HW12_soln

# HW12_soln - PHY 5667 Problem Set no 12 solution Problem 1 L...

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Unformatted text preview: PHY 5667 Problem Set no. 12 solution Problem 1 L = ( ∂ μ φ ∗ )( ∂ μ φ ) − m 2 φ ∗ φ + i ψ † L ¯ σ μ ∂ μ ψ L − i λφψ T L σ 2 ψ L + i λφ ∗ ψ † L σ 2 ψ ∗ L (i) The U(1) charge of ψ L has to be − 1 2 . The kinetic term of the lefthanded spinor remains unchanged with under any U(1)-charge (because it contains ψ L and ψ † L , the first contributes a factor e i αq and the second e − i αq , and the two cancel out for any q ), but the interaction terms require q = − 1 2 to be invariant: i λφψ T L σ 2 ψ L → i λe i α φe i αq ψ T L σ 2 e i αq ψ L = e i α (1+2 q ) i λφψ T L σ 2 ψ L ⇒ q = − 1 2 i λφ ∗ ψ † L σ 2 ψ ∗ L → i λe − i α φ ∗ e − i αq ψ † L σ 2 e − i αq ψ ∗ L = e − i α (1+2 q ) i λφ ∗ ψ † L σ 2 ψ ∗ L ⇒ q = − 1 2 (ii) When we promote α → α ( x ), extra terms in the Lagrangian can stem only from derivatives. In particular ( ∂ μ φ ∗ )( ∂ μ φ ) → [ ∂ μ ( e − i α ( x ) φ ∗ )][ ∂ μ ( e i α ( x ) φ )] = ( ∂ μ φ ∗ )( ∂ μ φ ) − i( ∂ μ α ( x ))[ φ ∗ ∂ μ φ − φ∂ μ φ ∗ ] i ψ † L ¯ σ μ ∂ μ ψ L → i e i α ( x ) / 2 ψ † L ¯ σ μ ∂ μ ( e − i α ( x ) / 2 ψ L ) = i ψ † L ¯ σ μ ∂ μ ψ L − i i 2 ( ∂ μ α ( x )) ψ † L ¯ σ μ ψ L So L [( e i α ( x ) φ ) , e − i α ( x ) / 2 ψ L ] − L [ φ, ψ L ] = − i( ∂ μ α ( x )) [ φ ∗ ∂ μ φ − φ∂ μ φ ∗ + i 2 ψ † L ¯ σ μ ψ L ] bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright J μ Thus, the Noether current is J μ = φ ∗ ∂ μ φ − φ∂ μ φ ∗ + i 2 ψ † L ¯ σ μ ψ L (iii) Let’s first get the equations of motion from the variation of the action. By the usual manipulations (in particular integration by parts and reshuffling of spinors, δψ T L σ 2 ψ L = ψ T L σ 2 δψ L and ψ † L σ 2 δψ ∗ L = δψ † L σ 2 ψ ∗ L ) we obtain δS = 0 = integraldisplay d 4 x parenleftBig [ − ∂ 2 φ − m 2 φ + i λψ † L σ 2 ψ ∗ L ] δφ ∗ + [ − ∂ 2 φ ∗ − m 2 φ ∗ − i λψ T L σ 2 ψ L ] δφ + [ − i( ∂ μ ψ † L )¯ σ μ − 2i λφψ T L σ 2 ] δψ L + δψ † L [i¯ σ μ ∂ μ ψ L + 2i λφ ∗ σ 2 ψ ∗ L ] parenrightBig 1 So the equations of motion are − ∂ 2 φ − m 2 φ + i λψ † L σ 2 ψ ∗ L = 0 − ∂ 2 φ ∗ − m 2 φ ∗ − i λψ T L σ 2 ψ L = 0 − i( ∂ μ ψ † L )¯ σ μ − 2i λφψ T L σ 2 = 0 i¯ σ μ ∂ μ ψ L + 2i λφ ∗ σ 2 ψ ∗ L = 0 Now ∂ μ J μ = ∂ μ [ φ ∗ ∂ μ φ − φ∂ μ φ ∗ + i 2 ψ † L ¯ σ μ ψ L ] = φ ∗ ∂ 2 φ − φ∂ 2 φ ∗ + i 2 ( ∂ μ ψ † L )¯ σ μ ψ L + i 2 ψ † L ¯ σ μ ( ∂ μ ψ L ) = φ ∗ [ − m 2 φ + i λψ † L σ 2 ψ ∗ L ] − φ [ − m 2 φ ∗ − i λψ T L σ 2 ψ L ] − i λφψ T L σ 2 ψ L − i λφ ∗ ψ † L σ 2 ψ ∗ L = 0 (iv) We need the interaction terms, which contain...
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HW12_soln - PHY 5667 Problem Set no 12 solution Problem 1 L...

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