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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 p6 p = 2 6 p (c) Note first that 6 p 6 q 6 p = 2 p q 6 p6 q 6 p 6 p = 2 p q 6 p6 qp 2 Now we can use part (b) to find 6 p 6 q 6 p = (2 p q 6 p6 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....
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This note was uploaded on 02/04/2012 for the course PHYS 253A at Harvard.
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