Quantized Electromagnetic Modes

A quantized electromagnetic mode is a classical radiation normal mode whose amplitude and conjugate momentum have been promoted to operators. For a lossless source-free field, every independent normal mode has the Hamiltonian of a harmonic oscillator. Its excitation number is the photon number in that mode.

The mode’s geometry and energy normalization determine its electric field per photon and its coupling to an atom. The derivation below starts with vacuum standing waves, then connects them to traveling pulses and explains why a lossy cavity resonance requires a different normalization.

Required background. Quantum Optics Overview introduces field states; Ladder-Operator Solution gives oscillator quantization; Mode Occupations gives bosonic number states; Minimal Coupling separates radiation coordinates from constrained Coulomb fields.

Transverse Radiation and Boundary Conditions

The parent problem on this page is a source-free electromagnetic field in a fixed, lossless environment. The cleanest derivation uses:

Ingredient	Assumption here	What changes outside it
charges and currents	absent from the quantization region	longitudinal fields and matter degrees of freedom enter
gauge	Coulomb gauge for transverse radiation	canonical variables and interaction terms are rearranged
boundaries	periodic or perfectly reflecting	radiation leakage requires ports, continua, or quasinormal modes
materials	vacuum first; lossless nondispersive dielectric later	dispersion and absorption change energy normalization
geometry	time independent	time-dependent modes can mix creation and annihilation operators

The underlying charge–field Hamiltonian and gauge dictionary live on Minimal Coupling and Gauge Choices in Light–Matter Physics. The present derivation isolates only the free transverse radiation sector.

Full relativistic quantum electrodynamics does more: it quantizes matter fields, maintains relativistic covariance, introduces propagators and interactions, and renormalizes ultraviolet-sensitive quantities. The few-mode construction here is the AMO entry point, not the whole theory.

Source-Free Transverse Field

In a vacuum region with no free charge or current, write the electromagnetic fields in terms of a vector potential. In Coulomb gauge,

$$\nabla\mathbin{\cdot}\mathbf A_{\mathrm T} = 0.$$

For the source-free transverse sector, the scalar potential can be taken to vanish and

$$\mathbf E_{\mathrm T} = - \frac{\partial\mathbf A_{\mathrm T}}{\partial t}, \qquad \mathbf B = \nabla\times\mathbf A_{\mathrm T}.$$

The vector potential obeys

$$\frac{\partial^2\mathbf A_{\mathrm T}}{\partial t^2} + c^2 \nabla\times\nabla\times\mathbf A_{\mathrm T} = 0.$$

For a transverse field, this is equivalent to the vector wave equation. The curl–curl form keeps the connection to boundary-value electromagnetism visible.

Why only the transverse field appears

Coulomb gauge separates the vector potential into physical transverse radiation coordinates while the longitudinal electric field is constrained by Gauss’s law and the charge distribution. In a matter-coupled problem, the Coulomb interaction and longitudinal sector have not disappeared; they are carried by the matter and scalar-potential terms of the full Hamiltonian.

Quantizing three unconstrained Cartesian components as independent scalar fields would overcount degrees of freedom. In free space, each nonzero wave vector has two transverse polarizations.

Mode Decomposition

Let the real mode functions $\mathbf u_\mu(\mathbf r)$ solve

$$\begin{aligned} \nabla\times\nabla\times\mathbf u_\mu &= \frac{\omega_\mu^2}{c^2} \mathbf u_\mu, \\ \nabla\mathbin{\cdot}\mathbf u_\mu &= 0, \end{aligned}$$

with the physical boundary conditions. For a perfectly conducting wall, the tangential electric field vanishes. In the simple standing-mode representation used here, the corresponding vector-potential mode functions can be chosen consistently with that condition.

Choose the vacuum normalization

$$\int_V d^3r\, \mathbf u_\mu(\mathbf r) \mathbin{\cdot} \mathbf u_\nu(\mathbf r) = \delta_{\mu\nu}.$$

This normalization gives $\mathbf u_\mu$ dimensions of inverse square-root volume. Expand the transverse vector potential as

$$\mathbf A_{\mathrm T}(\mathbf r,t) = \frac{1}{\sqrt{\epsilon_0}} \sum_\mu Q_\mu(t) \mathbf u_\mu(\mathbf r).$$

The factor $1/\sqrt{\epsilon_0}$ is a convention chosen so that $Q_\mu$ has the standard oscillator normalization.

