where$$p\cdot x = p_\mu x^\mu = Et-\mathbf{p}\cdot\mathbf{x}$$p⋅x=pμ​xμ=Et−p⋅x

so explicitly$$\phi(x) = e^{-i(Et-\mathbf{p}\cdot\mathbf{x})}$$ϕ(x)=e−i(Et−p⋅x)

Why look for energy and momentum functions as a trial solution?

Step 1: Think about the Schrödinger Equation

In ordinary quantum mechanics, if the Hamiltonian doesn’t depend on position,$$\hat H = \frac{\hat p^2}{2m}$$H^=2mp^​2​

then momentum is conserved.

The momentum operator is$$\hat p = -i\nabla$$p^​=−i∇

and its eigenfunctions satisfy$$\hat p \psi = p\psi.$$p^​ψ=pψ.

The solutions are$$\psi(\mathbf x) = e^{i\mathbf p\cdot\mathbf x}.$$ψ(x)=eip⋅x.

Why?

Because derivatives of exponentials reproduce the same exponential:$$-i\nabla e^{i\mathbf p\cdot\mathbf x} = \mathbf p\, e^{i\mathbf p\cdot\mathbf x}.$$−i∇eip⋅x=peip⋅x.

This makes exponentials the natural building blocks of the theory.

Step 2: Same Idea for the KG Equation

The KG equation is$$(\Box+m^2)\phi=0.$$(□+m2)ϕ=0.

Notice that its coefficients are constants.

There is no preferred location:$$x \rightarrow x+a.$$x→x+a.

Likewise there is no preferred time:$$t \rightarrow t+b.$$t→t+b.

Therefore:

• Momentum is conserved.
• Energy is conserved.

Whenever a differential equation has translation symmetry, its natural modes are eigenfunctions of the translation operators.

Step 3: What Generates Translations?

Suppose we shift space:$$x \rightarrow x+\epsilon.$$x→x+ϵ.

The generator of this transformation is$$\hat p = -i\partial_x.$$p^​=−i∂x​.

Likewise time translations are generated by$$\hat H = i\partial_t.$$H^=i∂t​.

Thus momentum and energy are literally the operators that describe spacetime translations.

Step 4: Find Simultaneous Eigenfunctions

We seek states satisfying$$\hat H \phi = E\phi$$H^ϕ=Eϕ

and$$\hat{\mathbf p}\phi = \mathbf p\phi.$$p^​ϕ=pϕ.

Using$$\hat H=i\partial_t, \qquad \hat{\mathbf p}=-i\nabla,$$H^=i∂t​,p^​=−i∇,

we get$$i\partial_t\phi=E\phi$$i∂t​ϕ=Eϕ

and$$-i\nabla\phi=\mathbf p\phi.$$−i∇ϕ=pϕ.

Solving these gives$$\phi(x) = e^{-iEt} e^{i\mathbf p\cdot\mathbf x} = e^{-ip\cdot x}.$$ϕ(x)=e−iEteip⋅x=e−ip⋅x.

So the plane wave is not a guess.

It is the unique simultaneous eigenfunction of energy and momentum.

Step 5: Why Are Plane Waves So Useful?

Because the KG equation is linear.

If$$\phi_1$$ϕ1​

and$$\phi_2$$ϕ2​

are solutions, then$$a\phi_1+b\phi_2$$aϕ1​+bϕ2​

is also a solution.

The plane waves form a complete basis.

Therefore any solution can be written as$$\phi(x) = \int d^3p\, A(\mathbf p)e^{-ip\cdot x}.$$ϕ(x)=∫d3pA(p)e−ip⋅x.

This is exactly analogous to a Fourier transform.

Step 6: The Deeper QFT View

In QFT, every momentum mode becomes an independent harmonic oscillator.

For a given momentum $\mathbf p$p,$$\phi_{\mathbf p}(t) \sim e^{-iE_{\mathbf p}t}.$$ϕp​(t)∼e−iEp​t.

where$$E_{\mathbf p} = \sqrt{\mathbf p^2+m^2}.$$Ep​=p2+m2​.

Thus the field can be decomposed into infinitely many oscillators labeled by momentum.

That is why Peskin and Schroeder immediately move to momentum eigenmodes.

The momentum basis diagonalizes the theory.

The Most Fundamental Reason

The deepest reason comes from Noether’s theorem.

The Klein-Gordon equation is invariant under both.

Therefore energy and momentum are the natural quantum numbers.

The plane waves$$e^{-ip\cdot x}$$e−ip⋅x

are precisely the states with definite values of those conserved quantities.

The post Why look for eigenfunctions of energy and momentum (KG Equation)? appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

In natural units ($\hbar=c=1$ℏ=c=1):$$(\Box + m^2)\phi(x)=0$$(□+m2)ϕ(x)=0

where$$\Box \equiv \partial_\mu\partial^\mu = \frac{\partial^2}{\partial t^2} -\nabla^2.$$□≡∂μ​∂μ=∂t2∂2​−∇2.

Explicitly,$$\left( \frac{\partial^2}{\partial t^2} -\nabla^2 +m^2 \right)\phi(x)=0.$$(∂t2∂2​−∇2+m2)ϕ(x)=0.

Step 1: Plane-Wave Solutions

We first look for solutions of the form$$\phi(x)=Ae^{-ip\cdot x}$$ϕ(x)=Ae−ip⋅x

where$$p\cdot x = Et-\mathbf p\cdot \mathbf x.$$p⋅x=Et−p⋅x.

Substituting into the KG equation gives$$(-E^2+\mathbf p^2+m^2)\phi=0.$$(−E2+p2+m2)ϕ=0.

Therefore$$E^2=\mathbf p^2+m^2.$$E2=p2+m2.

This is the relativistic energy-momentum relation.

Thus for every momentum $\mathbf p$p,$$E_p=\sqrt{\mathbf p^2+m^2}.$$Ep​=p2+m2​.

and we obtain two solutions$$e^{-iE_pt+i\mathbf p\cdot\mathbf x}$$e−iEp​t+ip⋅x

and$$e^{+iE_pt-i\mathbf p\cdot\mathbf x}.$$e+iEp​t−ip⋅x.

Step 2: General Solution

The most general solution is a superposition of all momentum modes:$$\boxed{ \phi(x)= \int \frac{d^3p}{(2\pi)^3} \left[ a(\mathbf p)e^{-ip\cdot x} + b(\mathbf p)e^{ip\cdot x} \right] }$$ϕ(x)=∫(2π)3d3p​[a(p)e−ip⋅x+b(p)eip⋅x]​

where$$p^\mu=(E_p,\mathbf p).$$pμ=(Ep​,p).

