Quantum Field Theory

The minus signs here are odd to say the least! For spacelike λ = 1, 2, 3, everythinglooks fine,[a λ ⃗p, a λ′ †⃗q] = δ λλ′ (2π) 3 δ (3) (⃗p − ⃗q) λ, λ ′ = 1, 2, 3 (6.46)But for the timelike annihilation and creation operators, we have[a 0 ⃗p , a0 †⃗q ] = −(2π)3 δ (3) (⃗p − ⃗q) (6.47)This is very odd! To see just how strange this is, we take the Lorentz invariant vacuum|0〉 defined byThen we can create one-particle states in the usual way,a λ ⃗p |0〉 = 0 (6.48)|⃗p, λ〉 = a λ †⃗p|0〉 (6.49)For spacelike polarization states, λ = 1, 2, 3, all seems well. But for the timelikepolarization λ = 0, the state |⃗p, 0〉 has negative norm,〈⃗p, 0|⃗q, 0〉 = 〈0|a 0 ⃗p a0 †⃗q|0〉 = −(2π) 3 δ (3) (⃗p − ⃗q) (6.50)Wtf? That’s very very strange. A Hilbert space with negative norm means negativeprobabilities which makes no sense at all. We can trace this negative norm back to thewrong sign of the kinetic term for A 0 in our original Lagrangian: L = + 1 ˙⃗A 2 − 1 A ˙22 2 0 +. . ..At this point we should remember our constraint equation, ∂ µ A µ = 0, which, untilnow, we’ve not imposed on our theory. This is going to come to our rescue. We will seethat it will remove the timelike, negative norm states, and cut the physical polarizationsdown to two. We work in the Heisenberg picture, so that∂ µ A µ = 0 (6.51)makes sense as an operator equation. Then we could try implementing the constraintin the quantum theory in a number of different ways. Let’s look at a number ofincreasingly weak ways to do this• We could ask that ∂ µ A µ = 0 is imposed as an equation on operators. But thiscan’t possibly work because the commutation relations (6.39) won’t be obeyedfor π 0 = −∂ µ A µ . We need some weaker condition.– 133 –