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Unformatted text preview: 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 ( k ) = + ~ 2 2 ( k k ) T M 1 ( k k ) , (1.1) where M 1 is the reciprocal e ff ective mass tensor. As can be seen from its definition in Eq. (12.29) in A&M, the reciprocal e ff ective mass tensor is symmetric. Near a band minimum or maximum, the eigenvalues 1 / m * i of M 1 have the same sign, which means that the surfaces of constant energy as defined by (1.1) are ellipsoids. (a) In the absence of a magnetic field we are free to choose the Cartesian axes to coincide with the principal axes of these ellipsoid. However, when a magnetic field is present, the direction of the magnetic field will usually be taken as the zaxis, following which the xyplane will also be fixed. Therefore, for cyclotron motion, we would in general need to find an area A ( , k z ) like the one shown below: k z k x k y k z k z (,29 A To go about calculating this area, let us get ready some useful mathematical machinery. First, consider the integral I = $ x 2 + y 2 + z 2 R 2 dx dy dz ( z z ) . (1.2) By dimensional arguments, [ dx ] = [ dy ] = [ dz ] = length, while [ ( z z )] = 1 / length, the integral has dimensions of length 2 , i.e. it has the dimensions of an area. Going to cylindrical coordinates, we can write the integral as I = Z dz Z ( R 2 z 2 ) 1 / 2 d Z 2 d ( z z ) dz . (1.3) Performing the integration over z , and noting that this partial integration of the delta function modifies the limits of the remaining integration over and , we find that I = 2 Z ( R 2 z 2 ) 1 / 2 d = 2 " 2 2 # ( R 2 z 2 ) 1 / 2 = ( R 2 z 2 ) , (1.4) 1 which is none other than the area of the (circular) section of the sphere x 2 + y 2 + z 2 = R 2 at a fixed z = z . Next suppose that the set of unprimed coordinates ( x , y , z ) are related to another set of primed coordi nates ( x , y , z ) by a linear transformation r = A r , such that x y z = A xx A xy A xz A yx A yy A yz A zx A zy A zz x y z = A xx x + A xy y + A xz z A yx x + A yy y + A yz z A zx x + A zy y + A zz z . (1.5) Then the volume element dx dy dz is related to the volume element dx dy dz by dx dy dz = x / x x / y x / z y / x y / y y / z z / x z / y z / z dx dy dz =  A  dx dy dz , (1.6) where  A  is the determinant of the matrix A ....
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This note was uploaded on 06/07/2010 for the course MECH 1122 taught by Professor Dont during the Spring '10 term at A.T. Still University.
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