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: " # Z a σ ∂y 2 T ∂y 2 L = dx − (1) 0 2 ∂t 2 ∂x Express y(x, t) as a Fourier series: r ∞ 2 X nπx y(x, t) = q (t) sin . (2) a n a n=1 The derivative of y(x, t) with respect to x and t is r ∞ ∂y(x, t) 2 X nπx = q˙ (t) sin (3) ∂t a n a n=1 r ∞ ∂y(x, t) 2 X nπ nπx = q (t) cos , (4) ∂x a a n a n=1 where we abbreviateq ˙ ≡ ∂q/∂t. Substituting Eqs. (3) and (4) in (1) gives 1 X Z a nπx mπx nmπ2 nπx mπx L = dx σq˙ (t)q ˙ (t) sin sin − T q (t)q (t) cos cos . (5) a n m a a a2 n m a a n,m 0 Using the orthonormality relations 2 Z a nπx mπx dx sin sin = δnm and (6) a 0 a a 2 Z a nπx mπx dx cos cos = δnm (7) a 0 a a we see that the terms with n 6= m vanish. We obtain: X Z a σ T nπ 2 L = dx q˙2 (t) − q2 (t) . (8) 2 n 2 a n n 0 The Euler-Lagrange equations are given by d ∂L ∂L − = 0, n = 0, 1, 2,.... (9) dt ∂q˙n ∂qn With ∂L = σq˙n and (10) ∂q˙n ∂L nπ 2 = −T qn (11) ∂qn a we obtain the equations of motion: T nπ 2 q¨ (t) + q (t) = 0, (12) n σ a n nπ q T where n = 0, 1, 2,.... These are the equations of motion of harmonic oscillators with frequency ωn = a σ . 1 Exercise 2 µ 0µ µ ν The scalar field φ(x) transforms under a Lorentz transformation x → x = Λ ν x as φ(x) → φ0(x0) = φ(x) = φ(Λ−1x0). 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 ∂ ∂xν ∂ ∂0 ≡ = = (Λ−1)ν ∂ν µ ∂x0µ ∂x0µ ∂xν µ Now, µ 2 0µ 0 0 0 2 0 0 ∂ ∂µφ(x) + m φ(x) → ∂ ∂µφ (x ) + m φ (x ) µν −1 ρ −1 κ 2 = g (Λ ) µ∂ρ(Λ ) ν ∂κφ(x) + m φ(x) ρκ 2 = g ∂ρ∂κφ(x) + m 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 gµν , −1 ρ µν −1 κ ρκ (Λ ) µg (Λ ) ν = g . Eq. (13) therefore demonstrates that if φ(x) fulfils the Klein-Gordon equation µ 2 ∂ ∂µφ(x) + m φ(x) = 0, then φ(Λ−1x) likewise fulfils it. Exercise 3 Lagrangian density: λ L = ∂ φ∗∂µφ − m2φ∗φ − (φ∗φ)2. (14) µ 2 Euler-Lagrange equation for φ∗: ∂L ∂L ∗ − ∂µ ∗ = 0. (15) ∂φ ∂(∂µφ ) With ∂L = −m2φ − λ(φ∗φ)φ and (16) ∂φ∗ ∂L = ∂µφ. (17) ∂(∂µφ) We obtain the equation of motion for φ∗: µ 2 ∗ ∂µ∂ φ + m 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) → eiαφ(x), and φ†(x) → eiαφ∗(x), (19) where α ∈ [0, 2π) is a constant. It is straightforward to check that the Lagrangian (14) is invariant under this transformation. This is an example of a global symmetry, i.e., the symmetry transformation acts in the same way on the fields at each point in space and time. The infinitesimal transformation of (19) is: φ → φ + δφ, where δφ = iαφ, and (20) φ∗ → φ∗ + δφ∗, δφ∗ = −iαφ. (21) We can check explicitly that the Lagrangian density is invariant under this transformation: ∗ µ ∗ µ 2 ∗ ∗ ∗ ∗ ∗ δL = ∂µ(δφ )∂ φ + ∂µφ ∂ (δφ) − m (δφ φ + φ δφ) − λ(φ φ)(φ δφ + δφ φ) ∗ µ ∗ µ 2 ∗ ∗ ∗ ∗ ∗ = −iα∂µφ ∂ φ + iα∂µφ ∂ φ + iαm (φ φ − φ φ) − iαλ(φ φ)(φ φ − φ φ) = 0. (22) 2 According to Noether’s theorem, a global symmetry implies the existence of a conserved current. If L → L + α∆L, where µ ∆L = ∂µJ , then the conserved Noether current is given by the formula (Schroeder & Peskin p. 18) µ ∂L µ j = ∆ φa − J , (23) ∂(∂µφa) where a labels all the fields in the theory. In this case there are two independent fields φ and φ∗, and ∆ φ = iφ , ∆ φ∗ = −iφ , ∆ L = 0. The conserved current is thus µ ∂L ∂L ∗ j = (iφ) + ∗ (−iφ ) ∂(∂µφ) ∂(∂µφ ) = −i {φ∗(∂µφ) − (∂µφ∗)φ} . (24) Notice that this quantity is real, and we can check explicitly that it is indeed conserved: µ ∗ µ µ ∗ ∗ µ µ ∗ ∂µj = −i (φ ∂µ(∂ φ) − (∂ φ )(∂µφ) + (∂µφ )(∂ φ) − ∂µ(∂ φ )φ) ∗ µ µ ∗ = −i (φ ∂µ(∂ φ) − ∂µ(∂ φ )φ) = −i φ∗ −m2φ − λ(φ∗φ)φ − −m2φ∗ − λ(φ∗φ)φ∗ φ = 0. (25) In the last step, we used the