Free Fields, Creation Operators, and Fock Space

A mode excitation is interpretable as a particle when the theory supplies a positive-norm one-particle space, a stable vacuum, and a physically preferred positive-energy evolution. Minkowski spacetime provides these structures through time-translation symmetry. This page derives the ladder algebra, Fock states, energy–momentum eigenvalues, relativistic normalization, and spacetime field commutator that make the interpretation precise. The oscillator algebra alone does not make it background independent.

Required background. Quantum Mechanics supplies Hilbert spaces, observables, and oscillator quantization; Classical Field Theory supplies the field action and canonical momentum; Special Relativity supplies Minkowski causality and four-vector transformations. Helpful background. Calculus and Linear Algebra supplies basis changes and inner products; Waves, Green Functions, and Geometric Optics supplies normal modes and Fourier superposition. The relativistic mass-shell relation and Fourier-mode dispersion are the two calculations used most directly below.

From modes to ladder operators

Start from the classical canonical variables developed on Classical Field Modes and Canonical Variables. Canonical quantization promotes them to operator-valued distributions obeying

$$[\hat\phi(t,\mathbf x),\hat\pi(t,\mathbf y)] =i\delta^{(3)}(\mathbf x-\mathbf y), \qquad [\hat\phi,\hat\phi]=[\hat\pi,\hat\pi]=0.$$

Operators at a point are singular idealizations. Smearing them against test functions produces genuine operators on a suitable domain.

To see where the ladder operators and their normalization come from, first put the field in a finite box and expand it in real orthonormal spatial modes. For each mode $A$ with $\omega_A\gt 0$, the mode operators obey

$$[\hat q_A,\hat p_B]=i\delta_{AB},$$

and the oscillator variables are

$$\hat a_A =\sqrt{\frac{\omega_A}{2}}\,\hat q_A +\frac{i}{\sqrt{2\omega_A}}\,\hat p_A, \qquad \hat a_A^\dagger =\sqrt{\frac{\omega_A}{2}}\,\hat q_A -\frac{i}{\sqrt{2\omega_A}}\,\hat p_A.$$

They satisfy

$$[\hat a_A,\hat a_B^\dagger]=\delta_{AB}, \qquad [\hat a_A,\hat a_B]=0.$$

Define the occupation-number operator $\hat N_A=\hat a_A^\dagger\hat a_A$. Then

$$\hat H=\sum_A\omega_A\left(\hat N_A+\frac12\right), \qquad [\hat N_A,\hat a_B^\dagger] =\delta_{AB}\hat a_B^\dagger, \qquad [\hat H,\hat a_B^\dagger] =\omega_B\hat a_B^\dagger.$$

If $\hat a_A|0\rangle=0$, the normalized states

$$|n_A\rangle =\frac{(\hat a_A^\dagger)^{n_A}}{\sqrt{n_A!}}|0\rangle$$

obey $\hat N_A|n_A\rangle=n_A|n_A\rangle$. One application of $\hat a_A^\dagger$ therefore raises both occupation number and energy by one quantum. The ladder coefficients are fixed simultaneously by the canonical commutator, the oscillator Hamiltonian, and state normalization. A genuine $\omega_A=0$ mode is a free canonical degree of freedom rather than an oscillator and must be handled separately. Tong 2006–2007, §2.1.1, printed pp. 21–23, PDF gives the corresponding oscillator construction.

For a periodic box of volume $V$, the continuum limit uses

$$\frac1V\sum_{\mathbf k}\longrightarrow \int\frac{d^3k}{(2\pi)^3}, \qquad \hat a_{\mathbf k}^{\mathrm{cont}} =\sqrt V\,\hat a_{\mathbf k}^{\mathrm{box}}, \qquad V\delta_{\mathbf k\mathbf q} \longrightarrow(2\pi)^3\delta^{(3)}(\mathbf k-\mathbf q).$$

