What it means to quantize a field.

The cleanest entry into quantum field theory; where fields become operators and particles emerge as their excitations.

This is the first of the QFT documents. The prerequisite is everything we’ve built: Lagrangian mechanics, special relativity and tensors, classical field theory (especially the Klein-Gordon theory), quantum mechanics, and a bit of statistical mechanics. If you’re fluent with the harmonic oscillator in QM and can write down the Klein-Gordon Lagrangian, you’re ready.

The goal of this document is modest but profound: take the classical Klein-Gordon field and quantize it, using the same canonical procedure that turns classical mechanics into quantum mechanics. What comes out is not a theory of a single particle with a wave function; it’s a theory of a field operator whose excitations are particles. This reframing is the conceptual foundation of all QFT.

Signature Convention

Switching back to the mostly-minus convention $\eta_{\mu\nu} = \text{diag}(+,-,-,-)$ for the QFT series, consistent with particle physics practice and the classical field theory document. (Recall the GR document used mostly-plus.)

We also set $\hbar = c = 1$. Factors restore if you need SI units.

Table of Contents

1. The Setup: Why Quantize a Field?
2. Canonical Quantization Recipe
3. The Classical Klein-Gordon Field in Review
4. Quantizing: Equal-Time Commutation Relations
5. Fourier Decomposition and the Oscillator Picture
6. Creation and Annihilation Operators
7. The Hamiltonian and Vacuum Energy
8. Fock Space and Particle States
9. Normalization and Lorentz Invariance
10. Propagators
11. The Complex Scalar Field and Antiparticles
12. Physical Content and the Casimir Effect
13. Appendix: Formulas and Identities

1. The Setup: Why Quantize a Field?

A Persistent Problem with “Relativistic Quantum Mechanics”

In ordinary quantum mechanics, a particle is described by a wave function $\psi(\vec x, t)$ obeying the Schrödinger equation. To make this relativistic, you might try to replace the Schrödinger equation with a relativistic wave equation; say, Klein-Gordon or Dirac.

This doesn’t work, for several related reasons:

Negative energies. The Klein-Gordon equation $(\Box + m^2)\phi = 0$ has plane-wave solutions $e^{-iEt + i\vec p \cdot \vec x}$ with $E^2 = |\vec p|^2 + m^2$; so $E = \pm\sqrt{|\vec p|^2 + m^2}$. Negative energy solutions seem unphysical.

Negative probabilities. The natural “probability density” for Klein-Gordon, $\rho = i(\phi^* \partial_t\phi - \phi \partial_t\phi^*)$, isn’t positive-definite. You can’t interpret it as a probability.

Particle number isn’t conserved. At sufficient energies, particles can be created and destroyed. $e^+e^- \to \gamma\gamma$ annihilates two particles and creates two. No single-particle wave function can describe this; the very dimension of the Hilbert space changes.

Localization is problematic. Trying to build a position-space wave function for a relativistic particle leads to pathologies: the “wave function” spreads superluminally, and attempts to confine a particle to a region smaller than its Compton wavelength produce enough energy to create new particles.

The fundamental insight: relativistic quantum mechanics is not a theory of a fixed number of particles. It’s a theory of fields, whose excitations happen to look like particles.

The Shift: Fields Are Primary

In QFT, the fundamental object is not a particle but a field $\phi(\vec x, t)$; a dynamical variable defined at every spacetime point. The field is quantized (promoted to an operator); particles emerge as discrete excitations.

This resolves all the problems above:

• Negative-energy classical solutions are reinterpreted as antiparticles
• Probabilities are defined over states of the field, not positions of particles; always positive
• Particle creation and annihilation are built in (creation/annihilation operators)
• Localization limits are natural; the field is defined at points, but extracting a single localized particle is what’s restricted

The Plan

1. Start with the classical Klein-Gordon Lagrangian
2. Identify the canonical variables (field and conjugate momentum)
3. Impose canonical commutation relations; promote them to operators
4. Decompose into Fourier modes; each mode becomes a quantum harmonic oscillator
5. Creation/annihilation operators build up a Hilbert space (“Fock space”) with states of any particle number
6. The “particles” are what you see when you examine the states

The procedure is mechanical. The interpretation is deep.

