Canonical quantization and the free scalar

The free real scalar is the simplest relativistic field whose quantization can be carried from an action all the way to particles and propagators without an approximation. The decisive step is not memorizing a mode expansion: it is choosing its normalization so that the oscillator algebra reproduces the equal-time field commutator. That same choice fixes the Hamiltonian, the one-particle norm, and the residue of the propagator.

Required background. You should be comfortable with Fourier transforms and delta distributions, harmonic-oscillator commutators, and the classical Klein–Gordon equation. Review quantum fields, states, and observables if the difference between an operator algebra, a state, and a particle excitation is not yet clear; use the readiness diagnostics if the Fourier or quantum-mechanics steps need repair.

The Klein–Gordon field is a continuum of oscillators

Consider a real field $\phi$ of mass $m>0$ on $d$-dimensional Minkowski spacetime. With the site’s $(+,-,\ldots,-)$ metric, its action is

$$S[\phi]=\frac12\int \mathrm d^d x\, \left(\partial_\mu\phi\,\partial^\mu\phi-m^2\phi^2\right).$$

Varying the action and discarding the boundary term gives

$$(\Box+m^2)\phi=0, \qquad \pi(x)\equiv\frac{\partial\mathcal L}{\partial\dot\phi(x)}=\dot\phi(x).$$

The Hamiltonian is

$$H=\frac12\int\mathrm d^{d-1}\mathbf x\, \left[\pi^2+(\boldsymbol\nabla\phi)^2+m^2\phi^2\right].$$

Write $n=d-1$, $E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}$, and $\int_{\mathbf p}\equiv \int\mathrm d^n\mathbf p/(2\pi)^n$. Each spatial Fourier mode then obeys $\ddot\phi_{\mathbf p}+E_{\mathbf p}^2\phi_{\mathbf p}=0$: the free field is a continuum of oscillators, one for each momentum. Reality pairs the $\mathbf p$ and $-\mathbf p$ coefficients rather than supplying two independent real fields.

The equal-time algebra fixes the mode normalization

Canonical quantization promotes the field and momentum to operator-valued distributions with

$$\begin{aligned} [\phi(t,\mathbf x),\pi(t,\mathbf y)] &=i\delta^{(n)}(\mathbf x-\mathbf y),\\ [\phi(t,\mathbf x),\phi(t,\mathbf y)] &=[\pi(t,\mathbf x),\pi(t,\mathbf y)]=0. \end{aligned}$$

These equations become ordinary operator statements after smearing in $\mathbf x$ and $\mathbf y$. Choose oscillator operators satisfying

$$[a(\mathbf p),a^\dagger(\mathbf q)] =(2\pi)^n\delta^{(n)}(\mathbf p-\mathbf q), \qquad [a,a]=[a^\dagger,a^\dagger]=0,$$

and start with an unknown real, even coefficient $c_{\mathbf p}$:

$$\phi(x)=\int_{\mathbf p}c_{\mathbf p} \left[a(\mathbf p)e^{-ip\cdot x} +a^\dagger(\mathbf p)e^{ip\cdot x}\right], \qquad p^0=E_{\mathbf p}.$$

Differentiation gives

$$\pi(x)=-i\int_{\mathbf p}E_{\mathbf p}c_{\mathbf p} \left[a(\mathbf p)e^{-ip\cdot x} -a^\dagger(\mathbf p)e^{ip\cdot x}\right].$$

At equal times, the two nonzero cross-commutators give

$$\begin{aligned} [\phi(t,\mathbf x),\pi(t,\mathbf y)] &=i\int_{\mathbf p}E_{\mathbf p}c_{\mathbf p}^{2} \left[e^{i\mathbf p\cdot(\mathbf x-\mathbf y)} +e^{-i\mathbf p\cdot(\mathbf x-\mathbf y)}\right]\\ &=i\int_{\mathbf p}2E_{\mathbf p}c_{\mathbf p}^{2} e^{i\mathbf p\cdot(\mathbf x-\mathbf y)}. \end{aligned}$$

