Michael E. Peskin & Daniel V. Schroeder, *An Introduction to Quantum Field Theory*, (Perseus Books, 1995) – Chapter 2.

Continuing our discussion of the derivation of the Klein-Gordon field by analogy with the harmonic oscillator, we arrived at the field and its conjugate momentum density:

The operators and are analogues of the raising and lowering operators and in the harmonic oscillator, with the extra condition that we have one pair of operators for each momentum . In the harmonic oscillator, the operators satisfied the commutation relation

When applied to the Klein-Gordon field, we assume that operators with different momenta commute, but those with the same momenta do not. The assumption is that

[These equations differ from those in Klauber’s book as discussed earlier in that the factors of turn up in different places. However, the final result for the commutator is the same, which is what matters.]

From here, we can work out the commutator by plugging the operator commutators into the integrals for and . This is similar to the derivation we did earlier, except here we’re assuming the commutators for and are as given above, rather than deriving them from the assumed commutator . This calculation proves to be somewhat easier than the previous one, in that we can throw away all integrals containing and . We get

This is the same result we got earlier using Klauber’s method. Remember that we’re still taking the field to be a real field.

P&S now derive the total Hamiltonian in their equation 2.31. The technique is very similar to that used by Klauber. The differences are (i) we take the momentum to be continuous rather than the discrete used by Klauber; (ii) the Klein-Gordon field is real, rather than the two complex fields used by Klauber; (iii) the Hamiltonian density has a factor of not found in Klabuer; and (iv) the factors of show up in different places.

P&S’s Hamiltonian density is

From 1 we have

Plugging this and 2 into 11 and integrating over gives the total Hamiltonian

where comes from the term in 11 and comes from :

Doing the integral first, we have

so we can set and eliminate the integral over . Further, using converts and to

Since we’re integrating over all we can replace by in the last term, and use 4 on the first term:

Since the commutator in the last term has two operators both with suffix , it is an infinite quantity, so its integral is infinite. This is swept under the carpet by saying that since this energy is present in all states and it’s only the difference between a given state and the ground state that can be measured, we can ignore it. In the harmonic oscillator, the ground state has energy so this infinite term can be thought of as the sum of this zero-point energy over all momentem states.

In field theory, a state is postulated such that for all . This is the vacuum state, which has the infinite zero-point energy. If we operate on with this produces an eigenstate of as we can show using the commutator 4 and ignoring the infinite energy term:

The operator acts on the vacuum to create an excited state with energy in the same way that the operator in the harmonic oscillator operates on the ground state to produce an oscillator in the next highest energy state. In field theory, this excitation is called a particle, so becomes a creation operator that creates a particle with energy . A similar calculation shows that , acting on a state containing a particle of energy , removes this particle from the state and produces an eigenstate with energy lowered by , so is called an annihilation operator. These operators can produce and destroy multiple particles within a single state.