This is the classical KG field.

Step 3: Meaning of Each Term

The integral

$$\int d^3p$$∫d3p

adds together waves of every possible momentum.

Just as a Fourier series adds sine waves of different frequencies.

The factor

$$e^{-ip\cdot x} = e^{-iE_pt+i\mathbf p\cdot\mathbf x}$$e−ip⋅x=e−iEp​t+ip⋅x

represents a positive-frequency mode.

The phase oscillates forward in time.

The factor

$$e^{+ip\cdot x} = e^{+iE_pt-i\mathbf p\cdot\mathbf x}$$e+ip⋅x=e+iEp​t−ip⋅x

represents a negative-frequency mode.

In relativistic quantum mechanics this was interpreted as a negative-energy solution.

In QFT it becomes an antiparticle mode.

The coefficients

$$a(\mathbf p)$$a(p)

tell us how much of momentum $\mathbf p$p exists in the positive-frequency part.

$$b(\mathbf p)$$b(p)

tell us how much of momentum $\mathbf p$p exists in the negative-frequency part.

They are determined by the initial conditions:$$\phi(\mathbf x,0)$$ϕ(x,0)

and$$\dot\phi(\mathbf x,0).$$ϕ˙​(x,0).

Step 4: Real Scalar Field

If the field is real,$$\phi^*(x)=\phi(x),$$ϕ∗(x)=ϕ(x),

then the coefficients cannot be independent.

Reality requires$$b(\mathbf p)=a^*(\mathbf p).$$b(p)=a∗(p).

Therefore$$\boxed{ \phi(x)= \int \frac{d^3p}{(2\pi)^3} \left[ a(\mathbf p)e^{-ip\cdot x} + a^*(\mathbf p)e^{ip\cdot x} \right] }$$ϕ(x)=∫(2π)3d3p​[a(p)e−ip⋅x+a∗(p)eip⋅x]​

The field contains positive and negative frequencies, but only one physical particle species.

Examples:

• Neutral pion (approximately)
• Higgs field

Step 5: Complex Scalar Field

For a complex field,$$\phi(x)\neq\phi^*(x),$$ϕ(x)=ϕ∗(x),

and $a$a and $b$b are independent.

The solution becomes$$\boxed{ \phi(x)= \int \frac{d^3p}{(2\pi)^3} \left[ a(\mathbf p)e^{-ip\cdot x} + b(\mathbf p)e^{ip\cdot x} \right] }$$ϕ(x)=∫(2π)3d3p​[a(p)e−ip⋅x+b(p)eip⋅x]​

Now there are two independent sets of excitations.

In QFT these become:

• particle operators
• antiparticle operators

respectively.

Step 6: QFT Interpretation

After quantization,$$a(\mathbf p) \rightarrow \hat a(\mathbf p)$$a(p)→a^(p)

and$$b(\mathbf p) \rightarrow \hat b^\dagger(\mathbf p).$$b(p)→b^†(p).

The field operator becomes$$\boxed{ \hat\phi(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_p}} \left[ \hat a(\mathbf p)e^{-ip\cdot x} + \hat b^\dagger(\mathbf p)e^{ip\cdot x} \right] }$$ϕ^​(x)=∫(2π)3d3p​2Ep​​1​[a^(p)e−ip⋅x+b^†(p)eip⋅x]​

where:

• $\hat a$a^ annihilates a particle
• $\hat a^\dagger$a^† creates a particle
• $\hat b$b^ annihilates an antiparticle
• $\hat b^\dagger$b^† creates an antiparticle

This is the modern interpretation of the KG solution.

Physical Picture

Think of the KG field as an infinite collection of relativistic harmonic oscillators.

For every momentum $\mathbf p$p, there are two oscillatory modes:$$e^{-iE_pt} \qquad\text{and}\qquad e^{+iE_pt}.$$e−iEp​tande+iEp​t.

In classical field theory they are simply Fourier components.

In QFT they become:$$\text{particle creation/annihilation modes}$$particle creation/annihilation modes

and$$\text{antiparticle creation/annihilation modes}.$$antiparticle creation/annihilation modes.

That reinterpretation is precisely what removes the “negative energy problem” you asked about earlier. The $e^{+iE_pt}$e+iEp​t solutions never disappear; they are reinterpreted as antiparticle degrees of freedom rather than physical states of negative energy.

The post Plane wave solutions to the Klein Gordon Equation appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

The short answer is:

In RQM, negative-energy solutions appear because we are trying to describe relativistic particles with a single-particle wavefunction. In QFT, those same solutions are reinterpreted as antiparticle creation operators, so physical states always have positive energy.

Let’s walk through it carefully.

1. The Problem Appears in the Dirac Equation

The free-particle Dirac equation is$$(i\gamma^\mu \partial_\mu – m)\psi = 0$$(iγμ∂μ​−m)ψ=0

Assume a plane-wave solution$$\psi(x)=u(p)e^{-ip\cdot x}.$$ψ(x)=u(p)e−ip⋅x.

Substituting gives$$(\gamma^\mu p_\mu-m)u(p)=0.$$(γμpμ​−m)u(p)=0.

Nontrivial solutions require$$\det(\gamma^\mu p_\mu-m)=0,$$det(γμpμ​−m)=0,

which yields$$E^2=p^2+m^2.$$E2=p2+m2.

Therefore$$E=\pm \sqrt{p^2+m^2}.$$E=±p2+m2​.

The Dirac equation naturally contains:

• Positive-energy solutions
• Negative-energy solutions

2. Why This Is a Disaster in Single-Particle Quantum Mechanics

In ordinary QM, energy eigenstates are physical particle states.

Suppose a particle occupies$$E=+10\ \text{MeV}.$$E=+10 MeV.

Then there are states with$$E=-10,\,-20,\,-100\ \text{MeV}$$E=−10,−20,−100 MeV

available.

The particle could continuously emit photons and fall to lower and lower energies.

There is no lowest energy state.

The vacuum would be unstable.

This is unacceptable.

3. Dirac’s Original Solution: The Dirac Sea

Paul Dirac proposed:

• Every negative-energy state is already occupied.
• Electrons obey the Pauli exclusion principle.
• A normal electron cannot fall into those states.