The delta function is recovered if and only if $2E_{\mathbf p}c_{\mathbf p}^{2}=1$. Taking the positive square root fixes the standard expansion:

$$\boxed{ \phi(x)=\int_{\mathbf p}\frac{1}{\sqrt{2E_{\mathbf p}}} \left[a(\mathbf p)e^{-ip\cdot x} +a^\dagger(\mathbf p)e^{ip\cdot x}\right]. }$$

The equal-time $[\phi,\phi]$ and $[\pi,\pi]$ commutators vanish because their remaining integrands are odd under $\mathbf p\mapsto-\mathbf p$. Thus the mode expansion and oscillator algebra reproduce the complete canonical field algebra. The same normalization is developed in Schwartz 2014, §§ 2.2–2.3, pp. 17–26 and Weinberg 1995, § 5.2, pp. 201–206.

The vacuum generates a positive-energy Fock space

Substitution into $H$ and use of $E_{\mathbf p}^2=\mathbf p^2+m^2$ gives

$$H=\frac12\int_{\mathbf p}E_{\mathbf p} \left[a^\dagger(\mathbf p)a(\mathbf p) +a(\mathbf p)a^\dagger(\mathbf p)\right] =:H:+E_0,$$

where

$$:H:\equiv\int_{\mathbf p}E_{\mathbf p} a^\dagger(\mathbf p)a(\mathbf p), \qquad E_0=\frac12(2\pi)^n\delta^{(n)}(0) \int_{\mathbf p}E_{\mathbf p}.$$

$E_0$ is the divergent infinite-volume zero-point term. A finite box and a momentum cutoff turn it into an ordinary half-quantum sum before either regulator is removed. Normal ordering subtracts this constant relative to the chosen free vacuum; it is not a general prescription for gravitational vacuum energy or interacting composite operators.

Select the Poincaré-invariant free vacuum by $a(\mathbf p)|0\rangle=0$. Since

$$[:H:,a^\dagger(\mathbf p)]=E_{\mathbf p}a^\dagger(\mathbf p),$$

each creation operator adds positive energy $E_{\mathbf p}$. The generalized one-particle state

$$|\mathbf p\rangle_{\mathrm{rel}} =\sqrt{2E_{\mathbf p}}\,a^\dagger(\mathbf p)|0\rangle$$

has Lorentz-invariant normalization

$${}_{\mathrm{rel}}\langle\mathbf p|\mathbf q\rangle_{\mathrm{rel}} =2E_{\mathbf p}(2\pi)^n\delta^{(n)}(\mathbf p-\mathbf q).$$

Plane-wave states are delta-normalized, not finite-norm vectors. Physical one-particle vectors are wave packets, and multiparticle states are the symmetrized products expected for bosons.

The two-point function carries spectrum and boundary data

The positive-frequency vacuum correlator follows immediately:

$$W^+(x-y)\equiv\langle0|\phi(x)\phi(y)|0\rangle =\int_{\mathbf p}\frac{e^{-ip\cdot(x-y)}}{2E_{\mathbf p}}.$$

Time ordering joins the positive- and negative-frequency boundary values:

$$\begin{aligned} D_F(x-y) &\equiv\langle0|\mathrm T\phi(x)\phi(y)|0\rangle\\ &=\theta(x^0-y^0)W^+(x-y) +\theta(y^0-x^0)W^+(y-x)\\ &=\int\frac{\mathrm d^d p}{(2\pi)^d} \frac{i\,e^{-ip\cdot(x-y)}}{p^2-m^2+i0}. \end{aligned}$$

The pole location records the mass spectrum, its residue reflects the field normalization, and $+i0$ specifies the vacuum Feynman boundary value. Acting distributionally gives the useful sign check

$$(\Box_x+m^2)D_F(x-y)=-i\delta^{(d)}(x-y).$$

The Feynman function need not vanish outside the light cone. Causality is instead tested by the field commutator

$$[\phi(x),\phi(y)] =\int_{\mathbf p}\frac{1}{2E_{\mathbf p}} \left[e^{-ip\cdot(x-y)}-e^{ip\cdot(x-y)}\right].$$

