Harmonic Oscillator to Fields

A free bosonic field is the continuum analogue of a system of coupled oscillators: diagonalize its quadratic Hamiltonian into normal modes, quantize each independent mode, and obtain one oscillator algebra per mode. Occupying a field mode then has the particle interpretation appropriate to that free field.

The word free is essential. Interactions couple modes, gauge fields contain constraints, and fermionic fields use anticommutation rather than oscillator commutation relations.

Quantum-Mechanics Starting Point

For one oscillator of mass $M$ and frequency $\omega$,

$$H =\frac{p^2}{2M} +\frac12M\omega^2q^2, \qquad [q,p]=i\hbar I.$$

Define

$$\begin{aligned} a &= \sqrt{\frac{M\omega}{2\hbar}}\,q +\frac{i}{\sqrt{2M\hbar\omega}}\,p, \\ a^\dagger &= \sqrt{\frac{M\omega}{2\hbar}}\,q -\frac{i}{\sqrt{2M\hbar\omega}}\,p. \end{aligned}$$

Then

$$[a,a^\dagger]=I, \qquad H=\hbar\omega \left( a^\dagger a+\frac12I \right).$$

The number states satisfy

$$a^\dagger a\lvert n\rangle =n\lvert n\rangle, \qquad E_n=\hbar\omega \left(n+\frac12\right).$$

For a material particle trapped in this potential, $a^\dagger$ raises the particle’s motional excitation. It does not create another material particle. The particle interpretation changes only when oscillator modes are the normal modes of a quantized field or many-body medium.

Classical Field as Normal Modes

Set $\hbar=c=1$ from this section onward. Consider a real scalar field in a finite spatial region with boundary conditions that make

$$\mathcal K=-\nabla^2+m^2$$

a positive self-adjoint spatial operator. The classical Hamiltonian is

$$H =\frac12 \int d^3x\, \left[ \pi^2 +(\nabla\phi)^2 +m^2\phi^2 \right],$$

where $\pi=\dot\phi$ for the standard scalar Lagrangian.

Choose a real orthonormal eigenbasis:

$$\mathcal K f_\alpha(\boldsymbol x) =\omega_\alpha^2 f_\alpha(\boldsymbol x), \qquad \int d^3x\, f_\alpha(\boldsymbol x)f_\beta(\boldsymbol x) =\delta_{\alpha\beta}.$$

Expand

$$\phi(t,\boldsymbol x) =\sum_\alpha q_\alpha(t)f_\alpha(\boldsymbol x), \qquad \pi(t,\boldsymbol x) =\sum_\alpha p_\alpha(t)f_\alpha(\boldsymbol x).$$

Integration by parts, together with the stated boundary conditions, gives

$$H =\frac12 \sum_\alpha \left( p_\alpha^2+\omega_\alpha^2q_\alpha^2 \right).$$

The normal modes are therefore independent classical oscillators. Using real modes avoids a common bookkeeping trap: in a plane-wave expansion of a real field, the coefficients at $\boldsymbol k$ and $-\boldsymbol k$ are complex conjugates and are not independent coordinates.

Quantizing the Modes

Impose the equal-time canonical relation

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

Orthonormality implies

$$[q_\alpha,p_\beta] =i\delta_{\alpha\beta}I, \qquad [q_\alpha,q_\beta] =[p_\alpha,p_\beta]=0.$$

For each mode define

$$a_\alpha = \sqrt{\frac{\omega_\alpha}{2}}\,q_\alpha +\frac{i}{\sqrt{2\omega_\alpha}}\,p_\alpha, \qquad [a_\alpha,a_\beta^\dagger] =\delta_{\alpha\beta}I.$$

The Hamiltonian becomes

$$H =\sum_\alpha \omega_\alpha \left( a_\alpha^\dagger a_\alpha+\frac12I \right).$$

