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# ELEM MATH PHYS PHYS 227

UW

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This 10 page Class Notes was uploaded by Dr. Simeon Wiza on Wednesday September 9, 2015. The Class Notes belongs to PHYS 227 at University of Washington taught by Stephen Ellis in Fall. Since its upload, it has received 42 views. For similar materials see /class/192436/phys-227-university-of-washington in Physics 2 at University of Washington.

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Date Created: 09/09/15

Lecture 8 Appendix B Some sample problems from Boas Here are some solutions to the sample problems assigned for Chapter 3 32 9 Solution We want think about the following set of equations in terms of matrices and their properties x y2251 22 x 5 2x 3y z 2 4c 2 3 l y 4 2x 2y 42 2 6 2 2 4 z 6 Thus we have M N 3 and can proceed to row reduce the relevant matrices We find 1 l 2 l l 2 l 0 l A 3 1 ga gizz ll 0 5 5 RHRluSRZE 0 5 5 2 2 4 0 0 0 0 0 0 M 22 l l 2 5 l l 2 5 14ng 2 3 1 4 m 0 5 5 6 2 2 4 6 mm 0 0 0 41 l 0 1 195 R1 R115R2 0 5 5 6 0 0 0 4 l 0 l 0 RHRl1920R3E 0 5 5 0 MAugz3 R2gtR2732R3 0 0 0 4 Since the ranks of the 2 matrices satisfy M lt M mg we conclude that the equations are inconsistent over constrained and that there is no consistent solution Physics 227 Lecture 8 Appendix B Autumn 2007 32 17 Solution We want to determine the rank of the following matrix using row reduction l l 4 3 l l 4 3 l l 4 3 3 l 10 7 2 O 6 4 2 O 6 4 4 2 14 10 j fi l 2 O 6 4 31132 O O O O I 2 O 6 4 2 O 6 4 O O O 0 We cannot increase the number of zeros by combining these 2 rows and conclude that the rank of this matriX is 2 In Mathematica this looks like Ex 3217 A1143311074214102064 ll4331107l4214102064 MatrixForm A 1 1 4 3 3 1 10 7 4 2 14 10 2 o 6 4 MatrixRank A 2 33 12 Solution Now we want to evaluate a determinant First simplify by combining the second and fourth rows and then proceed to find the determinant 0 1 2 1 0 1 2 1 O 1 1 1 0 3 0 0 0 4 0 1 2 3 0 1W4 3 0 1123412 3 I 1 0 1 0 1 0 1 0 1 41 11 241316 Physics 227 Lecture 8 Appendix B 2 Autumn 2007 33 16 Solution Here we use Cramer s rule to solve a set of 2 simultaneous equations with complex coefficients from quantum mechanics We have A B 2 n3 1 1 A 1 m4 KB m 39 K B ik 1 4 Azk KiKik Kik 1 1 Km K m 1k K gt 1 4 Bik m mk 2m 1 1 Km K m K Since A has the form of a phase the ratio of a complex number and its complex conjugate we have K 39k K 39k A Z 1 l 1 K z39k K z39k 38 25 Solution We want to determine when the following set of simultaneous equations has a nontriVial solution ie the eigenvalue condition We have Physics 227 Lecture 8 Appendix B 3 Autumn 2007 x 2 ly 0 3x 2 292 0 1 2 1 3 x 0 gt 1 2 2 0 y 0 3 0 2 2 2 0 In order for there to be a nontrivial solution the determinant of the coefficient matrix must vanish so that it has no inverse This leads to 1 3 1 3 1 2 2 0 12A 3 0 2 2 3 22 3 132 3 9 3 21 32311 3 2333 4 2 1 3 4 2 3 1l 3 l 3 2 So now for each eigenvalue we want to solve the equations find the eigenvectors We have 3 l 3 0 x0 i2 1 0 0 y 0 3 y 3z 3 0 0 z 0 2 l 3 0 x 5y l 3l 5 0 0 33x2 52 3 0 5 0 3yz x y z 5 l 3 x 0 x2y y 0 33x22 2 0 3yz Physics 227 Lecture 8 Appendix B 4 Autumn 2007 3115 Solution Now think about a 2D matrix as a rotation We make the identification C ltbfose sin9 y sin9 cos9 I So the matrix C can be identified as a passive rotation through 19 sin 1 1107 radians 634 degrees 311 12 Solution Here we practice finding eigenvalues