The Klein-Gordon (KG) equation is the relativistic wave equation for a spin-0 particle.

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.

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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.

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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.

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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.

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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.

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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.

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The coefficients

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

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

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

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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

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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.

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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.

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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.