Chapter 2: Canonical Quantization

Canonical quantization is the procedure by which we promote classical fields to quantum operators. By imposing commutation (or anticommutation) relations, we transform the classical field into a quantum field capable of creating and destroying particles. This reveals that particles are quantized excitations of underlying fields — the central insight of QFT.

Derivation 1: Quantizing the Scalar Field

In classical mechanics, quantization promotes phase space variables to operators:$q \to \hat{q}$, $p \to \hat{p}$, with $[\hat{q}, \hat{p}] = i\hbar$. For fields, we apply the same logic at each spatial point.

From Classical to Quantum Fields

For the real Klein-Gordon field, the classical canonical variables are:

$\text{Field:} \quad \phi(\mathbf{x}, t), \qquad \text{Conjugate momentum:} \quad \pi(\mathbf{x}, t) = \dot{\phi}(\mathbf{x}, t)$

We promote these to operators and impose the equal-time canonical commutation relations(in natural units $\hbar = 1$):

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

And the vanishing commutators:

$[\hat{\phi}(\mathbf{x}, t), \hat{\phi}(\mathbf{y}, t)] = 0, \qquad [\hat{\pi}(\mathbf{x}, t), \hat{\pi}(\mathbf{y}, t)] = 0$

The Dirac delta function $\delta^3(\mathbf{x} - \mathbf{y})$ is the continuum analog of the Kronecker delta in $[q_i, p_j] = i\delta_{ij}$. It reflects that field degrees of freedom at different spatial points are independent.

Why These Commutation Relations?

Physical motivation: The field at position $\mathbf{x}$plays the role of a generalized coordinate $q_\mathbf{x}$. In the continuum limit of a lattice with spacing $a$:

$[q_\mathbf{x}, p_\mathbf{y}] = i\delta_{\mathbf{x}\mathbf{y}} \xrightarrow{a \to 0} i\delta^3(\mathbf{x} - \mathbf{y})$

The Heisenberg Picture

In the Heisenberg picture, operators evolve in time while states are fixed:

$\hat{\phi}(\mathbf{x}, t) = e^{i\hat{H}t} \hat{\phi}(\mathbf{x}, 0) e^{-i\hat{H}t}$

The Heisenberg equation of motion $i\partial_t \hat{\phi} = [\hat{\phi}, \hat{H}]$ reproduces the Klein-Gordon equation as an operator equation, ensuring consistency between the classical and quantum dynamics.

Derivation 2: Creation and Annihilation Operators

Just as the quantum harmonic oscillator is solved by introducing ladder operators, we decompose the quantum field into momentum modes, each of which acts as an independent oscillator.

Mode Expansion of the Quantum Field

We expand the field operator in plane waves:

$\boxed{\hat{\phi}(x) = \int \frac{d^3k}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\mathbf{k}}}} \left[ \hat{a}(\mathbf{k}) \, e^{-ik \cdot x} + \hat{a}^\dagger(\mathbf{k}) \, e^{ik \cdot x} \right]}$

where $k \cdot x = \omega_\mathbf{k} t - \mathbf{k} \cdot \mathbf{x}$ with$\omega_\mathbf{k} = \sqrt{|\mathbf{k}|^2 + m^2}$. The conjugate momentum is:

$\hat{\pi}(x) = \dot{\hat{\phi}}(x) = -i \int \frac{d^3k}{(2\pi)^3} \sqrt{\frac{\omega_{\mathbf{k}}}{2}} \left[ \hat{a}(\mathbf{k}) \, e^{-ik \cdot x} - \hat{a}^\dagger(\mathbf{k}) \, e^{ik \cdot x} \right]$

Deriving the Commutation Relations

Substituting the mode expansion into the canonical commutation relation$[\hat{\phi}(\mathbf{x}), \hat{\pi}(\mathbf{y})] = i\delta^3(\mathbf{x} - \mathbf{y})$and using the orthogonality of plane waves, we find:

