Quantum Field Theory

• Obviously we can’t cope with bound states. For example, this formalism can’tdescribe the scattering of an electron and proton which collide, bind, and leaveas a Hydrogen atom. It’s possible to circumvent this objection since it turns outthat bound states show up as poles in the S-matrix.• More importantly, a single particle, a long way from its neighbors, is never alonein field theory. This is true even in classical electrodynamics, where the electronsources the electromagnetic field from which it can never escape. In quantumelectrodynamics (QED), a related fact is that there is a cloud of virtual photonssurrounding the electron. This line of thought gets us into the issues of renormalization— more on this next term in the “AQFT” course. Nevertheless, motivatedby this problem, after developing scattering theory using the assumption of noninteractingasymptotic states, we’ll mention a better way.3.2.1 An Example: Meson DecayConsider the relativistically normalized initial and final states,|i〉 = √ 2E ⃗p a † ⃗p |0〉|f〉 = √ 4E ⃗q1 E ⃗q2 b † ⃗q 1c † ⃗q 2|0〉 (3.27)The initial state contains a single meson of momentum p; the final state contains anucleon-anti-nucleon pair of momentum q 1 and q 2 . We may compute the amplitude forthe decay of a meson to a nucleon-anti-nucleon pair. To leading order in g, it is∫〈f|S |i〉 = −ig 〈f| d 4 xψ † (x)ψ(x)φ(x) |i〉 (3.28)Let’s go slowly. We first expand out φ ∼ a + a † using (2.84). (Remember that the φ inthis formula is in the interaction picture, which is the same as the Heisenberg pictureof the free theory). The a piece will turn |i〉 into something proportional to |0〉, whilethe a † piece will turn |i〉 into a two meson state. But the two meson state will havezero overlap with 〈f|, and there’s nothing in the ψ and ψ † operators that lie betweenthem to change this fact. So we have∫〈f|S |i〉 = −ig 〈f|∫= −ig 〈f|∫d 4 xψ † (x)ψ(x)d 3 k(2π) 3 √2E⃗p√2E⃗ka ⃗k a † ⃗p e−ik·x |0〉d 4 xψ † (x)ψ(x)e −ip·x |0〉 (3.29)where, in the second line, we’ve commuted a ⃗k past a † ⃗p , picking up a δ(3) (⃗p − ⃗ k) deltafunctionwhich kills the d 3 k integral. We now similarly expand out ψ ∼ b + c † and– 55 –