For spacelike $x-y$, choose an inertial frame in which $x^0=y^0$; the two terms then cancel after $\mathbf p\mapsto-\mathbf p$. Lorentz covariance therefore makes the commutator vanish at every spacelike separation. Its equal-time initial data also recover the canonical algebra:

$$[\phi(t,\mathbf x),\phi(t,\mathbf y)]=0, \qquad \partial_{x^0}[\phi(x),\phi(y)]\big|_{x^0=y^0} =-i\delta^{(n)}(\mathbf x-\mathbf y).$$

These distinctions among Wightman, Feynman, and commutator distributions are treated in Schwartz 2014, § 6.2, pp. 75–77 and Weinberg 1995, § 6.2, pp. 274–279.

What the construction establishes—and where it stops

This is an exact construction for a massive free scalar in Minkowski spacetime and the selected vacuum representation. It demonstrates how local commutators, positive energy, particle states, and time-ordered correlators fit together in one model. The more detailed treatments of real-scalar quantization and scalar propagators develop the regulator and distributional details.

The construction does not make particle number representation-independent. Interactions replace the free pole by a renormalized spectral structure; curved or time-dependent backgrounds may lack a preferred positive-frequency split; thermal states are not the vacuum; and the massless limit can be singular because of zero modes or long-distance behavior. Consult massless scalars and infrared limits before setting $m=0$ in finite volume or low dimension.

Common pitfalls

Mixing normalization conventions. With the oscillator commutator used above, the field coefficient is $1/\sqrt{2E_{\mathbf p}}$. If the commutator instead contains $2E_{\mathbf p}$, the field coefficient must change too; moving only one factor breaks the equal-time commutator.

Treating the Feynman function as a causal response. $D_F$ is a time-ordered vacuum correlator and is generally nonzero at spacelike separation. The commutator, or retarded propagator built from it, carries the causal support statement.

Calling a plane wave a normalizable state. $a^\dagger(\mathbf p)|0\rangle$ is distributionally normalized. Use a wave packet or a finite box whenever a finite norm or probability is required.

Exercises

1. Normalize a one-particle wave packet. In the oscillator convention used above, let

$$|f\rangle=\int_{\mathbf p}f(\mathbf p) a^\dagger(\mathbf p)|0\rangle.$$

Find its norm and its energy expectation when $\langle f|f\rangle=1$.

Solution

Using the oscillator commutator once gives

$$\langle f|f\rangle =\int_{\mathbf p}|f(\mathbf p)|^2.$$

The commutator $[:H:,a^\dagger(\mathbf p)] =E_{\mathbf p}a^\dagger(\mathbf p)$ then gives

$$\langle f|:H:|f\rangle =\int_{\mathbf p}E_{\mathbf p}|f(\mathbf p)|^2.$$

Thus a normalized packet has mean energy equal to the probability-weighted average of the positive on-shell energies. Its spread in momentum makes both the norm and the energy finite.

2. Translate to relativistically normalized oscillators. Define $b(\mathbf p)=\sqrt{2E_{\mathbf p}}\,a(\mathbf p)$. Find the $b$ commutator and rewrite $\phi(x)$. Verify that no physical normalization changes.

Solution

Because the energy factors are ordinary functions of momentum,

$$[b(\mathbf p),b^\dagger(\mathbf q)] =2E_{\mathbf p}(2\pi)^n\delta^{(n)}(\mathbf p-\mathbf q).$$

Substituting $a=b/\sqrt{2E}$ into the field gives

$$\phi(x)=\int_{\mathbf p}\frac{1}{2E_{\mathbf p}} \left[b(\mathbf p)e^{-ip\cdot x} +b^\dagger(\mathbf p)e^{ip\cdot x}\right].$$

The extra $2E$ in the commutator precisely cancels the extra $2E$ in the two mode coefficients when $[\phi,\pi]$ is computed. The field, propagator, and one-particle norm are unchanged; only the allocation of normalization factors has changed.

Previous: Quantum fields, states, and observables. Next: Functional integrals and correlators, which reconstructs the same free two-point function from a regulated Gaussian integral.

References

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge: Cambridge University Press, 2014. DOI.
• Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge: Cambridge University Press, 1995. DOI.