This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHY 5667 Problem Set no. 5 solution Problem 1 The multiplication table of S 3 (with the convention g table = g A g B ) reads: g A \ g B 1 g 12 g 23 g 31 g 123 g 132 1 1 g 12 g 23 g 31 g 123 g 132 g 12 g 12 1 g 123 g 132 g 23 g 31 g 23 g 23 g 132 1 g 123 g 31 g 12 g 31 g 31 g 123 g 132 1 g 12 g 23 g 123 g 123 g 31 g 12 g 23 g 132 1 g 132 g 132 g 23 g 31 g 12 1 g 123 When talking about permutations of three elements ABC under S 3 , actu ally all one has to do is to specify the position of two elements (say A and B) after the S 3transformation, since the third element (C) then will automati cally be in the one position not occupied by the first two. The mathematical interpretation is to see S 3 as the surface of x 1 + x 2 + x 3 = 0, where after specifying x 1 , x 2 we automatically know x 3 = x 1 x 2 . It suffices to define the S 3 transformations in the basis of { x 1 , x 2 } , i.e. by 2 × 2 matrices. The group elements in this space are: • 1 = parenleftbigg 1 0 0 1 parenrightbigg ⇒ x ′ 1 = x 1 x ′ 2 = x 2 x ′ 3 = ( x ′ 1 + x ′ 2 ) = ( x 1 + x 2 ) = x 3 • g 12 = parenleftbigg 0 1 1 0 parenrightbigg ⇒ x ′ 1 = x 2 x ′ 2 = x 1 x ′ 3 = ( x ′ 1 + x ′ 2 ) = ( x 2 + x 1 ) = x 3 • g 23 = parenleftbigg 1 1 1 parenrightbigg ⇒ x ′ 1 = x 1 x ′ 2 = ( x 1 + x 2 ) = x 3 x ′ 3 = ( x ′ 1 + x ′ 2 ) = ( x 1 + x 3 ) =...
View
Full
Document
This note was uploaded on 12/14/2011 for the course PHY 5667 taught by Professor Okui during the Fall '10 term at FSU.
 Fall '10
 Okui

Click to edit the document details