and eigenvectors We have for the eigenvalues 1 3 TrA3 2 1 A 3 A 211 2 2 detA2 422112 124 So we can determine the eigenvectors up to an arbitrary normalization and phase here we choose unit vectors to be 1 3 x x x 3y 2 x 1 3 A l 3 3 v1 2 2 y y 2x 2 y y J 2 l 3 x x x 3y 2 4x 1 l 12 4 4 3 3 v2 2 2 y y 2x 2 y 4y J5 1 Note that since the original matrix is not Hermitian the eigenvectors are not orthogonal Using Mathematica we find Ex 31112 A1322 l322ll MatrixForm A Physics 227 Lecture 8 Appendix B 5 Autumn 2007 l 3 2 2 Eigenvalues A 41 Eigenvectors A ll 32 311 15 Solution This exercise is like the preVious one but with a 3x3 matrix So now we evaluate the determinant directly and set it to zero We have 3 2 2 0 1 t2 t2 9 t 112 4t 5 0 0 1 111 2 l lt1t 50gt 224 13 5 Notice that it is helpful to keep any common factors factored out rather than multiplying through The eigenyectors are found from the corresponding simultaneous equations up to a normalization and phase here we choose unit vectors Physics 227 Lecture 8 Appendix B Autumn 2007 2 3 0 x x 2x3yx3x 3y 0 2121 3 2 0 y y 3x2yy3x yv1 0 0 0 1 z 2 22 1 2 3 0 x x 2x3y xx y 1 2213 2 0 y y 3x2y yx yv2i 1 0 0 1 z 2 22 23220 J 0 2 3 0 x x 2x3y5x3xy 1 232513 2 0 y 5 y 3x2y5yxyv3L 1 0 0 1 z 2 22523220 J5 0 Note that in this case the original matrix is Hermitian symmetric and real and the eigenvectors are orthogonal Using Mathematica we find Ex 31115 A230320001 112303201100lll MatrixForm A 2 3 o 3 2 o o o l Eigenvalues A 5 l l Eigenvectors A 11110ll000lll 31122 Solution Finally consider a more complicated but still Hermitian 3X3 matriX We have Physics 227 Lecture 8 Appendix B Autumn 2007 3 2 2 3 2 2 2 A 2 1 3 2 2 1 2 3 2 3 1 2 3 1 2 231 22 9 221 2 626 21 2 2322 2 8822 232 422822 2222 2 20 222 524 21 2 0 222 4 2 Now we can proceed to find the corresponding eigenvectors by solving the equations and checking that we get orthogonal vectors but specified only up to a normalization and phase here we choose unit vectors We find 3 2 2 x x 3x2y2z 2x32yzx 212 2 2 l 3 y 2y 32xy3z 2y33yz 2x 2 3 1 z z 2x3yz 2233yz 2x x0 1 1 3y23v1 1 3 2 2 x x 3x2y2z 4x34yz 2x 222 4 2 l 3 y 2 4 y 3 2xy3z 4y35y3z 2x 2 3 1 z z 2x3yz 4233ySZ 2x 4 2 yzz 2v 2 1 x4y 2 332 1 Physics 227 Lecture 8 Appendix B 8 Autumn 2007 5 y 3 2xy3z5y38y 6z4x 3 2 2 x x 3x2y225xyz4x l 3 y 3 1 z z 2x3yz52 6y8z4x UsingMathemalica we find Ex31122 PPM 322213231 i322213i2r3rlll MatrixForm A 3 2 2 213 231 Eigenvalues A 5 4 2 Eigenvectors A l224llOll 312 6 Solution This is a different application of the diagonalization problem finding the principal axes of a quadratic surface As usual the starting step is to find the eigenvalues and we have note that we construct the matrix assuming it is symmetric Physics 227 Lecture 8 Appendix B 9 Autumn 2007 l 2 l x x y z 2 l l y x2y2zz4xy2xz 2yzl2 l l 1 z 1 2 2 1 gt 2 1 2 1 1 21 22 1 221 21 2 1 2 1 1 1 2 1 222 22 5 4 233223z 9 z 322 3 A 3 2 32 J 2J 2 2 J 23 J So we know that expressed in terms of the principal axes the coefficient matrix for the equation of the surface will be diagonal with these eigenvalues down the diagonal although the choices of which axis corresponds to the specific axes x y or z is arbitrary as we can interchange any pair with a further rotation We have one choice as note that as expected the trace and determinant of the matrix are unchanged 3 0 0 x x yr Zr 0 0 yr 23x72J y72J 272 12 00 J Z Physics 227 Lecture 8 Appendix B 10 Autumn 2007

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