Orthogonality from the wave problem

For two modes, integration by parts gives

$$\begin{aligned} \mathcal I_{\mu\nu} \equiv{}& \int_V d^3r\, \mathbf u_\mu \mathbin{\cdot} \left( \nabla\times\nabla\times\mathbf u_\nu \right), \\ \mathcal I_{\mu\nu} ={}& \int_V d^3r\, \left( \nabla\times\mathbf u_\mu \right) \mathbin{\cdot} \left( \nabla\times\mathbf u_\nu \right) + \mathcal B_{\mu\nu}. \end{aligned}$$

For the periodic or perfectly conducting boundaries used here, the surface term $\mathcal B_{\mu\nu}$ vanishes and the Maxwell operator is self-adjoint on its constrained domain. Modes with distinct eigenfrequencies are orthogonal. Within a degenerate subspace, one chooses an orthonormal basis.

For a radiating resonator, an outgoing complex-frequency resonance is not an ordinary square-integrable eigenmode. The full conservative field plus its exterior can still have a self-adjoint description with a continuum; retaining only resonance poles changes the mathematical problem.

From Field Energy to Oscillators

The source-free field energy is

$$H_{\mathrm{field}} = \frac12 \int_V d^3r \left( \epsilon_0\mathbf E_{\mathrm T}^2 + \frac{\mathbf B^2}{\mu_0} \right).$$

Using the mode expansion,

$$\mathbf E_{\mathrm T} = - \frac{1}{\sqrt{\epsilon_0}} \sum_\mu \dot Q_\mu \mathbf u_\mu$$

and

$$\mathbf B = \frac{1}{\sqrt{\epsilon_0}} \sum_\mu Q_\mu \nabla\times\mathbf u_\mu.$$

Orthogonality gives the electric contribution

$$\frac{\epsilon_0}{2} \int_V d^3r\, \mathbf E_{\mathrm T}^2 = \frac12 \sum_\mu \dot Q_\mu^2.$$

The eigenvalue equation and $c^2=1/(\epsilon_0\mu_0)$ give the magnetic contribution

$$\frac{1}{2\mu_0} \int_V d^3r\, \mathbf B^2 = \frac12 \sum_\mu \omega_\mu^2Q_\mu^2.$$

Therefore

$$\begin{aligned} H_{\mathrm{field}} ={}& \frac12 \sum_\mu \left( P_\mu^2 + \omega_\mu^2Q_\mu^2 \right), \\ P_\mu ={}& \dot Q_\mu. \end{aligned}$$

The geometry has disappeared from the oscillator Hamiltonian only because it has already fixed $\omega_\mu$ and $\mathbf u_\mu(\mathbf r)$. Those data return when the field is evaluated at a detector or coupled to matter.

Canonical Quantization

Promote the normal coordinates to operators with

$$\begin{aligned} [Q_\mu,P_\nu] &= i\hbar\delta_{\mu\nu}, \\ [Q_\mu,Q_\nu] &= 0, \qquad [P_\mu,P_\nu] &= 0. \end{aligned}$$

Define

$$\begin{aligned} a_\mu &= \sqrt{ \frac{\omega_\mu}{2\hbar} } Q_\mu + \frac{i}{ \sqrt{2\hbar\omega_\mu} } P_\mu, \\ a_\mu^\dagger &= \sqrt{ \frac{\omega_\mu}{2\hbar} } Q_\mu - \frac{i}{ \sqrt{2\hbar\omega_\mu} } P_\mu. \end{aligned}$$

Then

$$[a_\mu,a_\nu^\dagger] = \delta_{\mu\nu}, \qquad [a_\mu,a_\nu] = 0,$$

and the inverse relations are

$$\begin{aligned} Q_\mu &= \sqrt{ \frac{\hbar}{2\omega_\mu} } \left( a_\mu+a_\mu^\dagger \right), \\ P_\mu &= -i \sqrt{ \frac{\hbar\omega_\mu}{2} } \left( a_\mu-a_\mu^\dagger \right). \end{aligned}$$

Substitution gives

$$H_{\mathrm{field}} = \sum_\mu \hbar\omega_\mu \left( a_\mu^\dagger a_\mu + \frac12 \right).$$

For finitely many retained modes this is a tensor product of oscillator Hilbert spaces. The vacuum-based infinite-mode construction is bosonic Fock space over the one-photon mode space; it is not an unrestricted infinite tensor product. See Fock Space and Mode Occupations.

Constrained field commutator

The canonical momentum density in this source-free Coulomb-gauge problem is

$$\boldsymbol\Pi_{\mathrm T} = \epsilon_0 \dot{\mathbf A}_{\mathrm T} = -\epsilon_0\mathbf E_{\mathrm T}.$$

The equal-time field commutator is transverse:

$$\left[ A_{{\mathrm T},i}(\mathbf r), \Pi_{{\mathrm T},j}(\mathbf r') \right] = i\hbar \delta_{ij}^{\mathrm T} (\mathbf r,\mathbf r').$$

The kernel $\delta^{\mathrm T}$ projects onto the allowed transverse mode space and incorporates the boundary conditions. Replacing it by an unconstrained three-component delta function would reintroduce unphysical longitudinal coordinates.

