Quantization of the Electromagnetic Field

Field quantization is the founding act of quantum optics, and it is almost entirely a borrowing. It takes the fully solved quantum harmonic oscillator from quantum mechanics — its ladder operators, its evenly spaced levels, its irreducible ground-state energy $\tfrac12\hbar\omega$ — and the normal-mode decomposition of the source-free electromagnetic field from Maxwell's equations, and notices that the two are the same mathematics in different clothes. Each plane-wave mode is an amplitude that oscillates at $\omega_k=c|\mathbf{k}|$ and stores energy quadratic in that amplitude: that is a harmonic oscillator. So the field, expanded in modes, is an infinite orchestra of independent oscillators, one per wavevector and polarization, and quantizing the field means quantizing every one. Classical electromagnetism cannot survive this scrutiny. A continuous field carrying arbitrarily small energy at a given frequency gives the ultraviolet catastrophe, cannot explain the photoelectric threshold, forbids spontaneous emission (an excited atom in the dark should never decay), and has no room for photon antibunching. Promoting the mode amplitudes to operators fixes all of it at once: $\hat a$ and $\hat a^\dagger$ remove and add one photon of a mode, $\hat n=\hat a^\dagger\hat a$ counts them, and $\hat{\mathbf{E}}$ and $\hat{\mathbf{B}}$ become non-commuting operators that jitter even in the vacuum. From this single construction flow Fock states, coherent states, squeezed light, and the correlations that separate quantum from classical light — with Maxwell's theory recovered as the many-photon coherent limit.

Key takeaways

The free electromagnetic field, expanded in normal modes, is identical to an infinite collection of independent harmonic oscillators — one per mode $(\mathbf{k},\lambda)$ .
Quantization promotes each amplitude to operators $\hat a_{\mathbf{k}\lambda},\hat a^\dagger_{\mathbf{k}\lambda}$ with $[\hat a_{\mathbf{k}\lambda},\hat a^\dagger_{\mathbf{k}'\lambda'}]=\delta_{\mathbf{k}\mathbf{k}'}\delta_{\lambda\lambda'}$ ; a photon is one quantum of one mode.
The Hamiltonian is $\hat H=\sum_{\mathbf{k}\lambda}\hbar\omega_k(\hat n_{\mathbf{k}\lambda}+\tfrac12)$ ; the $\tfrac12$ is the zero-point energy, and the vacuum $|0\rangle$ has zero photons yet nonzero energy and field variance.
Because $\hat{\mathbf{E}}$ and $\hat{\mathbf{B}}$ are built from non-commuting $\hat a,\hat a^\dagger$ , they cannot both be sharp; a Fock state $|n\rangle$ has $\langle\hat{\mathbf{E}}\rangle=0$ and is not a classical wave.
Classical Maxwell theory is the coherent-state limit: for large $|\alpha|$ , $\langle\hat{\mathbf{E}}\rangle$ oscillates classically and the fractional noise falls as $1/|\alpha|$ .
Key idea: A normal mode of the electromagnetic field is a harmonic oscillator; quantizing the field means quantizing every oscillator, so that $\hat a^\dagger$ and $\hat a$ add and remove single quanta — photons — from each mode.

Where this lives today

The most visible payoff is squeezed light in gravitational-wave detection. LIGO and Virgo are shot-noise limited: their sensitivity floor is the vacuum fluctuation entering the interferometer's dark port — the $\tfrac14$ quadrature variance of equation (7). Since 2019 both inject squeezed vacuum, engineered by parametric down-conversion to have $\Delta X_1<\tfrac12$ in the measured quadrature at the cost of extra phase noise, buying several decibels of noise reduction and a larger volume of space surveyed for black-hole mergers. There is no classical account of this; it is direct manipulation of the quantized vacuum. The framework is equally the operating system of cavity and circuit QED. In cavity QED a single atom couples to one mode whose field per photon (the $\mathcal{E}_k$ of Worked Example 1) is large enough for a lone photon and a lone atom to exchange excitation coherently. In circuit QED a superconducting qubit plays the atom and a microwave resonator the mode; because a few-GHz mode carries over a thousand thermal photons at room temperature (Worked Example 2), these processors run at millikelvin temperatures so the mode is genuinely in its vacuum. Both are, at bottom, one quantized oscillator talking to one two-level system.

