Intro

The key point is that constant coefficients mean the equation itself does not change when you shift the coordinates.

Let’s look at the KG equation carefully:(□+m2)ϕ(x)=0(\Box + m^2)\phi(x)=0(□+m2)ϕ(x)=0

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

Notice that the coefficients in front of the derivatives are just numbers:1,−1,m2.1,\quad -1,\quad m^2.1,−1,m2.

They do not depend on xxx or ttt.

Contrast with a non-translationally invariant equation

Suppose instead we had(∂2∂t2−∇2+x2)ϕ=0.\left( \frac{\partial^2}{\partial t^2} -\nabla^2 +x^2 \right)\phi=0.(∂t2∂2​−∇2+x2)ϕ=0.

Now perform a translationx→x+a.x \rightarrow x+a.x→x+a.

The equation becomes(∂2∂t2−∇2+(x+a)2)ϕ=0.\left( \frac{\partial^2}{\partial t^2} -\nabla^2 +(x+a)^2 \right)\phi=0.(∂t2∂2​−∇2+(x+a)2)ϕ=0.

Expanding:x2+2ax+a2.x^2+2ax+a^2.x2+2ax+a2.

The equation has changed!

Therefore the physics at x=0x=0x=0 differs from the physics at x=100x=100x=100.

There is a preferred location.

Translation symmetry is broken.

Now do the same for KG

Define a translated fieldϕ′(x)=ϕ(x−a).\phi'(x)=\phi(x-a).ϕ′(x)=ϕ(x−a).

Apply the KG operator:(□+m2)ϕ′(x)=(□+m2)ϕ(x−a).(\Box+m^2)\phi'(x) = (\Box+m^2)\phi(x-a).(□+m2)ϕ′(x)=(□+m2)ϕ(x−a).

Since derivatives commute with constant shifts,∂μϕ(x−a)=(∂μϕ)(x−a).\partial_\mu \phi(x-a) = (\partial_\mu \phi)(x-a).∂μ​ϕ(x−a)=(∂μ​ϕ)(x−a).

Therefore(□+m2)ϕ(x−a)=[(□+m2)ϕ](x−a).(\Box+m^2)\phi(x-a) = \big[(\Box+m^2)\phi\big](x-a).(□+m2)ϕ(x−a)=[(□+m2)ϕ](x−a).

But ϕ\phiϕ satisfies KG:(□+m2)ϕ=0.(\Box+m^2)\phi=0.(□+m2)ϕ=0.

Hence(□+m2)ϕ′(x)=0.(\Box+m^2)\phi'(x)=0.(□+m2)ϕ′(x)=0.

The translated solution is again a solution.

That is exactly what we mean by translation symmetry.

Time translations work identically

Takeϕ′(t,x)=ϕ(t−b,x).\phi'(t,\mathbf x) = \phi(t-b,\mathbf x).ϕ′(t,x)=ϕ(t−b,x).

Then(□+m2)ϕ′=0.(\Box+m^2)\phi’ = 0.(□+m2)ϕ′=0.

Again, the equation is unchanged.

No preferred time exists.

The deeper mathematical statement

A translation isxμ→xμ+aμ.x^\mu \rightarrow x^\mu+a^\mu.xμ→xμ+aμ.

The KG operator is□+m2.\Box+m^2.□+m2.

Notice that neither □\Box□ nor m2m^2m2 contains xμx^\muxμ.

Therefore[□+m2,  Pμ]=0,[\Box+m^2,\;P_\mu]=0,[□+m2,Pμ​]=0,

wherePμ=i∂μP_\mu=i\partial_\muPμ​=i∂μ​

is the generator of translations.

Because the translation generators commute with the equation, solutions can be chosen to be eigenfunctions of PμP_\muPμ​.

Those eigenfunctions satisfyPμϕ=pμϕ.P_\mu\phi=p_\mu\phi.Pμ​ϕ=pμ​ϕ.

Solving givesϕ(x)=e−ip⋅x.\phi(x)=e^{-ip\cdot x}.ϕ(x)=e−ip⋅x.

This is where the plane waves come from.

Physical intuition

Imagine a perfectly infinite ocean.

The wave equation is∂2ψ∂t2−c2∇2ψ=0.\frac{\partial^2\psi}{\partial t^2} -c^2\nabla^2\psi=0.∂t2∂2ψ​−c2∇2ψ=0.

Every point of the ocean looks identical.

A wave doesn’t care whether it is at:

  • x=0x=0x=0
  • x=1000x=1000x=1000
  • x=−106x=-10^6x=−106

The equation is the same everywhere.

The natural solutions are traveling wavesei(kx−ωt).e^{i(kx-\omega t)}.ei(kx−ωt).

The KG field is exactly the relativistic version of this idea.

If the coefficients depended on position, the medium would be inhomogeneous, like water whose density changes from place to place. Then momentum eigenstates would no longer be the natural modes.

The connection to Noether’s theorem

The chain of logic is:Constant coefficients⟹Translation symmetry⟹Conserved momentum and energy⟹Momentum/energy eigenfunctions are natural⟹e−ip⋅x.\text{Constant coefficients} \Longrightarrow \text{Translation symmetry} \Longrightarrow \text{Conserved momentum and energy} \Longrightarrow \text{Momentum/energy eigenfunctions are natural} \Longrightarrow e^{-ip\cdot x}.Constant coefficients⟹Translation symmetry⟹Conserved momentum and energy⟹Momentum/energy eigenfunctions are natural⟹e−ip⋅x.

This is why Peskin and Schroeder immediately look for plane-wave solutions. They are the normal modes associated with spacetime translation symmetry.