Canonical quantisation of the real Klein-Gordon field

SI units throughout, with x^μ = (ct, x) and the metric (+, −, −, −) of Unit 3. Topic 9.01's density, ℒ = ½[(1/c²)(∂φ/∂t)² − (∇φ)² − (mc/ħ)²φ²] in J m⁻³, gives φ the unit √(J m⁻¹) and π = ∂ℒ/∂(∂φ/∂t) = (1/c²)∂φ/∂t. To see what to impose, cut space into cells of side ℓ. The Lagrangian becomes Σⱼ ℓ³ℒⱼ, so cell j has coordinate φⱼ and momentum pⱼ = ℓ³πⱼ, and Quantum Mechanics I's rule for each pair, [φᵢ, pⱼ] = iħδᵢⱼ, reads [φᵢ, πⱼ] = iħδᵢⱼ/ℓ³. As ℓ → 0, δᵢⱼ/ℓ³ becomes δ³(x − y). That is a postulate, and an equal-time one: it says nothing about φ and π at different times. Dimensions agree, φπ and iħδ³ both in J s m⁻³. A consistency check follows: with H = ∫ℋ d³x and ℋ = ½[c²π² + (∇φ)² + (mc/ħ)²φ²], the Heisenberg equations give ∂φ/∂t = c²π and ∂π/∂t = ∇²φ − (mc/ħ)²φ, so the operator φ obeys Unit 3's Klein-Gordon equation, derived.

The field equation is linear, so its operator solution is a sum of plane waves. In a periodic cube of side L the allowed wavevectors are k = 2πn/L, n a vector of integers, and e^{(i(k⋅x − ωt))} is a solution only for ω = ±ωₖ with ωₖ = c√(k² + (mc/ħ)²). Each k therefore brings a positive-frequency wave with some operator Aₖ and a negative-frequency wave e^{(−i(k⋅x − ωₖt))} with an operator Bₖ. In Unit 3 the two were independent, and the second was a negative-energy state. Here φ is Hermitian, and taking the adjoint of the sum swaps the two waves, so Bₖ = Aₖ† for every k: one operator per mode, not two. Writing Aₖ = √(ħc²/(2ωₖV)) aₖ, a factor in √(J m⁻¹) chosen so that aₖ is dimensionless with unit commutator, gives the expansion on the card. For the π⁰ field in a cube of side 20.0 fm the spacing is 2π/L = 0.314 fm⁻¹: the k = 0 mode has ħωₖ = 134.98 MeV, the six with |n| = 1 have 148.5 MeV, and 147 modes have |k| ≤ 1 fm⁻¹, against the continuum estimate VK³/(6π²) = 135.

At t = 0, project the field onto one wave: ∫d³x e^(−ik⋅x)φ = √(ħc²V/(2ωₖ))(aₖ + a₋ₖ†), and ∫d³x e^(−ik⋅x)π = −i√(ħωₖV/(2c²))(aₖ − a₋ₖ†). Adding the two with the right weights isolates aₖ = (2ħV)^(−1/2)∫d³x e^(−ik⋅x)[(√ωₖ/c)φ + i(c/√ωₖ)π]. Now feed in the postulate. In [aₖ, aₖ'†] only the φπ and πφ cross terms survive, and each contributes ½ through ∫d³x e^{(−i(k − k')⋅x)} = Vδₖₖ', so [aₖ, aₖ'†] = δₖₖ', while [φ, φ] = [π, π] = 0 give [aₖ, aₖ'] = 0: derived. Once k' = k, the √ωₖ in each term's φ weight cancels the 1/√ωₖ in its π weight, which is why the mass never enters the algebra. Dimensions: both sides are pure numbers. The one-mode limit is Quantum Mechanics I's [a, a†] = 1, and the Heisenberg equation with the Hamiltonian of the next step gives aₖ(t) = aₖe^(−iωₖt): the time dependence written into the expansion is what the operators do of their own accord.

