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advanced33 min studyLesson 651 in Physics

Scalar field mode expansion and normalization

Construct a real Klein–Gordon field from positive and negative frequencies while keeping box, continuum, and relativistic state normalizations distinct.

Before you begin
On this page

The question for today

What you’ll understand

  • Fix a metric, Fourier phase, and positive-energy dispersion relation.
  • Translate between discrete and continuum ladder normalizations.
  • Use a normalized wavepacket rather than a squared momentum delta function.

A free field has infinitely many oscillatory degrees of freedom, but its particle interpretation depends on a precise normalization of those oscillations. A missing square root of energy can move between a mode expansion, a ladder algebra, and a one-particle state without changing physics, provided every part changes consistently. This lesson chooses one convention and checks it in finite volume before taking the continuum limit. The construction is exact for the free massive real scalar field, rather than a general solution of interacting quantum field theory.

Fix conventions before promoting amplitudes

Use natural units with hbar and light speed equal to one in four-dimensional Minkowski spacetime. Choose metric signature plus, minus, minus, minus, and a Fourier phase with positive energy multiplying negative imaginary time. The real Klein–Gordon field is Hermitian, so its positive-frequency annihilation part must be accompanied by its Hermitian conjugate. The negative-frequency term does not introduce negative-energy particles; its coefficient creates a positive-energy excitation of the same neutral species.

A Fourier mode satisfies an oscillator equation whose frequency is the positive square root of momentum squared plus mass squared. Reality relates the coefficients of opposite phases, so counting both as unrelated real fields doubles the degrees of freedom. Mass and momentum have energy dimension, while spacetime has inverse-energy dimension. The scalar field has energy dimension one in this spacetime dimension. Its mode operators have dimensions determined by the chosen integration measure and delta-function algebra.

ημν=diag⁡(1,−1,−1,−1),p⋅x=Ept−p⋅x,Ep=p2+m2>0\eta_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1),\quad p\cdot x=E_{\mathbf p}t-\mathbf p\cdot\mathbf x,\quad E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}>0
We use ℏ = c = 1 and m > 0. E and spatial momentum are in common energy units. The Fourier convention fixes all later signs; it is not the opposite-sign metric convention used by some references.↔ Scroll wide equations horizontally.

The integration measure belongs to the algebra

With a spatial momentum measure containing three factors of two pi, put an inverse square root of twice energy in the field expansion. The ladder commutator then contains that same Fourier volume factor times a momentum delta function. At fixed time, differentiating the phases supplies the energy factors needed to reproduce the canonical field–momentum commutator. Omitting the negative-frequency term would leave only one of the contributions and would also destroy Hermiticity.

A finite periodic cube replaces momentum integrals by sums and Dirac deltas by Kronecker deltas. Its expansion coefficient is the inverse square root of twice energy times volume, when each discrete ladder pair has unit commutator. Taking the large-volume limit changes the density of allowed momenta, rather than simply replacing the symbol for a sum by an integral. A finite box also makes explicit that normalization of a single oscillator is different from normalization of a continuum momentum eigenstate.

ϕ(x)=∫d3p(2π)32Ep(ape−ip⋅x+ap†eip⋅x),[ap,aq†]=(2π)3δ3(p−q)\phi(x)=\int\frac{d^3p}{(2\pi)^3\sqrt{2E_{\mathbf p}}}\left(a_{\mathbf p}e^{-ip\cdot x}+a_{\mathbf p}^{\dagger}e^{ip\cdot x}\right),\quad [a_{\mathbf p},a_{\mathbf q}^{\dagger}]=(2\pi)^3\delta^3(\mathbf p-\mathbf q)
The real free field is an operator-valued distribution. Equalities mean smeared operator identities on a suitable dense finite-particle domain. Continuum aₚ has energy dimension −3/2. In a periodic volume V the coefficient is 1/√(2EV), with dimensionless ladders and [aₙ,aₘ†] = δₙₘ.↔ Scroll wide equations horizontally.