The field-theory derivation of canonical field brackets is developed on From Phase Space to Canonical Quantization.

Electric and Magnetic Field Operators

For real standing-wave modes, the Schrödinger-picture field operators at a reference time are

$$\begin{aligned} \mathbf A_{\mathrm T}(\mathbf r) ={}& \sum_\mu \sqrt{ \frac{\hbar}{ 2\epsilon_0\omega_\mu } } \, \mathbf u_\mu(\mathbf r) \times \left( a_\mu+a_\mu^\dagger \right), \\ \mathbf E_{\mathrm T}(\mathbf r) ={}& i \sum_\mu \sqrt{ \frac{\hbar\omega_\mu}{ 2\epsilon_0 } } \, \mathbf u_\mu(\mathbf r) \times \left( a_\mu-a_\mu^\dagger \right), \\ \mathbf B(\mathbf r) ={}& \sum_\mu \sqrt{ \frac{\hbar}{ 2\epsilon_0\omega_\mu } } \, \nabla\times\mathbf u_\mu(\mathbf r) \times \left( a_\mu+a_\mu^\dagger \right). \end{aligned}$$

Hermiticity is explicit because the $\mathbf u_\mu$ are real. In the Heisenberg picture,

$$a_\mu(t) = a_\mu(0)e^{-i\omega_\mu t}.$$

It is then useful to split the field into positive- and negative-frequency parts:

$$\mathbf E_{\mathrm T} = \mathbf E^{(+)} + \mathbf E^{(-)}, \qquad \mathbf E^{(-)} = \left( \mathbf E^{(+)} \right)^\dagger.$$

For complex traveling-wave modes,

$$\mathbf E^{(+)}(\mathbf r,t) = i \sum_\mu \sqrt{ \frac{\hbar\omega_\mu}{ 2\epsilon_0 } } \, \mathbf u_\mu(\mathbf r) a_\mu e^{-i\omega_\mu t}.$$

The positive-frequency part contains annihilation operators. This convention is central to absorption photodetection and normally ordered optical correlations.

Unit check

Because $\int|\mathbf u_\mu|^2d^3r=1$, $|\mathbf u_\mu|\sim V^{-1/2}$. The coefficient

$$\sqrt{ \frac{\hbar\omega_\mu}{\epsilon_0} } \mathbf u_\mu$$

has units of electric field. A missing square root of volume or an inconsistent continuum delta function can therefore be detected dimensionally before any rate is calculated.

Plane Waves in a Periodic Box

Take a cubic quantization box of volume $V=L^3$ with periodic boundary conditions. The allowed wave vectors are

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

For each nonzero $\mathbf k$, choose two real orthonormal polarization vectors $\boldsymbol\epsilon_{\mathbf k\lambda}$ satisfying

$$\mathbf k \mathbin{\cdot} \boldsymbol\epsilon_{\mathbf k\lambda} = 0, \qquad \lambda=1,2.$$

A normalized complex mode is

$$\mathbf u_{\mathbf k\lambda}(\mathbf r) = \frac{ \boldsymbol\epsilon_{\mathbf k\lambda} }{ \sqrt V } e^{i\mathbf k\cdot\mathbf r}, \qquad \omega_{\mathbf k} = c|\mathbf k|.$$

The positive-frequency electric field is

$$\mathbf E^{(+)}(\mathbf r,t) ={} i \sum_{\mathbf k,\lambda} \sqrt{ \frac{ \hbar\omega_{\mathbf k} }{ 2\epsilon_0V } } \times \boldsymbol\epsilon_{\mathbf k\lambda} a_{\mathbf k\lambda} e^{i(\mathbf k\cdot\mathbf r-\omega_{\mathbf k}t)}.$$

Its Hermitian conjugate supplies $\mathbf E^{(-)}$. The Hamiltonian is

$$H_{\mathrm{field}} = \sum_{\mathbf k,\lambda} \hbar\omega_{\mathbf k} \left( a_{\mathbf k\lambda}^\dagger a_{\mathbf k\lambda} + \frac12 \right).$$

What the box means

The quantization box is usually a normalization device, not a physical cavity. It:

• discretizes the continuum;
• makes plane waves normalizable;
• assigns a temporary field amplitude proportional to $V^{-1/2}$;
• converts sums into integrals in the infinite-volume limit.

No prediction may depend on this arbitrary $V$. In free-space emission, for example, the squared coupling to one plane-wave mode scales as $1/V$, while the number of available modes in a frequency interval scales as $V$. The factors cancel.

The $\mathbf k=0$ coordinate is not a propagating transverse photon mode with positive frequency and is omitted from this radiation expansion.

