advanced27 min studyLesson 43 in Physics

Quantizing the scalar field

Build a Fock space from oscillator modes and distinguish particles, field operators, vacuum fluctuations, and classical waves.

On this page

What you’ll understand

  • Connect equal-time field commutators with creation and annihilation operators.
  • Calculate a mode’s occupation and energy above the vacuum.
  • Explain why a one-particle state is not a classical oscillating field.

Quantization does not mean painting tiny balls onto a classical wave. It replaces dynamical variables with operators on a space of states. The scalar field’s normal modes are harmonic oscillators, so their quantum mechanics gives a systematic starting point. Excitations of those modes behave as identical spin-zero particles. The construction is exact for a free field; interactions will require more careful definitions of observable particle states.

Start from the canonical pair

Use ℏ=c=1\hbar=c=1, η=diag⁡(1,−1,−1,−1)\eta=\operatorname{diag}(1,-1,-1,-1), and the stable massive real scalar action from the preceding lesson. Its canonical momentum is π=ϕ˙\pi=\dot\phi. At equal time we impose a commutator between the field and momentum, while fields commute with fields and momenta commute with momenta. The spatial delta function plays the role of the Kronecker delta for a continuum of oscillator coordinates.

A field at a single point is more singular than an ordinary operator matrix: quantum fields are operator-valued distributions. Smearing with a smooth test function produces the meaningful integrated observables used in a careful formulation. In introductory calculations we write pointwise expressions as shorthand, but products at coincident points can diverge. That warning is already present before interactions and explains why a regulator is a tool with a purpose, rather than cosmetic algebra.

[ϕ^(t,x),π^(t,y)]=iδ(3)(x−y),[a^p,a^q†]=(2π)3δ(3)(p−q)[\hat\phi(t,\mathbf x),\hat\pi(t,\mathbf y)]=i\delta^{(3)}(\mathbf x-\mathbf y),\qquad [\hat a_{\mathbf p},\hat a_{\mathbf q}^{\dagger}]=(2\pi)^3\delta^{(3)}(\mathbf p-\mathbf q)
These are equal-time canonical relations and one chosen continuum normalization of the mode operators. They imply bosonic occupation numbers.↔ Scroll wide equations horizontally.

The mode expansion fixes normalization

Choose p0=Ep=p2+m2p^0=E_{\mathbf p}=\sqrt{\mathbf p^2+m^2} and p⋅x=Ept−p⋅xp\cdot x=E_{\mathbf p}t-\mathbf p\cdot\mathbf x. The annihilation part carries e−ip⋅xe^{-ip\cdot x} and the creation part its conjugate. The factor 1/2Ep1/\sqrt{2E_{\mathbf p}} works with our commutator and integration measure. Other books distribute these factors differently; an isolated factor has no meaning until the state normalization is also stated.

The vacuum is annihilated by every a^p\hat a_{\mathbf p}. Applying creation operators builds Fock states, with unrestricted nonnegative occupation of each bosonic mode. Momentum eigenstates in infinite volume are delta-normalized rather than normalizable. A localized preparation is a wave packet built by integrating momentum modes with a square-integrable envelope. Particle number is well defined for this free theory, but need not be conserved once interactions are introduced.

ϕ^(x)=∫d3p(2π)32Ep(a^pe−ip⋅x+a^p†eip⋅x),Ep=p2+m2\hat\phi(x)=\int\frac{d^3p}{(2\pi)^3\sqrt{2E_{\mathbf p}}}\left(\hat a_{\mathbf p}e^{-ip\cdot x}+\hat a_{\mathbf p}^{\dagger}e^{ip\cdot x}\right),\qquad E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}
The real field is Hermitian. Positive and negative frequency terms are paired, but their coefficients are now operators rather than classical complex amplitudes.↔ Scroll wide equations horizontally.

Vacuum and particles are different kinds of statements

A number state has a sharp occupation, but its field expectation is zero: the field changes occupation by one, making the relevant states orthogonal. Its fluctuations are nevertheless nonzero. A coherent state, a superposition of different occupations with a definite phase relationship, can have an expectation value following a classical free-field solution. This is the bridge to a classical wave, rather than equating a one-particle wavefunction with a directly measured classical field.

The free Hamiltonian sums oscillator energies, including a formally divergent zero-point contribution. Normal ordering subtracts the free vacuum constant when computing excitation energies in nongravitational flat-space calculations. This does not solve the cosmological constant problem or erase all physically relevant vacuum effects. What we can calculate here without ambiguity is the energy difference between finite-occupation states with the same regulator and reference vacuum.

Two quanta in one box mode

Place the field in a periodic box so modes are discrete. Choose one mode of frequency ω=5\omega=5 and normalized state ∣2⟩=(a†)2∣0⟩/2!|2\rangle=(a^{\dagger})^2|0\rangle/\sqrt{2!}. Find its occupation, excitation energy, and field expectation.

  1. Use [a,a†]=1[a,a^{\dagger}]=1 for the discrete box mode. The number operator satisfies a†a∣2⟩=2∣2⟩a^{\dagger}a|2\rangle=2|2\rangle, confirming the normalization and occupation.

  2. The oscillator energy is ω(n+1/2)\omega(n+1/2). Subtract the energy of that same mode in its vacuum to obtain ΔE=2ω=10\Delta E=2\omega=10; the zero-point contribution cancels.

  3. The mode’s field coordinate is proportional to a+a†a+a^{\dagger}. Each term changes occupation, so its expectation in ∣2⟩|2\rangle vanishes by orthogonality. A zero mean does not imply zero energy or zero variance.

The result

Two particles can carry a definite nonzero excitation energy while producing no oscillating mean field. A classical-looking mean requires a different state preparation.

Explore the model

Interactive lab

Interactive lab · oscillator

Modes before particle language

Set the damping ratio to zero and vary the natural frequency of the unit-mass oscillator. Its moving classical coordinate represents the equation underlying one free mode. Compare this coordinate with the text’s distinction between a number state’s zero mean and a coherent state’s oscillating mean.

Modes before particle languageUnit 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 forThe lab solves a classical oscillator with fixed initial data, not a quantum measurement trajectory or a number-state probability distribution. Nonzero damping adds a phenomenological effect outside the isolated free-field model.

Check your understanding

Why can a free scalar one-particle number state have zero mean field?

Try it yourself

For one normalized oscillator mode, compute ⟨n∣(a+a†)2∣n⟩\langle n|(a+a^{\dagger})^2|n\rangle. What does it say about a zero mean?

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

Give me a hint

Expand the square without replacing aa†aa^{\dagger} by a†aa^{\dagger}a.

Compare with the solution

The a2a^2 and (a†)2(a^{\dagger})^2 terms change occupation by two and have zero expectation. The remaining terms give (n+1)+n=2n+1(n+1)+n=2n+1. Thus the mean of a+a†a+a^{\dagger} is zero, but its variance is positive, including in the vacuum. Multiplying by the mode’s coordinate normalization restores physical units.

Take it with you

  • Field modes quantize as oscillators with a precisely stated continuum normalization.
  • Free particles are mode excitations; a classical wave is associated with different quantum-state structure.
  • Vacuum subtraction fixes excitation energies without resolving every question about vacuum energy.

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