$\boxed{[\hat{a}(\mathbf{k}), \hat{a}^\dagger(\mathbf{k}')] = (2\pi)^3 \delta^3(\mathbf{k} - \mathbf{k}')}$

With all other commutators vanishing:

$[\hat{a}(\mathbf{k}), \hat{a}(\mathbf{k}')] = 0, \qquad [\hat{a}^\dagger(\mathbf{k}), \hat{a}^\dagger(\mathbf{k}')] = 0$

Proof of the Commutation Relation

Let us verify this explicitly. At equal time $t = 0$:

Compute $[\hat{\phi}(\mathbf{x}), \hat{\pi}(\mathbf{y})]$:

$= \int \frac{d^3k}{(2\pi)^3} \frac{1}{\sqrt{2\omega_k}} \int \frac{d^3k'}{(2\pi)^3} \sqrt{\frac{\omega_{k'}}{2}} \cdot (-i) \left[ -[\hat{a}(\mathbf{k}), \hat{a}^\dagger(\mathbf{k}')] e^{i\mathbf{k}\cdot\mathbf{x}} e^{-i\mathbf{k}'\cdot\mathbf{y}} - [\hat{a}^\dagger(\mathbf{k}), \hat{a}(\mathbf{k}')] e^{-i\mathbf{k}\cdot\mathbf{x}} e^{i\mathbf{k}'\cdot\mathbf{y}} \right]$

Using $[\hat{a}(\mathbf{k}), \hat{a}^\dagger(\mathbf{k}')] = (2\pi)^3\delta^3(\mathbf{k}-\mathbf{k}')$ and integrating over $\mathbf{k}'$:

$= i \int \frac{d^3k}{(2\pi)^3} \frac{1}{2} \left[ e^{i\mathbf{k}\cdot(\mathbf{x}-\mathbf{y})} + e^{-i\mathbf{k}\cdot(\mathbf{x}-\mathbf{y})} \right] = i\delta^3(\mathbf{x}-\mathbf{y})$

The factor $1/\sqrt{2\omega_\mathbf{k}}$ in the mode expansion is precisely what is needed to make this work. It also ensures Lorentz-invariant normalization of states.

Derivation 3: Fock Space

The Hilbert space of the quantum field theory — the Fock space — is built by acting with creation operators on the vacuum state. It describes states with arbitrary numbers of particles.

The Vacuum State

The vacuum $|0\rangle$ is defined as the state annihilated by all annihilation operators:

$\hat{a}(\mathbf{k}) |0\rangle = 0 \quad \text{for all } \mathbf{k}$

This is the state of lowest energy — no particles are present. However, as we shall see, it is not a state of zero energy.

One-Particle States

A single particle with momentum $\mathbf{k}$ is created by:

$|\mathbf{k}\rangle = \hat{a}^\dagger(\mathbf{k}) |0\rangle$

This state has energy $\omega_\mathbf{k} = \sqrt{|\mathbf{k}|^2 + m^2}$ and momentum $\mathbf{k}$. The normalization is Lorentz-invariant:

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

Multi-Particle States

An $n$-particle state is obtained by acting $n$ times:

$|\mathbf{k}_1, \mathbf{k}_2, \ldots, \mathbf{k}_n\rangle = \hat{a}^\dagger(\mathbf{k}_1) \hat{a}^\dagger(\mathbf{k}_2) \cdots \hat{a}^\dagger(\mathbf{k}_n) |0\rangle$

Since $[\hat{a}^\dagger(\mathbf{k}), \hat{a}^\dagger(\mathbf{k}')] = 0$, these states are symmetric under exchange of any two particles — these are bosons!

