrDu]

This is the so called Hubbard Model hamiltonian. For U=0 it reduces to the Hückel hamiltonian.
Note that t is usually negative. Hopping stabilizes the system. The limit U=infinity corresponds to valence bond theory.