Homework 6 # 4.1, 4.2, 4.10, 4.14, 4.15

Homework 6: # 4.1, 4.2, 4.10, 4.14, 4.15 4.1 Prove that matrix multiplication is associative. Show that the product of two orthogonal matrices is also orthogonal. Answer: Matrix associativity means A(BC ) = (AB )C The elements for any row i and column j, are X A(BC) = (AB )C = k Aik ( m ( X Bkm Cmj ) X m X Aik Bkm )Cmj k Both the elements are the same. They only differ in the order of addition. As long as the products are defined, and there are finite dimensions, matrix multiplication is associative. Orthogonality may be defined by σx σz = 1 0 = 0 −1 1 = 1 0 0 1 1 0 0 1 0 −1 1 0 =I   1 0 0 −1 = =I   0 1 1 0  1 0 σ˜z σz = 0 −1 The product of these two: 0 1 1 0  1 1 0  0  σ˜x σx =  AA= I e The Pauli spin matrices, σx , and σz are both orthogonal. ≡q is also orthogonal: ˜ qq= More generally, if 0 1 −1 0  0 1  −1 0 = 1 0 0 1  =I BB= 1 e e AA= 1 then both A, and B are orthogonal. We can look at gABAB= X k g )ik (AB )kj = (AB X k AB ki AB kj = aks bsi akr brj X k,s,r The elements are X k,s,r aks bsi akr brj = X bsi aks akr brj = X AA)sr brj bsi ( ˜ This is BBij = δij X g ABAB= bsi δsr brj = ˜ Therefore the whole matrix is I and the product ABAB= I g is orthogonal. 4.2 Prove the following properties of the transposed and adjoint matrices: A g AB= eB e (AB )† = B † A† Answer: For transposed matrices X X X gAB= gAB ij = ABji = ajs bsi = bsi ajs = e As for the complex conjugate, A e ij = eB e Bis eAsj = ( eB A) AB)∗ (AB )† = ( g From our above answer for transposed matrices we can say A eB e g AB= 2 And so we have A∗ = B † A† e ∗ = eB ∗ e (AB )† = ( gAB)∗ = ( eB A) 4.10 If B is a square matrix and A is the exponential of B, defined by the infinite series expansion of the exponential, A ≡ eB = 1 + B + 1 2 B 2 + ...+ Bn n! + ..., then prove the following properties: • eB eC = eB+C , providing B and C commute. • A−1 = e−B −1 • eCBC = CAC −1 • A is orthogonal if B is antisymmetric Answer: Providing that B and C commute; BC− CB= 0 BC= CB we can get an idea of what happens: (1+ B + B2 C2 C2 B2 +O(B 3 ))(1+ C + +O(C 3 )) = 1+ C + + B + BC+ +O(3) 2 2 2 2 This is 1+(B +C)+ 1 2 (C 2 +2BC+B 2 )+ O(3) = 1+(B +C)+ (B + C )2 2 +O(3) = eB+C Because BC= CBand where O(3) are higher order terms with products of 3 or more matrices. Looking at the kth order terms, we can provide a rigorous proof. Expanding the left hand side of eB eC = eB+C and looking at the kth order term, by using the expansion for expwe get, noting that i + j = k k X 0 k BiC j B k−j C j = X ( k − j )!j ! i!j! 0 3 and using the binomial expansion on the right hand side for the kth order term, (a proof of which is given in Riley, Hobsen, Bence): (B + C )k k! k = k! 1 B k−j C j = X k! 0 (k − j )!j! k X 0 B k−j C j (k − j )!j ! we get the same term. QED. To prove A−1 = e−B We remember that A−1 A = 1 and throw e −B on the right A−1 Ae−B = 1e−B A−1 eB e−B = e−B and from our above proof we know eB eC = eB+C so A−1 eB−B = e−B Presto, A−1 = e−B To prove −1 eCBC = CAC −1 its best to expand the exp ∞ −1 −1 1 −1 n −1 CBC CBC X n! (CBC ) = 1+CBC + 2 0 +....+ CBC −1 CBC −1 CBC −1 ... n! +... Do you see how the middle C −1 C terms cancel out? And how they cancel each out n times? So we are left with just the C and C −1 on the outside of the B’s. ∞ 1 −1 n X n! (CBC ) = 0 ∞ 1 n −1 B −1 X n! CB C = Ce C 0 Remember A = eB and we therefore have −1 eCBC = CAC −1 4 To prove A is orthogonal A = A−1 e if B is antisymmetric B −B = e We can look at the transpose of A ∞ e = A Xg 0 ∞ Bn = X n! ∞ Bn (−B )n = e−B e = X n! n! 0 0 But from our second proof, we know that e−B = A−1 , so A = A−1 e and we can happily say A is orthogonal. 4.14 • Verify that the permutation symbol satisfies the following identity in terms of Kronecker delta symbols: ǫijp ǫrmp = δir δjm − δim δjr • Show that ǫijp ǫijk = 2δpk Answer: To verify this first identity, all we have to do is look at the two sides of the equation, analyzing the possibilities, i.e. if the right hand side has i=r j= m 6= i we get +1. If i = m j= r 6= i we get −1. For any other set of i, j , r, and m we get 0. For the left hand side, lets match conditions, if i=r j= m 6= i then ǫijp = ǫrmp and whether or not ǫijp is ±1 the product of the two gives a +1. If i = m j= r 6= i 5 then ǫrmp = ǫjip = −ǫijp and whether or not ǫijp is ±1 the product is now equal to −1. These are the only nonzero values because for i,j,r,m, none can have the same value as p. Since there are only three values, that any of the