The physical picture

Think of the radiation field in a large box as a grand piano with infinitely many strings. Each string is a standing-wave mode with its own pitch $\omega_k$ , and the strings are orthogonal, so plucking one deposits no energy into any other. What quantum mechanics adds is that a string cannot be plucked by an arbitrary amount: its energy climbs a ladder in fixed steps of $\hbar\omega$ , one photon per step, and can never be brought fully to rest — the lowest rung sits at $\tfrac12\hbar\omega$ . A photon is not a little bullet; it is one rung of excitation of a delocalized mode. "How many photons" is a sharp question, the eigenvalue of $\hat n$ ; "where exactly is the photon" usually is not.

Figure

Energy ladder of one field mode at λ = 633 nm: rungs are evenly spaced by one photon ħω = 1.96 eV, the ground state sits at the zero-point energy ½ħω = 0.98 eV, and â† climbs one rung while â descends one

The electric field of a mode plays the role of the oscillator's position and the magnetic field its momentum. Because $\hat x$ and $\hat p$ cannot both be sharp, neither can $\hat{\mathbf{E}}$ and $\hat{\mathbf{B}}$ — the field trembles even in empty space, and that vacuum jitter tips an excited atom into spontaneous emission. The cleanest picture of this noise is the phase space of the two quadratures $\hat X_1,\hat X_2$ : the vacuum is not a point at the origin but a fuzzy disk of radius $\tfrac12$ , and a bright classical wave is that same disk displaced far out to $|\alpha|$ , so its relative noise shrinks as $1/|\alpha|$ .

Figure

Quadrature phase space of one mode: the vacuum is a noise disk of radius ΔX₁ = ΔX₂ = ½ at the origin, and a coherent state is the same disk displaced out to amplitude α, so ⟨n⟩ = α² = 9 here and the relative noise ΔX ⁄ α ≈ 0.5 ⁄ 3 ≈ 0.17 falls as 1 ⁄ α

The core equations

Everything begins with the field as a superposition of modes. In the Coulomb gauge the transverse vector potential carries the whole radiation field, and each mode's amplitude is promoted to an annihilation or creation operator, weighted by a normalization that makes them obey the standard oscillator algebra.

\hat{\mathbf{A}}(\mathbf{r},t)=\sum_{\mathbf{k}\lambda}\sqrt{\frac{\hbar}{2\varepsilon_0\omega_k V}}\,\boldsymbol{\epsilon}_{\mathbf{k}\lambda}\Big(\hat a_{\mathbf{k}\lambda}e^{i(\mathbf{k}\cdot\mathbf{r}-\omega_k t)}+\hat a^\dagger_{\mathbf{k}\lambda}e^{-i(\mathbf{k}\cdot\mathbf{r}-\omega_k t)}\Big) \qquad (1)

where: $\hat{\mathbf{A}}$ — vector potential operator, Coulomb gauge $\nabla\cdot\hat{\mathbf{A}}=0$ ; $\hat a_{\mathbf{k}\lambda},\hat a^\dagger_{\mathbf{k}\lambda}$ — annihilation and creation operators for mode $(\mathbf{k},\lambda)$ ; $\boldsymbol{\epsilon}_{\mathbf{k}\lambda}$ — unit polarization vector, transverse ( $\boldsymbol{\epsilon}_{\mathbf{k}\lambda}\cdot\mathbf{k}=0$ , two per $\mathbf{k}$ ); $V$ — quantization volume with $\mathbf{k}=2\pi\mathbf{n}/L$ (periodic boundary conditions); $\omega_k=c|\mathbf{k}|$ — mode frequency; $\varepsilon_0$ — vacuum permittivity. Reading: the field is a sum over independent oscillators, normalized so each obeys the standard oscillator algebra. The operators are fixed by demanding equal-time canonical commutation between the field and its conjugate momentum $\hat{\boldsymbol{\Pi}}=-\varepsilon_0\hat{\mathbf{E}}$ . After using mode orthogonality, what survives is the bosonic algebra — the most important structural fact of the theory.