Each occupation number $n_\alpha=a_\alpha^\dagger a_\alpha$ counts quanta in one normal mode. A many-mode basis state is

$$\lvert n_1,n_2,\ldots\rangle = \prod_\alpha \frac{(a_\alpha^\dagger)^{n_\alpha}} {\sqrt{n_\alpha!}} \lvert0\rangle.$$

The vacuum is characterized by

$$a_\alpha\lvert0\rangle=0 \qquad \text{for every mode }\alpha.$$

Plane-Wave Convention

In infinite Minkowski space, one common real-scalar convention is

$$\begin{aligned} \phi(t,\boldsymbol x) =\int\frac{d^3k}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\boldsymbol k}}} \Big[ &a_{\boldsymbol k} e^{-i\omega_{\boldsymbol k}t +i\boldsymbol k\cdot\boldsymbol x} \\ &+ a_{\boldsymbol k}^\dagger e^{i\omega_{\boldsymbol k}t -i\boldsymbol k\cdot\boldsymbol x} \Big], \end{aligned}$$

where

$$\omega_{\boldsymbol k} =\sqrt{\lVert\boldsymbol k\rVert^2+m^2}.$$

The mode algebra is

$$[a_{\boldsymbol k},a_{\boldsymbol k'}^\dagger] =(2\pi)^3 \delta^{(3)} (\boldsymbol k-\boldsymbol k')I,$$

with all annihilator–annihilator and creator–creator commutators zero.

After separating the formal vacuum contribution,

$$H = \int\frac{d^3k}{(2\pi)^3} \omega_{\boldsymbol k} a_{\boldsymbol k}^\dagger a_{\boldsymbol k} +E_0,$$

where

$$E_0 =\frac{V}{2} \int\frac{d^3k}{(2\pi)^3} \omega_{\boldsymbol k}$$

is the formal infinite-volume zero-point expression. It requires a regulator and a physical renormalization prescription before interpretation.

The one-quantum state

$$\lvert\boldsymbol k\rangle =a_{\boldsymbol k}^\dagger\lvert0\rangle$$

has the noncovariant normalization

$$\langle\boldsymbol k\vert\boldsymbol k'\rangle =(2\pi)^3 \delta^{(3)} (\boldsymbol k-\boldsymbol k').$$

Another common convention rescales states to obtain

$$\langle\boldsymbol k\vert\boldsymbol k'\rangle_{\mathrm{cov}} = 2\omega_{\boldsymbol k}(2\pi)^3 \delta^{(3)} (\boldsymbol k-\boldsymbol k').$$

Mode expansions, commutators, state normalization, and phase-space measures must be translated as one package.

Finite Box and Continuum

For a periodic cube of volume $V=L^3$,

$$\boldsymbol k =\frac{2\pi}{L}\boldsymbol n, \qquad \boldsymbol n\in\mathbb Z^3.$$

The useful conversion rules are

Finite periodic box	Infinite-volume continuum
$\sum_{\boldsymbol k}$	$V\int d^3k/(2\pi)^3$
$\delta_{\boldsymbol k,\boldsymbol k'}$	$(2\pi)^3\delta^{(3)}(\boldsymbol k-\boldsymbol k')/V$
$[a_{\boldsymbol k},a_{\boldsymbol k'}^\dagger]=\delta_{\boldsymbol k,\boldsymbol k'}$	$[a(\boldsymbol k),a^\dagger(\boldsymbol k')]=(2\pi)^3\delta^{(3)}(\boldsymbol k-\boldsymbol k')$
$a(\boldsymbol k)=\sqrt V\,a_{\boldsymbol k}$	$a_{\boldsymbol k}=a(\boldsymbol k)/\sqrt V$

Checking the volume dimension of each row is a reliable way to catch missing factors.