The Number Operator

The number operator counts the total number of particles:

$\hat{N} = \int \frac{d^3k}{(2\pi)^3} \hat{a}^\dagger(\mathbf{k}) \hat{a}(\mathbf{k})$

Verification: Using the commutation relations:

$\hat{N} |0\rangle = 0, \quad \hat{N} |\mathbf{k}\rangle = |\mathbf{k}\rangle, \quad \hat{N} |\mathbf{k}_1, \mathbf{k}_2\rangle = 2|\mathbf{k}_1, \mathbf{k}_2\rangle$

The full Fock space is the direct sum:

$\mathcal{F} = \mathcal{H}_0 \oplus \mathcal{H}_1 \oplus \mathcal{H}_2 \oplus \cdots$

where $\mathcal{H}_n$ is the $n$-particle Hilbert space. Unlike ordinary quantum mechanics, Fock space accommodates states with different particle numbers — essential for describing particle creation and annihilation processes.

Lorentz-Invariant Normalization

Our states use the relativistically invariant normalization:

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

An alternative, explicitly Lorentz-invariant convention uses $|k\rangle_\text{cov} = \sqrt{2\omega_\mathbf{k}} |\mathbf{k}\rangle$, giving $ {}_\text{cov}\langle k|k'\rangle_\text{cov} = 2\omega_\mathbf{k}(2\pi)^3\delta^3(\mathbf{k}-\mathbf{k}')$. The Lorentz-invariant integration measure is:

$\int \frac{d^3k}{(2\pi)^3 2\omega_\mathbf{k}} = \int \frac{d^4k}{(2\pi)^4} (2\pi)\delta(k^2 - m^2)\theta(k^0)$

This measure is invariant under Lorentz transformations, ensuring our quantum theory respects the symmetries of special relativity.

Derivation 4: Normal Ordering and Vacuum Energy

When we express the Hamiltonian in terms of creation and annihilation operators, we encounter the infinite vacuum energy — the most dramatic consequence of field quantization.

The Hamiltonian in Terms of Ladder Operators

The classical Hamiltonian for the Klein-Gordon field is:

$H = \int d^3x \left[ \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 \right]$

Substituting the mode expansion for $\hat{\phi}$ and $\hat{\pi}$ and using the orthogonality of plane waves:

$\hat{H} = \int \frac{d^3k}{(2\pi)^3} \frac{\omega_\mathbf{k}}{2} \left[ \hat{a}(\mathbf{k})\hat{a}^\dagger(\mathbf{k}) + \hat{a}^\dagger(\mathbf{k})\hat{a}(\mathbf{k}) \right]$

Using $\hat{a}\hat{a}^\dagger = \hat{a}^\dagger\hat{a} + (2\pi)^3\delta^3(0)$:

$\hat{H} = \int \frac{d^3k}{(2\pi)^3} \omega_\mathbf{k} \left[ \hat{a}^\dagger(\mathbf{k})\hat{a}(\mathbf{k}) + \frac{1}{2}(2\pi)^3\delta^3(0) \right]$

The Infinite Vacuum Energy

The vacuum expectation value of the Hamiltonian is:

$\langle 0 | \hat{H} | 0 \rangle = \frac{1}{2} \int \frac{d^3k}{(2\pi)^3} \omega_\mathbf{k} \cdot (2\pi)^3\delta^3(0) = \infty$

This is doubly infinite: the $\delta^3(0)$ is an infrared divergence (from infinite volume, regularized as $V = (2\pi)^3\delta^3(0)$), and the integral $\int d^3k \, \omega_\mathbf{k}$ diverges in the UV. The vacuum energy density is:

$\frac{E_\text{vac}}{V} = \int \frac{d^3k}{(2\pi)^3} \frac{\omega_\mathbf{k}}{2} \sim \frac{\Lambda^4}{16\pi^2}$

where $\Lambda$ is a UV cutoff. Each mode of the field contributes $\omega_\mathbf{k}/2$ of zero-point energy, just like a quantum harmonic oscillator.