Pairing the real box modes into complex plane waves then carries the oscillator coefficient $1/\sqrt{2\omega_{\mathbf k}}$ into the continuum expansion. With $\omega_{\mathbf k}=\sqrt{\mathbf k^2+m^2}$, the real field is therefore

$$\hat\phi(x)= \int\frac{d^3k}{(2\pi)^3\sqrt{2\omega_{\mathbf k}}} \left[ \hat a_{\mathbf k}e^{-i\omega_{\mathbf k}t+i\mathbf k\cdot\mathbf x} +\hat a_{\mathbf k}^{\dagger} e^{i\omega_{\mathbf k}t-i\mathbf k\cdot\mathbf x} \right].$$

Reality is the Hermiticity condition $\hat\phi^\dagger=\hat\phi$; it does not identify $\hat a_{\mathbf k}$ with $\hat a^\dagger_{\mathbf k}$. The canonical commutator is equivalent to

$$[\hat a_{\mathbf k},\hat a^\dagger_{\mathbf q}] =(2\pi)^3\delta^{(3)}(\mathbf k-\mathbf q), \qquad [\hat a_{\mathbf k},\hat a_{\mathbf q}]=0.$$

Substitution into the canonical Hamiltonian gives, before reordering,

$$\hat H =\frac12\int\frac{d^3k}{(2\pi)^3}\, \omega_{\mathbf k} \left( \hat a_{\mathbf k}^\dagger\hat a_{\mathbf k} +\hat a_{\mathbf k}\hat a_{\mathbf k}^\dagger \right).$$

The $\hat a\hat a$ and $\hat a^\dagger\hat a^\dagger$ terms cancel because their coefficient is $-\omega_{\mathbf k}^2+\mathbf k^2+m^2=0$. Likewise, substituting into $\hat{\mathbf P}=-\int d^3x\,\hat\pi\boldsymbol\nabla\hat\phi$ cancels the equal-creation and equal-annihilation terms pairwise under $\mathbf k\mapsto-\mathbf k$. The remaining mixed terms give the free translation generators

$$\hat H=\int\frac{d^3k}{(2\pi)^3}\, \omega_{\mathbf k}\hat a^\dagger_{\mathbf k}\hat a_{\mathbf k} +E_0, \qquad \hat{\mathbf P} =\int\frac{d^3k}{(2\pi)^3}\, \mathbf k\,\hat a^\dagger_{\mathbf k}\hat a_{\mathbf k},$$

where, formally,

$$E_0=\frac{V}{2}\int\frac{d^3k}{(2\pi)^3}\,\omega_{\mathbf k}, \qquad V=(2\pi)^3\delta^{(3)}(\mathbf0).$$

Apart from the vacuum constant, these expressions give

$$[\hat H,\hat a_{\mathbf k}^\dagger] =\omega_{\mathbf k}\hat a_{\mathbf k}^\dagger, \qquad [\hat{\mathbf P},\hat a_{\mathbf k}^\dagger] =\mathbf k\,\hat a_{\mathbf k}^\dagger.$$

Thus $\hat a^\dagger_{\mathbf k}$ creates one quantum of four-momentum $k^\mu=(\omega_{\mathbf k},\mathbf k)$. The formal zero-point term is divergent and proportional to spatial volume. In semiclassical gravity the source is $\langle\hat T_{\mu\nu}^{\mathrm{ren}}\rangle$, not the unregulated symbol $E_0$; in gravitational effective field theory, vacuum terms renormalize the cosmological coefficient. Normal ordering can define nongravitational energy differences relative to the chosen Minkowski vacuum, but it is state- and representation-dependent and does not solve the cosmological-constant problem.

The spacetime commutator and microcausality

The ladder algebra determines more than the equal-time canonical relation. Let $\Delta t=t_x-t_y$ and $\Delta\mathbf x=\mathbf x-\mathbf y$. Direct substitution of the mode expansion gives the c-number distribution

$$\begin{aligned} C(x-y) &\equiv[\hat\phi(x),\hat\phi(y)]\\ &=\int\frac{d^3k}{(2\pi)^3\,2\omega_{\mathbf k}} \left[ e^{-i\omega_{\mathbf k}\Delta t+i\mathbf k\cdot\Delta\mathbf x} -e^{i\omega_{\mathbf k}\Delta t-i\mathbf k\cdot\Delta\mathbf x} \right]. \end{aligned}$$

