Quantum Field Theory

Claim: The field commutation relations (5.3) are equivalent to[b r ⃗p , bs †⃗q ] = (2π)3 δ rs δ (3) (⃗p − ⃗q)[c r ⃗p , cs †⃗q ] = −(2π)3 δ rs δ (3) (⃗p − ⃗q) (5.5)with all other commutators vanishing. Note the strange minus sign in the [c, c † ] term.It’s not yet obvious that it’s bad, but we should be aware of it. For now, let’s just carryon.Proof: Let’s show that the [b, b † ] and [c, c † ] commutators reproduce the field commutators(5.3),[ψ(⃗x), ψ † (⃗y)] = ∑ ∫ d 3 p d 3 q 1(√ [b r(2π)⃗p, b s †6 ⃗q 4E⃗p E ]ur (⃗p)u s (⃗q) † e i(⃗x·⃗p−⃗y·⃗q)r,s⃗q)+ [c r †⃗p , cs ⃗q ]vr (⃗p)v s (⃗q) † e −i(⃗x·⃗p−⃗y·⃗q)= ∑ ∫d 3 p 1 (u s (⃗p)ū s (⃗p)γ 0 e i⃗p·(⃗x−⃗y) + v s (⃗p)¯v s (⃗p)γ 0 e −i⃗p·(⃗x−⃗y)) (5.6)(2π) 3 2Es⃗pAt this stage we use the outer product formulae (4.128) and (4.129) which tell us∑s us (⃗p)ū s (⃗p) = /p + m and ∑ s vs (⃗p)¯v s (⃗p) = /p − m, so that∫[ψ(⃗x), ψ † (⃗y)] =∫=d 3 p 1 (( /p + m)γ 0 e i⃗p·(⃗x−⃗y) + ( /p − m)γ 0 e −i⃗p·(⃗x−⃗y))(2π) 3 2E ⃗pd 3 p 1 ((p0 γ 0 + p(2π) 3 i γ i + m)γ 0 + (p 0 γ 0 − p i γ i − m)γ 0) e +i⃗p·(⃗x−⃗y)2E ⃗pwhere, in the second term, we’ve changed ⃗p → −⃗p under the integration sign. Now,using p 0 = E ⃗p we have∫[ψ(⃗x), ψ † d 3 p(⃗y)] =3e+i⃗p·(⃗x−⃗y) = δ (3) (⃗x − ⃗y) (5.7)(2π)as promised. Notice that it’s a little tricky in the middle there, making sure thatthe p i γ i terms cancel. This was the reason we needed the minus sign in the [c, c † ]commutator terms in (5.5).□5.1.1 The HamiltonianTo proceed, let’s construct the Hamiltonian for the theory. Using the momentumπ = iψ † , we haveH = π ˙ψ − L = ¯ψ(−iγ i ∂ i + m)ψ (5.8)– 107 –