Part II, Chapter 7

Photon Field Quantization

Quantizing the electromagnetic field: gauge freedom and physical photons

🔗Course Connections

📚Prerequisites

QFT Part I

Electromagnetic Field

Classical Maxwell theory and gauge invariance

→

QFT Part II

Free Scalar Quantization

Mode expansion and creation operators

→

▶️

Video Lecture

Lecture 20: Photon Field Quantization - MIT 8.323

Quantizing the electromagnetic field and electron-photon interactions (MIT QFT Course)

💡 Tip: Watch at 1.25x or 1.5x speed for efficient learning. Use YouTube's subtitle feature if available.

7.1 The Gauge Freedom Problem

The electromagnetic field A^μ(x) is a vector field (spin 1) described by the Lagrangian:

$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$

where F^μν = ∂^μA^ν - ∂^νA^μ is the field strength tensor.

⚠️ The Problem: Gauge Invariance

The Lagrangian is invariant under gauge transformations:

$$A^\mu(x) \to A^\mu(x) + \partial^\mu \chi(x)$$

This means A^μ has redundant degrees of freedom! We can't directly quantize it like we did for scalars—we'd be quantizing unphysical modes.

To quantize, we must:

Fix the gauge (choose a specific A^μ from each gauge equivalence class)
Identify the physical degrees of freedom
Quantize only the physical modes

7.2 Gauge Fixing Choices

There are several popular gauge choices. We'll focus on Coulomb gauge (radiation gauge).

Coulomb Gauge (Radiation Gauge)

Gauge Condition:

$$\nabla \cdot \mathbf{A} = 0 \quad \text{and} \quad A^0 = 0$$

This is called "radiation gauge" because it's natural for describing electromagnetic radiation (photons).

In Coulomb gauge:

A is purely transverse: A ⊥ k for plane waves
Electric field: E = -∂_tA (no scalar potential contribution)
Magnetic field: B = ∇ × A (as always)

Other Gauge Choices

Lorenz Gauge

∂_μA^μ = 0

Manifestly Lorentz covariant, used in covariant quantization

Temporal Gauge

A⁰ = 0

Simple but not manifestly covariant

Axial Gauge

n_μA^μ = 0 (n = fixed vector)

Useful for non-Abelian gauge theories

Light-Cone Gauge

A⁺ = 0

Natural for high-energy scattering

7.3 Mode Expansion in Coulomb Gauge

Expand the vector potential in plane waves:

$$\mathbf{A}(\mathbf{x}, t) = \int \frac{d^3k}{(2\pi)^3} \frac{1}{\sqrt{2\omega_k}} \sum_{\lambda=1}^2 \left[ \hat{a}_{\mathbf{k},\lambda} \boldsymbol{\epsilon}^{(\lambda)}(\mathbf{k}) e^{i\mathbf{k} \cdot \mathbf{x} - i\omega_k t} + \hat{a}_{\mathbf{k},\lambda}^\dagger \boldsymbol{\epsilon}^{(\lambda)*}(\mathbf{k}) e^{-i\mathbf{k} \cdot \mathbf{x} + i\omega_k t} \right]$$

where:

ω_k = |k| (photons are massless!)
λ = 1, 2 labels the two transverse polarizations
ε^(λ)(k) are polarization vectors

7.4 Polarization Vectors

The polarization vectors must satisfy:

\begin{align*} \mathbf{k} \cdot \boldsymbol{\epsilon}^{(\lambda)}(\mathbf{k}) &= 0 \quad \text{(transversality from } \nabla \cdot \mathbf{A} = 0\text{)} \\ \boldsymbol{\epsilon}^{(\lambda)}(\mathbf{k}) \cdot \boldsymbol{\epsilon}^{(\lambda')*}(\mathbf{k}) &= \delta^{\lambda\lambda'} \quad \text{(orthonormality)} \\ \sum_{\lambda=1}^2 \epsilon_i^{(\lambda)}(\mathbf{k}) \epsilon_j^{(\lambda)*}(\mathbf{k}) &= \delta_{ij} - \frac{k_i k_j}{|\mathbf{k}|^2} \quad \text{(completeness)} \end{align*}

💡Physical Meaning of Polarizations

For a photon traveling in the z-direction (k = k ẑ), the two polarization vectors are:

ε⁽¹⁾ = x̂ (linear polarization in x-direction)
ε⁽²⁾ = ŷ (linear polarization in y-direction)

Alternatively, we can use circular polarizations:

ε⁽⁺⁾ = (x̂ + iŷ)/√2 (right circular, helicity +1)
ε^(-) = (x̂ - iŷ)/√2 (left circular, helicity -1)

These correspond to the photon's spin projections along its direction of motion!