[\hat a_{\mathbf{k}\lambda},\hat a^\dagger_{\mathbf{k}'\lambda'}]=\delta_{\mathbf{k}\mathbf{k}'}\delta_{\lambda\lambda'},\qquad [\hat a_{\mathbf{k}\lambda},\hat a_{\mathbf{k}'\lambda'}]=0 \qquad (2)

where: $\delta_{\mathbf{k}\mathbf{k}'}$ — Kronecker delta over discrete mode labels. Reading: distinct modes commute (they are independent), while within one mode $[\hat a,\hat a^\dagger]=1$ is the seed of every quantum feature — zero-point energy, vacuum noise, and the impossibility of a sharp field. Substituting the field operators into the classical energy and doing the volume integral collapses everything to a sum of oscillator Hamiltonians; the $e^{\pm2i\omega_k t}$ cross terms integrate to zero and orthonormality ties each term to one mode.

\hat H=\sum_{\mathbf{k}\lambda}\hbar\omega_k\Big(\hat a^\dagger_{\mathbf{k}\lambda}\hat a_{\mathbf{k}\lambda}+\tfrac12\Big)=\sum_{\mathbf{k}\lambda}\hbar\omega_k\Big(\hat n_{\mathbf{k}\lambda}+\tfrac12\Big) \qquad (3)

where: $\hat n_{\mathbf{k}\lambda}=\hat a^\dagger_{\mathbf{k}\lambda}\hat a_{\mathbf{k}\lambda}$ — number operator, eigenvalues $n=0,1,2,\dots$ ; $\tfrac12\hbar\omega_k$ — zero-point energy of each mode. Reading: the field energy is the photon count in each mode times $\hbar\omega_k$ , plus an irremovable $\tfrac12$ per mode; the photon is the quantum of one mode. The electric field is the time derivative of $\hat{\mathbf{A}}$ , so it too is linear in the ladder operators, with a factor of $i$ and a relative minus sign that make it Hermitian. Its per-mode amplitude sets the "electric field of one photon".

\hat{\mathbf{E}}(\mathbf{r},t)=i\sum_{\mathbf{k}\lambda}\mathcal{E}_k\,\boldsymbol{\epsilon}_{\mathbf{k}\lambda}\Big(\hat a_{\mathbf{k}\lambda}e^{i(\mathbf{k}\cdot\mathbf{r}-\omega_k t)}-\hat a^\dagger_{\mathbf{k}\lambda}e^{-i(\mathbf{k}\cdot\mathbf{r}-\omega_k t)}\Big),\qquad \mathcal{E}_k=\sqrt{\frac{\hbar\omega_k}{2\varepsilon_0 V}} \qquad (4)

where: $\hat{\mathbf{E}}$ — Hermitian electric field operator; $\mathcal{E}_k$ — the electric field per photon, the amplitude scale that fixes atom–field coupling. Reading: because $\hat{\mathbf{E}}$ contains both $\hat a$ and $\hat a^\dagger$ , measuring it mixes adjacent photon-number states, which is why a definite- $n$ state has no definite field. The ladder operators act on the photon-number (Fock) states of a mode — the entire working algebra of quantum optics.

\hat a^\dagger|n\rangle=\sqrt{n+1}\,|n+1\rangle,\qquad \hat a|n\rangle=\sqrt{n}\,|n-1\rangle,\qquad \hat a|0\rangle=0 \qquad (5)

where: $|n\rangle$ — Fock state of a mode with exactly $n$ photons; $|0\rangle$ — vacuum, annihilated by every $\hat a_{\mathbf{k}\lambda}$ . Reading: the $\sqrt{n+1}$ and $\sqrt{n}$ factors are what make stimulated emission grow with photon number while spontaneous emission survives even at $n=0$ — the " $+1$ " is the vacuum term. For homodyne measurement and squeezing, resolve the field into two Hermitian quadratures — the analogues of position and momentum — whose commutator is fixed by the bosonic algebra.

\hat X_1=\tfrac12(\hat a+\hat a^\dagger),\quad \hat X_2=\tfrac{1}{2i}(\hat a-\hat a^\dagger),\qquad [\hat X_1,\hat X_2]=\tfrac{i}{2} \qquad (6)

where: $\hat X_1,\hat X_2$ — amplitude and phase quadratures. Reading: because they do not commute, no state can pin both, and the vacuum spreads its noise equally between them. The uncertainty relation that follows caps how sharp the quadratures can jointly be; the vacuum saturates it with equal noise in each — the shot-noise floor of every optical measurement.