Traveling waves and double counting

The labels $\mathbf k$ and $-\mathbf k$ denote distinct traveling directions and have independent annihilation operators. Hermiticity relates the positive- and negative-frequency pieces, not $a_{\mathbf k}$ to $a_{-\mathbf k}^\dagger$.

Confusion arises when formulas for real classical standing-wave coordinates are mixed with formulas for complex traveling-wave operators. Either basis is valid, but the independent degrees of freedom must be counted once.

Continuum Normalization

As $V\to\infty$,

$$\sum_{\mathbf k} \longrightarrow \frac{V}{(2\pi)^3} \int d^3k.$$

Define continuum operators by

$$a_\lambda(\mathbf k) = \sqrt{ \frac{V}{(2\pi)^3} } a_{\mathbf k\lambda}.$$

They satisfy

$$\left[ a_\lambda(\mathbf k), a_{\lambda'}^\dagger(\mathbf k') \right] = \delta_{\lambda\lambda'} \delta^{(3)}(\mathbf k-\mathbf k').$$

The electric field becomes

$$\mathbf E^{(+)}(\mathbf r,t) ={} i \sum_{\lambda=1}^{2} \int \frac{d^3k}{(2\pi)^{3/2}} \sqrt{ \frac{\hbar\omega_{\mathbf k}}{ 2\epsilon_0 } } \times \boldsymbol\epsilon_{\mathbf k\lambda} a_\lambda(\mathbf k) e^{i(\mathbf k\cdot\mathbf r-\omega_{\mathbf k}t)}.$$

Continuum annihilation operators carry dimensions inherited from the delta function. They are operator-valued distributions, not ordinary single-oscillator operators.

Normalized wave-packet modes

A physical pulse mode is built from a square-integrable spectral amplitude:

$$A_f^\dagger = \sum_\lambda \int d^3k\, f_\lambda(\mathbf k) a_\lambda^\dagger(\mathbf k),$$

with

$$\sum_\lambda \int d^3k\, |f_\lambda(\mathbf k)|^2 = 1.$$

Then

$$[A_f,A_f^\dagger] = 1.$$

The state $A_f^\dagger|0\rangle$ is a normalizable one-photon wave packet. A monochromatic plane wave in infinite volume is an idealized delta-normalized mode, not a normalizable pulse.

The commutator makes $A_f$ a bosonic mode operator, but a broadband pulse is not an energy eigenmode. With vacuum energy subtracted,

$$H A_f^\dagger|0\rangle =\sum_\lambda\int d^3k\,\hbar\omega_{\mathbf k} f_\lambda(\mathbf k)a_\lambda^\dagger(\mathbf k)|0\rangle.$$

Its mean energy is $\hbar\langle\omega\rangle_f$ and its energy variance is $\hbar^2(\langle\omega^2\rangle_f-\langle\omega\rangle_f^2)$. Free evolution changes the envelope to $f_\lambda(\mathbf k)e^{-i\omega_{\mathbf k}t}$. In a fixed wave-packet basis, the free Hamiltonian generally couples different packet modes; $\hbar\omega_f A_f^\dagger A_f$ alone is only a narrow-band approximation.

Photon Number

The number operator of mode $\mu$ is

$$N_\mu = a_\mu^\dagger a_\mu.$$

Its eigenstates obey

$$\begin{aligned} N_\mu|n_\mu\rangle &= n_\mu|n_\mu\rangle, \\ a_\mu^\dagger|n_\mu\rangle &= \sqrt{n_\mu+1} |n_\mu+1\rangle, \\ a_\mu|n_\mu\rangle &= \sqrt{n_\mu} |n_\mu-1\rangle. \end{aligned}$$

The energy in that mode is

$$E_{n_\mu} = \hbar\omega_\mu \left( n_\mu+\frac12 \right).$$

Photon Number States develops phase properties, source preparation, counting statistics, and nonclassicality. Here the key point is structural: photon number is an occupation number in a declared mode basis.

Basis dependence

For a passive unitary mode change,

$$c_j = \sum_\mu U_{j\mu}a_\mu, \qquad U^\dagger U=I,$$

the total occupation of the retained mode set is invariant:

$$\sum_j c_j^\dagger c_j = \sum_\mu a_\mu^\dagger a_\mu.$$

Individual occupations are not invariant. A photon in one path superposition can be a photon in a single output mode after a beam splitter. If the mixed frequencies differ, the free Hamiltonian is generally nondiagonal in the new basis. A static passive optical element mixes channels at the same frequency; driven frequency conversion can conserve photon number while exchanging energy with its pump.

A transformation mixing $a$ and $a^\dagger$ is different. Time-dependent boundaries, parametric amplification, and changes between inequivalent positive-frequency definitions can produce a Bogoliubov transformation and need not preserve the original photon number.

