Is the field strength coupling of physical significance? Consider the Lagrangian \begin{equation} L=\frac{1}{2}(\partial_{\mu}\phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4 \end{equation} I have arbitrarily taken the co-efficient of the $\frac{1}{2}(\partial_{\mu}\phi)^2$ to be $1$, and it seems this is often done. But if I were attempting to model a real physical statistical system, such as a real magnet, would I be modelling the system incorrectly if I took this co-efficient to be $1$? If not, then why is this co-efficient of no physical significance?
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Possible duplicate of What´s the importance of the normalization of the Kinetic term? – ohneVal May 17 '18 at 09:48
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In QFT the field strength is associated with the amplitude for the field operator to create a 1 particle state from the vacuum, see the discussion in Peskin&Schroeder part 7.1. – tsufli May 17 '18 at 13:41
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Your kinetic term can always be chosen to be $1$ by a rescaling of the field variable. – Prof. Legolasov May 18 '18 at 06:54