The vacuum becomes$$\text{Vacuum} = \text{all negative-energy states filled}.$$Vacuum=all negative-energy states filled.

A missing electron (a “hole”) behaves like a positively charged particle.

This predicted the positron.

Historically this was brilliant.

But it has problems:

• Works only for fermions.
• Requires an infinite sea of particles.
• Doesn’t generalize well.

QFT replaces it with something much cleaner.

4. The Key Idea in QFT

QFT quantizes the field, not the particle.

Instead of a wavefunction$$\psi(x),$$ψ(x),

we promote it to an operator field$$\hat\psi(x).$$ψ^​(x).

The field is expanded as$$\hat\psi(x) = \sum_s \int \frac{d^3p}{(2\pi)^3} \left[ b_s(p)u_s(p)e^{-ipx} + d_s^\dagger(p)v_s(p)e^{ipx} \right].$$ψ^​(x)=s∑​∫(2π)3d3p​[bs​(p)us​(p)e−ipx+ds†​(p)vs​(p)eipx].

Notice something important.

There is no negative-energy operator.

Instead:$$e^{-iEt}$$e−iEt

appears with an electron annihilation operator$$b.$$b.

while$$e^{+iEt}$$e+iEt

appears with an antiparticle creation operator$$d^\dagger.$$d†.

The “negative-energy solution” has been reinterpreted.

5. Where Did the Negative Energy Go?

In RQM we read$$v(p)e^{+iEt}$$v(p)e+iEt

as

a particle with energy $-E$−E.

In QFT we read exactly the same mathematical object as

creation of an antiparticle with energy $+E$+E.

The sign has moved from the energy to the operator interpretation.

This is the crucial step.

6. Hamiltonian in QFT

After quantization, the Hamiltonian becomes$$H= \sum_s \int d^3p\, E_p \left( b_s^\dagger b_s + d_s^\dagger d_s \right) + E_{\rm vac}.$$H=s∑​∫d3pEp​(bs†​bs​+ds†​ds​)+Evac​.

Every excitation contributes$$+E_p.$$+Ep​.

Electron:$$E_p>0.$$Ep​>0.

Positron:$$E_p>0.$$Ep​>0.

No negative-energy particles remain.

The only leftover infinity is the vacuum energy$$E_{\rm vac},$$Evac​,

which is handled separately by normal ordering or renormalization.

7. Feynman’s Interpretation

Richard Feynman provided another viewpoint.

A negative-energy electron moving forward in time can be reinterpreted as

a positive-energy positron moving backward in time.

Mathematically,$$e^{+iEt}$$e+iEt

can be viewed as either:

• negative-energy particle forward in time
• positive-energy antiparticle backward in time

Both descriptions are equivalent.

QFT adopts the positive-energy antiparticle interpretation.

8. Why the Klein-Gordon Equation Has the Same Issue

The Klein-Gordon equation also gives$$E=\pm \sqrt{p^2+m^2}.$$E=±p2+m2​.

So the problem is not unique to Dirac particles.

The issue arises whenever we combine:

• quantum mechanics
• special relativity

The cure is always the same:

Stop treating the object as a single particle and treat it as a quantum field.

The Deep Physical Lesson

The negative-energy states are not actually removed in QFT.

They are reinterpreted.

In that sense, the existence of antiparticles is not an additional prediction of QFT. It is the mechanism by which QFT resolves the negative-energy catastrophe of relativistic quantum mechanics. The negative-energy solutions never disappear from the mathematics—they are simply understood as antiparticle degrees of freedom rather than physical states of negative energy.

The post Negative energy states in Relativistic QM, but not in QFT appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

A relativistic particle is an object that obeys the relativistic energy-momentum relation

$E^2=p^2c^2+m^2c^4$E2=p2c2+m2c4

A relativistic field is a quantity defined at every point in spacetime whose dynamics are Lorentz invariant.

Particles are what you observe.

Fields are the underlying entities that produce those particles.

In modern QFT, fields are fundamental; particles are excitations of fields.

1. Relativistic Particle

In classical mechanics, a particle has:

• Position $x(t)$x(t)
• Momentum $p(t)$p(t)
• Energy $E(t)$E(t)

Special relativity modifies the relationship between momentum and energy:$$E=\gamma mc^2$$E=γmc2 $$p=\gamma mv$$p=γmv

where$$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$$γ=1−v2/c2​1​

The particle traces out a worldline in spacetime.

A relativistic particle is therefore:

A localized object moving through spacetime whose energy and momentum satisfy Einstein’s relativistic equations.

Examples:

• Electron
• Proton
• Photon

treated as individual objects.

2. Why Particles Alone Become Problematic

Suppose we try to quantize a relativistic particle.

We might write$$E \rightarrow i\hbar\frac{\partial}{\partial t}$$E→iℏ∂t∂​ $$p \rightarrow -i\hbar\nabla$$p→−iℏ∇

and substitute into$$E^2=p^2c^2+m^2c^4$$E2=p2c2+m2c4

giving the Klein-Gordon equation.

The problem:

The resulting theory predicts

• Negative-energy solutions
• Particle creation
• Particle annihilation

which cannot be described by a fixed number of particles.

Nature allows:$$\gamma \rightarrow e^- + e^+$$γ→e−+e+ $$e^- + e^+ \rightarrow \gamma+\gamma$$e−+e+→γ+γ

A particle-only description breaks down.

3. Relativistic Field

A field assigns a value to every spacetime point.

Examples:

Temperature field:$$T(x,y,z,t)$$T(x,y,z,t)

Electric field:$$\mathbf E(x,y,z,t)$$E(x,y,z,t)

Quantum field:$$\phi(x,t)$$ϕ(x,t)

or$$\psi(x,t)$$ψ(x,t)

Instead of tracking a particle trajectory, we describe the evolution of the entire field.

4. The Klein-Gordon Field

Consider a scalar field$$\phi(x)$$ϕ(x)

Its dynamics obey

$(\Box+m^2)\phi=0$(□+m2)ϕ=0

where$$\Box=\partial_\mu\partial^\mu$$□=∂μ​∂μ

This equation is Lorentz invariant.

Notice:

There is no particle anywhere in the equation.

Only a field.

5. Quantizing the Field

The crucial step:

Treat the field itself as an operator.

