Quantum Field Theory

from which we see thatδφ = −ω µ ν xν ∂ µ φ (1.52)By the same argument, the Lagrangian density transforms asδL = −ω µ νx ν ∂ µ L = −∂ µ (ω µ νx ν L) (1.53)where the last equality follows because ω µ µ = 0 due to anti-symmetry. Once again,the Lagrangian changes by a total derivative so we may apply Noether’s theorem (nowwith F µ = −ω µ ν xν L) to find the conserved currentj µ = −∂L∂(∂ µ φ) ωρ νx ν ∂ ρ φ + ω µ ν x ν L[ ]∂L= −ω ρ ν∂(∂ µ φ) xν ∂ ρ φ − δ µ ρ x ν L = −ω ρ ν T µ ρx ν (1.54)Unlike in the previous example, I’ve left the infinitesimal choice of ω µ ν in the expressionfor this current. But really, we should strip it out to give six different currents, i.e. onefor each choice of ω µ ν . We can write them as(J µ ) ρσ = x ρ T µσ − x σ T µρ (1.55)which satisfy ∂ µ (J µ ) ρσ = 0 and give rise to 6 conserved charges. For ρ, σ = 1, 2, 3,the Lorentz transformation is a rotation and the three conserved charges give the totalangular momentum of the field.∫Q ij = d 3 x (x i T 0j − x j T 0i ) (1.56)But what about the boosts? In this case, the conserved charges are∫Q 0i = d 3 x (x 0 T 0i − x i T 00 ) (1.57)The fact that these are conserved tells us that∫ ∫0 = dQ0i= d 3 x T 0i + t d 3 0i∂Tx − d ∫d 3 x x i T 00dt∂t dt= P i + t dP idt − d ∫d 3 x x i T 00 (1.58)dtBut we know that P i is conserved, so dP i /dt = 0, leaving us with the following consequenceof invariance under boosts:∫dd 3 x x i T 00 = constant (1.59)dtThis is the statement that the center of energy of the field travels with a constantvelocity. It’s kind of like a field theoretic version of Newton’s first law but, rathersurprisingly, appearing here as a conservation law.– 17 –