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.

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