Quantum Field Theory

Substituting into the above expressions we findwhile the Hamiltonian is given by[a, a † ] = 1 (2.11)H = 1 2 ω(aa† + a † a)= ω(a † a + 1 2 ) (2.12)One can easily confirm that the commutators between the Hamiltonian and the creationand annihilation operators are given by[H, a † ] = ωa † and [H, a] = −ωa (2.13)These relations ensure that a and a † take us between energy eigenstates. Let |E〉 bean eigenstate with energy E, so that H |E〉 = E |E〉. Then we can construct moreeigenstates by acting with a and a † ,Ha † |E〉 = (E + ω)a † |E〉 , Ha |E〉 = (E − ω)a |E〉 (2.14)So we find that the system has a ladder of states with energies. . .,E − ω, E, E + ω, E + 2ω, . . . (2.15)If the energy is bounded below, there must be a ground state |0〉 which satisfies a |0〉 = 0.This has ground state energy (also known as zero point energy),Excited states then arise from repeated application of a † ,H |0〉 = 1 ω |0〉 (2.16)2|n〉 = (a † ) n |0〉 with H |n〉 = (n + 1 )ω |n〉 (2.17)2where I’ve ignored the normalization of these states so, 〈n| n〉 ≠ 1.2.2 The Free Scalar FieldWe now apply the quantization of the harmonic oscillator to the free scalar field. Wewrite φ and π as a linear sum of an infinite number of creation and annihilation operatorsa † ⃗p and a ⃗p, indexed by the 3-momentum ⃗p,∫φ(⃗x) =∫π(⃗x) =d 3 p 1[]√ a(2π) 3 ⃗p e i⃗p·⃗x + a † e−i⃗p·⃗x ⃗p 2ω⃗p√d 3 p ω⃗p(2π) 3(−i) 2[a ⃗p e i⃗p·⃗x − a † ⃗p e−i⃗p·⃗x ](2.18)(2.19)– 23 –