Instead of$$\phi(x)$$ϕ(x)

we write$$\hat{\phi}(x)$$ϕ^​(x)

and expand it into modes:$$\hat{\phi}(x) = \sum_k \left( a_k e^{-ikx} + a_k^\dagger e^{ikx} \right)$$ϕ^​(x)=k∑​(ak​e−ikx+ak†​eikx)

The operators$$a_k^\dagger$$ak†​

create excitations.

The operators$$a_k$$ak​

destroy excitations.

Now particles appear naturally.

6. Particle in QFT

In QFT a particle is

A quantized excitation of a field mode.

For example:

Electron field:$$\psi(x)$$ψ(x)

One excitation:

electron.

Two excitations:

two electrons.

No excitations:

vacuum.

Likewise:

Photon field → photons

Gluon field → gluons

Higgs field → Higgs bosons

7. Key Philosophical Difference

Relativistic Particle View

Reality consists of particles.

Fields are mathematical tools.

Particle  → Fundamental
Field     → Secondary

This was roughly the view before QFT.

Relativistic Field View

Reality consists of fields.

Particles are excitations of fields.

Field      → Fundamental
Particle   → Emergent

This is the modern Standard Model viewpoint.

8. Example: Electron

Particle Picture

Electron is a tiny point object.

You ask:

• Where is it?
• How fast is it moving?

This works reasonably well at low energies.

Field Picture

There exists an electron field$$\psi(x)$$ψ(x)

throughout the universe.

What we call “an electron” is simply a localized excitation of that field.

The field is everywhere.

The particle is local.

9. Why Fields Are Necessary Relativistically

Special relativity implies$$E=mc^2$$E=mc2

which means energy can become matter.

Particles can be created and destroyed.

A fixed-particle theory cannot handle:

• Pair creation
• Pair annihilation
• Vacuum fluctuations
• Hawking radiation
• Particle decays

Fields can.

This is the deepest reason QFT replaces relativistic quantum mechanics.

10. Dirac’s View

Historically, Paul Dirac first wrote a relativistic wave equation for the electron:$$(i\gamma^\mu\partial_\mu-m)\psi=0$$(iγμ∂μ​−m)ψ=0

Initially it looked like a relativistic particle equation.

But the existence of antimatter and pair creation forced a reinterpretation:

The Dirac equation is actually the equation of a relativistic spinor field, not merely a single relativistic particle.

The one-sentence summary

A relativistic particle is a localized object obeying Einstein’s energy-momentum relation, while a relativistic field is a Lorentz-invariant entity spread throughout spacetime whose quantized excitations appear to us as particles; modern quantum field theory treats the field—not the particle—as fundamental.

The post Relativistic Particle versus Relativistic Field appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

Let’s do a few concrete Weyl-vector examples.

1. Expand $p_\mu\sigma^\mu$pμ​σμ

Use$$\sigma^\mu=(I,\sigma_x,\sigma_y,\sigma_z)$$σμ=(I,σx​,σy​,σz​)

and, with metric $(+,-,-,-)$(+,−,−,−),$$p_\mu=(E,-p_x,-p_y,-p_z)$$pμ​=(E,−px​,−py​,−pz​)

So$$p_\mu\sigma^\mu = E I – p_x\sigma_x – p_y\sigma_y – p_z\sigma_z$$pμ​σμ=EI−px​σx​−py​σy​−pz​σz​

Example: particle moving in the $z$z-direction:$$p^\mu=(E,0,0,p)$$pμ=(E,0,0,p)

Then$$p_\mu\sigma^\mu = E I-p\sigma_z$$pμ​σμ=EI−pσz​ $$= \begin{pmatrix} E-p & 0\\ 0 & E+p \end{pmatrix}$$=(E−p0​0E+p​)

Similarly,$$p_\mu\bar{\sigma}^\mu = E I+p\sigma_z = \begin{pmatrix} E+p & 0\\ 0 & E-p \end{pmatrix}$$pμ​σˉμ=EI+pσz​=(E+p0​0E−p​)

2. Massless right-handed Weyl equation

For a massless right-handed spinor,$$p_\mu\sigma^\mu \psi_R=0$$pμ​σμψR​=0

Using the $z$z-direction result:$$\begin{pmatrix} E-p & 0\\ 0 & E+p \end{pmatrix} \begin{pmatrix} a\\ b \end{pmatrix} =0$$(E−p0​0E+p​)(ab​)=0

For a massless particle,$$E=p$$E=p

so$$\begin{pmatrix} 0 & 0\\ 0 & 2E \end{pmatrix} \begin{pmatrix} a\\ b \end{pmatrix} =0$$(00​02E​)(ab​)=0

This gives$$2E b=0$$2Eb=0

so$$b=0$$b=0

Therefore$$\boxed{ \psi_R = \begin{pmatrix} 1\\ 0 \end{pmatrix} }$$ψR​=(10​)​

up to normalization.

This is spin-up along the direction of motion.

So a right-handed massless Weyl spinor has positive helicity:$$\boxed{ h=+\frac12}$$h=+21​​

3. Massless left-handed Weyl equation

For left-handed spinors,$$p_\mu\bar{\sigma}^\mu \psi_L=0$$pμ​σˉμψL​=0

Using$$p_\mu\bar{\sigma}^\mu = \begin{pmatrix} E+p & 0\\ 0 & E-p \end{pmatrix}$$pμ​σˉμ=(E+p0​0E−p​)

and $E=p$E=p,$$\begin{pmatrix} 2E & 0\\ 0 & 0 \end{pmatrix} \begin{pmatrix} a\\ b \end{pmatrix} =0$$(2E0​00​)(ab​)=0

This gives$$a=0$$a=0

so$$\boxed{ \psi_L = \begin{pmatrix} 0\\ 1 \end{pmatrix} }$$ψL​=(01​)​

This is spin-down along the direction of motion.

