Posted by Steve Simon on June 10, 2014, 11:25 pm, in reply to "Re: 2011 Question 7 (Nearly Free Electron Model)"
| Very good question. Yes, in principle k mixes with k+G, k+2G, k+3G etc etc etc. However, the reason we consider only k and k+G is because they are degenerate (E0(k) = E0(k+G)) , so they mix very strongly. The mixing with the other states is weak as it is suppressed by an energy denominator in perturbation theory. |
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