Why look for eigenfunctions of energy and momentum (KG Equation)?

Chapter 2 of Peskin and Schroder ‘An Intro to QFT’ contains something like this:

Just as in ordinary quantum mechanics, we look for eigenfunctions of momentum and energy:$$\phi(x)=e^{-ip\cdot x}$$ϕ(x)=e−ip⋅x

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)

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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.

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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.

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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.

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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.

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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.

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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.

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The Most Fundamental Reason

The deepest reason comes from Noether’s theorem.

Symmetry	Conserved Quantity
Time translation	Energy
Space translation	Momentum

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.