Every integrand lies on the mass shell, so

$$(\Box_x-m^2)C(x-y)=0$$

as a distribution. Its equal-time data are

$$C(0,\Delta\mathbf x)=0, \qquad \left.\partial_{t_x}C(x-y)\right|_{\Delta t=0} =-i\delta^{(3)}(\Delta\mathbf x).$$

The first identity follows by changing $\mathbf k\mapsto-\mathbf k$ in the second term; the second follows by differentiating before setting $\Delta t=0$. Since $\hat\pi(y)=\partial_{t_y}\hat\phi(y)$ and the commutator depends on the time difference, $[\hat\phi(x),\hat\pi(y)]=\partial_{t_y}C(x-y)=-\partial_{\Delta t}C(x-y)$. These data therefore recover

$$[\hat\phi(t,\mathbf x),\hat\pi(t,\mathbf y)] =i\delta^{(3)}(\mathbf x-\mathbf y).$$

If $x-y$ is spacelike, a proper orthochronous Lorentz transformation sends the separation to an equal-time one. A scalar field transforms covariantly, and the equal-time commutator above vanishes. Therefore

$$[\hat\phi(x),\hat\phi(y)]=0 \qquad\text{when}\qquad (x-y)^2\gt 0,$$

which is microcausality in the site’s mostly-plus convention. The commutator can be singular on the light cone without violating this statement. Tong 2006–2007, §2.6.1, printed pp. 36–38, PDF derives the same causal support from the free-field mode algebra.

Fock space and normalization

The Minkowski vacuum is defined by

$$\hat a_{\mathbf k}|0\rangle=0 \quad\text{for every }\mathbf k.$$

Repeated creation operators generate the symmetric Fock space

$$\mathcal F(\mathcal H_1) =\mathbb C\oplus\mathcal H_1 \oplus\operatorname{Sym}^2\mathcal H_1\oplus\cdots,$$

where $\mathcal H_1$ is the positive-frequency one-particle Hilbert space. With

$$|\mathbf k\rangle =\sqrt{2\omega_{\mathbf k}}\, \hat a^\dagger_{\mathbf k}|0\rangle,$$

the relativistic normalization is

$$\langle\mathbf q|\mathbf k\rangle =2\omega_{\mathbf k}(2\pi)^3 \delta^{(3)}(\mathbf q-\mathbf k).$$

Indeed, the ladder commutator first gives a factor $\sqrt{2\omega_{\mathbf q}2\omega_{\mathbf k}}$; on the support of the delta distribution this is $2\omega_{\mathbf k}$. With the vacuum translation eigenvalue subtracted in the usual Minkowski convention,

$$\hat P^\mu|\mathbf k\rangle =k^\mu|\mathbf k\rangle.$$

The corresponding invariant on-shell measure is

$$d\Pi_{\mathbf k} \equiv\frac{d^3k}{(2\pi)^3\,2\omega_{\mathbf k}} =\frac{d^4k}{(2\pi)^3} \,\theta(k^0)\delta(k^2+m^2).$$

The four-dimensional expression makes Lorentz invariance under the proper orthochronous group manifest. Momentum kets are distributions, not normalizable vectors. A genuine one-particle state is a wave packet

$$|f\rangle =\int d\Pi_{\mathbf k}\,f(\mathbf k)|\mathbf k\rangle, \qquad \langle f|f\rangle =\int d\Pi_{\mathbf k}\,|f(\mathbf k)|^2.$$

Other normalizations are equally valid if states, operators, delta functions, and phase-space factors are changed together. Tong 2006–2007, §§2.2–2.4, printed pp. 24–33, PDF follows this normalization through the vacuum, multiparticle states, and particle interpretation.

What makes the interpretation physical?

Minkowski time translations have a conserved generator bounded below, and the proper orthochronous Poincaré group preserves the future mass shell. Thus inertial observers agree on which nonzero on-shell momenta have positive energy, even though they assign different energies. The positive-frequency Klein–Gordon modes form $\mathcal H_1$; symmetrized tensor powers then describe arbitrary boson number.

The argument can fail in several ways:

• a time-dependent background can mix positive- and negative-frequency modes;
• inequivalent choices of time evolution can define different annihilation operators;
• an unstable quadratic Hamiltonian has no stationary lowest-energy state;
• interacting theories need not possess sharp one-particle states for unstable excitations; and
• localized detectors respond along worldlines and need not measure a global mode-number operator.