\Delta X_1\,\Delta X_2\ge\tfrac14,\qquad (\Delta X_1)^2_{\text{vac}}=(\Delta X_2)^2_{\text{vac}}=\tfrac14 \qquad (7)

where: $\Delta X_i$ — standard deviation of quadrature $i$ . Reading: a coherent state and the vacuum both sit at equality with balanced noise; squeezing pushes one quadrature below $\tfrac14$ at the cost of the other, without violating the bound. In thermal equilibrium the mean occupation is the Bose–Einstein (Planck) distribution — the result that historically forced field quantization on physics.

\bar n(\omega,T)=\frac{1}{e^{\hbar\omega/k_BT}-1} \qquad (8)

where: $\bar n$ — mean photon number of a mode at temperature $T$ ; $k_B$ — Boltzmann constant. Reading: for $\hbar\omega\gg k_BT$ the mode is frozen in its vacuum ( $\bar n\to0$ ), killing the ultraviolet catastrophe; for $\hbar\omega\ll k_BT$ it fills as $\bar n\to k_BT/\hbar\omega$ , recovering classical equipartition.

Derivation AA field mode is a harmonic oscillator

The goal is to prove the central claim mechanically: that the source-free Maxwell equations, expanded in normal modes, hand each mode amplitude an equation of motion and an energy identical to a simple harmonic oscillator's. If this holds, quantizing the field cannot mean anything new — it must be the ordinary quantization of an oscillator, done once per mode. We work in the Coulomb gauge with no sources, so the scalar potential vanishes and $\hat{\mathbf{A}}$ carries the full transverse field.

Step 1. Source-free Maxwell's equations combine into $\nabla^2\mathbf{A}-c^{-2}\ddot{\mathbf{A}}=0$ with $\nabla\cdot\mathbf{A}=0$ . Why: it is second order in time — the hallmark of oscillatory motion.

Step 2. Expand in transverse orthonormal modes $\mathbf{A}(\mathbf{r},t)=\sum_{\mathbf{k}\lambda}q_{\mathbf{k}\lambda}(t)\,\mathbf{u}_{\mathbf{k}\lambda}(\mathbf{r})$ , with $\mathbf{u}_{\mathbf{k}\lambda}=\boldsymbol{\epsilon}_{\mathbf{k}\lambda}e^{i\mathbf{k}\cdot\mathbf{r}}/\sqrt{V}$ obeying $\nabla^2\mathbf{u}=-k^2\mathbf{u}$ . Why: separating space from time turns one field equation into one ODE per amplitude $q_{\mathbf{k}\lambda}(t)$ .

Step 3. Because the $\mathbf{u}_{\mathbf{k}\lambda}$ are spatial eigenfunctions, the equation decouples into $\ddot q_{\mathbf{k}\lambda}+\omega_k^2\,q_{\mathbf{k}\lambda}=0$ with $\omega_k=ck$ . Why: this is literally a mass on a spring of frequency $\omega_k$ ; the modes are independent oscillators.

Step 4. The field energy $H=\tfrac12\int\!\mathrm{d}^3r\,(\varepsilon_0\mathbf{E}^2+\mathbf{B}^2/\mu_0)$ reduces by orthonormality to $H=\sum_{\mathbf{k}\lambda}\tfrac12(p_{\mathbf{k}\lambda}^2+\omega_k^2 q_{\mathbf{k}\lambda}^2)$ with $p=\dot q$ . Why: energy and equation of motion both match an oscillator's — the analogy is exact.

Step 5. Promote $q_{\mathbf{k}\lambda},p_{\mathbf{k}\lambda}$ to operators with $[\hat q,\hat p]=i\hbar$ and set $\hat a_{\mathbf{k}\lambda}=(\omega_k\hat q_{\mathbf{k}\lambda}+i\hat p_{\mathbf{k}\lambda})/\sqrt{2\hbar\omega_k}$ . Why: textbook oscillator quantization per mode, reproducing equations (2) and (3).