Vacuum and Zero-Point Fluctuations

The multimode vacuum satisfies

$$a_\mu|0\rangle = 0$$

for every retained mode. For one real mode, define the local electric zero-point amplitude vector

$$\boldsymbol{\mathcal E}_\mu(\mathbf r) = \sqrt{ \frac{\hbar\omega_\mu}{ 2\epsilon_0 } } \mathbf u_\mu(\mathbf r).$$

Its field contribution is

$$\mathbf E_\mu(\mathbf r) = i \boldsymbol{\mathcal E}_\mu(\mathbf r) \left( a_\mu-a_\mu^\dagger \right).$$

The vacuum mean vanishes,

$$\langle0| \mathbf E_\mu(\mathbf r) |0\rangle = 0,$$

but a Cartesian component has nonzero variance:

$$\left\langle0\left| E_{\mu,i}^2(\mathbf r) \right|0\right\rangle = |\mathcal E_{\mu,i}(\mathbf r)|^2.$$

In a number state,

$$\left\langle n\left| E_{\mu,i}^2(\mathbf r) \right|n\right\rangle = (2n+1) |\mathcal E_{\mu,i}(\mathbf r)|^2.$$

These are finite single-mode variances. Summing the unsmeared field variance over all continuum frequencies is ultraviolet divergent; real measurements specify bandwidth, spatial resolution, and coupling.

The $1$ is the zero-point contribution. It reflects the oscillator commutator and cannot be removed while preserving both field quadratures.

Ordering matters

An ideal absorption detector is governed by a normally ordered quantity such as

$$\langle \mathbf E^{(-)} \mathbin{\cdot} \mathbf E^{(+)} \rangle.$$

For the vacuum this is zero. A symmetrized field variance retains a half-quantum per mode. These facts are compatible because they are different operator orderings tied to different measurement models.

“Vacuum fluctuations” therefore do not mean that an ideal ground-state photodetector continually absorbs real photons from empty space. They mean that field quadratures have irreducible quantum variance and nontrivial commutators. Ordered spectra determine whether a system can absorb from or emit into the field. See Thermal and Vacuum Noise for the open-system frequency-domain dictionary.

Zero-point energy

The free-field vacuum energy is formally

$$E_{\mathrm{vac}} = \frac12 \sum_\mu \hbar\omega_\mu.$$

For an infinite continuum this diverges. The formal sum is not a finite absolute laboratory observable. Measurable boundary-dependent energies, radiative shifts, and forces require a consistent subtraction, regularization, renormalization, or effective-theory matching procedure.

It is equally misleading to say that every phenomenon called a vacuum effect is caused by an independently observable sea of zero-point energy. Vacuum fluctuations, field commutators, radiation reaction, and virtual-process contributions can be redistributed by representation while final observables remain invariant.

Mode Volume and Coupling Strength

For a plane wave in a box,

$$|\mathbf u_{\mathbf k\lambda}|^2 = \frac1V,$$

so the field amplitude per excitation scales as

$$E_{\mathrm{zpf}} \sim \sqrt{ \frac{\hbar\omega}{ 2\epsilon_0V } }.$$

For a standing cavity mode, the field is not spatially uniform. Take a real positive scalar $\epsilon_r(\mathbf r)$, $\mu_r=1$, and negligible dispersion and absorption over the relevant band. The generalized Coulomb condition is $\nabla\cdot(\epsilon_r\mathbf A)=0$, and the mode inner product becomes $\int\epsilon_r\mathbf u_\mu^*\cdot\mathbf u_\nu\,d^3r =\delta_{\mu\nu}$. This weighted normalization reduces the energy to the same oscillator form; the unweighted vacuum condition cannot simply be reused in an inhomogeneous dielectric.

For a real unit dipole direction $\mathbf e_d$, an effective mode volume at emitter position $\mathbf r_0$ is

$$V_{\mathrm{eff}} (\mathbf r_0,\mathbf e_d) = \frac{ \displaystyle \int d^3r\, \epsilon_r(\mathbf r) |\mathbf E_c(\mathbf r)|^2 }{ \displaystyle \epsilon_r(\mathbf r_0) | \mathbf e_d \mathbin{\cdot} \mathbf E_c(\mathbf r_0) |^2 }.$$

$\mathbf E_c$ is any consistently normalized classical mode profile; its overall amplitude cancels. Under the assumptions above, the projected zero-point scale in this macroscopic field model is

$$E_{\mathrm{zpf},d}(\mathbf r_0) = \sqrt{ \frac{ \hbar\omega }{ 2\epsilon_0 \epsilon_r(\mathbf r_0) V_{\mathrm{eff}} } }.$$

The corresponding electric-dipole coupling is

$$g = - \frac{ \mathbf d_{eg} \mathbin{\cdot} \mathbf E_{\mathrm{zpf}}(\mathbf r_0) }{ \hbar },$$

up to the phase convention used for the mode and atomic states. For a complex transition dipole, evaluate the complex bilinear contraction $\mathbf d_{eg}\cdot\mathbf E^{(+)}_{\rm mode}$ directly; replacing it by overlap of real polarization vectors can select the wrong helicity. Microscopic local-field corrections for an emitter embedded in a dielectric are additional physics, not included by the macroscopic mode volume alone.