2. Canonical Quantization Recipe

Remember the recipe for ordinary quantum mechanics:

1. Start with a classical system with generalized coordinates $q_i$ and conjugate momenta $p_i = \partial L/\partial\dot q_i$
2. Promote $q_i$ and $p_i$ to operators $\hat q_i$ and $\hat p_i$
3. Replace Poisson brackets with commutators via Dirac’s rule: $\{q_i, p_j\} = \delta_{ij} \to [\hat q_i, \hat p_j] = i\hbar\delta_{ij}$
4. Hamiltonians, observables, time evolution all follow

For a classical field theory, the canonical “coordinates” are the field values $\phi(\vec x)$ at each spatial point $\vec x$; a continuous infinity of them. The conjugate momentum is:

$\pi(\vec x) = \frac{\partial \mathcal{L}}{\partial \dot\phi(\vec x)}$

The canonical commutation relation becomes:

$[\hat\phi(\vec x, t), \hat\pi(\vec y, t)] = i\delta^3(\vec x - \vec y)$

The Kronecker $\delta_{ij}$ becomes a Dirac $\delta^3(\vec x - \vec y)$; the continuum analog.

Note the equal time; both operators at the same time $t$. This is a constraint imposed at one moment; time evolution is separate (governed by the Heisenberg equations of motion).

That’s it. The rest of this document is working out what this recipe produces for the Klein-Gordon field.

3. The Classical Klein-Gordon Field in Review

Quick review from the classical field theory document.

Lagrangian Density

$\mathcal{L} = \tfrac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi) - \tfrac{1}{2}m^2\phi^2$

Expanding $\partial_\mu\phi \partial^\mu \phi = \dot\phi^2 - (\nabla\phi)^2$ (with mostly-minus metric):

$\mathcal{L} = \tfrac{1}{2}\dot\phi^2 - \tfrac{1}{2}(\nabla\phi)^2 - \tfrac{1}{2}m^2\phi^2$

Equation of Motion (Klein-Gordon)

$(\Box + m^2)\phi = 0$

Conjugate Momentum

$\pi(\vec x, t) = \frac{\partial\mathcal{L}}{\partial\dot\phi} = \dot\phi(\vec x, t)$

Hamiltonian Density

$\mathcal{H} = \pi\dot\phi - \mathcal{L} = \tfrac{1}{2}\pi^2 + \tfrac{1}{2}(\nabla\phi)^2 + \tfrac{1}{2}m^2\phi^2$

Hamiltonian

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

Every term is positive; the classical field has positive energy density everywhere. Negative energies will reappear as a subtlety when we quantize, and their resolution is important.

Plane Wave Solutions

Classical solutions of KG:

$\phi(x) = \int \frac{d^3p}{(2\pi)^3 \, 2\omega_p}\left[a(\vec p)e^{-ip\cdot x} + a^*(\vec p)e^{+ip\cdot x}\right]$

where $p\cdot x = \omega_p t - \vec p\cdot \vec x$ and $\omega_p = \sqrt{|\vec p|^2 + m^2} > 0$ is always positive.

The two terms are positive-frequency ($e^{-i\omega_p t}$) and negative-frequency ($e^{+i\omega_p t}$) modes. The coefficient $a^*(\vec p)$ is the complex conjugate of $a(\vec p)$; enforced by $\phi$ being real.

Keep this expression close; after quantization, $a(\vec p)$ and $a^*(\vec p)$ become the annihilation and creation operators.

4. Quantizing: Equal-Time Commutation Relations

The Canonical Commutation Relations

Promote $\phi$ and $\pi$ to operators, imposing:

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

$[\hat\phi(\vec x, t), \hat\phi(\vec y, t)] = 0$

$[\hat\pi(\vec x, t), \hat\pi(\vec y, t)] = 0$

All at the same time $t$. Equivalently: the field values at different spatial points (at the same time) commute; it’s only the conjugate pair $(\phi, \pi)$ at the same point that has a non-trivial commutator.

From here on, I’ll drop the hats; every $\phi$ is now an operator.

Why $\delta^3$ Instead of $\delta_{ij}$?

In QM, $[\hat q_i, \hat p_j] = i\delta_{ij}$ with integer indices. Here, the “index” is a continuous position $\vec x$, and the Kronecker delta becomes a Dirac delta. Physically: the field values at distinct spatial points are independent degrees of freedom, so they commute; the “self-interaction” at each point is what has a non-trivial commutator.

