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HW6_soln

# HW6_soln - PHY 5667 Problem Set no 6 solution Problem 1 1...

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Unformatted text preview: PHY 5667 Problem Set no. 6 solution Problem 1 1 φ(x + a) = φ(x) + aµ ∂µ φ(x) + aµ aν ∂µ ∂ν φ(x) + . . . 2 = exp[aµ ∂µ ]φ(x) = exp[iaµ Tµ ]φ(x) (Taylor expansion) (by deﬁnition) Comparing the second and the third line gives Tµ = −i∂µ Problem 2 In agreement with the transformation of 4-vectors x, 4-momenta k transform under a Lorentz boost in x1 direction as: k ′0 k ′1 k ′2 k ′3 = k 0 cosh η + k 1 sinh η = k 0 sinh η + k 1 cosh η = k2 = k3 Thus, k ′0 δ (k ′ − q ′ ) = k ′0 δ (k ′1 − q ′1 )δ (k ′2 − q ′2 )δ (k ′3 − q ′3 ) = k ′0 δ (k ′1 − q ′1 )δ (k 2 − q 2 )δ (k 3 − q 3 ) 1 From δ (f (x)) = |f ′ (x)|−1 δ (x), one can derive directly that δ (f (x) − f (a)) = |f ′ (x)|−1 δ (x − a), so we can rewrite δ (k ′1 − q ′1 ) = = ∂ ′1 k ∂k 1 −1 δ (k 1 − q 1 ) ∂ k 1 cosh η + k 0 sinh η ∂k 1 ∂k 0 = cosh η + 1 sinh η ∂k = cosh η + −1 δ (k 1 − q 1 ) −1 k1 sinh η k0 δ (k 1 − q 1 ) −1 δ (k 1 − q 1 ) 1 0 k where in the last line we have used ∂k1 = k0 which follows directly from ∂k k 0 = (k 1 )2 + (k 2 )2 + (k 3 )2 − m2 . So we end up with k ′0 δ (k ′ − q′) =k ′0 −1 k1 cosh η + 0 sinh η k δ (k 2 − q 2 )δ (k 3 − q 3 ) k1 k1 sinh η cosh η + 0 sinh η k0 k 0 1 1 2 2 3 3 = k δ (k − q )δ (k − q )δ (k − q ) −1 = k 0 cosh η + δ (k 1 − q 1 )δ (k 2 − q 2 )δ (k 3 − q 3 ) = k 0 δ (k − q ) Problem 3 The Heisenberg equation of motion for any operator A reads i∂0 A = [A, H ]. Here we consider the Hamiltonian H= dx H = dx m2 2 12 1 Π + (∇φ)(∇φ) + φ. 2 2 2 2 Using the commutation relation speciﬁed in the problem, we ﬁnd that the Heisenberg equation of motion for φ(t, x) gives i∂0 φ(t, x) = = = 1 1 m2 2 dy φ(t, x), (Π(t, y )) + (∇y φ(t, y ))(∇y φ(t, y )) + (φ(t, y ))2 2 2 2 1 1 dy [φ(t, x), Π(t, y )] Π(t, y ) + Π(t, y ) [φ(t, x), Π(t, y )] 2 2 dy iδ (x − y )Π(t, y ) = iΠ(t, x) Therefore, we ﬁnd Π(t, x) = ∂0 φ(t, x) The Heisenberg equation of motion for Π(t, x) gives i∂0 Π(t, x) = 1 m2 1 (Π(t, y ))2 + (∇y φ(t, y ))(∇y φ(t, y )) + (φ(t, y ))2 2 2 2 1 1 [Π(t, x), (∇y φ(t, y )](∇y φ(t, y )) + (∇y φ(t, y ))[Π(t, x), (∇y φ(t, y )] 2 2 2 m m2 + [Π(t, x), φ(t, y )] φ(t, y ) + φ(t, y ) [Π(t, x), φ(t, y)] 2 2 1 1 [Π(t, x), φ(t, y] (−∇y ∇y φ(t, y )) + (−∇y ∇y φ(t, y )) [Π(t, x), φ(t, y] 2 2 2 m2 m [Π(t, x), φ(t, y )] φ(t, y ) + φ(t, y ) [Π(t, x), φ(t, y)] + 2 2 dy Π(t, x), = dy = dy = dy (−i)δ (x − y ) · −∇y ∇y + m2 φ(t, y ) = i ∇x ∇x − m2 φ(t, x) = i ∂i ∂i − m2 φ(t, x) where the sum over i = 1, 2, 3 is left implicit. Combining this with the previous result, we ﬁnd ∂0 Π = ∂0 ∂0 φ = ∂i ∂i − m2 φ ∂0 ∂0 − ∂i ∂i + m2 φ = 0 ∂µ ∂ µ + m2 φ = 0 3 Problem 4 We are given the action L ∞ S= dt 1 m2 2 κ 3 λ 4 (∂µ φ)(∂ µ φ) − φ− φ− φ + 2 2 3 4 dx 0 −∞ ∞ dt −∞ M2 φ (t, L) 2 from which we get the equations of motions by setting δ S to zero. L ∞ δS = ∞ µ dt 2 2 3 dx (∂µ δφ)(∂ φ) − m φ δφ − κφ δφ − λφ δφ + 0 −∞ dt M φ δφ(t, L) −∞ Now use that (∂µ δφ)(∂ µ φ) = (∂t δφ)(∂t φ) − (∂x δφ)(∂x φ) and ∞ ∞ dt(∂t δφ)(∂t φ) = δφ(∂t φ)|∞ − −∞ dt δφ(∂t ∂t φ) −∞ −∞ ∞ =− dt δφ(∂t ∂t φ) since δφ(t = ±∞) = 0 −∞ L dx (∂x δφ)(∂x φ) = δφ(∂x φ)|L 0 L − 0 dx δφ(∂x ∂x φ) 0 L = (∂x φ)|x=L − (∂x φ)|x=0 − dx δφ(∂x ∂x φ) 0 to show that L ∞ δS = dx δφ −∂t ∂t φ + ∂x ∂x φ − m2 φ − κφ2 − λφ3 dt 0 −∞ ∞ + ∞ dt δφ [M − (∂x φ)] |x=L + −∞ dt δφ(∂x φ)|x=0 −∞ (a) Vanishing of the boundary localized terms for arbitrary δφ requires (BC1) (BC2) ∂x φ|x=0 = 0 [M − ∂x ]φ|x=L = 0 (b) With the boundary conditions now imposed, vanishing of δ S for arbitrary δφ in the region 0 < x < L gives: (EOM) ∂t ∂t φ − ∂x ∂x φ + m2 φ + κφ2 + λφ3 = 0. 4 ...
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