Normal Ordering

We define normal ordering $:\hat{O}:$ as rearranging all creation operators to the left of annihilation operators:

$\boxed{:\hat{H}: = \int \frac{d^3k}{(2\pi)^3} \omega_\mathbf{k} \, \hat{a}^\dagger(\mathbf{k})\hat{a}(\mathbf{k})}$

This removes the infinite vacuum energy: $\langle 0 | :\hat{H}: | 0\rangle = 0$. The physical justification is that only energy differences are measurable (in the absence of gravity).

The Cosmological Constant Problem

The deepest puzzle: In general relativity, gravity couples to absolute energy, not just energy differences. If we take the vacuum energy seriously with $\Lambda \sim M_\text{Planck}$, the predicted cosmological constant is $\sim 10^{120}$ times larger than observed. This is the cosmological constant problem — arguably the worst prediction in physics. Its resolution may require a deeper understanding of quantum gravity.

Derivation 5: Dirac Field Quantization

Quantizing the Dirac field requires anticommutation relations instead of commutation relations. This is not a choice — it is a necessity demanded by consistency of the theory.

Mode Expansion of the Dirac Field

The Dirac field operator is expanded as:

$\hat{\psi}(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_\mathbf{p}}} \sum_{s=1}^{2} \left[ \hat{b}^s(\mathbf{p}) \, u^s(\mathbf{p}) \, e^{-ip \cdot x} + \hat{d}^{s\dagger}(\mathbf{p}) \, v^s(\mathbf{p}) \, e^{ip \cdot x} \right]$

Here $\hat{b}^s$ annihilates a particle (e.g., electron) with spin $s$, and $\hat{d}^{s\dagger}$ creates an antiparticle (positron) with spin $s$. The spinors $u^s$ and $v^s$ are the positive and negative energy solutions of the Dirac equation.

Why Anticommutation?

If we try to quantize with commutation relations, the Hamiltonian becomes:

$\hat{H}_\text{wrong} = \int \frac{d^3p}{(2\pi)^3} E_\mathbf{p} \sum_s \left[ \hat{b}^{s\dagger}\hat{b}^s - \hat{d}^{s\dagger}\hat{d}^s \right]$

The minus sign before $\hat{d}^\dagger\hat{d}$ means creating antiparticleslowers the energy — the spectrum is unbounded below! With anticommutationrelations, $\hat{d}\hat{d}^\dagger = 1 - \hat{d}^\dagger\hat{d}$ (up to delta functions), which flips the sign:

$\boxed{\{\hat{b}^r(\mathbf{p}), \hat{b}^{s\dagger}(\mathbf{q})\} = \{\hat{d}^r(\mathbf{p}), \hat{d}^{s\dagger}(\mathbf{q})\} = (2\pi)^3 \delta^{rs} \delta^3(\mathbf{p} - \mathbf{q})}$

All other anticommutators vanish. The correctly normal-ordered Hamiltonian is:

$:\hat{H}: = \int \frac{d^3p}{(2\pi)^3} E_\mathbf{p} \sum_s \left[ \hat{b}^{s\dagger}\hat{b}^s + \hat{d}^{s\dagger}\hat{d}^s \right]$

Now both particles and antiparticles contribute positive energy.

Fermi-Dirac Statistics

The anticommutation relations immediately give:

$(\hat{b}^{s\dagger}(\mathbf{p}))^2 = 0$

This is the Pauli exclusion principle: you cannot create two identical fermions in the same state! Multi-particle states are automatically antisymmetric:

$|\mathbf{p}_1, \mathbf{p}_2\rangle = -|\mathbf{p}_2, \mathbf{p}_1\rangle$

The Spin-Statistics Theorem

A deep result of relativistic QFT: Integer-spin fields (scalars, vectors) must be quantized with commutation relations (bosons), while half-integer-spin fields (spinors) must use anticommutation relations (fermions). This is the spin-statistics theorem, proven by Pauli in 1940. The proof requires Lorentz invariance, locality, and positive-definite energy — all fundamental principles of QFT. Violating spin-statistics leads to either negative-norm states (ghosts) or an energy spectrum unbounded below.

