Exercises

Newton's formulation describes motion through forces. For a particle of constant mass, Newton's second law gives

$${\mathbf{F}}_{\mathrm{net}}=m\mathbf{a}=m\frac{d^2\mathbf{x}}{d{t}^2}.$$

Once we know the force law and the particle's initial position and velocity, we solve this second-order differential equation to find its trajectory $\mathbf{x}(t)$. Integrating the acceleration once gives the velocity, and integrating it again gives the position.

The Lagrangian is a function of the system's generalized coordinates, velocities, and time:

$$L(q_i,{\dot{q}}_i,t)=T-V,$$

for an ordinary mechanical system with kinetic energy $T$ and potential energy $V$.

The action assigns a number to each possible path between two fixed times:

$$S[q]={\int }_{t_i}^{t_f}L(q_i,{\dot{q}}_i,t)dt.$$

The physical path makes the action stationary. For every small variation of the path that leaves its endpoints fixed, the first-order change in the action vanishes:

$$\delta S=0.$$

Stationary does not always mean smallest; the action can also have a maximum or another stationary value. Applying $\delta S=0$ gives the Euler–Lagrange equations,

$$\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{q}}_i}\right)-\frac{\partial L}{\partial q_i}=0,$$

which determine the system's motion.

We want to derive

$$\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{q}}_i}\right)-\frac{\partial L}{\partial q_i}=0.$$

Consider a small variation of the physical path,

$$q_i(t)\to q_i(t)+ϵ{\eta }_i(t),$$

where $ϵ$ is small and ${\eta }_i(t_i)={\eta }_i(t_f)=0$, so the endpoints remain fixed. The corresponding variation of the velocity is

$$\delta {\dot{q}}_i=\frac{d}{dt}(\delta q_i).$$

The action is

$$S[q]={\int }_{t_i}^{t_f}L(q_i,{\dot{q}}_i,t)dt.$$

Its first-order variation is

$$\delta S={\int }_{t_i}^{t_f}\left(\frac{\partial L}{\partial q_i}\delta q_i+\frac{\partial L}{\partial {\dot{q}}_i}\delta {\dot{q}}_i\right)dt,$$

where repeated $i$ indices are summed. Substitute $\delta {\dot{q}}_i=d(\delta q_i)/dt$ and integrate the second term by parts:

$$\delta S={\left[\frac{\partial L}{\partial {\dot{q}}_i}\delta q_i\right]}_{t_i}^{t_f}+{\int }_{t_i}^{t_f}\left[\frac{\partial L}{\partial q_i}-\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{q}}_i}\right)\right]\delta q_{i}dt.$$

The boundary term vanishes because $\delta q_i(t_i)=\delta q_i(t_f)=0$. Stationary action requires $\delta S=0$, so

$${\int }_{t_i}^{t_f}\left[\frac{\partial L}{\partial q_i}-\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{q}}_i}\right)\right]\delta q_{i}dt=0.$$

The variations $\delta q_i(t)$ are arbitrary between the endpoints. Therefore, their coefficient must vanish at every time:

$$\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{q}}_i}\right)-\frac{\partial L}{\partial q_i}=0.$$

These are the Euler–Lagrange equations.

Generalized coordinates $q_i$ are independent variables that specify a system's configuration. They need not be Cartesian positions. For example, a pendulum can be described by one generalized coordinate, its angle $q=\theta$, rather than by the bob's Cartesian coordinates subject to a length constraint. Their time derivatives ${\dot{q}}_i$ are the generalized velocities.

The canonical momentum conjugate to $q_i$ is defined from the Lagrangian by

$$p_i=\frac{\partial L}{\partial {\dot{q}}_i}.$$

Canonical momentum does not always equal mechanical momentum $m{\dot{q}}_i$. The two agree for a Cartesian coordinate with the Lagrangian $L=\frac{1}{2}m{\dot{q}}_i^2-V(q)$, but constraints or velocity-dependent interactions can change the relation.

$$H(q_i,p_i,t)=\sum _i{p}_i{\dot{q}}_i-L(q_i,{\dot{q}}_i,t),$$

where the velocities on the right must be expressed in terms of $q_i$, $p_i$, and $t$.

The Hamiltonian generates time evolution through Hamilton's equations,

$${\dot{q}}_i=\frac{\partial H}{\partial p_i},{\dot{p}}_i=-\frac{\partial H}{\partial q_i}.$$

For the common Lagrangian $L=T-V$ with Cartesian coordinates and a velocity-independent potential,

$$H=T+V,$$

so the Hamiltonian equals the system's total energy. This equality does not hold for every possible Lagrangian.

A quantum state contains all the information that quantum mechanics can use to predict the outcomes of measurements. We represent a pure state by a normalized vector $|\psi ⟩$ in a complex Hilbert space:

$$⟨\psi |\psi ⟩=1.$$

Two vectors that differ only by an overall phase, $|\psi ⟩$ and $e^{i\alpha }|\psi ⟩$, represent the same physical state. In the position basis, the state is represented by the wave function

$$\psi (x,t)=⟨x|\psi (t)⟩.$$

An observable is a measurable physical quantity, such as position, momentum, energy, or spin. Quantum mechanics represents observables by self-adjoint linear operators (Hermitian matrices in finite-dimensional Hilbert spaces). Self-adjoint operators have real eigenvalues, so their possible measured values are real.

An eigenstate of an observable $\hat{A}$ is a state whose direction does not change when the operator acts on it:

$$\hat{A}|a_n⟩=a_n|a_n⟩.$$

Here, $|a_n⟩$ is the eigenstate and $a_n$ is its eigenvalue. If the system is in $|a_n⟩$, measuring $A$ gives $a_n$ with certainty. For example, an energy eigenstate satisfies

$$\hat{H}|E_n⟩=E_n|E_n⟩.$$

If an observable has orthonormal eigenstates $|a_n⟩$, we can expand the state as

$$|\psi ⟩=\sum _n{c}_n|a_n⟩.$$

$$P(a_n)=|c_n{|}^2=|⟨a_n|\psi ⟩{|}^2$$

to obtaining the value $a_n$ in a measurement.

Between measurements, the state evolves according to the Schrödinger equation,

$$i\hslash \frac{d}{dt}|\psi (t)⟩=\hat{H}|\psi (t)⟩.$$

In the position representation, the position operator multiplies the wave function by $x$:

$$\hat{x}\psi (x)=x\psi (x).$$

The momentum operator differentiates the wave function:

$$\hat{p}\psi (x)=-i\hslash \frac{d\psi (x)}{dx}.$$

In three dimensions,

$$\hat{\mathbf{x}}=\mathbf{x},\hat{\mathbf{p}}=-i\hslash \nabla .$$

They satisfy the canonical commutation relation

$$[\hat{x},\hat{p}]=i\hslash ,$$

which leads to the position–momentum uncertainty relation

$$\Delta x\Delta p\ge \frac{\hslash }{2}.$$

A harmonic oscillator is a system whose restoring force is proportional to its displacement from equilibrium and points back toward equilibrium:

$$F=-kx.$$

Its classical equation of motion is

$$m\ddot{x}+kx=0,$$

so it oscillates sinusoidally with angular frequency

$$\omega =\sqrt{\frac{k}{m}}.$$

$$H=\frac{p^2}{2m}+\frac{1}{2}m{\omega }^2{x}^2.$$

$$E_n=\hslash \omega \left(n+\frac{1}{2}\right),n=0,1,2,\dots .$$

Its lowest state has the nonzero zero-point energy $E_0=\frac{1}{2}\hslash \omega$. The harmonic oscillator appears throughout physics because a smooth potential near a stable equilibrium is approximately quadratic.