7.5 Canonical Commutation Relations

The creation and annihilation operators satisfy:

\begin{align*} [\hat{a}_{\mathbf{k},\lambda}, \hat{a}_{\mathbf{k}',\lambda'}^\dagger] &= (2\pi)^3 \delta^3(\mathbf{k} - \mathbf{k}') \delta^{\lambda\lambda'} \\ [\hat{a}_{\mathbf{k},\lambda}, \hat{a}_{\mathbf{k}',\lambda'}] &= 0 \\ [\hat{a}_{\mathbf{k},\lambda}^\dagger, \hat{a}_{\mathbf{k}',\lambda'}^\dagger] &= 0 \end{align*}

These are bosonic commutation relations (photons are spin-1 bosons, as expected from spin-statistics theorem!).

7.6 Hamiltonian and Photon States

The Hamiltonian is:

$$\hat{H} = \int \frac{d^3k}{(2\pi)^3} \omega_k \sum_{\lambda=1}^2 \hat{a}_{\mathbf{k},\lambda}^\dagger \hat{a}_{\mathbf{k},\lambda}$$

(after normal ordering to remove infinite vacuum energy).

Photon states:

Vacuum: |0⟩ (no photons)
One-photon state: |k,λ⟩ = â^†_k,λ|0⟩
Energy: ω_k = |k|, momentum: k, polarization: λ
Two-photon state: |k₁,λ₁; k₂,λ₂⟩ = â^†_k₁,λ₁ â^†_k₂,λ₂|0⟩
Bosonic: can have multiple photons with same k, λ!
Coherent states: |α⟩ ∝ exp(αâ^† - α*â)|0⟩
Closest quantum analog to classical EM wave (laser light!)

7.7 Photon Propagator

The Feynman propagator for the photon field is:

$$D_{\mu\nu}^F(x - y) = \langle 0 | T\{\hat{A}_\mu(x) \hat{A}_\nu(y)\} | 0 \rangle$$

In momentum space (Feynman gauge):

$$\tilde{D}_{\mu\nu}^F(k) = \frac{-i g_{\mu\nu}}{k^2 + i\epsilon}$$

Note: This is a tensor (has μ, ν indices), not a scalar! It's the photon version of the scalar propagator i/(p² - m² + iε).

In Coulomb gauge, the propagator is more complicated (not manifestly covariant):

$$\tilde{D}_{ij}^{\text{Coulomb}}(k) = \frac{-i}{k^2 + i\epsilon} \left(\delta_{ij} - \frac{k_i k_j}{|\mathbf{k}|^2}\right)$$

The projection operator (δ_ij - k_ik_j/|k|²) ensures we're only propagating transverse modes!

7.8 Coupling to Matter: QED Lagrangian

To describe interactions with charged particles (like electrons), we use the QED Lagrangian:

$$\mathcal{L}_{\text{QED}} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \bar{\psi}(i\gamma^\mu D_\mu - m)\psi$$

where D_μ = ∂_μ - ieA_μ is the covariant derivative.

This gives the interaction term:

$$\mathcal{L}_{\text{int}} = e \bar{\psi} \gamma^\mu \psi A_\mu = e j^\mu A_\mu$$

This is the famous electromagnetic vertex: electrons couple to photons with strength e (the electric charge)!

💡The QED Vertex in Feynman Diagrams

In Feynman diagrams, the vertex where an electron line meets a photon line contributes a factor:

-ie γ^μ

This is the fundamental interaction of QED! Every electromagnetic process (scattering, emission, absorption) is built from this basic vertex.

7.9 Why Photons Are Massless

A crucial fact: photons have zero mass (m_γ = 0).

This is enforced by gauge invariance! If we tried to add a mass term:

$$\mathcal{L}_{\text{mass}} = \frac{1}{2}m^2 A_\mu A^\mu$$

This is NOT gauge invariant:

$$A_\mu \to A_\mu + \partial_\mu \chi \quad \Rightarrow \quad A_\mu A^\mu \to A_\mu A^\mu + 2A^\mu \partial_\mu \chi + (\partial\chi)^2 \quad \text{(not invariant!)}$$

Therefore, gauge invariance forbids photon mass!

