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

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Phys 253a 1 Problem Set 7 Solutions November 22, 2010 1. (a) In the Majorana representation, γ 0 = 0 σ 2 σ 2 0 ! γ 1 = 3 0 0 3 ! γ 2 = 0 - σ 2 σ 2 0 ! γ 3 = - 1 0 0 - 1 ! It’s easy enough to compute the commutators and get S 01 = - i 2 0 σ 1 σ 1 0 ! S 02 = i 2 1 0 0 - 1 ! S 03 = - i 2 0 σ 3 σ 3 0 ! S 31 = 1 2 σ 2 0 0 σ 2 ! S 12 = i 2 0 - σ 1 σ 1 0 ! S 23 = i 2 0 σ 3 - σ 3 0 ! These are purely imaginary, as expected. (b) We have γ 5 = 0 γ 1 γ 2 γ 3 = i σ 2 σ 3 σ 2 σ 1 0 0 - σ 2 σ 3 σ 2 σ 1 ! Now, we can use the relation σ i σ j = δ ij + i ijk σ k to simplify the product σ 2 σ 3 σ 2 σ 1 = 1 σ 2 σ 1 = - σ 3 σ 1 = - 2 , giving γ 5 = σ 2 0 0 - σ 2 !
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Phys 253a 2 2. (a) The Clifford relation { γ μ , γ ν } = 2 η μν means that γ 2 0 = 1 , γ 2 i = - 1 , and γ μ anticommutes with γ ν when ν 6 = μ . Thus, γ 2 5 = i 2 γ 0 γ 1 γ 2 γ 3 γ 0 γ 1 γ 2 γ 3 = - i 2 γ 2 0 γ 1 γ 2 γ 3 γ 1 γ 2 γ 3 = - i 2 γ 2 0 γ 2 1 γ 2 γ 3 γ 2 γ 3 = i 2 γ 2 0 γ 2 1 γ 2 2 γ 2 3 = ( - 1)( 1 )( - 1 ) 3 = 1 In the future, we’ll be less careful and write 1 instead of 1 . (b) γ μ 6 μ = p ν γ μ γ ν γ μ = p ν (2 δ ν μ - γ ν γ μ ) γ μ = 2 6 p -6 μ μ = - 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 γ μ 6 p 6 q 6 μ = γ μ (2 p · q 6 p -6 qp 2 ) γ μ = - 4 p · q 6 p + 2 p 2 6 q (d) Suppose μ = 0. Then { γ 5 , γ 0 } = i ( γ 0 γ 1 γ 2 γ 3 γ 0 + γ 2 0 γ 1 γ 2 γ 3 ) = i ( - γ 2 0 γ 1 γ 2 γ 3 + γ 2 0 γ 1 γ 2 γ 3 ) = 0 Why did this happen? Well, we had to move one γ 0 past 3 γ 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 we’d taken μ = 1, we would have had to move one γ 1 past a single γ and the other past two γ ’s, again giving an opposite sign, causing the anticommutator to vanish. The same thing happens for μ = 2 , 3, so we get { γ 5 , γ μ } = 0
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Phys 253a 3 This is consistent with γ 5 being like a fifth gamma matrix. Indeed, if we let γ 4 = 5 , and let a, b run from 0 to 4, then we find { γ a , γ b } = 2 η ab , where η ab = diag(1 , - 1 , - 1 , - 1 , - 1) is the Minkowski metric in 5 dimensions. (e) There are two ways to approach this problem. The first is to simply use the Dirac algebra repeatedly to get γ α to the right hand side of the trace, and then use the cyclic property of the trace to get back what one started with, in addition to all terms from anticommutators. This will reduce the trace of four γ matrices to a sum of traces of pairs of γ matrices, which we can compute in turn.
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