The quantum harmonic oscillator is the quantum description of a particle moving in the quadratic potential

$$V(x)=\frac{1}{2}m{\omega }^2{x}^2.$$

$$\hat{H}=\frac{{\hat{p}}^2}{2m}+\frac{1}{2}m{\omega }^2{\hat{x}}^2.$$

$$\hat{H}|n⟩=E_n|n⟩,$$

gives discrete, equally spaced energy levels:

$$E_n=\hslash \omega \left(n+\frac{1}{2}\right),n=0,1,2,\dots .$$

The ground state has energy $E_0=\frac{1}{2}\hslash \omega$, so the oscillator never has zero energy. Ladder operators move between adjacent energy states:

$${\hat{a}}^{†}|n⟩=\sqrt{n+1}|n+1⟩,\hat{a}|n⟩=\sqrt{n}|n-1⟩.$$

The model matters because many systems behave approximately like harmonic oscillators near stable equilibria. Each mode of a free quantum field also behaves like a quantum harmonic oscillator.

The frequency $\omega$ can be any positive value because it is a parameter that defines the particular oscillator:

$$\omega =\sqrt{\frac{k}{m}}.$$

Once $m$, $k$, and therefore $\omega$ are fixed, that oscillator can have only the discrete energies

$$E_n=\hslash \omega \left(n+\frac{1}{2}\right),n=0,1,2,\dots .$$

Across different oscillators, the energy scale can vary continuously with $\omega$. For one fixed oscillator, however, the allowed energies remain discrete and adjacent levels differ by $\hslash \omega$. Quantization restricts the level number $n$, not the externally specified parameter $\omega$.

The principle of relativity states that the laws of physics have the same form in every inertial frame. The invariance of the speed of light states that every inertial observer measures the same vacuum light speed $c$, regardless of the source's motion.

Galilean transformations preserve absolute time and use ordinary velocity addition, which would make different observers measure different light speeds. Lorentz transformations preserve $c$ by mixing space and time:

$$t^{\prime }=\gamma \left(t-\frac{vx}{c^2}\right),x^{\prime }=\gamma (x-vt),\gamma =\frac{1}{\sqrt{1-v^2/c^2}}.$$

Observers can disagree about simultaneity, elapsed time, and spatial distance, but they agree on the spacetime interval

$$s^2=c^2\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2.$$

The sign of $s^2$ classifies event separations as timelike, lightlike, or spacelike and determines whether one event can causally influence another.

Four-vectors combine time and space components into objects that transform consistently between inertial frames. With the Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(+1,-1,-1,-1)$, contracting an upper and lower index produces a Lorentz scalar, whose value every inertial observer agrees on.

The four-momentum is

$$p^{\mu }=\left(\frac{E}{c},\mathbf{p}\right)$$

and has the invariant norm $p^{\mu }{p}_{\mu }=m^2{c}^2$. Therefore,

$$E^2=(m{c}^2{)}^2+(pc{)}^2.$$

This relation reduces to $E=m{c}^2$ for a massive particle at rest and to $E=pc$ for a massless particle.

Relativistic equations must retain their form under Lorentz transformations. Maxwell's equations meet this requirement when written using the four-current $J^{\mu }$ and electromagnetic field tensor $F^{\mu \nu }$:

$${\partial }_{\mu }{F}^{\mu \nu }=J^{\nu }.$$

Electric and magnetic fields form parts of one relativistic field and can mix under a boost.

The Schrödinger equation treats time and space derivatives differently, so its form changes when a Lorentz boost mixes space and time. It also uses the non-relativistic relation $E={\mathbf{p}}^2/(2m)$. Starting instead from the exact relativistic energy–momentum relation leads toward the Klein–Gordon and Dirac equations.

A free particle with definite momentum $\mathbf{p}$ and energy $E$ has the plane-wave form

$$\psi (\mathbf{x},t)=e^{i(\mathbf{p}\cdot \mathbf{x}-Et)/\hslash }.$$

Substituting this form into a wave equation determines the equation's relation between energy and momentum, called its dispersion relation. An exact plane wave extends across all space, so a localized particle requires a wave packet built from many momenta.

The Schrödinger equation uses the non-relativistic kinetic energy $E={\mathbf{p}}^2/(2m)$. Special relativity instead gives the exact relation

$$E^2={\mathbf{p}}^2+m^2$$

in units where $c=1$. A relativistic wave equation must reproduce this relation and retain the same form under Lorentz transformations.

Replace energy and momentum in the relativistic relation with their quantum operators:

$$E\to i\hslash {\partial }_t,\mathbf{p}\to -i\hslash \nabla .$$

This gives

$$\left(\square +\frac{m^2}{{\hslash }^2}\right)\psi =0,\square ={\partial }_t^2-{\nabla }^2.$$

The d'Alembertian $\square ={\partial }_{\mu }{\partial }^{\mu }$ is a Lorentz scalar, so the Klein–Gordon equation has the same form in every inertial frame.

A plane wave satisfies

$$E=\pm \sqrt{{\mathbf{p}}^2+m^2}.$$

If the negative branch described ordinary one-particle energies, the spectrum would have no lower bound. The equation is also second order in time, so an initial state requires both $\psi$ and ${\partial }_t\psi$. In contrast, the first-order Schrödinger equation determines the future state from $\psi$ alone.

The conserved Klein–Gordon current is

$$j^{\mu }=i\left({\psi }^{\ast }{\partial }^{\mu }\psi -\psi {\partial }^{\mu }{\psi }^{\ast }\right),{\partial }_{\mu }{j}^{\mu }=0.$$

For a plane wave, its time component is

$$j^0=\frac{2E}{\hslash }|\psi {|}^2.$$

This density changes sign with $E$, so it cannot represent a nonnegative probability for all solutions. In quantum field theory, the conserved quantity instead becomes charge, which can have either sign.

Dirac sought an equation that is first order in both time and space:

$$i\hslash {\partial }_t\psi =\left(-i\hslash \mathbf{\alpha }\cdot \nabla +\beta m\right)\psi .$$

Requiring its square to reproduce $E^2={\mathbf{p}}^2+m^2$ gives

$${\beta }^2=1,\{{\alpha }_i,\beta \}=0,\{{\alpha }_i,{\alpha }_j\}=2{\delta }_{ij}.$$

Ordinary numbers cannot satisfy these relations. The coefficients must be matrices, and in three spatial dimensions the wave function becomes a four-component spinor. Defining ${\gamma }^0=\beta$ and ${\gamma }^i=\beta {\alpha }_i$ gives the covariant equation

$$(i\hslash {\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0.$$

The Dirac equation has the conserved current

$$j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi ,$$

whose time component is

$$j^0={\psi }^{†}\psi \ge 0.$$

It therefore provides a nonnegative single-particle density and naturally introduces spin-$\frac{1}{2}$. However, it still has positive- and negative-energy solutions. Quantum field theory completes the interpretation by treating negative-frequency modes as antiparticles with positive excitation energy and by allowing particles to be created and destroyed.