So a left-handed massless Weyl spinor has negative helicity:$$\boxed{ h=-\frac12}$$h=−21​​

4. Massive Dirac case: left and right are coupled

The chiral Dirac equations are$$p_\mu\sigma^\mu\psi_R=m\psi_L$$pμ​σμψR​=mψL​ $$p_\mu\bar{\sigma}^\mu\psi_L=m\psi_R$$pμ​σˉμψL​=mψR​

Again take motion along $z$z:$$p_\mu\sigma^\mu = \begin{pmatrix} E-p&0\\ 0&E+p \end{pmatrix}$$pμ​σμ=(E−p0​0E+p​) $$p_\mu\bar{\sigma}^\mu = \begin{pmatrix} E+p&0\\ 0&E-p \end{pmatrix}$$pμ​σˉμ=(E+p0​0E−p​)

Suppose spin-up, so use$$\psi_R= \begin{pmatrix} A\\ 0 \end{pmatrix}, \qquad \psi_L= \begin{pmatrix} B\\ 0 \end{pmatrix}$$ψR​=(A0​),ψL​=(B0​)

Then the first equation gives$$(E-p)A=mB$$(E−p)A=mB

The second gives$$(E+p)B=mA$$(E+p)B=mA

From the second,$$B=\frac{m}{E+p}A$$B=E+pm​A

Using$$E^2-p^2=m^2$$E2−p2=m2

we also have$$\frac{m}{E+p} = \frac{E-p}{m}$$E+pm​=mE−p​

So$$\boxed{ \psi_L= \frac{m}{E+p}\psi_R }$$ψL​=E+pm​ψR​​

For very high energy,$$E\gg m$$E≫m

so$$\frac{m}{E+p}\approx \frac{m}{2E}\ll 1$$E+pm​≈2Em​≪1

Therefore a high-energy spin-up massive fermion is mostly right-handed, with a small left-handed component.

That is why chirality and helicity become nearly the same at high energy.

Does This Mean QFT Is Linear Like QM?

Partly yes, but mostly no.

Free QFT is linear

The free Klein–Gordon equation is linear:$$(\Box+m^2)\phi=0$$(□+m2)ϕ=0

The free Dirac equation is linear:$$(i\gamma^\mu\partial_\mu-m)\psi=0$$(iγμ∂μ​−m)ψ=0

The free Weyl equation is linear:$$i\sigma^\mu\partial_\mu\psi_R=0$$iσμ∂μ​ψR​=0

So for free particles, QFT resembles ordinary quantum mechanics: superpositions work cleanly.

But interacting QFT is generally nonlinear

Once interactions are included, the equations are no longer simple linear wave equations.

Example scalar interaction:$$\mathcal L = \frac12(\partial\phi)^2 – \frac12m^2\phi^2 – \frac{\lambda}{4!}\phi^4$$L=21​(∂ϕ)2−21​m2ϕ2−4!λ​ϕ4

The equation of motion becomes$$(\Box+m^2)\phi + \frac{\lambda}{3!}\phi^3 =0$$(□+m2)ϕ+3!λ​ϕ3=0

That $\phi^3$ϕ3 term makes it nonlinear.

QED is also interacting

For the electron field interacting with electromagnetism:$$(i\gamma^\mu D_\mu-m)\psi=0$$(iγμDμ​−m)ψ=0

where$$D_\mu=\partial_\mu+ieA_\mu$$Dμ​=∂μ​+ieAμ​

So the electron field couples to the photon field.

The Maxwell equation also gets a source term:$$\partial_\mu F^{\mu\nu}=e\bar{\psi}\gamma^\nu\psi$$∂μ​Fμν=eψˉ​γνψ

That source contains products of fields.

So the full coupled theory is nonlinear.

The clean answer

$$\boxed{ \text{Quantum states still evolve linearly in Hilbert space.} }$$Quantum states still evolve linearly in Hilbert space.​

But:$$\boxed{ \text{the field dynamics of interacting QFT is not linear in the fields.} }$$the field dynamics of interacting QFT is not linear in the fields.​

So QFT keeps the linear superposition principle of quantum mechanics, but interactions make the field equations and scattering structure much richer.

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showing that the Klein–Gordon scalar field transforms consistently under Lorentz transformations and that the theory is Lorentz invariant.

The key idea is:$$\boxed{ \text{Physics must look identical in all inertial frames.} }$$Physics must look identical in all inertial frames.​

For scalar fields, this turns out to be beautifully simple.

1. Start with the Klein–Gordon Equation

The scalar field satisfies:$$(\partial_\mu\partial^\mu + m^2)\phi(x)=0$$(∂μ​∂μ+m2)ϕ(x)=0

or equivalently$$(\Box + m^2)\phi(x)=0$$(□+m2)ϕ(x)=0

where$$\Box \equiv \partial_\mu\partial^\mu$$□≡∂μ​∂μ

is the d’Alembertian operator.

Using metric signature $(+,-,-,-)$(+,−,−,−),$$\Box = \frac{\partial^2}{\partial t^2} – \nabla^2$$□=∂t2∂2​−∇2

2. Lorentz Transformations

A Lorentz transformation changes coordinates:$$x^\mu \rightarrow x’^\mu$$xμ→x′μ

with$$x’^\mu = \Lambda^\mu_{\ \nu}x^\nu$$x′μ=Λ νμ​xν

The matrix $\Lambda$Λ satisfies$$\Lambda^T \eta \Lambda = \eta$$ΛTηΛ=η

which preserves the spacetime interval:$$x_\mu x^\mu = x’_\mu x’^\mu$$xμ​xμ=xμ′​x′μ

3. What Is a Scalar Field?

A scalar field is defined by:$$\boxed{ \phi'(x’)=\phi(x) }$$ϕ′(x′)=ϕ(x)​

This means:

• the numerical value of the field is unchanged
• only the coordinates labeling spacetime points change

This is the defining property of a Lorentz scalar.

4. Transforming Derivatives

Now derive how derivatives transform.

Using the chain rule:$$\frac{\partial}{\partial x’^\mu} = \frac{\partial x^\nu}{\partial x’^\mu} \frac{\partial}{\partial x^\nu}$$∂x′μ∂​=∂x′μ∂xν​∂xν∂​

Since$$x’^\mu = \Lambda^\mu_{\ \nu}x^\nu$$x′μ=Λ νμ​xν

the inverse transformation is$$x^\nu = (\Lambda^{-1})^\nu_{\ \mu}x’^\mu$$xν=(Λ−1) μν​x′μ

Therefore:$$\partial’_\mu = (\Lambda^{-1})^\nu_{\ \mu}\partial_\nu$$∂μ′​=(Λ−1) μν​∂ν​

or equivalently$$\partial’^\mu = \Lambda^\mu_{\ \nu}\partial^\nu$$∂′μ=Λ νμ​∂ν

So derivatives transform like four-vectors.