Substitute the expansion into H = ∫ℋ d³x. The c²π² term gives (ħωₖ/4)(aₖaₖ† + aₖ†aₖ − aₖa₋ₖe^{(−2iωₖt)} − h.c.) for each k, and the gradient and mass terms together give (ħc²(k² + m²c²/ħ²)/(4ωₖ))(aₖaₖ† + aₖ†aₖ + aₖa₋ₖe^{(−2iωₖt)} + h.c.). The second prefactor equals ħωₖ/4 only because ωₖ² = c²(k² + m²c²/ħ²), so the terms that would make or destroy pairs at ±k cancel, and H = Σₖ ħωₖ(aₖ†aₖ + ½), derived. Quantum Mechanics I's norm argument, quoted, makes each nₖ = 0, 1, 2, …, and topic 9.01's Noether momentum, P = −∫π∇φ d³x, becomes Σₖ ħk aₖ†aₖ. So aₖ†|0⟩ has energy ħωₖ = √(ħ²k²c² + m²c⁴) and momentum ħk: a free relativistic particle. Limits: m = 0 gives ħωₖ = ħc|k|; for k ≪ mc/ħ, ħωₖ ≈ mc² + ½ħ²k²/m, and at k = 0.1 fm⁻¹ for the π⁰ the two give 136.412 and 136.419 MeV. Measured: the π⁰, chargeless and its own antiparticle as a real field's quantum must be, has mc² = 134.977 MeV.

The algebra gives [H, aₖ†] = ħωₖaₖ† and [P, aₖ†] = ħk aₖ†: acting on any state, aₖ† adds energy +ħωₖ and momentum +ħk, although in φ it multiplies e^{(−i(k⋅x − ωₖt))}, which Unit 3 would have read as energy −ħωₖ and momentum −ħk. The sign of a frequency in the field is not the energy of any state. It records whether the operator beside it removes a quantum or adds one, and aₖ†(t) = aₖ†e^(+iωₖt) is how every raising operator evolves in the Heisenberg picture. Every state has energy at least E₀, so there is nothing below to cascade into. The one-quantum wavefunction, ⟨0|φ(x, t)|1ₖ⟩ = √(ħc²/(2ωₖV)) e^{(i(k⋅x − ωₖt))}, has positive frequency only: the negative-frequency solutions have been given a job, not thrown away. What a real field cannot supply is a second species. With one operator per mode the quantum aₖ† adds is the one aₖ removes, chargeless and its own antiparticle; distinct antiparticles need a complex field, topic 9.03.

The vacuum, aₖ|0⟩ = 0 for every k, has energy E₀ = Σₖ½ħωₖ. With V/(2π)³ modes per unit k-space volume and |k| ≤ K, E₀/V = (ħc/(4π²))∫₀^K k²√(k² + (mc/ħ)²) dk ≈ ħcK⁴/(16π²): for K = 10 fm⁻¹, 12.5 GeV fm⁻³ or 2.0 × 10³⁶ J m⁻³. Normal ordering, every aₖ† moved left of every aₖ, replaces H by :H: = Σₖħωₖaₖ†aₖ, which differs from H only by the constant E₀ and so leaves every energy difference unchanged. Only gravity responds to an absolute energy, and the dark-energy density cosmology measures, about 5 × 10⁻¹⁰ J m⁻³, is some 4 × 10⁴⁵ times smaller: a mismatch this course leaves open. When the constant's change with boundaries is measurable is Unit 11's question. Symmetry needs no repair. Creation operators commute, so aₖ₁†aₖ₂†|0⟩ = aₖ₂†aₖ₁†|0⟩, derived: the two-quantum wavefunction ⟨0|φ(x₁)φ(x₂)|1ₖ₁1ₖ₂⟩ is symmetric in x₁ and x₂, and (aₖ†)²|0⟩ = √2|2ₖ⟩ is allowed. The quanta are bosons, and the excess of identical-pion pairs at small relative momentum, measured in collisions since 1960, is that symmetry seen.