advanced33 min studyLesson 652 in Physics

Canonical field commutators and Hamiltonians

Recover the free scalar dynamics from a canonical bracket and a positive quadratic Hamiltonian while distinguishing a formal density from its regulated operator.

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What you’ll understand

  • Obtain canonical momentum and Hamiltonian density from a chosen signature.
  • Use smeared canonical commutators without treating δ³(0) as a number.
  • Check the Heisenberg equation and the energy of a finite-particle state.

Canonical quantization connects the equal-time algebra of a field to the generator of its time evolution. For a real Klein–Gordon field, the canonical momentum is the time derivative and the Hamiltonian is a sum of three positive quadratic contributions. Promoting these expressions to operators introduces products at coincident points, so an algebraic derivation needs a regulator or a suitable definition of those products. The free theory provides a controlled construction against which later interacting prescriptions can be checked.

The Legendre transform fixes the signs

Retain natural units and the plus, minus, minus, minus metric. Expanding the Lorentz contraction gives a positive time-derivative square and negative spatial-gradient square in the Lagrangian. Differentiating with respect to the time derivative gives the canonical momentum. The Legendre transform reverses the spatial and mass contributions to produce a positive Hamiltonian density when mass squared is nonnegative. This sign check is essential before assigning a particle spectrum.

A boundary condition is part of the derivation. Integrations by parts assume periodic boundaries or sufficiently rapid decay, so no unaccounted surface work changes the conserved energy. A finite spatial lattice or truncated mode set gives a concrete quadratic system before continuum limits. Negative mass squared would change the stability of the chosen background rather than merely rename the oscillator frequency; the positive-mass free Fock construction used here would no longer describe that background in the same way.

L=12ϕ˙2−12(∇ϕ)2−12m2ϕ2,π=ϕ˙,H=12π2+12(∇ϕ)2+12m2ϕ2\mathcal L=\frac12\dot\phi^2-\frac12(\nabla\phi)^2-\frac12m^2\phi^2,\quad \pi=\dot\phi,\quad \mathcal H=\frac12\pi^2+\frac12(\nabla\phi)^2+\frac12m^2\phi^2
In 3+1 dimensions with ℏ = c = 1, φ has energy dimension one, π dimension two, and both densities dimension four. The integrated H has energy dimension one. We assume m > 0 and suitable spatial boundary conditions.↔ Scroll wide equations horizontally.

An equal-time delta is a kernel

Canonical field and momentum at the same time obey a commutator proportional to a spatial delta function, while field–field and momentum–momentum commutators vanish. This is not an instruction to assign an infinite value to a pointwise uncertainty. Smearing over space turns the delta into an overlap of test functions. Operators smeared over disjoint spatial supports have zero canonical overlap, and the degree of overlap determines the noncommutativity for intersecting probes.

At a regulator scale the delta kernel is represented by the identity on the retained modes or by a lattice Kronecker delta with its volume factor. Truncation generally changes exact localization: a finite Fourier sum cannot reproduce a compact spatial delta. The canonical relation should therefore be checked at the level appropriate to the regulator. Passing to the continuum means convergence of smeared expressions on a common domain, rather than substituting a divergent coincidence value into ordinary arithmetic.

[ϕ(t,x),π(t,y)]=iδ3(x−y),[ϕ(f),π(g)]=i∫d3x f(x)g(x)[\phi(t,\mathbf x),\pi(t,\mathbf y)]=i\delta^3(\mathbf x-\mathbf y),\qquad [\phi(f),\pi(g)]=i\int d^3x\,f(\mathbf x)g(\mathbf x)
Here φ(f) = ∫d³x fφ and π(g) = ∫d³x gπ at one fixed time, with real smooth compactly supported test functions. Their dimensions depend on the chosen test functions. The relation holds on an appropriate dense domain.↔ Scroll wide equations horizontally.

The generator reproduces the wave equation

The Heisenberg rule multiplies a commutator with the Hamiltonian by positive imaginary one. Using the canonical bracket makes the field derivative equal its momentum. Applying the rule again gives the spatial Laplacian minus mass squared times the field, after integration by parts. Combining the two equations recovers the Klein–Gordon equation and checks both the canonical normalization and the Hamiltonian's relative coefficients.

In free momentum modes the normal-ordered Hamiltonian counts quanta with positive energies. Creation raises energy by the corresponding positive frequency, and the vacuum is annihilated by every annihilation operator. Adding a constant to the Hamiltonian changes neither these commutators nor energy gaps. An interacting local polynomial may formally be added to the density, but coincident products then require renormalization and do not automatically define a self-adjoint continuum Hamiltonian on the free Fock space.

ϕ˙=i[H,ϕ]=π,π˙=i[H,π]=∇2ϕ−m2ϕ,:H:=∫d3p(2π)3Epap†ap\dot\phi=i[H,\phi]=\pi,\quad \dot\pi=i[H,\pi]=\nabla^2\phi-m^2\phi,\quad :H:=\int\frac{d^3p}{(2\pi)^3}E_{\mathbf p}a_{\mathbf p}^{\dagger}a_{\mathbf p}
The free Hamiltonian is defined after vacuum subtraction, for example first in a box with a mode cutoff. The displayed continuum expressions describe operators on the free-field domain; an arbitrary interacting density is not established by the same calculation.↔ Scroll wide equations horizontally.

Count energy in two retained modes

A regulated free scalar theory retains two independent oscillator modes with energies 2 and 5 in natural units. The state has occupation numbers 3 and 1. Compute its normal-ordered energy, its unsubtracted oscillator energy, and the energy gained by adding one quantum to the second mode.

  1. The normal-ordered energy is 3(2)+1(5)=113(2)+1(5)=11. Occupations count quanta in orthogonal modes.

  2. The two zero-point contributions add (2+5)/2=3.5(2+5)/2=3.5, giving unsubtracted energy 14.5.

  3. Adding one second-mode quantum raises either energy by 5. Subtracting the vacuum constant changes the zero but leaves that gap unchanged.

The result

The two energy conventions give 11 and 14.5; the same added quantum costs 5 in both.

Explore the model

Interactive lab

Interactive lab · function

Differentiate a quadratic contribution

Reset and set BOTH Coefficient a to 0.5 and Position x to 1; predict slope 1. Then Reset and set BOTH a to 0.5 and x to 2; predict slope 2. This exact scalar derivative illustrates a quadratic energy gradient. It has no canonical operator bracket, field commutator, regulator, or Hamiltonian evolution.

Differentiate a quadratic contributionThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forA quadratic energy has a linear derivative, but the canonical algebra supplies its dynamical meaning.

Check your understanding

With the Heisenberg convention dO/dt = i[H,O], which canonical identity gives the scalar velocity?

Try it yourself

At equal time, real test functions f and g have ∫fg d³x = 3/4 in declared test-function units. Let F = 2φ(f) and G = 4π(g). Find [F,G]. What happens if the supports of f and g are disjoint?

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

Give me a hint

Pull numerical factors outside the bracket and integrate the delta kernel.

Compare with the solution

The bracket is [F,G]=8i∫fg d3x=6i[F,G]=8i\int fg\,d^3x=6i in the corresponding product units. If supports are disjoint the overlap integral vanishes and the bracket is zero. This does not set the unsmeared delta at coincidence to a finite value; it computes a well-defined overlap for the specified probes.

Take it with you

  • The Legendre transform and metric signature determine the free Hamiltonian’s relative signs.
  • Smeared canonical brackets replace a delta kernel with a finite spatial overlap.
  • Free-field dynamics and energy gaps survive vacuum subtraction, while interactions require additional operator definitions.

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