The Dirac equation is a relativistic wave equation for spin-$\frac{1}{2}$ particles:

$$(i\hslash {\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0.$$

The wave function $\psi$ is a four-component spinor, and the gamma matrices satisfy

$$\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }1.$$

This algebra ensures that solutions obey the relativistic energy–momentum relation $E^2={\mathbf{p}}^2+m^2$ in units where $c=1$.

For the free Dirac Hamiltonian

$$H=\mathbf{\alpha }\cdot \hat{\mathbf{p}}+\beta m,$$

orbital angular momentum $\mathbf{L}$ is not conserved by itself. The Dirac components carry intrinsic angular momentum described by

$$\mathbf{S}=\frac{\hslash }{2}\mathbf{\Sigma },{\Sigma }_i=\mathrm{diag}({\sigma }_i,{\sigma }_i).$$

The total angular momentum $\mathbf{J}=\mathbf{L}+\mathbf{S}$ is conserved because $[H,\mathbf{J}]=0$. Since $S_z$ has eigenvalues $\pm \hslash /2$, the Dirac equation describes spin-$\frac{1}{2}$ particles.

For an operator $A$ with no explicit time dependence, the Heisenberg equation gives

$$\frac{dA}{dt}=\frac{i}{\hslash }[H,A].$$

An observable is conserved in every state when it commutes with the Hamiltonian. In the Dirac theory, $[H,\mathbf{L}]\ne 0$, but $[H,\mathbf{L}+\mathbf{S}]=0$, so spin and orbital angular momentum can change separately while their sum remains constant.

Minimal coupling replaces momentum by

$$\hat{\mathbf{p}}\to \hat{\mathbf{p}}-q\mathbf{A}.$$

Taking the non-relativistic limit then produces the Pauli equation and the magnetic moment

$$\mathbf{\mu }=\frac{q}{m}\mathbf{S}.$$

Comparing this with $\mathbf{\mu }=gq\mathbf{S}/(2m)$ gives $g=2$. For an electron, $q=-e$, so $\mathbf{\mu }=-(e/m)\mathbf{S}$.

For each momentum, the free Dirac equation has energies

$$E=\pm E_p,E_p=\sqrt{{\mathbf{p}}^2+m^2}.$$

Each energy sign has two independent spin states. At rest in the standard representation, the positive-energy solutions occupy the upper pair of spinor components, while the negative-energy solutions occupy the lower pair. Motion mixes the upper and lower components. We denote the positive-energy spinors by $u_s(p)$ and the negative-energy spinors by $v_s(p)$.

Charge conjugation maps a particle solution to an antiparticle solution:

$${\psi }_c=i{\gamma }^2{\psi }^{\ast }.$$

Complex conjugation reverses the sign of the electromagnetic charge, so the transformed solution has the same mass and opposite charge. For the electron, this antiparticle is the positron.

A one-particle theory cannot consistently treat negative-energy solutions as ordinary electron states because the energy would have no lower bound. Quantum field theory instead interprets the negative-frequency terms as operators that create positive-energy antiparticles.

When coordinates transform as $x^{\prime }=\Lambda x$, a Dirac spinor transforms as

$${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x),S(\Lambda )=\exp \left(-\frac{i}{4}{\omega }_{\mu \nu }{\sigma }^{\mu \nu }\right),$$

where

$${\sigma }^{\mu \nu }=\frac{i}{2}[{\gamma }^{\mu },{\gamma }^{\nu }].$$

A $2\pi$ rotation changes the sign of a spinor, while a $4\pi$ rotation returns it to its original value. The four-component Dirac spinor can also be separated into left- and right-handed Weyl spinors, which transform alike under rotations and oppositely under boosts.

The general solution superposes every momentum, both spin states, and both frequency signs:

$$\psi (x)=\sum _{s=1}^2\int \frac{d^{3}p}{(2\pi \hslash {)}^3}\frac{1}{\sqrt{2E_p}}\left[a_s(p)u_s(p)e^{-ip\cdot x/\hslash }+b_s^{\ast }(p)v_s(p)e^{+ip\cdot x/\hslash }\right].$$

In one-particle quantum mechanics, $a_s$ and $b_s^{\ast }$ are numerical amplitudes fixed by the initial state. Quantum field theory promotes them to operators: $a_s$ annihilates a particle, while $b_s^{\ast }\to {\hat{b}}_s^{†}$ creates an antiparticle.

A classical field assigns a physical value to every point in spacetime:

$$\varphi :x^{\mu }\mapsto \varphi (x).$$

A temperature field assigns a number to each point, while the electromagnetic field assigns electric and magnetic vectors. A classical field has a definite configuration, follows classical field equations, and is not an operator.

A quantum field is an operator-valued field defined throughout spacetime:

$$x^{\mu }\mapsto \hat{\varphi }(x).$$

Unlike a classical field, it does not assign one definite physical value to each point. Instead, it acts on quantum states.

Informally, a quantum field is an operator-valued field. More precisely, it is an operator-valued distribution, not an ordinary operator assigned independently to each exact point.

The notation $\hat{\varphi }(x)$ treats the field at $x$ like an operator, but it can be singular at an exact spacetime point. Averaging, or smearing, it with a smooth function $f(x)$ gives a well-defined operator:

$$\hat{\varphi }(f)=\int d^{4}xf(x)\hat{\varphi }(x).$$

This smeared field acts on states in Fock space.

In QFT, a state also belongs to a Hilbert space. For free fields and perturbative particle descriptions, this is usually a Fock space:

$$|\Psi ⟩\in \mathcal{F},\mathcal{F}={\mathcal{H}}_0\oplus {\mathcal{H}}_1\oplus {\mathcal{H}}_2\oplus \cdots .$$

Here, ${\mathcal{H}}_0$ contains the vacuum, ${\mathcal{H}}_1$ contains one-particle states, and ${\mathcal{H}}_2$ contains two-particle states. Creation and annihilation operators move states between these sectors:

$$a^{†}:{\mathcal{H}}_n\to {\mathcal{H}}_{n+1},a:{\mathcal{H}}_n\to {\mathcal{H}}_{n-1}.$$

Thus, QM and QFT both use Hilbert spaces, but QFT needs a state space that allows particle number to change. A general interacting QFT still uses a Hilbert space, although it may not decompose cleanly into free-particle sectors.

Quantum field theory assigns a field to every point in spacetime. A field can be a scalar, vector, or spinor depending on how its values transform. After quantization, each momentum mode behaves like a quantum oscillator, and applying creation operators produces particle excitations. Particles are therefore discrete quanta of fields, while the field remains the fundamental object.

