1. Free field theory: Why is it said that different Fourier modes in case of a free field (say, real Klein-Gordon field) are independent of each other?

  2. Interacting field theory: How exactly does the addition of non-linear term in the Lagrangian make the Fourier modes couple to each other?


1 Answer 1


In the case of a real, scalar field, we start from the action

$$S=\int d^4x\ \eta^{\mu\nu}\partial_\mu\phi\partial_\nu\phi+m^2\phi^2$$

The equation of motion is the Klein-Gordon equation

$$(-\eta^{\mu\nu}\partial_\mu\partial^\nu+m^2)\phi=\ddot{\phi}-\nabla^2 \phi+m^2\phi=0$$

Now, we introduce the Fourier modes $\phi_k$: $$\phi=\int \frac{d^3k}{(2\pi)^3}e^{i\vec{k}\cdot\vec{x}}\phi_k $$ In terms of the Fourier modes, the equation of motion becomes $$\ddot{\phi}_k +\omega^2\phi_k=0,\hspace{2cm}\omega\equiv \sqrt{k^2+m^2}$$ As you see, this is just an independent harmonic oscillator equation for each value of $k$. This is what it means to say that the different Fourier modes are independent.

When we introduce an interaction term proportional to $\phi^n$ where $n\geq 3$ (e.g. the canonical $\lambda \phi^4$), we will have terms proportional to $\phi^{n-1}$ ($4\lambda\phi^3$ in my example) in the equation of motion. When going to Fourier space, these powers of $\phi$ all have 'their own label', so we obtain a term something like $\phi_{k_1}\phi_{k_2}\dots\phi_{k_{n-1}}$. As you see, the equation of motion no longer depends only on a single Fourier mode! The Fourier modes are now coupled.


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