Heisenberg Equations of Motion

Under canonical quantization, time evolution of operators is governed by:

$i\frac{d\hat O}{dt} = [\hat O, H]$

Applied to $\phi$ and $\pi$ with our Hamiltonian, this reproduces the Klein-Gordon equation; but now as an operator equation:

$(\Box + m^2)\phi = 0$

(Check: $i\dot\phi = [\phi, H] = i\pi$ gives $\dot\phi = \pi$. Then $i\dot\pi = [\pi, H]$ works out to $\dot\pi = \nabla^2\phi - m^2\phi$. Combining: $\ddot\phi = \nabla^2\phi - m^2\phi$, which is KG.) The classical equation is preserved; we’re quantizing, not changing the dynamics.

What Does a “Field Operator” Mean?

A field operator $\phi(\vec x, t)$ is an operator that acts on states in the Hilbert space of the theory. It has different value-operators at every spacetime point. Specific matrix elements $\langle \alpha | \phi(\vec x, t) | \beta\rangle$ are ordinary complex numbers depending on $\vec x, t$.

This is not a wave function. The wave function concept is gone. $\phi(\vec x, t)$ is an observable; a measurable field; now promoted to an operator.

5. Fourier Decomposition and the Oscillator Picture

Commutation relations and equations of motion are all well and good, but to make progress we need to solve them. The trick: decompose $\phi$ into modes.

The Ansatz

Substitute the classical plane-wave solution form, but with $a(\vec p), a^*(\vec p)$ now promoted to operators $a_{\vec p}, a^\dagger_{\vec p}$ (with $\dagger$ because we’re in Hilbert space; complex conjugate becomes adjoint):

$\boxed{\phi(\vec x, t) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left[a_{\vec p}e^{-ip\cdot x} + a^\dagger_{\vec p} e^{+ip\cdot x}\right]}$

with $\omega_p = \sqrt{|\vec p|^2 + m^2}$ and $p\cdot x = \omega_p t - \vec p\cdot \vec x$.

The factor $1/\sqrt{2\omega_p}$ is a conventional normalization choice. Some texts use $1/(2\omega_p)$; the factor only affects what commutation relations the $a_{\vec p}$ satisfy. Our choice will give the cleanest answer.

The Conjugate Momentum

$\pi = \dot\phi$, so differentiating the expansion:

$\pi(\vec x, t) = \int\frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_p}{2}}\left[a_{\vec p} e^{-ip\cdot x} - a^\dagger_{\vec p} e^{+ip\cdot x}\right]$

Commutation Relations on $a_{\vec p}$

Now we translate the canonical commutation relation $[\phi(\vec x, t), \pi(\vec y, t)] = i\delta^3(\vec x - \vec y)$ into a relation on the $a_{\vec p}$.

Compute the commutator using the expansions:

$[\phi(\vec x, t), \pi(\vec y, t)] = \int\frac{d^3p\, d^3q}{(2\pi)^6}\frac{-i\sqrt{\omega_q/\omega_p}}{2}\left\{[a_{\vec p}, a^\dagger_{\vec q}]e^{-i(p-q)\cdot(x-y)} - [a^\dagger_{\vec p}, a_{\vec q}]e^{+i(p-q)\cdot(x-y)}\right\}$

(with other terms; cross terms $[a,a]$ and $[a^\dagger, a^\dagger]$ set to zero by symmetry/hermiticity).

Setting this equal to $i\delta^3(\vec x - \vec y)$ and working through requires:

$\boxed{[a_{\vec p}, a^\dagger_{\vec q}] = (2\pi)^3 \delta^3(\vec p - \vec q)}$

$[a_{\vec p}, a_{\vec q}] = 0, \quad [a^\dagger_{\vec p}, a^\dagger_{\vec q}] = 0$

These are exactly the commutation relations of a continuum of independent harmonic oscillators, one for each $\vec p$. The Klein-Gordon field, after quantization, is infinitely many harmonic oscillators coupled only by their shared Hamiltonian.

This is the key insight: the quantum field is a continuum of harmonic oscillators labeled by momentum $\vec p$.