They differ in the type of value assigned to each spacetime point and in how that value transforms. A scalar has one Lorentz-invariant component, a vector has components that mix like spacetime coordinates, and a spinor follows the spin-$\frac{1}{2}$ transformation law. These properties determine which Lorentz-invariant terms can appear in a field theory.

Spacetime labels where a field is evaluated; the value space specifies what lives at that point. A Lorentz transformation changes spacetime coordinates and also acts on vector or spinor components through the appropriate representation. Keeping these actions distinct prevents confusion between moving a point and transforming a field value.

In the Heisenberg picture, quantum states stay fixed while operators evolve with time:

$$|{\psi }_H⟩=\text{constant},{\hat{A}}_H(t)=e^{i\hat{H}t/\hslash }{\hat{A}}_H(0)e^{-i\hat{H}t/\hslash }.$$

Their evolution obeys

$$\frac{d{\hat{A}}_H}{dt}=\frac{i}{\hslash }[\hat{H},{\hat{A}}_H]+\frac{\partial {\hat{A}}_H}{\partial t}.$$

In QFT, the field operator $\hat{\varphi }(x)$ evolves and obeys the field equation while the state remains fixed. In the Schrödinger picture, operators are usually fixed and states obey $i\hslash d|{\psi }_S⟩/dt=\hat{H}|{\psi }_S⟩$. Both pictures give the same measurement predictions.

A field theory uses a Lagrangian density $\mathcal{L}(\varphi ,{\partial }_{\mu }\varphi )$ and action

$$S=\int d^{4}x\mathcal{L}.$$

Requiring stationary action gives the field Euler–Lagrange equation

$${\partial }_{\mu }\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\varphi )}-\frac{\partial \mathcal{L}}{\partial \varphi }=0.$$

Choosing a Lorentz-scalar density makes the resulting equations covariant.

For a real scalar field,

$$\mathcal{L}=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2.$$

The field Euler–Lagrange equation gives $(\square +m^2)\varphi =0$. The derivative term governs propagation, while the quadratic term sets the particle mass.

For a particle, the unknown is a path such as $q(t)$. Test the path by changing it slightly:

$$q(t)\to q(t)+\delta q(t).$$

For a field, the unknown is a value at every spacetime point, $\varphi (x)$. Change the entire field slightly:

$$\varphi (x)\to \varphi (x)+\delta \varphi (x).$$

Keep $\delta \varphi =0$ at the boundary because the starting and ending configurations are fixed. Requiring the action to be stationary for every allowed change gives

$${\partial }_{\mu }\left(\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }\varphi )}\right)-\frac{\partial \mathcal{L}}{\partial \varphi }=0.$$

The main difference is that we vary an entire function $\varphi (x)$ rather than a finite list of coordinates.

The density $\mathcal{L}$ is local in spacetime. Integrating it over space gives the Lagrangian, $L=\int d^{3}x\mathcal{L}$, and integrating that over time gives the action, $S=\int d^{4}x\mathcal{L}$.

If $\mathcal{L}$ is a Lorentz scalar, every inertial observer assigns the same value to the action. Its stationary configurations then obey field equations with the same form in every inertial frame.

Define

$$\pi =\frac{\partial \mathcal{L}}{\partial \dot{\varphi }},\mathcal{H}=\pi \dot{\varphi }-\mathcal{L},H=\int d^{3}x\mathcal{H}.$$

After quantization, $\hat{H}$ generates time evolution.

A state on one time slice is the pair of functions $(\varphi (\mathbf{x}),\pi (\mathbf{x}))$. Because it contains one conjugate pair at every spatial point, the state space is infinite-dimensional.

It assigns energy to a field configuration and generates its time evolution. When the action has time-translation symmetry, Noether's theorem connects these roles by making the energy conserved.

For the electromagnetic potential,

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

Varying $A_{\nu }$ gives ${\partial }_{\mu }{F}^{\mu \nu }=0$. The absence of an $A_{\mu }{A}^{\mu }$ term keeps the photon massless.

The density ${\psi }^{†}\psi$ is the time component of a current, not a Lorentz scalar. Defining $\bar{\psi }={\psi }^{†}{\gamma }^0$ makes $\bar{\psi }\psi$ invariant, so

$$\mathcal{L}=\bar{\psi }(i\hslash {\gamma }^{\mu }{\partial }_{\mu }-m)\psi$$

is a Lorentz scalar.

Treat $\psi$ and $\bar{\psi }$ as independent variables. Varying $\bar{\psi }$ gives the Dirac equation, while varying $\psi$ gives the adjoint Dirac equation. The derivative is first order, so specifying $\psi$ on one time slice is sufficient initial data.

Start with a Lorentz vector field $A_{\mu }$. For a massless spin-1 gauge field, require invariance under

$$A_{\mu }\to A_{\mu }+{\partial }_{\mu }\alpha .$$

The simplest object made from one derivative of $A_{\mu }$ that is unchanged by this transformation is the antisymmetric field strength

$$F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }.$$

The simplest Lorentz scalar quadratic in the field strength is therefore

$$\boxed{{\mathcal{L}}_{\mathrm{Maxwell}}=-\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }}.$$

Varying $A_{\nu }$ gives

$${\partial }_{\mu }{F}^{\mu \nu }=0,$$

the source-free Maxwell equation. The factor $-1/4$ is a normalization chosen so the components have canonical normalization. This construction is guided by Lorentz symmetry, locality, a simple quadratic free theory, and gauge invariance; symmetry alone does not make the Lagrangian mathematically unique.

In QED and the weak theory before the Higgs mechanism, the gauge symmetry selects the massless Maxwell or Yang–Mills form. The Higgs mechanism later produces an effective vector mass from a gauge-invariant scalar kinetic term.

Canonical quantization starts with a classical field $\varphi (\mathbf{x},t)$ and its conjugate momentum

$$\pi (\mathbf{x},t)=\frac{\partial \mathcal{L}}{\partial \dot{\varphi }(\mathbf{x},t)}.$$

Replace them with operators,

$$\varphi \to \hat{\varphi },\pi \to \hat{\pi },$$

and impose the equal-time commutator

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

Expanding a free field into momentum modes turns the mode coefficients into creation and annihilation operators:

$$\hat{\varphi }(x)\sim \int d^{3}p\left({\hat{a}}_{\mathbf{p}}{e}^{-ip\cdot x}+{\hat{a}}_{\mathbf{p}}^{†}{e}^{ip\cdot x}\right).$$

Each mode behaves like a quantum harmonic oscillator. The operator ${\hat{a}}_{\mathbf{p}}^{†}$ creates a particle, while ${\hat{a}}_{\mathbf{p}}$ removes one. Fermion fields use anticommutation relations instead of commutation relations. Path-integral quantization provides another route to the quantum theory.

