PHYS 635 Condensed Matter Physics
Assignment 4 (Nov 9, 2004) Solutions
1. A&M Problem 12.2. For electrons near a band minimum or maximum, we have
2
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PHYS 635 Solid State Physics Take home exam 1
Gregory Eremeev Fall 2004 Submitted: November 8, 2004
Problem 1:Ashcroft & Mermin, Ch.10, p.189, prob.2 a) Lets prove xx = yy = zz = xx = = Now 0 = xx yy = dr (x2 y 2 ) (r)2 U (r) (3) dr (r) x (r) U (r) = x
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Homework 6, due December 6, 1996
Problem 1 Part a) At x ? a (x) = AeiK x + (Ar + B t)e?iKx 2 At x a (x) = (At + Br)eiKx + Be?iKx a 2 2 From (8.68) ( a ) = eika (? a ) a= eika(Ae?aiK 2 + (Ar + Bt)eiK a ) 2 2 which is equal to (At + Br)eiK 2 + Be?iK 2 which
Problem A& M 6.1 First make sure you understand Equation 6.12 in A&M (pp 103104). It shows that the planes corresponding to the smallest reciprocal lattice vector yield the smallest angle for the ring. Thus we know that the angle at which the j th diract
P613 HW # 2 Solutions
JM Densmore
1. The Free and Independent Electron Gas in Two Dimensions (a) In a 2D system with periodic BC you can solve the SE with momentum kx = 2/Lnx and ky = 2/Lny . The number of allowed values of k space inside a volume will be
PHYSICS 880.06 (Fall 2005)
(1.1) A&M Problem 1.4
Proble Set 1 Solution
dp p p = e E + H , dt mc H = Hz z, E(t) = Re(E( )eit ). (a) Seek steadystate solutoin of this form p(t) = i p( ) = Re(p( )eit ), p( ) p( ) e E( ) + H . mc 1 py ( )Hz , mc 1 px ( )Hz ,
PHYSICS 880.06 (Fall 2004)
Problem Set 3 Solution
3.1
A&M Chapter 4 Problem 1
(a)
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