In relation to Feynman Close coupling procedure, how is the relation E(k)=E0-2Acos(kx).?

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In relation to Feynman Close coupling procedure, how is the relation E(k)=E0-2Acos(kx).?

The relation is for the superposition of two waves, the expression gives the energy level difference.....

2006-07-26 11:05:34 · 2 answers · asked by RobLough 3 in Science & Mathematics ➔ Physics

2 answers

It's a standard result in static perturbation theory for a periodic lattice in one dimension. Start with free wavefunctions (exp (+/-ikx)) find the Hamiltonian in this basis, and diagonalize it.

2006-07-26 15:13:49 · answer #1 · answered by Benjamin N 4 · 4⤊ 2⤋

That's funny - I read "Feynman Close Copulating Procedure" and nearly fell off my chair.
Hah!
Sorry I don't have an answer to your question.
(snicker)

2006-07-26 18:36:02 · answer #2 · answered by Anonymous · 0⤊ 0⤋