6. Creation and Annihilation Operators

Physical Interpretation

For each mode $\vec p$, we have an oscillator with:

• Annihilation operator $a_{\vec p}$
• Creation operator $a^\dagger_{\vec p}$
• Number operator $N_{\vec p} = a^\dagger_{\vec p} a_{\vec p}$

By analogy with the QM harmonic oscillator:

• $a_{\vec p}$ destroys one “quantum” of momentum $\vec p$
• $a^\dagger_{\vec p}$ creates one “quantum” of momentum $\vec p$
• $N_{\vec p}$ counts how many quanta of momentum $\vec p$ are present

These quanta are the particles. A “particle of momentum $\vec p$” is a single excitation of the $\vec p$-mode of the field.

The Vacuum

The vacuum state $|0\rangle$ is the state with no particles; no excitations in any mode:

$a_{\vec p}|0\rangle = 0 \quad \text{for all } \vec p$

It’s the ground state of all the oscillators simultaneously.

One-Particle States

$|\vec p\rangle = \sqrt{2\omega_p}\, a^\dagger_{\vec p}|0\rangle$

The factor $\sqrt{2\omega_p}$ is chosen so that this state has Lorentz-invariant normalization (see section 9).

This represents one particle of momentum $\vec p$. We can verify:

• $a_{\vec q}|\vec p\rangle \propto \delta^3(\vec p - \vec q)|0\rangle$: annihilating the same momentum returns vacuum; different momentum gives zero.

Multi-Particle States

$|\vec p_1, \vec p_2\rangle = \sqrt{2\omega_{p_1} \cdot 2\omega_{p_2}}\, a^\dagger_{\vec p_1} a^\dagger_{\vec p_2}|0\rangle$

Since $[a^\dagger_{\vec p_1}, a^\dagger_{\vec p_2}] = 0$:

$|\vec p_1, \vec p_2\rangle = |\vec p_2, \vec p_1\rangle$

The state is symmetric under exchange. Klein-Gordon field quanta are automatically bosons; this is the spin-statistics theorem in action for spin 0. (Fermionic fields, next document, will use anticommutators instead.)

7. The Hamiltonian and Vacuum Energy

Computing $H$

Substitute the mode expansions into:

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

This is a long computation (products of two mode expansions, integrating over space gives $\delta^3$‘s, etc.). The result:

$H = \int\frac{d^3p}{(2\pi)^3}\,\omega_p\left[a^\dagger_{\vec p}a_{\vec p} + \tfrac{1}{2}(2\pi)^3\delta^3(0)\right]$

The Infinite Vacuum Energy

The second term is a problem. $\delta^3(0)$ is infinite (formally; it represents the volume of space). So the vacuum energy $\langle 0 | H | 0\rangle = \tfrac{1}{2}\int d^3p\, \omega_p/(2\pi)^3 \cdot V$; integral of $\omega_p$ over all momenta, times the volume of space.

This is doubly infinite: infinite volume (IR divergence), and divergent integral over momentum ($\omega_p \to \infty$ as $|\vec p| \to \infty$; UV divergence).

Normal Ordering

The standard resolution: redefine the Hamiltonian by normal ordering; put all annihilation operators to the right of all creation operators, dropping the zero-point energy:

$:H: \equiv \int\frac{d^3p}{(2\pi)^3}\, \omega_p\, a^\dagger_{\vec p} a_{\vec p}$

Justification: only energy differences are observable in a non-gravitational theory. Subtracting an infinite constant doesn’t change anything physical.

After normal ordering:

$:H:|0\rangle = 0$

$:H:|\vec p\rangle = \omega_p |\vec p\rangle$

$:H:|\vec p_1, \vec p_2\rangle = (\omega_{p_1} + \omega_{p_2})|\vec p_1, \vec p_2\rangle$

One-particle states have energy $\omega_p = \sqrt{|\vec p|^2 + m^2}$; exactly the relativistic energy of a particle with mass $m$ and momentum $\vec p$. $\checkmark$

Multi-particle states have the sum of single-particle energies; particles don’t interact in this free theory. $\checkmark$

A Caveat

The vacuum energy isn’t really zero; gravitational physics cares about energy absolutely (all energy gravitates). The vacuum energy of quantum fields should contribute to the cosmological constant $\Lambda$; and famously, a naive estimate gives a value off from the observed value by ~$10^{120}$. This is the cosmological constant problem, arguably the worst prediction in physics. Normal ordering sidesteps the issue in non-gravitational QFT, but it doesn’t make the problem go away.