5. Transforming the d’Alembertian

Now examine$$\Box’ = \partial’_\mu\partial’^\mu$$□′=∂μ′​∂′μ

Substitute the transformed derivatives:$$\Box’ = (\Lambda^{-1})^\nu_{\ \mu}\partial_\nu \Lambda^\mu_{\ \rho}\partial^\rho$$□′=(Λ−1) μν​∂ν​Λ ρμ​∂ρ

Using$$(\Lambda^{-1})^\nu_{\ \mu}\Lambda^\mu_{\ \rho} = \delta^\nu_{\rho}$$(Λ−1) μν​Λ ρμ​=δρν​

we get$$\Box’ = \partial_\nu\partial^\nu = \Box$$□′=∂ν​∂ν=□

Thus:$$\boxed{ \Box \text{ is Lorentz invariant} }$$□ is Lorentz invariant​

This is the central result.

6. Lorentz Invariance of the KG Equation

Now apply this to the field equation:$$(\Box + m^2)\phi(x)=0$$(□+m2)ϕ(x)=0

Under Lorentz transformation:$$(\Box’ + m^2)\phi'(x’) = (\Box + m^2)\phi(x)$$(□′+m2)ϕ′(x′)=(□+m2)ϕ(x)

Since the RHS is zero,$$(\Box’ + m^2)\phi'(x’)=0$$(□′+m2)ϕ′(x′)=0

Therefore the Klein–Gordon equation has exactly the same form in every inertial frame.

So:$$\boxed{ \text{The KG equation is Lorentz invariant} }$$The KG equation is Lorentz invariant​

7. Lorentz Invariance of the Action

The action is$$S = \int d^4x\, \mathcal L$$S=∫d4xL

with$$\mathcal L = \frac12\partial_\mu\phi\partial^\mu\phi – \frac12m^2\phi^2$$L=21​∂μ​ϕ∂μϕ−21​m2ϕ2

Now check each term.

Kinetic Term

$$\partial_\mu\phi\partial^\mu\phi$$∂μ​ϕ∂μϕ

is a Lorentz scalar because it contracts two four-vectors.

Like:$$A_\mu A^\mu$$Aμ​Aμ

Mass Term

$$\phi^2$$ϕ2

is also scalar since $\phi$ϕ itself is scalar.

Volume Element

Lorentz transformations preserve spacetime volume:$$d^4x’ = d^4x$$d4x′=d4x

for proper Lorentz transformations.

Therefore:$$S’ = S$$S′=S

Thus the action is Lorentz invariant.

8. Physical Interpretation

The scalar field has:

• no direction in spacetime
• no spin index
• no vector structure

It behaves like temperature distributed through spacetime.

All observers agree on the field value at the same spacetime event.

9. Contrast with Spinors and Vectors

Scalar:$$\phi'(x’)=\phi(x)$$ϕ′(x′)=ϕ(x)

Vector:$$A’^\mu(x’) = \Lambda^\mu_{\ \nu}A^\nu(x)$$A′μ(x′)=Λ νμ​Aν(x)

Spinor:$$\psi'(x’) = S(\Lambda)\psi(x)$$ψ′(x′)=S(Λ)ψ(x)

where $S(\Lambda)$S(Λ) is the spinor representation of the Lorentz group.

Spinors transform much more subtly.

10. The Deep Idea

The entire structure of relativistic QFT comes from demanding:$$\boxed{ \text{the action be Lorentz invariant} }$$the action be Lorentz invariant​

That requirement strongly constrains:

• allowed fields
• allowed interactions
• allowed dynamics

For the Klein–Gordon field, Lorentz invariance emerges because:$$\boxed{ \partial_\mu\partial^\mu }$$∂μ​∂μ​

is a spacetime scalar operator.

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unless you are in a special approximation. The relativistic treatment changes the framework.

The main issue is this:$$\boxed{ \text{Relativity treats space and time together, but ordinary Hamiltonians single out time.} }$$Relativity treats space and time together, but ordinary Hamiltonians single out time.​

So the more natural object is the action or the field theory Lagrangian, not a nonrelativistic two-particle Hamiltonian.

For two relativistic particles/fields, one writes something like$$S = \int d^4x \, \mathcal L_A + \int d^4x \, \mathcal L_B + \int d^4x \, \mathcal L_{\text{int}}$$S=∫d4xLA​+∫d4xLB​+∫d4xLint​

or, schematically,$$\mathcal L = \mathcal L_A + \mathcal L_B + \mathcal L_{\text{int}}$$L=LA​+LB​+Lint​

For two scalar fields,$$\mathcal L_A = \frac12 \partial_\mu \phi_A \partial^\mu \phi_A – \frac12 m_A^2\phi_A^2$$LA​=21​∂μ​ϕA​∂μϕA​−21​mA2​ϕA2​ $$\mathcal L_B = \frac12 \partial_\mu \phi_B \partial^\mu \phi_B – \frac12 m_B^2\phi_B^2$$LB​=21​∂μ​ϕB​∂μϕB​−21​mB2​ϕB2​

and an interaction might be$$\mathcal L_{\text{int}} = -g \phi_A^2 \phi_B^2$$Lint​=−gϕA2​ϕB2​

Then the relativistic dynamics comes from$$\delta S = 0$$δS=0

rather than from a simple two-body Schrödinger Hamiltonian.

For quantum theory, the state lives in a tensor product of field Hilbert spaces:$$\mathcal H = \mathcal H_A \otimes \mathcal H_B$$H=HA​⊗HB​

but the particles are excitations of fields, not little objects with fixed wavefunctions.

A bipartite relativistic state might look like$$|\Psi\rangle = \sum_{p,q} C(p,q) \,a_A^\dagger(p) a_B^\dagger(q) |0\rangle$$∣Ψ⟩=p,q∑​C(p,q)aA†​(p)aB†​(q)∣0⟩

where $a_A^\dagger(p)$aA†​(p) creates particle $A$A with four-momentum $p$p, and $a_B^\dagger(q)$aB†​(q) creates particle $B$B with four-momentum $q$q.

The relativistic energy is$$E_{\mathbf p} = \sqrt{\mathbf p^2 + m^2}$$Ep​=p2+m2​

so the free Hamiltonian is roughly$$H_0 = \int \frac{d^3p}{(2\pi)^3} E_{\mathbf p} \,a^\dagger_{\mathbf p}a_{\mathbf p}$$H0​=∫(2π)3d3p​Ep​ap†​ap​

For two species,$$H_0 = \int \frac{d^3p}{(2\pi)^3} E^A_{\mathbf p} a_A^\dagger(\mathbf p)a_A(\mathbf p) + \int \frac{d^3q}{(2\pi)^3} E^B_{\mathbf q} a_B^\dagger(\mathbf q)a_B(\mathbf q)$$H0​=∫(2π)3d3p​EpA​aA†​(p)aA​(p)+∫(2π)3d3q​EqB​aB†​(q)aB​(q)

Then add an interaction Hamiltonian:$$H = H_A + H_B + H_{\text{int}}$$H=HA​+HB​+Hint​

But now $H_{\text{int}}$Hint​ comes from a Lorentz-invariant field interaction.