Quantization volume is not effective mode volume

These two volumes answer different questions:

• Quantization volume $V$ is an arbitrary box used to normalize continuum plane waves. It cancels from physical free-space predictions.
• Effective mode volume $V_{\mathrm{eff}}$ characterizes the spatial concentration and polarization overlap of a physical resonator mode. It affects measurable coupling strengths.

Substituting one for the other is a common source of erroneous cavity and spontaneous-emission rates.

When the textbook formula fails

The simple integral above is not universal.

Dispersive media. Electromagnetic energy density involves frequency derivatives of material response, not merely $\epsilon|\mathbf E|^2$. The material degrees of freedom carrying dispersion must be represented consistently.

Absorptive media. A field-only Hermitian normal-mode expansion is generally inadequate. Macroscopic QED introduces reservoir-assisted noise operators tied to the absorptive response and electromagnetic Green tensor.

Open resonators. Outgoing-wave resonances have complex frequencies and spatially divergent quasinormal profiles. Their normalization, mode volume, and completeness require an open-system or scattering formulation; a naive finite integral of $|\mathbf E|^2$ is not valid.

Strongly nonlocal or microscopic media. A local scalar permittivity may itself fail. Mode volume cannot repair an inadequate material model.

Degenerate or multimode systems. A single scalar $V_{\mathrm{eff}}$ can hide polarization, interference, and nonorthogonality. The Green tensor or a validated multimode model is often the safer object.

Worked Example: One-Dimensional Cavity

Consider an ideal one-dimensional cavity of length $L$ with one polarization and perfect mirrors at $x=0,L$. Use periodic transverse boundaries of normalization area $A$, neglecting diffraction; this is not a rectangular metal box with the same field uniform along all its side walls. A mode is

$$\mathbf u_m(\mathbf r) = \mathbf e_y \sqrt{ \frac{2}{AL} } \sin\left( \frac{m\pi x}{L} \right),$$

with

$$\omega_m = \frac{m\pi c}{L}, \qquad m=1,2,\ldots.$$

The normalization follows from

$$\int_0^L \sin^2\left( \frac{m\pi x}{L} \right) dx = \frac L2.$$

At an electric antinode,

$$|\mathbf u_m|_{\max} = \sqrt{ \frac{2}{AL} }.$$

The local zero-point electric amplitude is therefore

$$E_{\mathrm{zpf}}^{\max} = \sqrt{ \frac{\hbar\omega_m}{ \epsilon_0AL } }.$$

The effective mode volume at the antinode is

$$V_{\mathrm{eff}} = \frac{AL}{2},$$

so the same result follows from

$$E_{\mathrm{zpf}}^{\max} = \sqrt{ \frac{\hbar\omega_m}{ 2\epsilon_0V_{\mathrm{eff}} } }.$$

The factor of two is not a contradiction. It records that the standing-wave intensity is concentrated at antinodes rather than uniform throughout the geometric volume $AL$.

For $m=2$, the upper panel shows the signed mode profile and the electric vacuum RMS field, normalized to their antinode values. The lower panel shows the electric and magnetic vacuum energy densities divided by $\hbar\omega/(2AL)$. Their integrals give $\hbar\omega/4$ each, adding to the mode’s $\hbar\omega/2$ vacuum energy. A node of electric coupling is not a node of the total field energy density.

Worked Example: Free-Space Mode Density

In a large periodic box, the number of wave vectors in a shell $[k,k+dk]$ is

$$dN_k = 2 \frac{V}{(2\pi)^3} 4\pi k^2dk,$$

where the factor of two counts transverse polarizations. With $\omega=ck$,

$$\rho(\omega) \equiv \frac{dN}{d\omega} = \frac{V\omega^2}{ \pi^2c^3 }.$$

The density per unit volume is

$$\frac{\rho(\omega)}{V} = \frac{\omega^2}{ \pi^2c^3 }.$$

This density of states combines with the $V^{-1/2}$ field amplitude in transition rates. It also supplies the mode counting behind blackbody radiation once each mode is thermally populated.

Boundary geometry, dimensionality, waveguides, cavities, and material dispersion all modify the density of states. Replacing it by the free-space formula in a structured environment defeats the purpose of modeling that environment.

Harmonic Oscillator to Fields extends the oscillator construction to other free fields. The Bridge to QFT Roadmap continues to covariant constraints, matter fields, propagators, and renormalization.