subscripts may take, the only non-zero values are the ones above. (not all four subscripts may be equal because then it would be ǫ = 0 as if i = j or r = m). To show that ǫijp ǫijk = 2δpk we can use our previous identity, cast in a different form: ǫijk ǫimp = δjm δkp − δjp δkm This is equivalent because the product of two Levi-Civita symbols is found from the deteriment of a matrix of delta’s, that is ǫijk ǫrmp = δir δjm δkp + δim δjp δkr + δip δjr δkm − δim δjr δkp − δir δjp δkm − δip δjm δkr For our different form, we set i = r. If we also set j = m, this is called ‘contracting’ we get ǫijk ǫijp = δjj δkp − δjp δkj Using the summation convention, δjj = 3, ǫijk ǫijp = 3δkp − δkp ǫijk ǫijp = 2δkp 4.15 Show that the components of the angular velocity along the space set of axes are given in terms of the Euler angles by ωx = θ̇ cos φ + ψ̇ sin θ sin φ ωy = θ̇ sin φ − ψ̇ sin θ cos φ ωz = ψ̇ cos θ + φ˙ Answer: Using the same analysis that Goldstein gives to find the angular velocity along the body axes (x′ ,y ′ ,z ′ ) we can find the angular velocity along the space axes (x,y,z). To make a drawing easier, its helpful to label the axes of rotation for θ,˙ ψ̇ and φ.˙ 6 θ̇ → L.O.N. ψ̇ → z ′ φ̇ → z We want ωx = ˙θx + ψ̇x + φ̇x ωy = ˙θy + ψ̇y + φ̇y ωz = ˙θz + ψ̇z + φ̇z Lets start with ωz first to be different. If we look at the diagram carefully on ˙ along the line of nodes, that is θ revolves around page 152, we can see thatθ is the line of nodes. Therefore because the line of nodes is perpendicular to the z space axis there is no component of θ contributing to angular velocity around the z space axis. θ˙z = 0. What about ˙ψz ? Well, ψ revolves around z ′ . So there is a component along z due to a changing ψ. That component depends on how much angle there is between z ′ and z, which is θ. Does this makes sense? We find the z part, which is the adjacent side to θ. Thus we have ψz ˙= ψ̇ cos θ. Now lets look atφz˙. We can see that φ just revolves around z in the first place! Right? So there is no need to make any ‘transformation’ or make any changes. Lets take φ̇z = φ̇. Add them all up for our total ωz . ωz = θ̇z + ψ̇z + φ̇z = 0 + ψ̇ cos θ + φ̇ Now lets do the harder ones. Try ωx . What isθx˙? Well, θ̇ is along the line of nodes, that is, θ changes and revolves around the line of nodes axis. To find the x component of that, we just see that the angle between the line of nodes and the x axis is only φ, because they both lie in the same xyplane. Yes? So θ̇x = θ̇ cos φ. The adjacent side to φ with θ ˙as the hypotenuse. Lets look at φ˙x . See how φ revolves around the z axis? Well, the z axis is perpendicular to the ˙ x axis there for there is no component ofφ that contributes to the x space axis. φ˙x = 0. Now look at ˙ψx . We can see that ψ˙ is along the z ′ body axis, that is, it is in a whole different plane than x. We first have to find the component in the same xyplane, then find the component of the x direction. So to get into the xyplane we can take ψ̇x,y = ˙ψ sin θ. Now its in the same plane. But where is it facing in this plane? We can see that depends on the angle φ. If φ = 0 we would have projected it right on top of the y − axis! So we can make sure that if φ = 0 we have a zero component for x by multiplying by sin φ. So we get after two projections, ψ˙x = ˙ψ sin θ sin φ. Add these all up for our total ωx , angular velocity in the x space axis. ωx = ˙θx + ψ̇x + φ̇x = ˙θ cos φ + ψ˙ sin θ sin φ + 0 I’ll explain ωy for kicks, even though the process is exactly the same. Look for θ˙y . θ˙ is along the line of nodes. Its y component depends on the angle φ. So 7 project it to the y axis. θy˙ = θ̇ sin φ. Look for ψ˙y . Its in a different plane again, so two projections are necessary to find its component. Project down to the xy ˙ = ˙ψ sin θ and now we remember that if φ = 0 we plane like we did before,ψx,y would have exactly placed it on top of the y axis. Thats good! So lets make it if φ = 0 we have the fullψ ˙sin θ, (ie multiply by cos φ because cos 0 = 1). But we also have projected it in the opposite direction of the positive y direction, (throw in a negative). So we haveψy˙ = − ˙ψ sin θ cos φ. Forφy˙ we note that φ revolves around the z axis, completely perpendicular to y. Therefore no component in the y direction. φ˙y = 0. Add them all up ωy = θ̇y + ψ̇y + φ̇y = ˙θ sin φ − ψ̇ sin θ cos φ + 0 Here is all the ω’s together ωx = θ̇ cos φ + ψ̇ sin θ sin φ ωy = θ̇ sin φ − ψ̇ sin θ cos φ ωz = ψ̇ cos θ + φ̇ 8