A classical field behaves like an infinite collection of coupled classical coordinates, with roughly one coordinate $\varphi (\mathbf{x})$ at every point in space. Ordinary canonical quantization replaces a coordinate and its conjugate momentum with operators satisfying

$$[\hat{q},\hat{p}]=i\hslash .$$

The field version applies the same rule at every spatial point:

$$[\hat{\varphi }(\mathbf{x}),\hat{\pi }(\mathbf{y})]=i\hslash {\delta }^3(\mathbf{x}-\mathbf{y}).$$

The delta function says that $\hat{\varphi }(\mathbf{x})$ is conjugate to $\hat{\pi }(\mathbf{x})$, while variables at different points are independent at equal times. This prescription reproduces quantum mechanics for each field mode, gives the correct Heisenberg equations and particle spectrum, and produces bosonic statistics.

Canonical quantization is a construction rule, not something derived from classical physics alone. We test the resulting theory for mathematical consistency and agreement with experiment. Fermionic fields require anticommutators, while constrained and gauge theories need additional care.

Relativistic interactions can create and destroy particles, so a state with a fixed particle number is insufficient. A quantum field contains creation and annihilation operators and can connect sectors with different particle numbers while preserving relativistic locality.

Expand the classical field into momentum modes and promote each mode's amplitudes to creation and annihilation operators. For a scalar field, canonical quantization imposes equal-time commutators such as

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

The resulting ladder operators create and remove particles of definite momentum. Fermion fields use anticommutators, which enforce the exclusion principle.

A mode is one independent pattern in which a field can oscillate. A Fourier transform decomposes a free field into plane-wave modes labeled by momentum $\mathbf{p}$:

$$\varphi (x)=\int d^{3}p\left[a(\mathbf{p})e^{-ip\cdot x}+a^{\ast }(\mathbf{p})e^{ip\cdot x}\right].$$

Each mode has a definite momentum, frequency ${\omega }_{\mathbf{p}}=E_{\mathbf{p}}/\hslash$, and amplitude. The full field is the sum of all its modes, like a vibrating string built from musical harmonics. In a free quantum field, a mode's occupation number counts particles with that momentum. Interactions can couple different modes.

A free spin-0 field obeys the Klein–Gordon equation:

$$\left(\square +\frac{m^2}{{\hslash }^2}\right)\varphi (x)=0.$$

Its basic solutions are $e^{-ip\cdot x/\hslash }$ and $e^{+ip\cdot x/\hslash }$, with on-shell momentum

$$p^2=m^2,E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}.$$

For a real scalar field, the general classical solution is

$$\varphi (x)=\int \frac{d^{3}p}{(2\pi \hslash {)}^3}\frac{\hslash }{\sqrt{2E_{\mathbf{p}}}}\left[a(\mathbf{p})e^{-ip\cdot x/\hslash }+a^{\ast }(\mathbf{p})e^{+ip\cdot x/\hslash }\right].$$

The two terms are complex conjugates, which keeps $\varphi$ real. After quantization, $a\to \hat{a}$ and $a^{\ast }\to {\hat{a}}^{†}$. The negative-frequency term then creates a positive-energy particle rather than describing a physical negative-energy particle.

A free spin-$\frac{1}{2}$ field obeys the Dirac equation:

$$(i\hslash {\gamma }^{\mu }{\partial }_{\mu }-m)\psi (x)=0.$$

Its plane-wave spinors satisfy

$$({\gamma }^{\mu }{p}_{\mu }-m)u_s(p)=0,({\gamma }^{\mu }{p}_{\mu }+m)v_s(p)=0,$$

with $p^2=m^2$ and $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$. The general quantized free field is

$$\hat{\psi }(x)=\sum _{s=1}^2\int \frac{d^{3}p}{(2\pi \hslash {)}^3}\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left[{\hat{a}}_s(p)u_s(p)e^{-ip\cdot x/\hslash }+{\hat{b}}_s^{†}(p)v_s(p)e^{+ip\cdot x/\hslash }\right].$$

The operator ${\hat{a}}_s$ annihilates a particle, while ${\hat{b}}_s^{†}$ creates its antiparticle. The label $s=1,2$ identifies the two spin states.

Fourier expansion separates a free field into independent momentum modes. Each mode has a quadratic Hamiltonian with the same form as a harmonic oscillator. The operator ${\hat{a}}_{\mathbf{p}}^{†}$ adds one quantum of momentum $\mathbf{p}$ and energy $E_{\mathbf{p}}$, while ${\hat{a}}_{\mathbf{p}}$ removes one.

For a Lagrangian density $\mathcal{L}$, the conjugate momentum is

$$\pi (x)=\frac{\partial \mathcal{L}}{\partial \dot{\varphi }(x)}.$$

For the free real scalar field, $\pi =\dot{\varphi }$. It is the field analogue of $p=\partial L/\partial \dot{q}$.

Canonical commutation implies

$$[a_{\mathbf{p}},a_{\mathbf{q}}^{†}]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q}),$$

up to the normalization convention, while two creation operators or two annihilation operators commute. Different momentum modes are independent oscillators.

The vacuum obeys $a_{\mathbf{p}}|0⟩=0$ for every momentum. Acting once with a creation operator gives $|\mathbf{p}⟩=a_{\mathbf{p}}^{†}|0⟩$, a one-particle momentum eigenstate. Repeated action builds multiparticle bosonic states.

After mode expansion, the Hamiltonian is a sum of oscillator energies. Normal ordering removes the formal vacuum constant and leaves

$$H=\int d^{3}p{E}_{\mathbf{p}}{a}_{\mathbf{p}}^{†}{a}_{\mathbf{p}},$$

so each occupied mode contributes its relativistic energy.

Its particle and antiparticle amplitudes become operators $a_s$ and $b_s$ that obey anticommutation relations. The field contains $a_s{u}_s{e}^{-ipx}$ and $b_s^{†}{v}_s{e}^{ipx}$ terms. Anticommutation restricts each fermionic mode to occupation zero or one and makes both particle and antiparticle excitation energies positive.

The electric four-current describes the density and flow of electric charge:

$$J^{\mu }=(\rho ,\mathbf{j}),$$

where $\rho$ is the charge density and $\mathbf{j}$ is the three-dimensional current density. For a Dirac field $\psi$ with electric charge $q$, the QED current is

$$J^{\mu }=q\bar{\psi }{\gamma }^{\mu }\psi .$$

Its time component gives the charge density, $J^0=q{\psi }^{†}\psi$, while its spatial components give the charge flow. The current is conserved:

$${\partial }_{\mu }{J}^{\mu }=0,$$

which is the relativistic form of charge conservation.

The electromagnetic field strength is the antisymmetric tensor

$$F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu },$$

built from the gauge potential $A_{\mu }$. Its components contain the electric and magnetic fields. With the convention ${\eta }_{\mu \nu }=\mathrm{diag}(+1,-1,-1,-1)$ used here, the raised tensor is

$$F^{\mu \nu }=\left(\begin{array}{cccc}0& -E_x& -E_y& -E_z\\ E_x& 0& -B_z& B_y\\ E_y& B_z& 0& -B_x\\ E_z& -B_y& B_x& 0\end{array}\right).$$

The time–space components contain $\mathbf{E}$, while the space–space components contain $\mathbf{B}$. Antisymmetry, $F^{\mu \nu }=-F^{\nu \mu }$, leaves six independent components: three electric and three magnetic.

