From my naive understanding of the symmetry principle, in inertial frames the space is uniform and homogeneous, so the action must not depend explicitly on coordinates (or fields). Thus the action must only depend on magnitude of velocity or rate of change of the field. But still, quadratic term is only one way to satisfy this. Is it necessarily the only possible term?
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$\begingroup$ Related: physics.stackexchange.com/q/811056/2451 $\endgroup$– Qmechanic ♦Commented Apr 20 at 6:18
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Well, this answer might not be totally satisfying to you, but a free theory is a theory who is described by linear partial differential equations by definition. This is most commonly achieved through the well known Klein-Gordon or Dirac equations for fields of spin $0$ and $1/2$, respectively, which are quadratic.
Now, it is also possible to do a field redefinition, that renders your Lagrangian "more complicated", while still being a free QFT. This is the case of the Lagrangian $$ \mathcal{L} = \dfrac{1}{2}\partial_\mu \phi \partial^\mu\phi + \lambda\phi\partial_\mu \phi \partial^\mu\phi,\tag{1} $$ which turns out to be free, as seen by doing a field redefinition such that $$ \dfrac{1}{2}\partial_\mu \chi \partial^\mu\chi = \dfrac{1}{2}\partial_\mu \phi \partial^\mu\phi + \lambda\phi\partial_\mu \phi \partial^\mu\phi, \tag{2} $$ or, $$ \chi = \dfrac{1}{3\lambda} \left(1 + 2 \lambda\phi \right)^{3/2}. \tag{3} $$ So free Lagrangians do not have to be quadratic.
As a side note, field redefinition (3) also changes the metric of the path integral, which turns out to vanish after dimensional regularization.
answered Apr 20 at 10:35