Go deeper · Derivation B · 1 minThe field Hamiltonian and its zero-point energy

Derivation BThe field Hamiltonian and its zero-point energy

The goal is to see where the ground-state energy comes from, because that $\tfrac12\hbar\omega$ per mode is not an optional constant — it is forced by operator ordering and has measurable consequences. We insert the operator expansions into the classical field energy and track how the non-commutativity of $\hat a$ and $\hat a^\dagger$ survives the volume integral to leave energy in a state with no photons.

Step 1. Write $\hat{\mathbf{E}}=-\dot{\hat{\mathbf{A}}}$ and $\hat{\mathbf{B}}=\nabla\times\hat{\mathbf{A}}$ and substitute into $\hat H=\tfrac12\int\!\mathrm{d}^3r\,(\varepsilon_0\hat{\mathbf{E}}^2+\hat{\mathbf{B}}^2/\mu_0)$ . Why: the energy is quadratic in the field, so products $\hat a\hat a$ , $\hat a^\dagger\hat a^\dagger$ , $\hat a\hat a^\dagger$ , $\hat a^\dagger\hat a$ appear and their ordering must be kept.

Step 2. Do the volume integral: the two- $\hat a$ and two- $\hat a^\dagger$ terms carry $e^{\pm2i\omega_k t}$ and cancel between $\hat{\mathbf{E}}^2$ and $\hat{\mathbf{B}}^2$ , while $\int\mathrm{d}^3r\,e^{i(\mathbf{k}-\mathbf{k}')\cdot\mathbf{r}}=V\delta_{\mathbf{k}\mathbf{k}'}$ . Why: only same-mode, number-conserving products $\hat a\hat a^\dagger+\hat a^\dagger\hat a$ survive, decoupling the modes.

Step 3. This leaves the symmetrized form $\hat H=\sum_{\mathbf{k}\lambda}\tfrac12\hbar\omega_k(\hat a\hat a^\dagger+\hat a^\dagger\hat a)$ . Why: the symmetric ordering comes straight from adding the electric and magnetic parts, and it is what forbids a vanishing ground state.

Step 4. Apply $\hat a\hat a^\dagger=\hat a^\dagger\hat a+1$ to get $\hat H=\sum_{\mathbf{k}\lambda}\hbar\omega_k(\hat a^\dagger\hat a+\tfrac12)$ , so $\hat H|0\rangle=\sum_{\mathbf{k}\lambda}\tfrac12\hbar\omega_k\,|0\rangle$ . Why: the $+\tfrac12$ is the commutator's fingerprint — zero photons yet nonzero energy, whose boundary dependence is the Casimir effect.

Limiting cases

Case	Result	Meaning
Large coherent amplitude	$\vert\alpha\vert\to\infty$	$\langle\hat{\mathbf{E}}\rangle$ reproduces a classical Maxwell wave; noise $\Delta n/\langle n\rangle\sim\vert\alpha\vert^{-1}\to0$ . Classical electrodynamics is the many-photon coherent limit.
Single mode	$\hat H=\hbar\omega(\hat a^\dagger\hat a+\tfrac12)$	The field collapses onto one harmonic oscillator — the arena of cavity and circuit QED.
Vacuum	$\hat a\vert0\rangle=0,\ E_0=\tfrac12\hbar\omega$	Zero photons but nonzero energy and variance $\langle0\vert\hat{\mathbf{E}}^2\vert0\rangle\ne0$ ; drives spontaneous emission, Lamb shift, Casimir force.
High-temperature	$k_BT\gg\hbar\omega$	$\bar n\to k_BT/\hbar\omega$ , recovering Rayleigh–Jeans; photon discreteness invisible.
Optical / low- $T$	$\hbar\omega\gg k_BT$	$\bar n\to e^{-\hbar\omega/k_BT}\to0$ ; mode frozen in its vacuum, so no ultraviolet catastrophe.