8. Fock Space and Particle States

The Structure of the Hilbert Space

The Hilbert space of the free scalar field theory has a characteristic structure: Fock space.

$\mathcal{F} = \mathbb{C} \oplus \mathcal{H}_1 \oplus (\mathcal{H}_1 \otimes_S \mathcal{H}_1) \oplus (\mathcal{H}_1 \otimes_S \mathcal{H}_1 \otimes_S \mathcal{H}_1) \oplus \cdots$

Components:

• $\mathbb{C}$: the 0-particle sector (vacuum only)
• $\mathcal{H}_1$: the 1-particle sector (single-particle states of any momentum)
• $\mathcal{H}_1 \otimes_S \mathcal{H}_1$: the 2-particle sector (symmetrized tensor product)
• etc.

Every state in the free theory is a superposition of multi-particle states of definite (though unrestricted) particle number.

Notice: the dimension is infinite; not because the single-particle Hilbert space is infinite-dimensional (it is), but because you can have any number of particles. This could not be described by ordinary quantum mechanics, which fixes particle number.

General States

A general state:

$|\Psi\rangle = c_0|0\rangle + \int\frac{d^3p}{(2\pi)^3}\,\psi_1(\vec p)|\vec p\rangle + \int\frac{d^3p_1 d^3p_2}{(2\pi)^6}\,\psi_2(\vec p_1, \vec p_2)|\vec p_1, \vec p_2\rangle + \cdots$

with coefficients that are wave functions in momentum space.

The Vacuum Is Not Empty

Quantum mechanically, the vacuum state $|0\rangle$ is highly nontrivial. Consider:

• Vacuum fluctuations: $\langle 0|\phi^2|0\rangle \neq 0$
• Zero-point energy (before normal ordering): infinite
• Virtual particle processes in interacting theories

The vacuum is the ground state of a system of infinitely many oscillators, each of which has nonzero zero-point motion. It’s “empty” in the sense of containing no excitations, but quantum fluctuations mean the field values are never exactly zero.

This is observable: vacuum fluctuations produce the Casimir effect, contribute to the Lamb shift, create Hawking radiation near black holes, and many other things.

9. Normalization and Lorentz Invariance

The Problem

In nonrelativistic QM, we normalize states with $\langle \vec p | \vec q\rangle = (2\pi)^3 \delta^3(\vec p - \vec q)$. In QFT we want Lorentz-invariant normalization; the inner product shouldn’t depend on which frame you’re in.

But $\delta^3(\vec p)$ is not Lorentz-invariant (three-momentum transforms nontrivially).

The Solution

With the normalization $|\vec p\rangle = \sqrt{2\omega_p}\, a^\dagger_{\vec p}|0\rangle$:

$\langle \vec p | \vec q\rangle = 2\omega_p (2\pi)^3 \delta^3(\vec p - \vec q)$

This factor $2\omega_p$ is chosen precisely so that under Lorentz transformations, the combination $d^3p/(2\omega_p)$ is invariant:

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

(The right side is manifestly Lorentz-invariant if you restrict to positive energy on the mass shell.)

Lorentz-Invariant Measure

$\int\frac{d^3p}{(2\pi)^3 2\omega_p}$

This is the natural measure on single-particle momentum space in QFT. You’ll see it constantly.

Final Forms

For comparison, textbooks differ on conventions. Common variants:

• Peskin & Schroeder: $|\vec p\rangle = \sqrt{2\omega_p} a^\dagger_{\vec p}|0\rangle$, $\langle \vec p|\vec q\rangle = 2\omega_p(2\pi)^3\delta^3(\vec p - \vec q)$ (our convention)
• Srednicki: uses natural units but slightly different conventions
• Weinberg: relativistic normalization, explicit $\sqrt{2\omega_p}$

Stay consistent within one set; translate when needed.

10. Propagators

The propagator is the two-point correlation function; a central object in QFT. It tells you the amplitude for a field disturbance to propagate from one spacetime point to another.

