By Clarkson R., McKeon D.G.C.

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92)), 0 = iγ µ aλµ ∂ ∂x λ − m S −1 (a)ψ (x ) → multiply on left by S(a). 93) 42 CHAPTER 2. GRASSMANN VARIABLES But change of variables should leave us with an equation in the same form. 97) Now take S(a) = S(I + ∆w) i = I − σµν ∆wµν 4 → Need σµν . 100) → must hold to order ∆w. 3. QUANTIZATION OF THE SPINNING PARTICLE. 105) If ∆wµν = ∆w(I)µν → ∆w = scalar, (I)µν = 4 × 4 matrix characterizing the Lorentz transformation. ) In this case, 0 −1 0 +1 0 0 ∆wµν = ∆w 0 0 0 0 0 0 (where rows/columns = [x, y, z, t]) i Thus S(a) = exp − σµν wµν 4 = e−iσµν w/2 0 0 0 0 44 CHAPTER 2.

2n+1 ] = Tr[γµ1 . . γµ2n+1 γ5 γ5 ] ; γ5 γ5 = 1 Tr[γ5 γµ1 . . γµ2n+1 γ5 ] (Using property of traces) (−1)2n+1 Tr[γ5 γµ1 . . e. 3) eimt ψ40 Using Boosts: i S = exp − σµν ω µν → Can express in closed form. 4. 6) i ∴ S = exp − ωγ0 γ1 2 ω 0 σ1 = exp − 2 σ1 0 Recall, σ12 0 σ1 σ1 0 =1 → 2 = 1 0 0 1 2 ω 0 σ1 1 ω 1 ω 2 0 σ1 − + σ1 0 2 σ1 0 2! 2 3! 8) But tanh(ω) = −v, and recall the trig identities, sinh(x + y) = sinh(x) cosh(y) + cosh(x) sinh(y) ; ∴ sinh(2θ) = 2 sinh(θ) cosh(θ) cosh(2θ) = cosh2 (θ) + sinh2 (θ) 3 x=y=θ 52 CHAPTER 2.

105) If ∆wµν = ∆w(I)µν → ∆w = scalar, (I)µν = 4 × 4 matrix characterizing the Lorentz transformation. ) In this case, 0 −1 0 +1 0 0 ∆wµν = ∆w 0 0 0 0 0 0 (where rows/columns = [x, y, z, t]) i Thus S(a) = exp − σµν wµν 4 = e−iσµν w/2 0 0 0 0 44 CHAPTER 2. 108) If w = 2π (1 revolution) S(w = 2π) = exp iπ σ3 0 0 σ3 ; σ32 = 1 We know, (a · σ)2 + ... 2! But → (a · σ)2 = a2 [σi σj = δij + ijk σk ] |a|3 a·σ a2 (a2 )2 |a| + + + ... + + ... 1+ 2! 4! |a| 3! σ·a cosh |a| + sinh |a| |a| σ3 0 exp iπ 0 σ3 cosh(iπ) + sinh(iπ) σ3 ea·σ = 1 + (a · σ) + = = and so S(w = 2π) = = =−1 =0 = −1 Thus if w = 2π, ψ (x ) = −ψ(x), and if w = 4π, ψ (x ) = +ψ(x) ∴ All physical quantities need an even # of ψ’s.