Write the four-potential as $A^{\mu }=(\phi ,\mathbf{A})$ in units where $c=1$. Take its antisymmetrized derivative:

$$F^{\mu \nu }={\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu }.$$

The same calculation in three-vector notation gives

$$\mathbf{E}=-\nabla \phi -{\partial }_t\mathbf{A},\mathbf{B}=\nabla \times \mathbf{A}.$$

Thus,

$$F^{0i}=-E_i,F^{ij}=-{\epsilon }^{ijk}{B}_k.$$

For example, $F^{01}={\partial }_t{A}_x+{\partial }_x\phi =-E_x$ and $F^{12}={\partial }_y{A}_x-{\partial }_x{A}_y=-B_z$. Computing the other index pairs fills the field-strength matrix.

No. The gauge transformation

$$A_{\mu }\to A_{\mu }-{\partial }_{\mu }\alpha$$

leaves the field strength unchanged because the extra second derivatives cancel:

$$F_{\mu \nu }^{\prime }=F_{\mu \nu }.$$

Different potentials related this way describe the same electric and magnetic fields. An arbitrary change $A_{\mu }\to A_{\mu }+B_{\mu }$ generally changes the physics; it is locally a pure gauge only when ${\partial }_{\mu }{B}_{\nu }-{\partial }_{\nu }{B}_{\mu }=0$.

The potential $A_{\mu }$ is a dynamical field. Given a four-current $J^{\nu }$, solve Maxwell's equation

$${\partial }_{\mu }{F}^{\mu \nu }=J^{\nu },F^{\mu \nu }={\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu }.$$

Substitution gives

$$\square A^{\nu }-{\partial }^{\nu }({\partial }_{\mu }{A}^{\mu })=J^{\nu }.$$

Because the solution is not unique, choose a gauge. In Lorenz gauge, ${\partial }_{\mu }{A}^{\mu }=0$, the equation becomes

$$\square A^{\nu }=J^{\nu }.$$

The sources, boundary or initial conditions, and gauge choice then determine a representative $A_{\mu }$. The function $\alpha (x)$ does not generate the physical field; it only changes which gauge-equivalent representative describes the same field.

Maxwell's four equations combine into two tensor equations:

$${\partial }_{\mu }{F}^{\mu \nu }=J^{\nu },$$

$${\partial }_{\lambda }{F}_{\mu \nu }+{\partial }_{\mu }{F}_{\nu \lambda }+{\partial }_{\nu }{F}_{\lambda \mu }=0.$$

In the first equation, choosing $\nu =0$ gives Gauss's electric law,

$$\nabla \cdot \mathbf{E}=\rho ,$$

while $\nu =1,2,3$ gives the Ampère–Maxwell law,

$$\nabla \times \mathbf{B}={\partial }_t\mathbf{E}+\mathbf{j}.$$

In the second equation, choosing three spatial indices gives $\nabla \cdot \mathbf{B}=0$. Choosing one time index gives Faraday's law,

$$\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}.$$

The second tensor equation follows automatically from $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ because partial derivatives commute. The first follows from varying the electromagnetic action, including its coupling to the four-current $J^{\mu }$.

A global symmetry is a transformation applied in the same way at every point in spacetime. For example, the Dirac field has the global U(1) phase symmetry

$$\psi (x)\to e^{i\alpha }\psi (x),$$

where $\alpha$ is the same constant everywhere. This changes the field's mathematical phase but leaves the Lagrangian and physical predictions unchanged. By Noether's theorem, every continuous global symmetry has a conserved quantity. For this U(1) symmetry, that quantity is electric charge.

A spacetime-dependent phase transformation of the Dirac field breaks invariance of the free derivative. Introducing the gauge field $A_{\mu }$ and covariant derivative

$$D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$$

restores local U(1) symmetry. The QED Lagrangian is

$${\mathcal{L}}_{\mathrm{QED}}=\bar{\psi }(i{\gamma }^{\mu }{D}_{\mu }-m)\psi -\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }.$$

Its interaction term couples charged fermions to photons.

Making the charged field's U(1) phase symmetry local requires a connection $A_{\mu }$ with the transformation

$$A_{\mu }\to A_{\mu }-{\partial }_{\mu }\alpha .$$

This is exactly the gauge freedom of the electromagnetic four-potential. Its gauge-invariant curvature

$$F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$$

contains the electric and magnetic fields. Giving $A_{\mu }$ the kinetic term $-F_{\mu \nu }{F}^{\mu \nu }/4$ produces Maxwell's equations, while the covariant derivative produces the coupling $-e\bar{\psi }{\gamma }^{\mu }{A}_{\mu }\psi$ to electric current.

Thus the new U(1) gauge field has the transformation law, dynamics, and coupling of electromagnetism. Experiment confirms this identification, and its quantum excitations are photons.

If $\psi (x)\to e^{ie\alpha (x)}\psi (x)$, covariance requires

$$A_{\mu }\to A_{\mu }-{\partial }_{\mu }\alpha .$$

This cancels the derivative of the local phase in $D_{\mu }\psi$. The field strength $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ remains unchanged, so the electric and magnetic fields are gauge invariant.

The QED Lagrangian is

$${\mathcal{L}}_{\mathrm{QED}}=\bar{\psi }(i{\gamma }^{\mu }{D}_{\mu }-m)\psi -\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu },$$

where

$$D_{\mu }={\partial }_{\mu }+ie{A}_{\mu },F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }.$$

The first term describes the charged Dirac field, including its kinetic and mass terms. The second describes the photon field. Expanding the covariant derivative gives the interaction term

$$-e\bar{\psi }{\gamma }^{\mu }{A}_{\mu }\psi ,$$

which describes the coupling between the charged fermion and the photon.

Expand $A_{\mu }$ into momentum and polarization modes and promote their coefficients to ladder operators. Gauge freedom removes unphysical redundancies, leaving two transverse polarizations for an on-shell photon. Creation operators add photon quanta to the Fock state.

A plane wave $A^{\mu }={\epsilon }^{\mu }{e}^{-ik\cdot x}$ starts with four polarization components. Maxwell's equation imposes $k\cdot \epsilon =0$, removing one. Gauge freedom ${\epsilon }^{\mu }\to {\epsilon }^{\mu }+\alpha k^{\mu }$ removes another, leaving two directions transverse to the momentum.

A polarization proportional to its momentum, ${\epsilon }^{\mu }=C{k}^{\mu }$, gives

$$F^{\mu \nu }\propto k^{\mu }{k}^{\nu }-k^{\nu }{k}^{\mu }=0.$$

The potential oscillates, but it produces no electric or magnetic field, so it is not a physical photon.

The Maxwell Lagrangian contains no time derivative of $A_0$. Its conjugate momentum is therefore

$${\pi }^0=-F^{00}=0.$$

This prevents imposing an ordinary canonical commutator for $A_0$. The component $A_0$ instead enforces Gauss's-law constraint.