For a finite collection of oscillators, two linear canonical mode choices are unitarily related. For infinitely many modes, implementability can fail. In the common case where a discrete Bogoliubov transformation is diagonal in the mode label, the condition is $\sum_{\mathbf k}|\beta_{\mathbf k}|^2\lt \infty$; otherwise the transformed vacuum would contain an infinite expected number of the original quanta. Jacobson 2004, §5.3.2, printed p. 27, PDF derives this criterion. Wald 1995, printed pp. 6–10, especially pp. 8–9, PDF constructs the one-particle and Fock spaces and explains why a generic nonstationary spacetime need not select a preferred representation. The curved-spacetime consequences are developed on Fock Space and Field Operators, Positive Frequency and Particle Number, and Why Curved Spacetime Has No Preferred Vacuum.

Exercises

Recover the canonical relation

Insert the mode expansion into the equal-time commutator and recover $i\delta^{(3)}(\mathbf x-\mathbf y)$.

Solution

The conjugate momentum is $\hat\pi=\partial_t\hat\phi$. Only the $[\hat a,\hat a^\dagger]$ terms survive. At equal time they give

$$[\hat\phi(t,\mathbf x),\hat\pi(t,\mathbf y)] =\frac{i}{2}\int\frac{d^3k}{(2\pi)^3} \left[ e^{i\mathbf k\cdot(\mathbf x-\mathbf y)} +e^{-i\mathbf k\cdot(\mathbf x-\mathbf y)} \right].$$

Changing $\mathbf k\mapsto-\mathbf k$ in the second term makes the two integrals equal, leaving $i\delta^{(3)}(\mathbf x-\mathbf y)$. The factor $1/\sqrt{2\omega_{\mathbf k}}$ is therefore fixed by the canonical normalization.

Check a one-particle state

Using $|\mathbf k\rangle=\sqrt{2\omega_{\mathbf k}}\,\hat a_{\mathbf k}^\dagger|0\rangle$, derive its four-momentum eigenvalue and relativistic delta-function normalization. Then verify the norm of the wave packet $|f\rangle=\int d\Pi_{\mathbf k}\,f(\mathbf k)|\mathbf k\rangle$.

Solution

Let the vacuum translation eigenvalue be zero after the conventional Minkowski subtraction. The commutator with the translation generator gives

$$\begin{aligned} \hat P^\mu|\mathbf k\rangle &=\sqrt{2\omega_{\mathbf k}} \left( [\hat P^\mu,\hat a_{\mathbf k}^\dagger] +\hat a_{\mathbf k}^\dagger\hat P^\mu \right)|0\rangle\\ &=k^\mu|\mathbf k\rangle. \end{aligned}$$

The ladder algebra gives

$$\langle\mathbf q|\mathbf k\rangle =\sqrt{2\omega_{\mathbf q}2\omega_{\mathbf k}} (2\pi)^3\delta^{(3)}(\mathbf q-\mathbf k) =2\omega_{\mathbf k}(2\pi)^3 \delta^{(3)}(\mathbf q-\mathbf k).$$

Finally, substituting the two wave-packet integrals and using this delta function cancels one invariant measure:

$$\langle f|f\rangle =\int d\Pi_{\mathbf k}\,|f(\mathbf k)|^2.$$

The result is finite when $f$ is square-integrable with respect to $d\Pi_{\mathbf k}$.

Next, Correlation Functions and Propagators replaces mode occupation by two-point distributions that diagnose fluctuations, propagation, and response.

References

• Ted Jacobson, “Introduction to Quantum Fields in Curved Spacetime and the Hawking Effect,” arXiv:gr-qc/0308048v3 (2004), §5.3.2, printed p. 27. Author-manuscript PDF.
• David Tong, Quantum Field Theory, University of Cambridge Part III lecture notes, Michaelmas Terms 2006 and 2007, §§2.1–2.6.1, printed pp. 21–38. Official author PDF.
• Robert M. Wald, “Quantum Field Theory in Curved Spacetime,” arXiv:gr-qc/9509057 (1995), printed pp. 6–10. Author-manuscript PDF.