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Unformatted text preview: Chemistry 120A Problem Set 9 (due April 26, 2010) 1. In lecture, we discussed a procedure for determining the energy levels of an electron in a ring of N tight binding states. The site energy was α , and the nearest neighbor hopping element was β . With these parameters, the energy was α- 2 | β | cos( kb ), where k = 2 πn/Nb, n = 1 , 2 ,...,N . In this problem, consider not a ring, but a line of M sites that begins with site 1 at position x 1 = b and ends with site M at x M = Mb . To do it, use a trick of constructing a suitable linear combination of states for a ring with N sites, N M . Here’s how: Let ψ k = exp( ikx n ) denote a stationary state of the N site chain. Since ψ- k is degenerate with ψ k , the state Φ k = ψ- k- ψ k is also an eigen solution. (a) Verify this last statement, and further show that Φ k ( x ) is zero at x = 0, which means that Φ k ( x n ) satisfies one of the two boundary conditions for the stationary states of an electron on the line. (b) Show that the other boundary condition, Φ k ( b + Mb ) = 0 is satisfied if k = π ( M + 1) b s, s = 1 , 2 ,...,M . Note that k is now positive only. This is because Φ- k is the same as Φ k to within a constant, so switching k to- k does not change the state. By enforcing the boundary conditions, the electron is trapped between the two points x = 0 and x = ( M +1) b , and thus this analysis determines the stationary state wave...
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This note was uploaded on 05/14/2010 for the course CHEM 120ACHEM taught by Professor Chandler during the Spring '10 term at Berkeley.
- Spring '10