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The classical Klein-Gordon theory for a real scalar field is called a relativistic free theory.

It is called a free theory because the dynamics of the degrees of freedom in the momentum space of the Klein-Gordon theory are decoupled from each other. In other words, the Klein-Gordon theory is devoid of interactions among the excitations of the Klein-Gordon field in the momentum space.

Is the Klein-Gordon theory also devoid of interactions among the excitations of the Klein-Gordon field in the position space?

Why is the Klein-Gordon theory called a relativistic theory? Is it because the Klein-Gordon Lagrangian and the Klein-Gordon equation of motion are Lorentz invariant?

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Yes. You are correct. A non-relativistic theory would be invariant under the Galilean group. Lorentz invariance (specifically, invariance under Lorentz boosts) is what defines a relativistic theory.

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  • $\begingroup$ Thanks! I was also wondering if there the lack of interactions among the excitations of the Klein-Gordon theory in the momentum space of the theory necessarily imply a lack of interactions among the excitations of the theory in the position space of the theory as well. $\endgroup$ – nightmarish Apr 26 '16 at 16:21
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And regarding why it's called a "free" theory, it's not specific to a momentum-space formulation. It's "free" because the Lagrangian is quadratic in the fields, and therefore the equations of motion (what you get from plugging the Lagrangian into the Euler-Lagrange equation) are linear in the fields. Therefore you can superpose different classical solutions to the EOM and get another solution. If you have two solutions consisting of one wave packet propagating in one direction and another propagating in a different direction, then to "combine" the two solutions you simply add them together and you don't get any kind of scattering - the particles just pass right through each other.

Whether the field has interactions in position space depends on exactly what you mean. The derivative terms relate the value of the field at one point to its values at nearby points and could therefore be considered a "position-space interaction." If you think of a free scalar field as a network of coupled simple harmonic oscillators (springs), then the springs are indeed coupled together - that's why it's very energetically unfavorable to have two nearby points have completely different values (because then there would have to be a large derivative connecting them). Also, in statistical mechanics, the Klein-Gordon equation often arises as a low-energy approximate description of an interacting system, such as the Ising model. The underlying degrees of freedom (the "springs" or the Ising spins) are coupled together and therefore interact, but the excitations (the "ripples" or oscillations in the field) do not interact. When people say that the Klein-Gordon equation is "noninteracting," they're referring to the excitations, not the underlying degrees of freedom (like the field values themselves). And whether or not you're interacting is independent of whether you're working in position space or momentum space, because the Fourier transform of zero is zero and the FT of something nonzero is nonzero. Regarding the excitations, anything quadratic in the fields (or their derivatives) is noninteracting and anything higher than quadratic is interacting, because those terms will contribute nonlinearly to the EOM.

Finally, to clarify Prahar's answer: for a system to be relativistic, it must be invariant under the entire Lorentz group, not just boosts. But local rotational invariance is often the case in practical applications - boost invariance, not always.

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