# Interpretation of the four-vector $k$ in scalar QFT

I'm studying the canonical quantization of the Klein-Gordon real scalar quantum field theory, given by the classical Lagrangian density $$\mathscr L = {1\over 2}\partial_\mu\phi\partial^\mu\phi-{1\over 2}m^2\phi^2.$$

The plane wave solutions to the Euler-Lagrange equation (which becomes the KG equation) are, of course, of the form $$\phi(t,\vec x)\backsim e^{-i\vec k\cdot\vec x}$$

where $\vec k\cdot\vec x = k_\mu x^\mu$ and $k_0\equiv\omega = \sqrt{k^2+m^2}$. To find arbitrary solutions you take superpositions over the three spatial dimensions of $k$, and thus most of your integrals start with something looking like

$$\int \frac{d^3 k}{(2\pi)^3 2\omega}...$$

Indeed, the Hamiltonian and the creation and annihilation operators have such integral forms, and thus when you add bosons to the vacuum the parameter is this mysterious vector $\vec k$. Is there any sort of intuitive physical meaning to this (maybe having to do with the bosons for which it is an initial parameter) or is it simply a byproduct of abstract mathematics?

• What do you get if you apply the momentum operator to $\phi(t, \vec x) = \exp(-i\vec k \cdot \vec x)$? – Robin Ekman Dec 8 '14 at 0:42

This is simply a Fourier transform. The wave equation, Hamiltonian, and Green's functions are all simpler in momentum space than in position space. This is because, for the free, theory, the different momenta decouple, and you can create a particle with momentum $k$ with $a^\dagger_k \left|0\right\rangle$, and that state will evolve in a nice way. This is dependent on translation invariance of the Lagrangian, among other things.
When you act $\phi(t,\vec{x})$ on the vacuum, you get a particle at $(\vec{x},t)$, that is $|\vec{x},t>=\phi(t,\vec{x})|0>=\int \frac{d^3 k}{(2\pi)^3}e^{-ikx}a^\dagger _k|0>$. It's a Fourier transformation. And it means that if a particle wants to be in $(\vec{x},t)$ then it should have all kinds of momenta, which is actually the uncertainty principle. To @lionelbrits, the Lagrangian is not translation invariant, but the action is.