Homework 3 for Quantum Field Theory - Advanced Quantum Mechanics I | PHYSICS 513, Assignments of Quantum Mechanics

Download Homework 3 for Quantum Field Theory - Advanced Quantum Mechanics I | PHYSICS 513 and more Assignments Quantum Mechanics in PDF only on Docsity! PHYSICS 513: QUANTUM FIELD THEORY HOMEWORK 3 1 Physics 513, Quantum Field Theory Homework 3 Due Tuesday, 23rd September 2003 Jacob Lewis Bourjaily 1. a) We are given complex scalar Lagrangian, L = ∂µφ∗∂µφ−m2φ∗φ. It is clear that the canonical momenta of the field are π = ∂L ∂(∂0φ) = ∂0φ∗; π∗ = ∂L ∂(∂0φ∗) = ∂0φ. The canonical commutation relations are then [φ(x), ∂0φ∗(y)] = [φ∗(x), ∂0φ(y)] = iδ(3)(x− y), with all other combinations commuting. As in Homework 2, the Hamiltonian can be directly computed, H = ∫ d3xH = ∫ d3x (π∂0φ− L), = ∫ d3x ( π∗π − 1/2π∗π + 1/2∇φ∗∇φ + 1/2m2φ∗φ), = 1 2 ∫ d3x ( π∗π +∇φ∗∇φ + m2φ∗φ). We can use this expression for the Hamiltonian to find the Heisenberg equation of motion. We have i∂0φ(x) = [ φ(x), 1 2 ∫ d3y ( π∗(y)π(y) +∇φ∗(y)∇φ(y) + m2φ∗(y)φ(y)) ] , = 1 2 ∫ d3y [φ(x), π(y)]π∗(y), = i 2 ∫ d3y δ(3)(x− y)π∗(y), = i 2 π∗(x). Analogously, i∂0φ∗(x) = i2π(x). Notice that this derivation used the fact that φ commutes with everything in H except for π. Before we compute the commutator of π∗(x) with the Hamiltonian, we should re-write H as PS did so that our conclusion will be more lucid. We have from above that H = 1 2 ∫ d3x ( π∗π +∇φ∗∇φ + m2φ∗φ). We can evaluate the middle term in H using Green’s Theorem (essentially integration by parts). We will assume that the surface term vanishes at infinity because the fields must. This allows us to write the Hamiltonian as, H = 1 2 ∫ d3x ( π∗π + φ∗(−∇2 + m2)φ). 2 JACOB LEWIS BOURJAILY Commuting this with π∗(x), we conclude that i∂0π ∗(x) = 1 2 ∫ d3y [π∗(x), φ∗(y)](−∇2 + m2)φ(y), = − i 2 ∫ d3y (−∇2 + m2)φ(y)δ(3)(x− y), = − i 2 φ(x). Combining the two results, it is clear that ∂20φ(x) = (∇)2 −m2)φ(x), =⇒ (∂µ∂µ + m2)φ = 0. This is just the Klein-Gordon equation. The result is the same for the complex conjugate field. b) Because the field is no longer purely real, we cannot assume that the coefficient of eip·x in the ladder-operator Fourier expansion is the adjoint of the coefficient of e−ip·x. Therefore we will use the operator b. The expansion of the fields are then φ(xµ) = ∫ d3p (2π)3 1√ 2ωp ( ape −ipµxµ + b†pe ipµx µ ) ; φ∗(xµ) = ∫ d3q (2π)3 1√ 2ωq ( a†qe iqµx µ + bqe−iqµx µ ) . It is easy to show that these allow us to define π and π∗ in terms of a and b operators as well. These become, π(xµ) = ∂0φ∗(xµ) = ∫ d3q (2π)3 i √ ωq 2 ( a†qe iqµx µ − bqe−iqµx µ ) ; π∗(xµ) = ∂0φ(xµ) = ∫ d3p (2π)3 i √ ωp 2 ( −ape−ipµx µ + b†pe ipµx µ ) . These allow us to directly demonstrate that [φ(xµ), π(yµ)] = ∫ d3pd3q (2π)6 −i 2 √ ωq ωp ([ ap, a † q ] e−i(pµx µ−qµxµ) − [b†p, bq ] ei(pµx µ−qµxµ) ) , = iδ(3)(x− y), while noting that [ ap, a † q ] = [ bp, b † q ] = (2π)3δ(3)(p− q), and all other terms commute. This implies that there are in fact two entirely different sets of particles with the same mass: those created by b† and those created by a†. c) I computed the conserved Noether charge in Homework 2 as jµ = i (φ∂µφ∗ − φ∗∂µφ) . We integrate this over all space to see the conserved current in the 0 component. When expressing phi and pi in terms of ladder operators, we can evaluate this directly. Q = i 2 ∫ dx(φ∗(x)π∗(x)− π(x)φ(x)), = i 2 ∫ d3xd3pd3q (2π)6 ( apa † qe ixµ(qµ−pµ) − apbqe−ix µ(pµ+qµ) + b†pa † qe ixµ(pµ+qµ) − b†pbqeix µ(qµ−pµ) ) − c.c., = i 2 ∫ d3pd3q (2π)3 ( apa † qδ (3)(p− q)− apbqδ(3)(p + q) + b†pa†qδ(3)(p + q)− b†pbqδ(3)(p− q) ) − c.c., = i 2 ∫ d3p (2π)3 ( apa † p − apb−p + b†pa†−p − b†pbp ) − c.c., = i ∫ d3p (2π)3 ( a†pap − b†pbp ) . ‘óπ²ρ ’²́δ²ι δ²Äιξαι