Feynman Propagator

The most important propagator:

$D_F(x - y) = \langle 0 | T\{\phi(x)\phi(y)\}|0\rangle$

where $T$ is the time-ordering operator:

$T\{\phi(x)\phi(y)\} = \theta(x^0 - y^0)\phi(x)\phi(y) + \theta(y^0 - x^0)\phi(y)\phi(x)$

Computing the Feynman Propagator

Substitute the mode expansion, use the vacuum conditions ($a_{\vec p}|0\rangle = 0$, $\langle 0|a^\dagger_{\vec p} = 0$), and work through. The result can be written as an integral:

$\boxed{D_F(x - y) = \int\frac{d^4p}{(2\pi)^4}\,\frac{i}{p^2 - m^2 + i\epsilon}\,e^{-ip\cdot(x-y)}}$

The $+i\epsilon$ prescription (infinitesimal positive imaginary part) handles the poles at $p^0 = \pm\omega_p$ in a specific way; this is what gives time ordering.

Alternative Propagators

Different contour prescriptions give different propagators:

The Feynman propagator is what appears in Feynman diagrams. The retarded/advanced are used for classical-like calculations.

Meaning

In position space, $D_F(x - y)$ gives the quantum amplitude for a disturbance at $y$ to be detected at $x$ (with time ordering). For spacelike separations (no causal contact classically), $D_F$ is still nonzero but small; leading to the “virtual particle” picture of off-shell propagation.

Why It Matters

Every Feynman diagram has a propagator for each internal line. Computing scattering amplitudes, decay rates, cross-sections; all require propagators. Once we introduce interactions (next document), the propagator becomes the bread and butter of calculations.

11. The Complex Scalar Field and Antiparticles

For a real scalar field, there’s just one type of particle. But many physical fields are complex; leading to particles and antiparticles.

Complex Scalar Lagrangian

$\mathcal{L} = (\partial_\mu\phi^*)(\partial^\mu\phi) - m^2\phi^*\phi$

Treat $\phi$ and $\phi^*$ as independent fields. The Euler-Lagrange equations give Klein-Gordon for each.

Mode Expansion

$\phi(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left[a_{\vec p}e^{-ip\cdot x} + b^\dagger_{\vec p}e^{+ip\cdot x}\right]$

$\phi^\dagger(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left[b_{\vec p}e^{-ip\cdot x} + a^\dagger_{\vec p}e^{+ip\cdot x}\right]$

Note: $\phi^\dagger$ has $b_{\vec p}$ where $\phi$ has $a_{\vec p}$. These are different particles.

Commutation Relations

Two independent sets of oscillators:

$[a_{\vec p}, a^\dagger_{\vec q}] = (2\pi)^3\delta^3(\vec p - \vec q)$

$[b_{\vec p}, b^\dagger_{\vec q}] = (2\pi)^3\delta^3(\vec p - \vec q)$

$[a_{\vec p}, b_{\vec q}] = [a_{\vec p}, b^\dagger_{\vec q}] = 0$

Particles and Antiparticles

$a^\dagger_{\vec p}|0\rangle$ creates a particle of momentum $\vec p$.

$b^\dagger_{\vec p}|0\rangle$ creates an antiparticle of momentum $\vec p$.

They have the same mass (same $\omega_p = \sqrt{|\vec p|^2 + m^2}$) but opposite charge under the $U(1)$ symmetry.

The Conserved Charge

The $U(1)$ symmetry $\phi \to e^{i\alpha}\phi$ gives (by Noether) a conserved current:

$j^\mu = i(\phi^\dagger \partial^\mu \phi - (\partial^\mu \phi^\dagger)\phi)$

The total charge $Q = \int d^3x\, j^0$ evaluates to:

$Q = \int\frac{d^3p}{(2\pi)^3}\left[a^\dagger_{\vec p}a_{\vec p} - b^\dagger_{\vec p} b_{\vec p}\right]$

Particles contribute +1 per particle; antiparticles contribute −1. If we identify $Q$ with electric charge (up to a factor $e$), we see that:

• The complex scalar describes electrically charged bosons
• Each particle type comes with a corresponding antiparticle of opposite charge

This is how antimatter enters QFT: not as a postulate, but as the conjugate half of complex fields.

Example: Pions

The $\pi^+$ and $\pi^-$ are (approximately) described by a complex scalar field: $\pi^+$ is the particle, $\pi^-$ the antiparticle. They have the same mass (139.6 MeV) but opposite charge. The $\pi^0$, by contrast, is described by a real scalar; it’s its own antiparticle.

12. Physical Content and the Casimir Effect

What We’ve Achieved

After all this machinery, what have we gotten?

