PS7-solutions

PS7-solutions - Phys 253a 1 Problem Set 7 Solutions...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Phys 253a 1 Problem Set 7 Solutions November 22, 2010 1. (a) In the Majorana representation, = 2 2 ! 1 = i 3 i 3 ! 2 =- 2 2 ! 3 =- i 1- i 1 ! Its easy enough to compute the commutators and get S 01 =- i 2 1 1 ! S 02 = i 2 1- 1 ! S 03 =- i 2 3 3 ! S 31 = 1 2 2 2 ! S 12 = i 2- 1 1 ! S 23 = i 2 3- 3 ! These are purely imaginary, as expected. (b) We have 5 = i 1 2 3 = i 2 3 2 1- 2 3 2 1 ! Now, we can use the relation i j = ij + i ijk k to simplify the product 2 3 2 1 = i 1 2 1 =- 3 1 =- i 2 , giving 5 = 2- 2 ! Phys 253a 2 2. (a) The Clifford relation { , } = 2 means that 2 = 1 , 2 i =- 1 , and anticommutes with when 6 = . Thus, 2 5 = i 2 1 2 3 1 2 3 =- i 2 2 1 2 3 1 2 3 =- i 2 2 2 1 2 3 2 3 = i 2 2 2 1 2 2 2 3 = (- 1)( 1 )(- 1 ) 3 = 1 In the future, well be less careful and write 1 instead of 1 . (b) 6 p = p = p (2 - ) = 2 6 p-6 p =- 2 6 p (c) Note first that 6 p 6 q 6 p = 2 p q 6 p-6 q 6 p 6 p = 2 p q 6 p-6 qp 2 Now we can use part (b) to find 6 p 6 q 6 p = (2 p q 6 p-6 qp 2 ) =- 4 p q 6 p + 2 p 2 6 q (d) Suppose = 0. Then { 5 , } = i ( 1 2 3 + 2 1 2 3 ) = i (- 2 1 2 3 + 2 1 2 3 ) = 0 Why did this happen? Well, we had to move one past 3 i s to the left, and the other not at all. Thus, the two terms in the anticommutator had an opposite sign, and cancelled to give zero. If wed taken = 1, we would have had to move one 1 past a single and the other past two s, again giving an opposite sign, causing the anticommutator to vanish. The same thing happens for = 2 , 3, so we get { 5 , } = 0 Phys 253a 3 This is consistent with 5 being like a fifth gamma matrix. Indeed, if we let 4 = i 5 , and let a,b run from 0 to 4, then we find { a , b } = 2 ab , where ab = diag(1 ,- 1 ,- 1 ,- 1 ,- 1) is the Minkowski metric in 5 dimensions....
View Full Document

This note was uploaded on 02/04/2012 for the course PHYS 253A at Harvard.

Page1 / 8

PS7-solutions - Phys 253a 1 Problem Set 7 Solutions...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online