Oscillator-to-Field Dictionary

One oscillator	Free bosonic field
Coordinate $q$	Normal-mode amplitude $q_\alpha$
Momentum $p$	Mode momentum $p_\alpha$
Frequency $\omega$	Dispersion value $\omega_\alpha$ or $\omega_{\boldsymbol k}$
Ground state $\lvert0\rangle$	Joint vacuum annihilated by every $a_\alpha$
Level number $n$	Occupation number of a field mode
$a^\dagger$ raises motional energy	$a_\alpha^\dagger$ creates one free mode quantum
$\frac12\omega$ zero-point energy	One formal zero-point contribution per mode
Coherent state	Classical-like excitation of one mode or a multimode field profile

The analogy is algebraic and dynamical for a quadratic bosonic theory. It does not say that a field quantum is a tiny classical oscillator located at a point.

Examples Beyond a Scalar Field

Cavity photons

Boundary conditions select discrete electromagnetic normal modes. After gauge constraints are handled, each physical polarization mode has an oscillator Hamiltonian. The mode frequency, spatial profile, and polarization are part of the label created by $a_\alpha^\dagger$. See Quantized Electromagnetic Modes.

Phonons

Diagonalizing a harmonic lattice produces normal coordinates and frequencies. Quantizing them gives phonons. They are collective excitations of the lattice, not additional microscopic atoms. See Phonons as Many-Body Excitations.

Complex scalar fields

A complex scalar field has independent particle and antiparticle mode operators, commonly written $a_{\boldsymbol k}$ and $b_{\boldsymbol k}$. The real-field expansion above uses one family because Hermiticity relates its positive- and negative-frequency parts.

Fermionic fields

Free fermionic modes are not bosonic oscillators. Their operators satisfy canonical anticommutation relations, and one mode has occupation zero or one. The broader mode-decomposition strategy survives, but the Hilbert-space and sign structure changes.

Two-Point Function

The mode expansion immediately determines vacuum correlations. For the free real scalar field,

$$\langle0\vert \phi(x)\phi(y) \vert0\rangle = \int\frac{d^3k}{(2\pi)^3} \frac{ e^{-i\omega_{\boldsymbol k}(x^0-y^0) +i\boldsymbol k\cdot(\boldsymbol x-\boldsymbol y)} }{2\omega_{\boldsymbol k}}.$$

The same factor $1/(2\omega_{\boldsymbol k})$ that appears in the mode normalization appears in the correlation function. The Green Functions bridge distinguishes this Wightman function from time-ordered, retarded, and other propagators.

What Interactions Change

Add, for example,

$$H_{\mathrm{int}} =\frac{\lambda}{4!} \int d^3x\,\phi^4(\boldsymbol x).$$

In momentum space this term contains products of four creation or annihilation operators and a momentum-conserving delta function. It therefore couples different modes:

$$\boldsymbol k_1+\boldsymbol k_2 +\boldsymbol k_3+\boldsymbol k_4=0$$

for the all-incoming momentum labels in a representative term. The free oscillator states remain a useful basis for perturbation theory, but they are not eigenstates of the full interacting Hamiltonian.

Interactions also modify the relation between field operators and observable particles. Stable particles are identified through asymptotic states and poles of correlation functions rather than by naively applying a bare creation operator at finite time.

Where the Particle Picture Becomes Subtle

• In an interacting theory, a field operator can overlap with multiparticle states as well as a one-particle state.
• In curved spacetime or for accelerated observers, positive frequency and hence the particle decomposition need not be unique.
• In media, quasiparticles are excitations relative to a material ground state and may decay.
• Gauge theories require constraints and physical-state conditions before all oscillator-like variables can be interpreted.
• In an infinite interacting theory, the exact representation need not be unitarily equivalent to the free Fock representation used for perturbative bookkeeping.

These caveats qualify the bridge; they do not erase the oscillator’s central role in free fields and perturbation theory.

How to Use This Bridge

1. Diagonalize the classical quadratic Hamiltonian into independent real normal modes.
2. Quantize each independent canonical pair.
3. state the finite-volume or continuum normalization.
4. Identify the vacuum and occupation-number basis.
5. Determine which mode quanta have a particle or quasiparticle interpretation.
6. Add interactions only after recording which free basis and normal ordering convention are being used.

