# cha13_5 - PHYS 635 Condensed Matter Physics Assignment 4...

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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 z-axis, following which the xy-plane 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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cha13_5 - PHYS 635 Condensed Matter Physics Assignment 4...

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