Common trapA Fock state of n photons is a classical wave with n photons' worth of field

It is not. A number state $|n\rangle$ has definite energy and photon number, but its mean electric field is exactly zero for every $n$ , including huge $n$ : $\langle n|\hat{\mathbf{E}}|n\rangle=0$ . The field operator is linear in $\hat a$ and $\hat a^\dagger$ , which shift photon number by $\pm1$ , and $|n\rangle$ is orthogonal to $|n\pm1\rangle$ , so the average vanishes. Sharp photon number means completely undefined phase, and an oscillating field with random phase averages to nothing. A classical-looking wave — a definite oscillating $\langle\hat{\mathbf{E}}(t)\rangle$ — requires a coherent superposition of adjacent number states, which is exactly what a coherent state $|\alpha\rangle$ is. Definite number and definite field are complementary.

Worked examples

1 · The electric field of a single photon (confidence builder)

A mode of a small optical cavity has $\lambda=633$ nm (a HeNe line) and volume $V=1000\ \mu\mathrm{m}^3=1.0\times10^{-15}\ \mathrm{m}^3$ . Find its photon energy, its zero-point energy, and the electric field per photon $\mathcal{E}_k=\sqrt{\hbar\omega/2\varepsilon_0V}$ .

Step 1. Frequency and photon energy: $\omega=2\pi c/\lambda=2.976\times10^{15}\ \mathrm{rad/s}$ , so $\hbar\omega=(1.0546\times10^{-34})(2.976\times10^{15})=3.14\times10^{-19}\ \mathrm{J}=\mathbf{1.96\ eV}$ — the energy of one 633 nm photon.

Step 2. The zero-point energy is half of that: $\tfrac12\hbar\omega=\mathbf{0.98\ eV}$ per mode — present even with no photons, the residual jitter no cooling can remove.

Step 3. Field per photon: $\mathcal{E}_k=\sqrt{(3.14\times10^{-19})/(2\times8.854\times10^{-12}\times1.0\times10^{-15})}$ . The denominator is $1.77\times10^{-26}$ , the ratio $1.77\times10^{7}$ , and the root $\mathbf{4.2\times10^{3}\ V/m}\approx4.2\ \mathrm{kV/m}$ . A single photon presents an atom with several kV/m — small next to a laser, but enough in a high-finesse cavity, where a photon crosses the atom many times, to reach strong coupling. ħω = 1.96 eV, zero-point ½ħω = 0.98 eV, ℰₖ ≈ 4.2 kV/m per photon.

2 · When is a mode thermally populated? (optical vs microwave)

At room temperature $T=300$ K, how many thermal photons occupy (a) the 633 nm optical mode above, and (b) a $5$ GHz microwave mode of a superconducting circuit? Below what frequency does a mode carry more than one thermal photon?

Step 1. Set the scale: $k_BT=(1.381\times10^{-23})(300)=4.14\times10^{-21}\ \mathrm{J}=0.02585\ \mathrm{eV}$ . Everything hinges on $\hbar\omega/k_BT$ in equation (8).

Step 2. Optical mode: $\hbar\omega/k_BT=1.96/0.02585=75.8$ , so $\bar n=(e^{75.8}-1)^{-1}\approx e^{-75.8}\approx\mathbf{1\times10^{-33}}$ . Optical modes are, to spectacular precision, empty of thermal photons at room temperature — frozen in their vacuum.

Step 3. Microwave mode: $\hbar\omega=\hbar(2\pi\times5\times10^{9})=3.31\times10^{-24}\ \mathrm{J}$ , so $\hbar\omega/k_BT=8.0\times10^{-4}$ and $\bar n=(e^{0.0008}-1)^{-1}\approx\mathbf{1.25\times10^{3}}$ . A 5 GHz mode holds about 1250 thermal photons at 300 K — exactly why superconducting qubits run in dilution refrigerators near 10 mK to reach $\bar n\ll1$ .

Step 4. The crossover $\bar n=1$ needs $e^{\hbar\omega/k_BT}=2$ , i.e. $\nu=k_BT\ln2/h=\mathbf{4.3\ THz}$ : below it a mode carries more than one thermal photon, above it the vacuum dominates. A coherent-state check on the classical limit: a $1$ mW beam over $1\ \mu\mathrm{s}$ carries $\langle n\rangle=P\tau/\hbar\omega=3.2\times10^{9}$ , so $|\alpha|=5.6\times10^{4}$ and the fractional noise $1/|\alpha|=1.8\times10^{-5}$ is negligible — a laser is a classical Maxwell wave. Optical n̄ ≈ 10⁻³³, microwave n̄ ≈ 1250, crossover ≈ 4.3 THz.