Physical packets have a finite norm

A sharp momentum state has a delta-function normalization, so its formal norm is not a finite probability. Smearing creation operators with a square-integrable momentum envelope gives a normalizable one-particle packet. The norm is the envelope's squared integral in the selected measure. This is also the correct setting for asking whether a proposed mode coefficient has an extra energy factor: the answer must agree with the normalization of the packet and its expected energy.

Relativistically normalized sharp states can instead absorb a square root of twice energy into their definition. Their completeness measure then contains its inverse. Those conventions are equivalent but cannot be mixed. Local fields themselves require spacetime smearing with smooth test functions; a pointwise field acting on the vacuum is not an ordinary finite-norm vector. Finite-volume and momentum cutoffs are useful intermediate regulators, but neither justifies ignoring the distributional nature of the continuum limit.

∣f⟩=∫d3p(2π)3f(p)ap†∣0⟩,⟨f∣f⟩=∫d3p(2π)3∣f(p)∣2|f\rangle=\int\frac{d^3p}{(2\pi)^3}f(\mathbf p)a_{\mathbf p}^{\dagger}|0\rangle,\qquad \langle f|f\rangle=\int\frac{d^3p}{(2\pi)^3}|f(\mathbf p)|^2
The packet has unit norm when the displayed integral is one. With |p⟩rel = √(2Ep)aₚ†|0⟩ the one-particle completeness measure is d³p/[(2π)³2Ep]. These are free-field Fock-space statements.↔ Scroll wide equations horizontally.

Work it through

Normalize one oscillator in a box

In natural units, a periodic box has V = 8 in inverse-energy cubed units. A scalar mode has m = 3 and momentum magnitude 4 in common energy units. Find E, its coefficient 1/√(2EV), and the coefficient after multiplying box volume by four at fixed mode energy.

  1. The positive frequency is E=42+32=5E=\sqrt{4^2+3^2}=5. It is energy, not momentum magnitude alone.

  2. The field coefficient is 1/2(5)(8)=1/80≃0.11181/\sqrt{2(5)(8)}=1/\sqrt{80}\simeq0.1118 in its corresponding field units.

  3. At V = 32 the denominator becomes √320, twice as large. The coefficient halves to approximately 0.0559, while the discrete ladder commutator remains one.

The result

The mode energy is 5 and the coefficient is approximately 0.1118; quadrupling volume halves that coefficient at fixed E.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

A mode has an oscillator frequency

Reset and set BOTH Damping ratio ζ to 0 and Natural frequency ω₀ to 1; predict undamped motion. Then Reset and set BOTH ζ to 0 and ω₀ to 2; predict half the period. This positive-stiffness classical oscillator starts at x = 1 and zero velocity. It illustrates the time dependence of one free mode only, without a quantum ladder algebra, vacuum fluctuations, or continuum normalization.

A mode has an oscillator frequencyUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forOscillatory time dependence does not by itself specify a field’s quantum normalization.

Pause & reason

Check your understanding

With the displayed continuum ladder algebra, which factor belongs in the scalar field expansion?

Your turn

Try it yourself

Two orthogonal normalized box modes have packet coefficients c₁ = 1/2 and c₂ = √3/2, and energies 2 and 6. Check the one-particle norm and compute its mean free energy.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Orthogonality removes cross terms; the squared amplitudes give the weights.

Compare with the solution

The norm is ∣c1∣2+∣c2∣2=1/4+3/4=1|c_1|^2+|c_2|^2=1/4+3/4=1. The mean energy is (1/4)(2)+(3/4)(6)=5(1/4)(2)+(3/4)(6)=5. This packet is normalized but is not an energy eigenstate because both distinct energies have nonzero weight. There is exactly one quantum in the packet, rather than two particles with fractional populations.

Keep the idea

Take it with you

  • Fourier phases, integration measures, and ladder normalizations form one consistent convention.
  • The negative-frequency term creates positive-energy neutral particles.
  • Wavepackets have finite norms; sharp momentum states and local fields need distributional treatment.

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