• A quantum theory of a relativistic scalar field
• Particles emerge as excitations of the field
• Multi-particle states exist naturally in Fock space
• Particle creation/annihilation operators
• Propagators encode virtual-particle physics
• Antiparticles appear for complex fields

The classical Klein-Gordon theory, quantized, is a quantum field theory; a theory of particles. The field operator $\phi(x)$ is the primary object; particles are what the theory describes.

The Casimir Effect: Real Consequences of “Empty” Vacuum

One concrete physical prediction of QFT, even at the level we’ve developed: the Casimir effect.

Consider two parallel conducting plates in vacuum, separated by distance $d$. QFT predicts an attractive force between them:

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

Tiny; about $10^{-3}$ N/m² at 1 micron separation; but measurable.

Origin: The vacuum of the electromagnetic field has zero-point fluctuations of all modes. Conducting plates constrain which modes are allowed (boundary conditions). The constrained modes have different zero-point energy than unconstrained modes, and the energy depends on $d$. Differentiating gives a force.

Hendrik Casimir predicted this in 1948. It’s been measured with high precision.

The Casimir effect is the simplest experimental demonstration that vacuum fluctuations are real; not an artifact of our mathematical formalism, but a physical consequence.

A Glimpse Ahead

This document covered the free Klein-Gordon field. No interactions. Scattering is trivial (particles fly by each other without interacting). Real physics requires interactions.

The next steps:

1. Dirac field (next doc): quantize a fermionic field. Same procedure, but anticommutators. Gives fermions, antifermions, and the spin-statistics theorem.
2. Electromagnetic field: quantize the gauge field. Gauge redundancy complicates things.
3. Interactions: add couplings like $\phi^4$ or $\bar\psi\gamma^\mu\psi A_\mu$. Perturbation theory. Feynman diagrams. Actual scattering calculations.

Each of these builds on the machinery we’ve developed here. The pattern; classical Lagrangian → canonical commutators → mode expansion → creation/annihilation operators → Fock space; repeats.

Appendix: Formulas and Identities

Conventions

• Metric: $\eta_{\mu\nu} = \text{diag}(+,-,-,-)$
• Natural units: $\hbar = c = 1$
• $p \cdot x = p^\mu x_\mu = \omega_p t - \vec p \cdot \vec x$
• $\omega_p = \sqrt{|\vec p|^2 + m^2}$

Key Expressions

Lagrangian: $\mathcal{L} = \tfrac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi) - \tfrac{1}{2}m^2\phi^2$

Conjugate momentum: $\pi = \dot\phi$

Hamiltonian: $H = \int d^3x[\tfrac{1}{2}\pi^2 + \tfrac{1}{2}(\nabla\phi)^2 + \tfrac{1}{2}m^2\phi^2]$

Equal-time commutation relations: $[\phi(\vec x), \pi(\vec y)] = i\delta^3(\vec x - \vec y)$ $[\phi(\vec x), \phi(\vec y)] = [\pi(\vec x), \pi(\vec y)] = 0$

Field expansion: $\phi(x) = \int\frac{d^3p}{(2\pi)^3\sqrt{2\omega_p}}[a_{\vec p}e^{-ip\cdot x} + a^\dagger_{\vec p}e^{+ip\cdot x}]$

Creation/annihilation commutators: $[a_{\vec p}, a^\dagger_{\vec q}] = (2\pi)^3\delta^3(\vec p - \vec q)$ $[a_{\vec p}, a_{\vec q}] = [a^\dagger_{\vec p}, a^\dagger_{\vec q}] = 0$

Normal-ordered Hamiltonian: $:H: = \int\frac{d^3p}{(2\pi)^3}\omega_p a^\dagger_{\vec p}a_{\vec p}$

Single-particle states: $|\vec p\rangle = \sqrt{2\omega_p}a^\dagger_{\vec p}|0\rangle$

Inner product: $\langle\vec p|\vec q\rangle = 2\omega_p(2\pi)^3\delta^3(\vec p - \vec q)$

Lorentz-invariant measure: $\int\frac{d^3p}{(2\pi)^3 2\omega_p}$

Feynman propagator: $D_F(x - y) = \int\frac{d^4p}{(2\pi)^4}\frac{i}{p^2 - m^2 + i\epsilon}e^{-ip\cdot(x-y)}$

Checklist: Have I Understood?

