QFT Example Sheet 1 Solutions - Quantum Field Theory...

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Contemporary Abstract Algebra
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Chapter 3 / Exercise 15
Contemporary Abstract Algebra
Gallian
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Quantum Field Theory Example Sheet 1 Michelmas Term 2011 Solutions by: Johannes Hofmann [email protected] Laurence Perreault Levasseur [email protected] David Morris [email protected] Marcel Schmittfull [email protected] Note: In the conventions of this course, the Minkowski metric is g = diag (1 , - 1 , - 1 , - 1), or ‘mostly minus.’ Exercise 1 Lagrangian: L = a 0 dx σ 2 ∂y ∂t 2 - T 2 ∂y ∂x 2 (1) Express y ( x, t ) as a Fourier series: y ( x, t ) = 2 a n =1 q n ( t ) sin nπx a . (2) The derivative of y ( x, t ) with respect to x and t is ∂y ( x, t ) ∂t = 2 a n =1 ˙ q n ( t ) sin nπx a (3) ∂y ( x, t ) ∂x = 2 a n =1 a q n ( t ) cos nπx a , (4) where we abbreviate ˙ q ∂q/∂t . Substituting Eqs. ( 3 ) and ( 4 ) in ( 1 ) gives L = 1 a n,m a 0 dx σ ˙ q n ( t ) ˙ q m ( t ) sin nπx a sin mπx a - T nmπ 2 a 2 q n ( t ) q m ( t ) cos nπx a cos mπx a . (5) Using the orthonormality relations 2 a a 0 dx sin nπx a sin mπx a = δ nm and (6) 2 a a 0 dx cos nπx a cos mπx a = δ nm (7) we see that the terms with n = m vanish. We obtain: L = n a 0 dx σ 2 ˙ q 2 n ( t ) - T 2 a 2 q 2 n ( t ) . (8) The Euler-Lagrange equations are given by d dt ∂L ˙ q n - ∂L ∂q n = 0 , n = 0 , 1 , 2 , . . . . (9) With ∂L ˙ q n = σ ˙ q n and (10) ∂L ∂q n = - T a 2 q n (11) we obtain the equations of motion: ¨ q n ( t ) + T σ a 2 q n ( t ) = 0 , (12) where n = 0 , 1 , 2 , . . . . These are the equations of motion of harmonic oscillators with frequency ω n = a T σ . 1
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Contemporary Abstract Algebra
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Chapter 3 / Exercise 15
Contemporary Abstract Algebra
Gallian
Expert Verified
Exercise 2 The scalar field φ ( x ) transforms under a Lorentz transformation x μ x μ = Λ μ ν x ν as φ ( x ) φ ( x ) = φ ( x ) = φ - 1 x ) . Note that this is an active transformation, i.e. the new field at the new coordinate equals the old field at the old coordinate. Using μ ∂x μ = ∂x ν ∂x μ ∂x ν = (Λ - 1 ) ν μ ν Now, μ μ φ ( x ) + m 2 φ ( x ) μ μ φ ( x ) + m 2 φ ( x ) = g μν - 1 ) ρ μ ρ - 1 ) κ ν κ φ ( x ) + m 2 φ ( x ) = g ρκ ρ κ φ ( x ) + m 2 φ ( x ) . (13) In the last step we used that Λ is a Lorentz transformation and so its inverse Λ - 1 preserves the inverse Minkowski metric g μν , - 1 ) ρ μ g μν - 1 ) κ ν = g ρκ . Eq. ( 13 ) therefore demonstrates that if φ ( x ) fulfils the Klein-Gordon equation μ μ φ ( x ) + m 2 φ ( x ) = 0 , then φ - 1 x ) likewise fulfils it. Exercise 3 Lagrangian density: L = μ φ * μ φ - m 2 φ * φ - λ 2 ( φ * φ ) 2 . (14) Euler-Lagrange equation for φ * : L ∂φ * - μ L ( μ φ * ) = 0 . (15) With L ∂φ * = - m 2 φ - λ ( φ * φ ) φ and (16) L ( μ φ ) = μ φ. (17) We obtain the equation of motion for φ * : μ μ φ + m 2 φ + λ ( φ * φ ) φ = 0 . (18) Similarly, we can calculate the field equation for φ . The result is the complex conjugate of Eq. ( 18 ). Consider the U(1) transformation φ ( x ) e φ ( x ) , and φ ( x ) e φ * ( x ) , (19) where α [0 , 2 π ) is a constant. It is straightforward to check that the Lagrangian ( 14 ) is invariant under this transformation.

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