Exercises

1. Locate electric and magnetic vacuum energy

For the standing mode in the worked example, calculate $u_E(x)=\epsilon_0\langle E_y^2\rangle/2$ and $u_B(x)=\langle B_z^2\rangle/(2\mu_0)$ in the vacuum. Integrate over the cavity. Is vanishing electric field at a mirror enough to make the local total energy density vanish?

Solution

With $k=m\pi/L$, the field operators give

$$u_E(x)=\frac{\hbar\omega}{2AL}\sin^2(kx),\qquad u_B(x)=\frac{\hbar\omega}{2AL}\cos^2(kx).$$

Each integrates to $\hbar\omega/4$. Their sum is uniform in this ideal one-dimensional single-mode model and integrates to $\hbar\omega/2$. At an electric node the magnetic variance remains nonzero. The full multimode vacuum requires the regularization discussed above.

2. Follow a pulse mode with two frequencies

Let $A^\dagger=(a_1^\dagger+a_2^\dagger)/\sqrt2$ and $B^\dagger=(a_1^\dagger-a_2^\dagger)/\sqrt2$, where $\omega_1\ne\omega_2$. Starting from $A^\dagger|0\rangle$, find the probability of occupying the fixed mode $B$ after time $t$, and the energy variance. Does one photon remain one photon?

Solution

Set $\bar\omega=(\omega_1+\omega_2)/2$ and $\delta\omega=\omega_1-\omega_2$. The evolved state is

$$e^{-i\bar\omega t}\left[ \cos(\delta\omega t/2)|1_A,0_B\rangle -i\sin(\delta\omega t/2)|0_A,1_B\rangle\right].$$

Thus $P_B=\sin^2(\delta\omega t/2)$ and $\operatorname{Var}H=\hbar^2(\delta\omega)^2/4$. Total photon number remains exactly one. It is the fixed packet profile, not photon number, that changes under free evolution.

3. Field variance in a number state

For one field component,

$$E = i\mathcal E_0 (a-a^\dagger),$$

show that $\langle n|E|n\rangle=0$ and

$$\langle n|E^2|n\rangle = (2n+1)\mathcal E_0^2.$$

Solution

$E$ changes photon number by one, so its diagonal matrix element in a number state vanishes:

$$\langle n|E|n\rangle = 0.$$

Squaring gives

$$E^2 = -\mathcal E_0^2 \left( a^2 -aa^\dagger -a^\dagger a +a^{\dagger2} \right).$$

The $a^2$ and $a^{\dagger2}$ terms have zero diagonal expectation. Using

$$\langle n|aa^\dagger|n\rangle = n+1$$

and

$$\langle n|a^\dagger a|n\rangle = n,$$

one finds

$$\langle n|E^2|n\rangle = (2n+1)\mathcal E_0^2.$$

At $n=0$, the remaining $\mathcal E_0^2$ is the vacuum quadrature variance.

4. Vacuum variance versus absorption

For one mode,

$$E^{(+)} = i\mathcal E_0a, \qquad E^{(-)} = -i\mathcal E_0a^\dagger.$$

Evaluate in the vacuum:

$$\langle E^{(-)}E^{(+)}\rangle$$

and

$$\frac12 \langle E^{(-)}E^{(+)} + E^{(+)}E^{(-)} \rangle.$$

Interpret the difference.

Solution

Normal ordering gives

$$\langle0| E^{(-)}E^{(+)} |0\rangle = \mathcal E_0^2 \langle0|a^\dagger a|0\rangle = 0.$$

For the symmetrized product,

$$\begin{aligned} \mathcal S_E \equiv{}& E^{(-)}E^{(+)} + E^{(+)}E^{(-)}, \\ \mathcal V_E \equiv{}& \frac12 \langle0| \mathcal S_E |0\rangle, \\ \mathcal V_E ={}& \frac{\mathcal E_0^2}{2} \langle0| a^\dagger a+aa^\dagger |0\rangle \\ ={}& \frac{\mathcal E_0^2}{2}. \end{aligned}$$

The numerical factor differs from the full real-quadrature variance because the symmetrized expression here uses the positive- and negative-frequency pieces. The normal-ordered absorption signal vanishes in vacuum, while the symmetrized quadrature noise retains a half-quantum. A detector model selects which ordering is operationally relevant.

5. Box-to-continuum rescaling

Given

$$[a_{\mathbf k\lambda}, a_{\mathbf k'\lambda'}^\dagger] = \delta_{\mathbf k\mathbf k'} \delta_{\lambda\lambda'},$$

show that

$$a_\lambda(\mathbf k) = \sqrt{ \frac{V}{(2\pi)^3} } a_{\mathbf k\lambda}$$

has the continuum commutator stated on this page.