Physics of the portrait

Field quantization is where the photon stopped being a heuristic and became a theorem. Planck and Einstein quantized the energy exchanged with the field; it took quantizing the field itself to derive spontaneous emission — one 1927 paper.

Origins · 1 minPaul Dirac and how this was discovered

Paul Dirac

1902–1984 · founder of quantum electrodynamics and field quantization

Paul Dirac's 1927 paper "The Quantum Theory of the Emission and Absorption of Radiation" is the birth certificate of quantum optics and quantum field theory alike. Before it the photon was an accounting device: Planck (1900) had quantized the oscillators radiating into a blackbody cavity, and Einstein (1905, 1917) had introduced the light quantum and the $A$ and $B$ coefficients that demanded a spontaneous-emission term — but nobody could derive that $A$ coefficient, because a classical field cannot make an isolated excited atom decay. Dirac quantized each mode of the radiation field as a harmonic oscillator with operators obeying $[\hat a,\hat a^\dagger]=1$ — the construction on this page — and spontaneous emission fell out automatically: the " $+1$ " in $\hat a^\dagger|n\rangle=\sqrt{n+1}\,|n+1\rangle$ let an atom emit into a mode with zero photons, and one Hamiltonian unified spontaneous emission, stimulated emission (the $\sqrt{n}$ scaling), and absorption — Einstein's three coefficients at once. His method — promote fields to operators, impose commutators, read off particles as quanta — became the template for all of quantum field theory; the infinities it exposed took Bethe, Feynman, Schwinger, and Tomonaga two decades to tame by renormalization.

History

1900

— Planck quantizes oscillator energy to fit the blackbody spectrum

1905

— Einstein proposes the light quantum to explain the photoelectric effect

1927

— Dirac quantizes the electromagnetic field and derives spontaneous emission from first principles

1948

— Casimir predicts an attractive force between plates from the change in zero-point energy

1963

— Glauber builds the quantum theory of optical coherence and the coherent states

Going further

Texts. Loudon, The Quantum Theory of Light, 3rd ed., Ch. 4 (canonical quantization of the field) · Foot, Atomic Physics, Ch. 7 and appendix · Cohen-Tannoudji, Dupont-Roc and Grynberg, Photons and Atoms, Ch. III · Scully and Zubairy, Quantum Optics, Ch. 1 · Saleh and Teich, Fundamentals of Photonics, 3rd ed., Ch. 12 · Gerry and Knight, Introductory Quantum Optics, Ch. 2. The key checkable results in this article are: (i) the mode wave equation reduces to $\ddot q+\omega^2 q=0$ with $\omega_k=ck$ ; (ii) the Hamiltonian $\sum\hbar\omega(\hat n+\tfrac12)$ follows from the field-energy integral and commutator ordering; (iii) the quadrature commutator is $[\hat X_1,\hat X_2]=i/2$ ; (iv) the single-mode vacuum variance is $\mathcal{E}_k^2=\hbar\omega/2\varepsilon_0V$ ; (v) the Bose–Einstein occupation follows from summing $e^{-\beta\hbar\omega n}$ over $n$ . The worked numbers follow from standard SI constants: for $\lambda=633$ nm, $\hbar\omega=1.96$ eV, $\tfrac12\hbar\omega=0.98$ eV, and $\mathcal{E}_k=4.2\ \mathrm{kV/m}$ at $V=10^{-15}\ \mathrm{m}^3$ ; at $T=300$ K, optical $\bar n\approx10^{-33}$ , a 5 GHz mode $\bar n\approx1.25\times10^{3}$ , and the $\bar n=1$ crossover falls at $\nu\approx4.3$ THz.

Prerequisites. Quantum Harmonic Oscillator · Maxwell's Equations · Canonical Quantization · Fourier / Normal-Mode Expansion. Leads to. Fock States and the Photon Number Basis · Vacuum State and Zero-Point Energy · Quadrature Operators and the Uncertainty Relation · Coherent States · Squeezed States of Light · Spontaneous Parametric Down-Conversion.