Experimental Limit:

If photons had mass, Coulomb's law would be modified to a Yukawa potential. Experiments constrain:

m_γ < 10^-18 eV

Consistent with m_γ = 0 exactly!

Practice Problems

📝 Practice Problems

Progress: 0/2 (0%)

⭐⭐ Medium

Show that in Coulomb gauge (∇·A = 0), the electromagnetic Lagrangian density becomes:

ℒ = -½(E² - B²) where E = -∂_tA

⭐⭐⭐ Hard

Explain why the photon has only 2 polarization states despite A^μ having 4 components.

📊 Problem Set Statistics

Total Problems

Attempted

Completion

⚠️Common Mistakes to Avoid

❌

Mistake:

Thinking the photon has 4 polarization states (one for each A^μ component)

🤔

Why it's wrong:

Gauge freedom and constraints eliminate 2 of the 4 components of A^μ.

✅

Correct approach:

After gauge fixing, the photon has only 2 physical polarization states (transverse modes). The timelike and longitudinal components are pure gauge!

❌

Mistake:

Confusing gauge transformations with physical changes

🤔

Why it's wrong:

Gauge transformations are redundancies in our mathematical description, not physical processes.

✅

Correct approach:

Gauge transformations A^μ → A^μ + ∂^μχ don't change any physical observables (E, B fields). They're redundancy in our description, not real physics!

❌

Mistake:

Forgetting that A^0 is not a dynamical variable

🤔

Why it's wrong:

A^0 has no conjugate momentum—it appears without time derivatives in the Lagrangian.

✅

Correct approach:

A^0 has no time derivative in the Lagrangian! It's not an independent degree of freedom—it's a constraint equation (Gauss's law).

🎯 Key Takeaways

Electromagnetic field A^μ has gauge redundancy A^μ → A^μ + ∂^μχ
Must fix gauge before quantization (e.g., Coulomb gauge ∇·A = 0)
Photon has 2 transverse polarizations (4 components - 1 gauge - 1 constraint)
Polarization vectors: ε^(λ) with k·ε = 0
Bosonic commutators: [â, â^†] = 1 (photons are spin-1 bosons)
Hamiltonian: Ĥ = ∫d³k ω_k Σ_λ â^†_k,λâ_k,λ
Photon propagator: D_μν(k) = -ig_μν/(k² + iε) (Feynman gauge)
QED vertex: eψ̄γ^μψA_μ (electron-photon coupling)
Gauge invariance → m_γ = 0 (photons massless!)
Part 2 complete! Next: Path integral formulation!

🎉 Congratulations!

You've completed Part II: Canonical Quantization!

You now understand:

How to promote classical fields to quantum operators
Creation and annihilation operators for bosons and fermions
Fock space and variable particle number
Propagators and their role in scattering amplitudes
The spin-statistics theorem and why it's unavoidable
Gauge theories and photon quantization

Next up: Part III - Path Integral Formulation, where we'll develop a completely different (and often more powerful) approach to QFT!

Quantum Field Theory Course

Photon Field Quantization

🔗Course Connections

📚Prerequisites

Video Lecture

7.1 The Gauge Freedom Problem

⚠️ The Problem: Gauge Invariance

7.2 Gauge Fixing Choices

Coulomb Gauge (Radiation Gauge)

Gauge Condition:

Other Gauge Choices

Lorenz Gauge

Temporal Gauge

Axial Gauge

Light-Cone Gauge

7.3 Mode Expansion in Coulomb Gauge

7.4 Polarization Vectors

💡Physical Meaning of Polarizations

7.5 Canonical Commutation Relations

7.6 Hamiltonian and Photon States

7.7 Photon Propagator

7.8 Coupling to Matter: QED Lagrangian

💡The QED Vertex in Feynman Diagrams

7.9 Why Photons Are Massless

Practice Problems

📝 Practice Problems

📊 Problem Set Statistics

⚠️Common Mistakes to Avoid

Mistake:

Why it's wrong:

Correct approach:

Mistake:

Why it's wrong:

Correct approach:

Mistake:

Why it's wrong:

Correct approach:

🎯 Key Takeaways

🎉 Congratulations!