By the end of this document, you should be able to:

• Write down the Lagrangian, Hamiltonian, and canonical commutators for the free scalar field
• Expand the field in terms of creation and annihilation operators
• Show that $[a_{\vec p}, a^\dagger_{\vec q}] = (2\pi)^3 \delta^3(\vec p - \vec q)$ follows from the canonical commutation relations
• Construct multi-particle states and explain why they’re symmetric
• Compute $H|\vec p\rangle = \omega_p|\vec p\rangle$
• Explain what normal ordering does and why
• Write down the Feynman propagator in momentum space
• Describe antiparticles as quanta of the complex conjugate mode coefficients

If any of these feel shaky, go back to that section before moving on.

Further Reading

• Peskin & Schroeder, An Introduction to Quantum Field Theory, Chapter 2: the standard treatment. Does everything we’ve done with full rigor.
• Schwartz, Quantum Field Theory and the Standard Model, Chapter 2: cleaner modern presentation.
• Srednicki, Quantum Field Theory, Chapters 1–3: pedagogically clean.
• Weinberg, The Quantum Theory of Fields, Vol. 1, Chapter 4 onward: rigorous, beautiful, hard.
• Coleman, Lectures on Quantum Field Theory: Sidney Coleman’s legendary course, written up by his students. Worth reading whenever you have time.

Problems to Work (Highly Recommended)

1. Derive the expansion of $\pi(\vec x, t)$ from that of $\phi$, and verify that the coefficients come out as claimed.
2. Compute $[\phi(\vec x, t), \pi(\vec y, t)]$ directly from the mode expansions, and verify that it gives $i\delta^3(\vec x - \vec y)$.
3. Verify that $H|\vec p\rangle = \omega_p|\vec p\rangle$ using the normal-ordered Hamiltonian.
4. Compute $\langle 0|\phi(x)\phi(y)|0\rangle$ directly from the mode expansion (without time ordering first). This gives the Wightman function; the Feynman propagator is a time-ordered average of Wightman functions.
5. For the complex scalar, verify that $Q|\vec p\rangle_a = +|\vec p\rangle_a$ and $Q|\vec p\rangle_b = -|\vec p\rangle_b$ (particles and antiparticles carry opposite charge).

These problems are not optional if you want to really internalize the material. Working through them converts “I read this” into “I understand this.”

Closing Note

What we’ve just done is arguably the most important conceptual step in 20th-century physics. Classical field theory was a beautiful framework for electromagnetism and (less obviously) Newton’s gravity. Quantum mechanics was a revolutionary theory of particles. Putting them together properly; quantizing a field; produces something neither was alone: a theory where particles are emergent, where their number isn’t fixed, where creation and annihilation are natural, and where antimatter falls out of the math.

This single idea; that nature is fundamentally described by quantum fields, with particles as their excitations; is the foundation of the Standard Model, of essentially all particle physics, of much of condensed matter theory, and of our best understanding of matter at its most fundamental.

There are a few things to emphasize as we close:

The classical-quantum pairing is tight. Everything in this document had a direct classical counterpart. Canonical commutation → Poisson brackets. Field expansion → classical plane waves. Hamiltonian → classical Hamiltonian. Mode energies → classical oscillator energies. QFT at this stage is not weird; it’s quantum mechanics applied to a particular classical system.

The weirdness is the Fock space structure. States with arbitrary particle numbers. Vacuum as nontrivial. Operators that change particle count. These are what’s new about QFT; and they trace back to the fact that a field has infinitely many degrees of freedom.

Normal ordering is a stopgap. We dropped an infinite zero-point energy with a wave of the hand. In gravitational contexts (cosmological constant), this is a disaster. It’s a clue that something deeper is missing, possibly related to whatever quantum gravity is.

Everything we built here is just the free theory. We haven’t turned on any interactions. Free field theories are simple and solvable. Interacting theories are the real challenge, and they’re where Feynman diagrams, perturbation theory, and renormalization enter.

What’s Next

The plan we laid out is on track. Next document: quantizing the Dirac field. Fermions. Anticommutators. The spin-statistics connection. Antiparticles from a slightly different angle. After that, the photon field, then interactions, then Feynman diagrams, then renormalization. The machine builds.

But take your time here first. Work the problems. Make sure the Fock space structure is second nature. The whole rest of QFT builds on understanding what we’ve done today.

You’re officially doing quantum field theory. Well done.