Ask the tutor: Ask me to walk through how the " $+1$ " in $\hat a^\dagger|n\rangle=\sqrt{n+1}\,|n+1\rangle$ turns into Einstein's spontaneous-emission $A$ coefficient once you couple a single mode to a two-level atom.

Why is a field mode "the same thing" as a harmonic oscillator, and not just similar to one?

Because both the equation of motion and the energy match exactly, not approximately. Expanded in normal modes, each amplitude obeys $\ddot q+\omega_k^2 q=0$ — the defining equation of a harmonic oscillator — and the field energy becomes $\sum\tfrac12(p^2+\omega_k^2q^2)$ with $p=\dot q$ . No leftover term, no coupling between modes, no approximation. So quantizing the field is oscillator quantization once per mode, which is why the whole toolkit — ladder operators, evenly spaced levels, zero-point energy — transfers to light verbatim.

If the vacuum has infinite zero-point energy, why doesn't it have infinite mass or gravitate wildly?

The divergent sum $\sum_{\mathbf{k}\lambda}\tfrac12\hbar\omega_k$ is a real embarrassment, and quantum optics sidesteps rather than solves it. In every laboratory prediction only changes in the zero-point energy matter: the Casimir force is how it shifts when you move two plates, the Lamb shift how a bound electron's mode coupling differs from a free one's, spontaneous emission the coupling near one transition — all finite. The absolute vacuum energy only becomes a genuine problem once gravity enters (the cosmological-constant problem), which lies outside quantum optics and remains unsolved.

What actually is a photon, if it isn't a little particle at a location?

A photon is one quantum of excitation of a single field mode — one rung up the ladder of one oscillator. The mode is spread over all of space, so the photon has no sharp position: asking "where is it" is like asking which point of a vibrating string the note is at. What is sharp is the photon number $\hat n=\hat a^\dagger\hat a$ and the mode's frequency, polarization, and wavevector. You can build localized wavepackets by superposing many modes, but a single monochromatic photon is inherently delocalized. The particle-like clicks in a detector come from the field depositing energy in lumps of $\hbar\omega$ , not from a tiny object arriving.

Why does a definite photon number force an undefined phase?

Photon number and field phase are conjugate variables, roughly like $n$ and the oscillator's phase-space angle, and conjugate variables obey an uncertainty relation. A Fock state $|n\rangle$ has zero number uncertainty, so its phase is uniformly random over $2\pi$ . In the second figure $|n\rangle$ is a ring of radius $\sqrt{n}$ about the origin, not a localized blob; averaging the oscillating field over that ring gives $\langle\hat{\mathbf{E}}\rangle=0$ . A coherent state $|\alpha\rangle$ trades number certainty for a definite phase, becoming a localized blob with an oscillating field. You cannot have both sharp number and sharp phase, which is why a "wave of exactly $n$ photons" is not a thing.

The electric field per photon in Worked Example 1 was only a few kV/m — how can that drive strong atom–light coupling?

Two effects amplify it. First, $\mathcal{E}_k\propto1/\sqrt{V}$ , so shrinking the mode volume raises the per-photon field steeply — microcavities and photonic-crystal modes reach far higher values than the femtolitre example. Second, in a high-finesse cavity a photon bounces thousands of times before leaking out, so the atom feels the field over and over and coherent coupling accumulates. Strong coupling requires the single-photon rate $g$ (proportional to $\mathcal{E}_k$ times the atomic dipole) to exceed both cavity decay and atomic linewidth — a race even a modest field can win if the cavity stores the photon long enough. That is the whole art of cavity QED.

Is the factor of ½ in the Hamiltonian ever physically removable, or is it always real?

You can subtract the constant $\sum\tfrac12\hbar\omega_k$ freely as long as the mode structure is fixed, because an overall offset never affects dynamics — this is "normal ordering". What you cannot do is pretend the zero-point energy has no physics: it depends on the boundary conditions that set which modes exist. Move a mirror, insert a dielectric, or change the geometry, and the sum shifts by a finite, measurable amount — the Casimir force and cavity-modified emission rates. So the $\tfrac12$ is like the sea level of a landscape: set the zero anywhere, but the slopes that respond to the environment are absolutely real.