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How do I understand that the action for the free relativistic scalar field theory is non linear? What will be the associated interaction potential of that equation?

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The action of a field theory itself is almost always nonlinear, i.e. not first-order in the fields. If actions were just first-order in fields, the equations of motion would pretty much say that the fields are equal to zero.

Simplest meaningful actions of field theories start with quadratic – bilinear – forms. When actions (or, equivalently, Hamiltonians) are bilinear, the equations of motion derived from these actions (or Hamiltonians) are linear. That's what we call "free field theory" because the solutions to these linear equations of motion are arbitrary superpositions of plane waves that freely propagate through each other, without interactions or other types of resistance or obstructions (violations of "freedom" of the wave packets).

The actions (or Hamiltonians) have a higher order (by one, in simplest cases) than the equations of motion because the equations of motion are obtained by differentiating the action (or Hamiltonians) with respect to fields (or other degrees of freedom) and the derivative of $\phi^n$ is $n\phi^{n-1}$, where the exponent grew by one. The bilinear/quadratic actions are "natural" because they are what is minimized or maximized in linear regression and similar situations, too.

Usually there are also higher-than-second-order terms, the interaction terms, that are responsible for the interactions between the fields (or a field with itself). There are many choices what these interactions may be. In quantum field theory, we usually consider "renormalizable" interactions which means in $d=4$ "at most fourth-order", roughly speaking.

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