Solution

In the large-box limit, the discrete Kronecker delta corresponds to

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

Therefore

$$\begin{aligned} \left[ a_\lambda(\mathbf k), a_{\lambda'}^\dagger(\mathbf k') \right] ={}& \frac{V}{(2\pi)^3} \delta_{\lambda\lambda'} \delta_{\mathbf k\mathbf k'} \\ \longrightarrow{}& \delta_{\lambda\lambda'} \delta^{(3)} (\mathbf k-\mathbf k'). \end{aligned}$$

This rescaling is also what removes the explicit box volume from the continuum field integral.

6. Move an emitter away from an antinode

In the one-dimensional cavity, a real dipole is at position $x_0$ and makes angle $\beta$ with $\mathbf e_y$. Find its effective mode volume and its coupling relative to a parallel dipole at an antinode. Interpret a divergent mode volume.

Solution

The squared projected normalized mode is $2\sin^2(m\pi x_0/L)\cos^2\beta/(AL)$. Consequently

$$V_{\rm eff}=\frac{AL}{2\sin^2(m\pi x_0/L)\cos^2\beta},\qquad \frac{g(x_0,\beta)}{g_{\rm antinode}}= \sin(m\pi x_0/L)\cos\beta.$$

The sign of $g$ depends on the mode phase; the splitting scale uses $|g|$. At an electric node or orthogonal dipole orientation, the mode volume diverges because this transition does not couple to that mode. The cavity has not become physically infinite.

7. Passive mode mixing

Let

$$c = \frac{a+b}{\sqrt2}, \qquad d = \frac{a-b}{\sqrt2}.$$

Show that $c^\dagger c+d^\dagger d=a^\dagger a+b^\dagger b$. Express $a^\dagger|0\rangle$ in the $c,d$ basis and interpret the result.

Solution

Direct substitution gives

$$\begin{aligned} c^\dagger c+d^\dagger d ={}& \frac12 \left( a^\dagger+b^\dagger \right) \left( a+b \right) + \frac12 \left( a^\dagger-b^\dagger \right) \left( a-b \right) \\ ={}& a^\dagger a+b^\dagger b. \end{aligned}$$

Inverting the transformation,

$$a^\dagger = \frac{ c^\dagger+d^\dagger }{ \sqrt2 }.$$

Therefore

$$a^\dagger|0\rangle = \frac{ |1_c,0_d\rangle + |0_c,1_d\rangle }{ \sqrt2 }.$$

The state has one photon in total in either basis, but definite occupation of mode $a$ becomes a superposition of occupations in the $c,d$ basis.

References

• E. S. C. Ching, P. T. Leung, A. Maassen van den Brink, W. M. Suen, S. S. Tong, and K. Young, “Quasinormal-Mode Expansion for Waves in Open Systems,” Reviews of Modern Physics 70, 1545–1554 (1998). Mathematical structure, completeness, and limitations of open-system resonant modes.
• C. Cohen-Tannoudji, J. Dupont-Roc, and G. Grynberg, Photons and Atoms: Introduction to Quantum Electrodynamics, Wiley (1989; online edition 2007). Detailed canonical and multipolar formulations, constraints, and mode quantization.
• P. A. M. Dirac, “The Quantum Theory of the Emission and Absorption of Radiation,” Proceedings of the Royal Society A 114, 243–265 (1927). Foundational operator treatment of radiation emission and absorption.
• G. Grynberg, A. Aspect, and C. Fabre, “Quantization of Free Radiation,” in Introduction to Quantum Optics, Cambridge University Press (2010). Pedagogical derivation from classical normal coordinates to radiation operators.
• R. Loudon, The Quantum Theory of Light, 3rd ed., Oxford University Press (2000). Standard treatment of field quantization, photon states, and light–matter processes.
• L. Mandel and E. Wolf, “Quantization of the Free Electromagnetic Field,” in Optical Coherence and Quantum Optics, Cambridge University Press (1995). Authoritative free-field and optical-coherence conventions.
• P. W. Milonni, The Quantum Vacuum: An Introduction to Quantum Electrodynamics, Academic Press (1994). Zero-point fluctuations, operator ordering, radiative effects, and vacuum-energy cautions.
• C. Sauvan, J. P. Hugonin, I. S. Maksymov, and P. Lalanne, “Theory of the Spontaneous Optical Emission of Nanosize Photonic and Plasmon Resonators,” Physical Review Letters 110, 237401 (2013). Open, lossy, dispersive quasinormal-mode normalization and generalized mode-volume cautions.
• S. Scheel and S. Y. Buhmann, “Macroscopic QED: Concepts and Applications,” Acta Physica Slovaca 58, 675–809 (2008). Quantization in dispersive and absorbing media using reservoir fields and Green tensors.