- #1
Hixy
- 6
- 0
1. The problem statement, all variables and givenknown data
We have a system of three atoms arranged in a circular arrangements. They each have a valence electron that can tunnel to the nearest neighbor. For a tunneling rate -A/\hbar we have the Hamiltonian (shifted by an energy E_a on the diagonal)
<br /> H = \begin{pmatrix}<br /> 0 & -A & -A\\<br /> -A & 0 & -A\\<br /> -A & -A & 0<br /> \end{pmatrix}<br />
which eigenvalues -2|A| corresponding to the eigenstate (1,1,1) and degenerated eigenvalues A corresponding to eigenstates (-1,0,1) and (-1,1,0). Since we have translational invariance, p is a good quantum number, hence [H,p]=0 and p and H can be diagonalized simultaneously. Assuming plane wave solutions \psi_n \sim \mathrm{e}^{ikn} we get from boundary conditions that k = 2/3 \pi N for N \in \mathbb{Z}. Now, the three lowest cases are k=0 and k= \pm 2/3 \pi. k=0 corresponds to the eigenvalue -2|A| since for k=0 we have \psi_1=\psi_2=\psi_3=1, i.e. the electrons are evenly distributed over the three atoms. The other two cases give the following linear combinations:
k=2/3 \pi: |\psi_2 \rangle + (-1/2 -i \sqrt{3}/2)|\psi_3 \rangle
k=-2/3 \pi: |\psi_2 \rangle + (-1/2 +i \sqrt{3}/2)|\psi_3 \rangle
My question is now: How do I find the momentum values p for each of the 3 simultaneous eigenstates?
\hat{p} | \psi_n \rangle = p | \psi_n \rangle
I'm really at a loss here. How can we find the momentum values for this discrete system with only assumed plane wave solutions?
We have a system of three atoms arranged in a circular arrangements. They each have a valence electron that can tunnel to the nearest neighbor. For a tunneling rate -A/\hbar we have the Hamiltonian (shifted by an energy E_a on the diagonal)
<br /> H = \begin{pmatrix}<br /> 0 & -A & -A\\<br /> -A & 0 & -A\\<br /> -A & -A & 0<br /> \end{pmatrix}<br />
which eigenvalues -2|A| corresponding to the eigenstate (1,1,1) and degenerated eigenvalues A corresponding to eigenstates (-1,0,1) and (-1,1,0). Since we have translational invariance, p is a good quantum number, hence [H,p]=0 and p and H can be diagonalized simultaneously. Assuming plane wave solutions \psi_n \sim \mathrm{e}^{ikn} we get from boundary conditions that k = 2/3 \pi N for N \in \mathbb{Z}. Now, the three lowest cases are k=0 and k= \pm 2/3 \pi. k=0 corresponds to the eigenvalue -2|A| since for k=0 we have \psi_1=\psi_2=\psi_3=1, i.e. the electrons are evenly distributed over the three atoms. The other two cases give the following linear combinations:
k=2/3 \pi: |\psi_2 \rangle + (-1/2 -i \sqrt{3}/2)|\psi_3 \rangle
k=-2/3 \pi: |\psi_2 \rangle + (-1/2 +i \sqrt{3}/2)|\psi_3 \rangle
My question is now: How do I find the momentum values p for each of the 3 simultaneous eigenstates?
Homework Equations
\hat{p} | \psi_n \rangle = p | \psi_n \rangle
The Attempt at a Solution
I'm really at a loss here. How can we find the momentum values for this discrete system with only assumed plane wave solutions?