Common Mistakes

• Saying the oscillator’s $a^\dagger$ always creates another material particle.
• Counting $\boldsymbol k$ and $-\boldsymbol k$ as independent complex coordinates for a real classical field.
• Writing the continuum Hamiltonian with a bare “one half per mode” without identifying its volume and regulator.
• Mixing covariant and noncovariant one-particle normalization.
• Treating gauge-field time and longitudinal components as unconstrained physical oscillators.
• Applying bosonic commutators to fermionic modes.
• Calling an interacting field a set of independent oscillators.
• Assuming a free-field creation operator creates an exact interacting eigenstate.
• Treating the particle concept as observer-independent in every background.

Exercises

Exercise 1: Diagonalizing a quadratic field

Starting from the real orthonormal eigenfunctions $f_\alpha$, derive the mode Hamiltonian from the field Hamiltonian.

Solution

Insert

$$\phi=\sum_\alpha q_\alpha f_\alpha, \qquad \pi=\sum_\alpha p_\alpha f_\alpha.$$

Orthonormality gives

$$\int d^3x\,\pi^2 =\sum_\alpha p_\alpha^2.$$

Using integration by parts and the boundary conditions,

$$\begin{aligned} \int d^3x\, \left[ (\nabla\phi)^2+m^2\phi^2 \right] &= \int d^3x\,\phi\mathcal K\phi \\ &= \sum_{\alpha,\beta} q_\alpha q_\beta\omega_\beta^2 \int d^3x\,f_\alpha f_\beta \\ &= \sum_\alpha \omega_\alpha^2q_\alpha^2. \end{aligned}$$

Therefore

$$H =\frac12 \sum_\alpha \left( p_\alpha^2+\omega_\alpha^2q_\alpha^2 \right).$$

Exercise 2: Box normalization

Show that $a(\boldsymbol k)=\sqrt V\,a_{\boldsymbol k}$ converts the box commutator into the continuum commutator.

Solution

In the box,

$$[a_{\boldsymbol k},a_{\boldsymbol k'}^\dagger] =\delta_{\boldsymbol k,\boldsymbol k'}.$$

The continuum limit replaces

$$\delta_{\boldsymbol k,\boldsymbol k'} \longrightarrow \frac{(2\pi)^3}{V} \delta^{(3)} (\boldsymbol k-\boldsymbol k').$$

Hence

$$\begin{aligned} [a(\boldsymbol k),a^\dagger(\boldsymbol k')] &= V[a_{\boldsymbol k},a_{\boldsymbol k'}^\dagger] \\ &\longrightarrow (2\pi)^3 \delta^{(3)} (\boldsymbol k-\boldsymbol k'). \end{aligned}$$

The same rescaling makes the discrete field expansion approach the stated continuum expansion.

Exercise 3: Mode coupling from an interaction

Why does a quartic interaction destroy independent mode evolution even though the free Hamiltonian remains diagonal?

Solution

Each field factor is a sum or integral over modes. Multiplying four factors produces terms containing four mode operators. Integrating over position gives a delta function imposing total momentum conservation, but it does not force all four momenta to be equal. The interaction therefore converts combinations of occupied modes into other combinations. The free Hamiltonian still defines the convenient oscillator basis; the full Hamiltonian mixes those basis states.

Canonical Links

• Quantum Harmonic Oscillator
• Ladder-Operator Solution
• Zero-Point Energy
• Oscillator as a Universal Local Model
• Bosonic Fock Space
• Creation and Annihilation Operators
• Bosonic Commutation Relations
• From Phase Space to Canonical Quantization
• Second Quantization
• Coherent-State Dynamics

Continue in Field Theory

Continue with free scalar-field quantization, mode expansions and relativistic normalization, vacuum correlation functions, Wick’s theorem, interacting fields, and renormalization at QFT.org.

References

• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.
• S. Weinberg, The Quantum Theory of Fields, Vol. I, Cambridge University Press, 1995.
• M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014.
• A. Zee, Quantum Field Theory in a Nutshell, 2nd ed., Princeton University Press, 2010.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.