A gauge-fixing term is an extra term added to the QED Lagrangian to control the redundant descriptions related by gauge transformations. It is not added to the gauge potential $A_{\mu }$. A common choice is

$${\mathcal{L}}_{\mathrm{QED}}⟶{\mathcal{L}}_{\mathrm{QED}}+{\mathcal{L}}_{\mathrm{gf}},{\mathcal{L}}_{\mathrm{gf}}=-\frac{1}{2\xi }({\partial }_{\mu }{A}^{\mu }{)}^2.$$

It changes the equations for the redundant components used during quantization without changing gauge-invariant physical predictions.

The Feynman-gauge term

$${\mathcal{L}}_{\mathrm{gf}}=-\frac{1}{2}({\partial }_{\mu }{A}^{\mu }{)}^2$$

supplies the missing time derivative of $A_0$ and makes all four components obey $\square A_{\mu }=0$. We can then quantize them uniformly, but we must remove the extra unphysical states afterward.

Quantizing all four components introduces two transverse modes, one longitudinal mode, and one timelike mode. The timelike mode has negative norm because the Minkowski metric gives its oscillator the opposite sign. Longitudinal and timelike excitations cannot represent physical photons.

After adding the Feynman-gauge term, the four components decouple. The momentum-space propagator is

$$D_{\mu \nu }(k)=\frac{-i{g}_{\mu \nu }}{k^2+iϵ}.$$

Unphysical components may appear on internal lines, but the physical-state conditions and gauge symmetry remove them from observable predictions.

The interaction Lagrangian determines the transition amplitude $\mathcal{M}$ through the S-matrix. To predict a scattering rate, calculate $|\mathcal{M}{|}^2$, sum and average over unobserved quantum states, integrate over final-state phase space, and divide by the incoming flux. The resulting cross section can be compared directly with event counts in an experiment.

Square the amplitude, average over unobserved initial states, sum over final states, impose four-momentum conservation, and integrate over Lorentz-invariant phase space. Dividing by the incoming flux gives $d\sigma$. This separates the dynamics in $\mathcal{M}$ from the kinematics of the initial and final particles.

It prepares incoming beams with known momenta and quantum numbers, then counts specified outgoing particles in detector bins. Luminosity measures the opportunity for collisions, and $N={\mathcal{L}}_{\mathrm{int}}\sigma$ relates an event count to the cross section, before efficiencies and backgrounds are included.

For each final particle it supplies $d^{3}p/[(2\pi {)}^{3}2E]$, and a delta function enforces total four-momentum conservation. Integrating this measure sums over all final configurations accepted by the measurement.

Flux and phase space follow from kinematics once the external particles are specified. The theory-dependent information lies in $\mathcal{M}$, which encodes couplings, exchanged particles, spin, and interference. Perturbation theory and Feynman rules provide a systematic way to compute it.

We want the transition amplitude between chosen initial and final states, $⟨f|S|i⟩$. Its interacting part defines the invariant amplitude $\mathcal{M}$:

$$⟨f|iT|i⟩=(2\pi {)}^4{\delta }^4(P_f-P_i)i\mathcal{M}.$$

Perturbation theory approximates $\mathcal{M}$ by expanding the S-matrix in powers of the interaction strength. We then use $|\mathcal{M}{|}^2$ to calculate measurable cross sections and decay rates.

Write

$$\mathcal{L}={\mathcal{L}}_0+{\mathcal{L}}_{\mathrm{int}}.$$

The quadratic terms form the exactly solvable free Lagrangian ${\mathcal{L}}_0$. The remaining terms couple fields and form ${\mathcal{L}}_{\mathrm{int}}$. We quantize ${\mathcal{L}}_0$ exactly and expand the S-matrix in powers of ${\mathcal{L}}_{\mathrm{int}}$ and its coupling constants. We split the Lagrangian; we do not expand the Lagrangian itself.

For a real scalar field $\varphi$,

$${\mathcal{L}}_{\mathrm{int}}=-\frac{g}{3!}{\varphi }^3.$$

The coupling $g$ sets the interaction strength. The factor $3!$ compensates for the equivalent ways of attaching three identical scalar lines to a vertex. Each insertion of ${\mathcal{L}}_{\mathrm{int}}$ in the Dyson series produces one three-legged vertex with factor $-ig$.

Yes. Cubic scalar theory has

$${\mathcal{L}}_{\mathrm{int}}=-\frac{g}{3!}{\varphi }^3,$$

which couples three identical scalar fields. QED instead has

$${\mathcal{L}}_{\mathrm{int}}=-q\bar{\psi }{\gamma }^{\mu }\psi A_{\mu },$$

which joins two fermion lines and one photon line and gives the vertex factor $-iq{\gamma }^{\mu }$. The perturbative method is the same, but the fields, propagators, spinor structure, and fermionic signs differ.

The S-matrix maps an incoming free-particle state in the distant past to an outgoing free-particle state in the distant future. After splitting

$$\mathcal{L}={\mathcal{L}}_0+{\mathcal{L}}_{\mathrm{int}},$$

${\mathcal{L}}_0$ defines the free fields and asymptotic particle states, while ${\mathcal{L}}_{\mathrm{int}}$ generates transitions between them. In the interaction picture,

$$S=\mathcal{T}\exp \left(i\int d^{4}x{\mathcal{L}}_{\mathrm{int}}(x)\right),$$

where $\mathcal{T}$ orders later interactions to the left. Expanding the exponential gives

$$\begin{array}{rl}S=& 1+i\int d^{4}x{\mathcal{L}}_{\mathrm{int}}(x)\\ & +\frac{i^2}{2!}\int d^{4}x{d}^{4}y\mathcal{T}[{\mathcal{L}}_{\mathrm{int}}(x){\mathcal{L}}_{\mathrm{int}}(y)]+\cdots .\end{array}$$

Each insertion of ${\mathcal{L}}_{\mathrm{int}}$ adds one interaction vertex. Writing $S=1+iT$ separates free passage from scattering, and matrix elements of $T$ define the invariant amplitude $\mathcal{M}$.

Start from the interaction Lagrangian and expand the S-matrix:

$$S=\mathcal{T}\exp \left(i\int d^{4}x{\mathcal{L}}_{\mathrm{int}}(x)\right)=1+iT.$$

The Dyson expansion therefore determines $iT=S-1$. In practice, we usually calculate only the matrix element needed for a chosen process:

$$⟨f|iT|i⟩=(2\pi {)}^4{\delta }^4(P_f-P_i)i\mathcal{M}.$$

Choose the initial and final states, expand to the required order in the coupling, and evaluate the result with Wick's theorem or the equivalent Feynman rules. Here $T$ is the transition operator, while $\mathcal{T}$ is the time-ordering operator.

Expand the interaction-picture S-matrix,

$$S=T\exp \left[-\frac{i}{\hslash }\int d^{4}x{\mathcal{H}}_I(x)\right],$$

in powers of the coupling. Wick's theorem rewrites time-ordered products as normal-ordered fields plus contractions. Contractions become propagators, interaction terms become vertices, and the remaining external fields become external lines. This dictionary turns each term in the series into a Feynman diagram and amplitude.