So the relativistic version of the bipartite Hamiltonian idea is:$$\boxed{ H_A \otimes I + I \otimes H_B + H_{\text{int}} }$$HA​⊗I+I⊗HB​+Hint​​

but interpreted in QFT, where $H_A$HA​, $H_B$HB​, and $H_{\text{int}}$Hint​ act on field Fock spaces.

The clean hierarchy is:$$\boxed{ \text{Nonrelativistic QM: two-particle Hamiltonian} }$$Nonrelativistic QM: two-particle Hamiltonian​ $$\boxed{ \text{Relativistic QM: constrained and delicate} }$$Relativistic QM: constrained and delicate​ $$\boxed{ \text{Relativistic QFT: fields, Fock space, Lorentz-invariant action} }$$Relativistic QFT: fields, Fock space, Lorentz-invariant action​

So the safest answer is:$$\boxed{ \text{Treat the bipartite system as two interacting quantum fields.} }$$Treat the bipartite system as two interacting quantum fields.​

Not as two particles with an ordinary potential $V(\mathbf x_A-\mathbf x_B)$V(xA​−xB​), because instantaneous potentials generally conflict with relativity.

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A spinor field is a field describing spin-$\frac12$21​ particles:

• electrons
• quarks
• neutrinos

The Dirac field is the standard example:$$\psi(x)$$ψ(x)

Unlike scalar or vector fields, spinors transform differently under rotations and Lorentz transformations.

2. Classical Macroscopic Fields Exist for Bosons

Examples:

• electromagnetic field $A_\mu(x)$Aμ​(x)
• classical light waves
• laser beams
• superfluids
• Bose condensates

Why?

Because bosons can pile into the same quantum state.

The occupation number can become enormous:$$N \gg 1$$N≫1

This allows the quantum field operator to behave approximately like a classical field.

For example:$$\hat A_\mu(x) \rightarrow A_\mu^{\text{classical}}(x)$$A^μ​(x)→Aμclassical​(x)

3. Fermions Cannot Do This

Spinor fields describe fermions.

Fermions obey:$$\boxed{ \text{Pauli exclusion principle} }$$Pauli exclusion principle​

No two identical fermions can occupy the same quantum state.

Mathematically:$$\{a_p,a_q^\dagger\}=\delta_{pq}$${ap​,aq†​}=δpq​

instead of bosonic commutators:$$[a_p,a_q^\dagger]=\delta_{pq}$$[ap​,aq†​]=δpq​

This changes everything.

4. Grassmann Nature of Fermionic Fields

In QFT, fermionic fields are not ordinary numbers.

They are Grassmann-valued:$$\psi_1\psi_2=-\psi_2\psi_1$$ψ1​ψ2​=−ψ2​ψ1​

In particular:$$\boxed{ \psi^2=0 }$$ψ2=0​

This nilpotent property means you cannot build arbitrarily large coherent amplitudes from a fermion field.

That prevents a classical macroscopic field interpretation.

5. Why Electromagnetic Waves Exist but Electron Waves Do Not

For photons:

many photons can occupy one mode:$$|N\rangle$$∣N⟩

with arbitrarily large $N$N.

This produces classical EM waves.

But for electrons:

occupation is only:$$0 \text{ or } 1$$0 or 1

per quantum state.

So there is no analog of a huge coherent classical electron field.

6. Macroscopic Matter Is NOT a Macroscopic Spinor Field

A metal contains enormous numbers of electrons.

But:$$\boxed{ \text{the electron field itself is still quantum} }$$the electron field itself is still quantum​

The matter behaves macroscopically because:

• enormous statistical averages emerge
• collective variables become classical
• densities and currents become classical observables

NOT because the spinor field becomes classical.

7. Contrast with Bose–Einstein Condensates

Bosonic field:$$\phi(x)$$ϕ(x)

can acquire a macroscopic expectation value:$$\langle \phi(x)\rangle \neq 0$$⟨ϕ(x)⟩=0

This is a true classical order parameter.

For fermions:$$\langle \psi(x)\rangle =0$$⟨ψ(x)⟩=0

in ordinary macroscopic systems.

Instead, fermions form composite bosons:

Examples:

• Cooper pairs
• helium-4 atoms
• mesons

Then THOSE bosonic composites can condense macroscopically.

8. Superconductivity Is the Key Example

Electrons themselves cannot form a classical spinor field.

But paired electrons:$$(e^- e^-)$$(e−e−)

form Cooper pairs with integer spin.

Those pairs behave bosonically.

Then you get a macroscopic condensate:$$\langle \Psi_{\text{Cooper pair}}\rangle \neq 0$$⟨ΨCooper pair​⟩=0

leading to superconductivity.

9. Deep Geometric View

Bosonic classical fields correspond to smooth amplitudes over spacetime:$$\phi(x)\in \mathbb R \text{ or } \mathbb C$$ϕ(x)∈R or C

Fermionic fields instead live in Grassmann algebra.

Grassmann quantities do not have ordinary macroscopic classical limits.

So:$$\boxed{ \text{fermionic fields are fundamentally quantum objects} }$$fermionic fields are fundamentally quantum objects​

while bosonic fields can emerge into classicality.

10. Important Caveat

People sometimes write “classical Dirac fields.”

This means:

• using the Dirac equation classically
• before quantization

But these are mathematical tools.

Physical macroscopic spinor fields are not observed in nature the way classical EM fields are.

The short version:$$\boxed{ \text{Bosons stack. Fermions exclude.} }$$Bosons stack. Fermions exclude.​

and therefore:$$\boxed{ \text{bosonic fields can become classical macroscopic fields,} }$$bosonic fields can become classical macroscopic fields,​

while$$\boxed{ \text{fermionic spinor fields cannot.} }$$fermionic spinor fields cannot.​

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This is one of the most important conceptual distinctions in field theory.

The symbols $\partial$∂ and $\delta$δ represent completely different operations.

1. The Partial Derivative $\partial$∂

$\partial$∂ means:

“How does a function change as spacetime coordinates change?”

