Goldstein_4.1, 4.2, 4.10, 4.14, 4.15

# Goldstein_4.1, 4.2, 4.10, 4.14, 4.15 - Homework 6: # 4.1,...

This preview shows pages 1–4. Sign up to view the full content.

Homework 6: # 4.1, 4.2, 4.10, 4.14, 4.15 Michael Good Oct 4, 2004 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 A ( BC ) = X k A ik ( X m B km C mj ) ( AB ) C = X m ( X k A ik B km ) C mj 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 e AA = I The Pauli spin matrices, σ x , and σ z are both orthogonal. ˜ σ x σ x = ± 0 1 1 0 ²± 0 1 1 0 ² = ± 1 0 0 1 ² = I ˜ σ z σ z = ± 1 0 0 - 1 ²± 1 0 0 - 1 ² = ± 1 0 0 1 ² = I The product of these two: σ x σ z = ± 0 1 1 0 ²± 1 0 0 - 1 ² = ± 0 - 1 1 0 ² q 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
is also orthogonal: ˜ qq = ± 0 1 - 1 0 ²± 0 - 1 1 0 ² = ± 1 0 0 1 ² = I More generally, if e AA = 1 e BB = 1 then both A , and B are orthogonal. We can look at g ABAB = X k ( g AB ) ik ( AB ) kj = X k AB ki AB kj = X k,s,r a ks b si a kr b rj The elements are X k,s,r a ks b si a kr b rj = X b si a ks a kr b rj = X b si ( ˜ AA ) sr b rj This is g ABAB = X b si δ sr b rj = ˜ BB ij = δ ij Therefore the whole matrix is I and the product g ABAB = I is orthogonal. 4.2 Prove the following properties of the transposed and adjoint matrices: g AB = e B e A ( AB ) = B A Answer: For transposed matrices g AB = g AB ij = AB ji = X a js b si = X b si a js = X e B is e A sj = ( e B e A ) ij = e B e A As for the complex conjugate, ( AB ) = ( g AB ) * From our above answer for transposed matrices we can say g AB = e B e A 2
And so we have ( AB ) = ( g AB ) * = ( e B e A ) * = e B * e A * = B 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 e B = 1 + B + 1 2 B 2 + ... + B n n ! + ..., then prove the following properties: e B e C = e B + C , providing B and C commute. A - 1 = e - B e CBC - 1 = 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 + B 2 2 + O ( B 3 ))(1+ C + C 2 2 + O ( C 3 )) = 1+ C + C 2 2 + B + BC + B 2 2 + O (3) This is 1+( B + C )+ 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## Goldstein_4.1, 4.2, 4.10, 4.14, 4.15 - Homework 6: # 4.1,...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online