Simulation: Mode Decomposition & Vacuum Fluctuations

This simulation visualizes the scalar field's mode decomposition showing how individual plane-wave modes combine to form the total field configuration, the time evolution of a single mode, the vacuum fluctuation power spectrum $\langle 0|\hat{\phi}_k\hat{\phi}_{-k}|0\rangle = 1/(2\omega_k)$, and the divergent vacuum energy as a function of the UV cutoff.

Canonical Quantization: Modes & Vacuum Fluctuations

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Causality and Microcausality

A fundamental requirement of relativistic QFT is microcausality: measurements at spacelike-separated points cannot influence each other. This translates to the vanishing of commutators at spacelike separation:

$[\hat{\phi}(x), \hat{\phi}(y)] = 0 \quad \text{for} \quad (x-y)^2 < 0 \text{ (spacelike)}$

This can be verified by explicit computation. The commutator for a free scalar field is:

$[\hat{\phi}(x), \hat{\phi}(y)] = \int \frac{d^3k}{(2\pi)^3 2\omega_\mathbf{k}} \left(e^{-ik(x-y)} - e^{ik(x-y)}\right) \equiv i\Delta(x-y)$

The function $\Delta(x-y)$ is Lorentz-invariant. At equal times ($x^0 = y^0$), the two exponentials differ only in the sign of $\mathbf{k}$, and the integral vanishes by symmetry. By Lorentz invariance, $\Delta(x-y) = 0$ for all spacelike separations.

Deep significance: Microcausality is essential for consistency with special relativity. It ensures that the theory respects the light-cone structure of spacetime. For fermionic fields, causality requires the anticommutator$\{\hat{\psi}(x), \hat{\bar{\psi}}(y)\}$ to vanish at spacelike separation. This is automatically guaranteed by the anticommutation relations — another instance of the spin-statistics connection.

The Feynman propagator, by contrast, does not vanish at spacelike separation:

$\Delta_F(x-y) \sim e^{-m|x-y|} \neq 0 \quad \text{for spacelike } (x-y)$

This exponential tail outside the light cone reflects virtual particle contributions. It does not violate causality because the propagator appears only as an intermediate quantity in calculations — observable amplitudes respect the light cone.

The Casimir Effect: Vacuum Energy Made Observable

While the absolute vacuum energy is removed by normal ordering, changes in vacuum energy due to boundary conditions are physical and measurable. The Casimir effectis the paradigmatic example.

Consider two parallel conducting plates separated by distance $d$. The allowed modes between the plates are quantized: $k_n = n\pi/d$. The vacuum energy per unit area between the plates, minus the vacuum energy without plates, gives a finite attractive force:

$\frac{F}{A} = -\frac{\pi^2}{240 d^4}$

The Casimir force — measured experimentally to within 1% accuracy

This effect has been measured with high precision, confirming that the quantum vacuum is not empty but seethes with zero-point fluctuations whose effects are physically real.

Summary: From Fields to Particles

Quantization Procedure

Promote fields and conjugate momenta to operators satisfying canonical commutation (or anticommutation) relations at equal times. This is the field-theoretic generalization of $[\hat{q}, \hat{p}] = i$.

Particles as Excitations

The mode expansion introduces creation ($\hat{a}^\dagger$) and annihilation ($\hat{a}$) operators. Particles are quantized excitations: $|\mathbf{k}\rangle = \hat{a}^\dagger(\mathbf{k})|0\rangle$.

Vacuum Energy and Normal Ordering

Zero-point fluctuations give an infinite vacuum energy. Normal ordering removes this, but the question of whether vacuum energy gravitates remains one of the deepest open problems.

Spin-Statistics Connection

Consistency demands bosonic fields use commutators and fermionic fields use anticommutators, giving rise to Bose-Einstein and Fermi-Dirac statistics respectively.