For a field$$\phi(x) = \phi(t,\mathbf x)$$ϕ(x)=ϕ(t,x)

the derivative$$\partial_\mu \phi$$∂μ​ϕ

describes how the field changes from point to point in spacetime.

Examples:$$\partial_0 \phi = \frac{\partial \phi}{\partial t}$$∂0​ϕ=∂t∂ϕ​

(time variation)

and$$\partial_i \phi = \frac{\partial \phi}{\partial x^i}$$∂i​ϕ=∂xi∂ϕ​

(spatial variation)

So:$$\boxed{ \partial = \text{ordinary spacetime derivative} }$$∂=ordinary spacetime derivative​

It acts inside spacetime.

2. The Variation $\delta$δ

$\delta$δ means:

“Suppose I slightly deform the entire field configuration.”

This is not motion through spacetime.

Instead:$$\phi(x) \rightarrow \phi(x)+\delta\phi(x)$$ϕ(x)→ϕ(x)+δϕ(x)

means:

• keep spacetime point $x$x fixed
• slightly change the value of the field there

So $\delta\phi$δϕ is an infinitesimal “test deformation” of the whole function.

Visual Intuition

Think of the field as a rubber sheet over spacetime.

• $\partial\phi$∂ϕ:
  measures the slope of the sheet from point to point
• $\delta\phi$δϕ:
  slightly lifts or perturbs the sheet itself

Why Both Appear Together

The Lagrangian depends on:$$\mathcal L(\phi,\partial_\mu\phi)$$L(ϕ,∂μ​ϕ)

meaning:

• the field value
• and its spacetime gradients

When varying the action:$$\delta S = \delta \int d^4x\,\mathcal L$$δS=δ∫d4xL

you must vary BOTH:$$\delta\phi$$δϕ

and$$\delta(\partial_\mu\phi)$$δ(∂μ​ϕ)

Since differentiation and variation commute:$$\boxed{ \delta(\partial_\mu\phi) = \partial_\mu(\delta\phi) }$$δ(∂μ​ϕ)=∂μ​(δϕ)​

This relation is central to the derivation.

Deep Geometric Interpretation

$\partial$∂ lives within spacetime.

$\delta$δ lives in the infinite-dimensional space of all possible field configurations.

So:

Analogy with Classical Mechanics

For a particle trajectory:$$x(t)$$x(t)

we distinguish:

Velocity:$$\frac{dx}{dt}$$dtdx​

vs variation of the path:$$x(t)\rightarrow x(t)+\delta x(t)$$x(t)→x(t)+δx(t)

The first measures motion along the path.

The second compares neighboring possible paths.

Field theory is exactly the same idea, but with fields instead of trajectories.

The Key Philosophical Idea

The action principle says:

Nature compares all nearby possible field configurations and selects the one where$$\boxed{ \delta S=0 }$$δS=0​

That does NOT mean the action is zero.

It means:$$\text{first-order change in action vanishes}$$first-order change in action vanishes

which is analogous to:$$\frac{df}{dx}=0$$dxdf​=0

for extrema in ordinary calculus.

The field equation emerges because the true field configuration is a stationary point in the space of all possible field configurations.

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The physical field configuration is the one for which a small variation of the field,$$\phi(x)\rightarrow \phi(x)+\delta\phi(x)$$ϕ(x)→ϕ(x)+δϕ(x)

does not change the action to first order:$$\delta S=0$$δS=0

Now vary the action:$$\delta S = \int d^4x \left[ \frac{\partial \mathcal{L}}{\partial \phi}\delta\phi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \delta(\partial_\mu\phi) \right]$$δS=∫d4x[∂ϕ∂L​δϕ+∂(∂μ​ϕ)∂L​δ(∂μ​ϕ)]

Since variation and differentiation commute,$$\delta(\partial_\mu\phi) = \partial_\mu(\delta\phi)$$δ(∂μ​ϕ)=∂μ​(δϕ)

so$$\delta S = \int d^4x \left[ \frac{\partial \mathcal{L}}{\partial \phi}\delta\phi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \partial_\mu(\delta\phi) \right]$$δS=∫d4x[∂ϕ∂L​δϕ+∂(∂μ​ϕ)∂L​∂μ​(δϕ)]

Now integrate the second term by parts:$$\int d^4x\, \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \partial_\mu(\delta\phi) = – \int d^4x\, \partial_\mu \left[ \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \right] \delta\phi$$∫d4x∂(∂μ​ϕ)∂L​∂μ​(δϕ)=−∫d4x∂μ​[∂(∂μ​ϕ)∂L​]δϕ

The boundary term vanishes because we assume$$\delta\phi=0$$δϕ=0

on the boundary of spacetime.

Therefore,$$\delta S = \int d^4x \left[ \frac{\partial \mathcal{L}}{\partial \phi} – \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \right) \right]\delta\phi$$δS=∫d4x[∂ϕ∂L​−∂μ​(∂(∂μ​ϕ)∂L​)]δϕ

For the action to be stationary for arbitrary $\delta\phi$δϕ, the bracket must vanish:$$\boxed{ \frac{\partial \mathcal{L}}{\partial \phi} – \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \right) =0 }$$∂ϕ∂L​−∂μ​(∂(∂μ​ϕ)∂L​)=0​

This is the Euler–Lagrange equation for a field.

For the scalar field,$$\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi – \frac{1}{2}m^2\phi^2$$L=21​∂μ​ϕ∂μϕ−21​m2ϕ2

we get$$\frac{\partial \mathcal{L}}{\partial \phi} = -m^2\phi$$∂ϕ∂L​=−m2ϕ

and$$\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\phi$$∂(∂μ​ϕ)∂L​=∂μϕ

So the Euler–Lagrange equation becomes$$-m^2\phi-\partial_\mu\partial^\mu\phi=0$$−m2ϕ−∂μ​∂μϕ=0

or$$\boxed{ (\partial_\mu\partial^\mu+m^2)\phi=0 }$$(∂μ​∂μ+m2)ϕ=0​

That is the Klein–Gordon equation.

In plain English:$$\boxed{ \delta S=0 \quad\Rightarrow\quad \text{field chooses stationary action} \quad\Rightarrow\quad \text{Euler–Lagrange equation} \quad\Rightarrow\quad \text{Klein–Gordon equation} }$$δS=0⇒field chooses stationary action⇒Euler–Lagrange equation⇒Klein–Gordon equation

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