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Discrete 3-particle system - Condensed Matter Pysics

  • Thread starter Hixy
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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 [itex]-A/\hbar[/itex] we have the Hamiltonian (shifted by an energy [itex]E_a[/itex] on the diagonal)
[tex]
H = \begin{pmatrix}
0 & -A & -A\\
-A & 0 & -A\\
-A & -A & 0
\end{pmatrix}
[/tex]
which eigenvalues [itex]-2|A|[/itex] corresponding to the eigenstate (1,1,1) and degenerated eigenvalues [itex]A[/itex] corresponding to eigenstates (-1,0,1) and (-1,1,0). Since we have translational invariance, [itex]p[/itex] is a good quantum number, hence [itex][H,p]=0[/itex] and [itex]p[/itex] and [itex]H[/itex] can be diagonalized simultaneously. Assuming plane wave solutions [itex]\psi_n \sim \mathrm{e}^{ikn}[/itex] we get from boundary conditions that [itex]k = 2/3 \pi N[/itex] for [itex]N \in \mathbb{Z}[/itex]. Now, the three lowest cases are [itex]k=0[/itex] and [itex]k= \pm 2/3 \pi[/itex]. [itex]k=0[/itex] corresponds to the eigenvalue [itex]-2|A|[/itex] since for [itex]k=0[/itex] we have [itex]\psi_1=\psi_2=\psi_3=1[/itex], i.e. the electrons are evenly distributed over the three atoms. The other two cases give the following linear combinations:
[itex]k=2/3 \pi[/itex]: [tex]|\psi_2 \rangle + (-1/2 -i \sqrt{3}/2)|\psi_3 \rangle[/tex]
[itex]k=-2/3 \pi[/itex]: [tex]|\psi_2 \rangle + (-1/2 +i \sqrt{3}/2)|\psi_3 \rangle[/tex]

My question is now: How do I find the momentum values [itex]p[/itex] for each of the 3 simultaneous eigenstates?

Homework Equations


[itex]\hat{p} | \psi_n \rangle = p | \psi_n \rangle[/itex]


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?
 

Answers and Replies

  • #2
fzero
Science Advisor
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Since we have translational invariance, [itex]p[/itex] is a good quantum number, hence [itex][H,p]=0[/itex] and [itex]p[/itex] and [itex]H[/itex] can be diagonalized simultaneously. Assuming plane wave solutions [itex]\psi_n \sim \mathrm{e}^{ikn}[/itex] we get from boundary conditions that [itex]k = 2/3 \pi N[/itex] for [itex]N \in \mathbb{Z}[/itex].
You should think about how we arrive at this form of the plane-wave. What do #k# and #n# have to do with position and momentum?
 

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