A contraction of two free fields gives a propagator, which connects two spacetime points or carries momentum internally. Each factor from the interaction Lagrangian gives a vertex and coupling constant. The expansion order counts vertices, so higher powers of a small coupling usually give smaller corrections.

Expanding the time-ordered exponential gives terms with zero, one, two, and more insertions of the interaction Hamiltonian. A process begins at the lowest order capable of connecting its external states; higher orders add vertices and loop corrections.

Wick's theorem rewrites a time-ordered product of free fields as its normal-ordered product plus every possible contraction. For two fields,

$$\mathcal{T}[\varphi (x)\varphi (y)]=:\varphi (x)\varphi (y):+⟨0|\mathcal{T}[\varphi (x)\varphi (y)]|0⟩.$$

Each contraction is a propagator. In a vacuum expectation value, terms with unpaired field operators vanish, so the surviving contractions map directly to Feynman diagrams.

A propagator is a time-ordered two-point function. In momentum space, the scalar propagator $i/(p^2-m^2+iϵ)$ is $i$ times the inverse of the free quadratic operator. Its pole structure encodes free-particle propagation, and the $iϵ$ specifies the time-ordering prescription.

Different contractions can produce the same diagram. Factorials from the exponential and interaction Lagrangian cancel many duplicated contractions; any remaining multiplicity becomes the diagram's symmetry factor. It prevents identical configurations from being overcounted.

The logic is unchanged, but contractions use Dirac and photon propagators, external fermions carry spinors, and each interaction insertion supplies $-ie{\gamma }^{\mu }$. Anticommuting fermion fields also introduce signs when their order changes.

Attach a spinor or polarization factor to each external line, a propagator to each internal line, and a factor $-ie{\gamma }^{\mu }$ to each electron–photon vertex. Conserve four-momentum at every vertex and integrate over independent loop momenta. Tree diagrams give the leading approximation; loops supply quantum corrections and can contain divergences that require renormalization.

Every distinct allowed contraction or internal routing contributes an amplitude. Add these amplitudes before taking the absolute square, so their cross terms produce quantum interference. Compton scattering, for example, has two tree-level electron-exchange diagrams with different vertex orderings.

External fermions contribute $u,\bar{u},v,$ or $\bar{v}$ spinors, external photons contribute polarization vectors, internal lines contribute propagators, and each vertex contributes $-ie{\gamma }^{\mu }$ with four-momentum conserved.

Electron–photon scattering has two tree diagrams because the photons can attach in two orders. Add both amplitudes before squaring; the internal electron momentum and propagator differ between the two terms.

Square the summed amplitude, average over initial spins and polarizations, and sum over final ones. Completeness relations turn spin sums into gamma-matrix traces and polarization sums into tensor contractions.

Tree diagrams contain no closed momentum cycles and usually give the leading result. Each loop introduces an unconstrained four-momentum integral, an additional coupling order, and a genuine quantum correction.

Their integrals include arbitrarily large virtual momenta and can diverge. Regularization exposes the divergence, and renormalization absorbs it into measured parameters so finite predictions remain.

Beta-minus (${\beta }^{-}$) decay is a radioactive process in which a neutron changes into a proton, an electron, and an electron antineutrino:

$$n⟶p+e^{-}+{\bar{\nu }}_e.$$

At the quark level, a down quark changes into an up quark through a virtual $W^{-}$ boson:

$$d⟶u+W^{-},W^{-}⟶e^{-}+{\bar{\nu }}_e.$$

The proton, electron, and antineutrino share the neutron's initial energy and momentum. In the neutron rest frame,

$${\mathbf{p}}_p+{\mathbf{p}}_e+{\mathbf{p}}_{\bar{\nu }}=0,$$

and

$$m_n=E_p+E_e+E_{\bar{\nu }}.$$

Because the three final particles can share the energy and momentum in many different ways, the electron has a continuous range of energies. The antineutrino is essential: without it, two-body kinematics would fix the electron energy to one value.

At the nuclear level,

$$0=(+e)+(-e)+0,$$

for $n\to p+e^{-}+{\bar{\nu }}_e$. At the quark level,

$$-\frac{e}{3}=+\frac{2e}{3}-e+0,$$

for $d\to u+e^{-}+{\bar{\nu }}_e$. The virtual $W^{-}$ carries charge $-e$ between the two vertices.

Parity is the spatial-inversion transformation

$$\mathcal{P}:(t,\mathbf{x})⟶(t,-\mathbf{x}).$$

It is the transformation associated with comparing a process with its mirror-image version. A parity-conserving interaction gives the same probabilities for the original and parity-transformed processes.

Position and momentum are polar vectors, so they reverse:

$$\mathbf{x}⟶-\mathbf{x},\mathbf{p}⟶-\mathbf{p}.$$

Spin is an axial vector, like angular momentum, so it does not reverse:

$$\mathbf{S}⟶\mathbf{S}.$$

Consequently, $\mathbf{S}\cdot \mathbf{p}$ changes sign under parity.

Wu and her collaborators cooled cobalt-60 nuclei to low temperature and aligned their spins with a magnetic field. They counted beta electrons emitted along and opposite to the nuclear-spin direction, then reversed the magnetic field and repeated the measurement. The electrons were emitted asymmetrically: more came out opposite to the nuclear spin. Since parity reverses momentum but not spin, this asymmetry differs from its parity-reflected version and shows that the weak interaction violates parity.

Apply a chiral projector to the Dirac field $\psi$:

$${\psi }_L=P_L\psi ,{\psi }_R=P_R\psi .$$

The full field is the sum of its two parts, $\psi ={\psi }_L+{\psi }_R$.

They are defined by

$$P_L=\frac{1-{\gamma }^5}{2},P_R=\frac{1+{\gamma }^5}{2}.$$

$P_L$ keeps the left-chiral component and removes the right-chiral component; $P_R$ does the reverse. They satisfy

$$P_L^2=P_L,P_R^2=P_R,P_L{P}_R=0,P_L+P_R=1.$$

The left-chiral lepton current is

$$j_{\ell }^{\mu }={\bar{e}}_L{\gamma }^{\mu }{\nu }_{eL}=\bar{e}{\gamma }^{\mu }{P}_L{\nu }_e=\frac{1}{2}\bar{e}{\gamma }^{\mu }(1-{\gamma }^5){\nu }_e.$$

Its field content allows the charged weak interaction to create an electron and an electron antineutrino. Its Hermitian-conjugate current describes the reverse process.

The corresponding left-chiral quark current is

$$j_q^{\mu }={\bar{u}}_L{\gamma }^{\mu }{d}_L=\bar{u}{\gamma }^{\mu }{P}_{L}d.$$

It converts a down quark into an up quark at the charged-current vertex. Together, the quark and lepton currents describe beta decay through charged $W$ exchange.

The weak interaction acts on left-handed fermion doublets and violates parity. Making their SU(2) symmetry local introduces three gauge fields $W_{\mu }^a$ with self-interactions. The combinations $W^{\pm }$ mediate charged-current processes such as beta decay, while $W^3$ contributes to the neutral interaction. At low energy, massive-W exchange reduces to Fermi's effective four-fermion interaction.