equation of motion for φ and φ∗, Eq. (18). Exercise 4 The Lagrangian will certainly be invariant under SO(3) rotations of the fields φa, a = 1, 2, 3, since only the ‘length’ φaφa enters the Lagrangian. Infinitesimally, consider the transformation of φaφa under a rotation of the fields φa by an infinitesimal angle θ φa → φa + θabcnbφc, i.e. ∆ φa = abcnbφc (26) where na is a constant unit vector. φaφa transforms as φaφa → (φa + θabcnbφc)(φa + θadendφe) 2 = φaφa + θabcnbφcφa + θadendφeφa + O(θ ) 2 = φaφa + O(θ ). In the last step we used that abc is skew-symmetric under exchange of c and a whereas φcφa is symmetric under this exchange, and therefore the sum vanishes. The same applies to the other term linear in θ. Dropping the θ2 term we find µ that φaφa is indeed invariant under (26). Similarly one can also show that ∂µφa∂ φa is invariant under (26). Therefore L is invariant under (26). To obtain the associated Noether current, we refer to the formula (23) above. In our case J = 0 and ∆φa = abcnbφc, so that µ µ j = (∂ φa)abcnbφc. The Noether current jµ implies a conserved charge Z Z 3 0 3 ˙ Q ≡ d x j = d x φaabcnbφc. This is conserved (Q˙ = 0) for any unit vector n. If we choose nb = δbd for d = 1, 2, 3 (i.e. n is any of the standard basis vectors of R3), we obtain three linearly independent charges Z Z 3 ˙ 3 ˙ Qd = d x adcφaφc = − d x dacφaφc. Let us finally show explicitly that these charges are conserved using the equations of motion. These are given by µ 2 ¨ 2 2 ∂µ∂ φa + m φa = φa − ∇ φa + m φa = 0. Then we get d d Z Q = − d3x φ˙ φ dt a dt abc b c Z 3 ¨ ˙ ˙ = − d x abc(φbφc + φbφc) Z 3 ¨ = − d x abc(φbφc) Z 3 2 2 = − d x abc ∇ φb φc − m φbφc Z 3 = d x abc∇φb · ∇φc = 0. 3 Here we used the skew-symmetry of the tensor several times and integrated by parts to go to the last line. We assume that the field falls off sufficiently fast so that we can neglect the boundary term of the partial integration. Exercise 5 Under a Lorentz transformation, the vector xµ transforms as: µ 0µ µ ν x → x = Λ ν x ; (27) and such transformations leave invariant the form µ ν 0µ 0ν gµν x x = gµν x x . (28) 0µ µ Replacing the x s by their definition in terms of Λ ν , 0µ 0ν µ α ν β µ ν α β α β µ ν gµν x x = gµν (Λ αx ) Λ βx = gµν Λ αΛ β x x = gαβΛ µΛ ν x x . (29) In the last step, we have interchanged the indices α and β with µ and ν. Now, equating (28) with the last step in (29), it follows straightforwardly that: α β gµν = gαβΛ µΛ ν (30) Now, let us consider an infinitesimal Lorentz transformation of the form: µ µ µ Λ ν = δ ν + ω ν ; |ω| 1 (31) Using (30), we find that, keeping only the terms up to first order in ω: α α β β 2 gµν = δ µ + ω µ gαβ δ ν + ω ν = gµν + ωµν + ωνµ + O ω , (32) which requires that ωµν is an antisymmetric tensor for the above transformation to indeed be a Lorentz transformation. For those of you who are interested in such matters, we introduce an alternative, more mathematically sound means of thinking about the concept of an ‘infinitesimal’ Lorentz transform. Consider a one-parameter family of Lorentz transfor- mations, that is a differentiable function Λ : R → SO(3, 1), i.e. for each value of the parameter t,Λ(t) defines a Lorentz transformation. Take the family so-defined to be such that Λ(0) = I, the identity. Then for all t, α β gµν = gαβ Λ(t) µ Λ(t) ν . (33) Looking ‘infinitesimally’ corresponds to differentiating at the identity, i.e. consider d α β 0 α β α 0 β gαβ Λ(t) µ Λ(t) ν = gαβΛ (0) µΛ(0) ν + gαβΛ(0) µΛ (0) ν dt t=0 0 α 0 β = gαν Λ (0) µ + gµβΛ (0) ν .