Timeline for What to do with a $\phi$ term in a Lagrangian?

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Dec 10, 2013 at 20:20	comment	added	Neuneck		@JerrySchirmer What Trimok is trying to say is that the interpretation in terms of creation and anihiliation of particles is only valid if you have a harmonic potential and you expand around the minimum. Moving to that minimum is exactly what completing the square does. Interpretation as interaction term means that you expand around the minimum, disregarding the shift and taking its effects in only at higher orders in perturbation theory.
Dec 10, 2013 at 19:13	comment	added	Trimok		@JerrySchirmer : The spatial Fourier transform would give : $\frac{\partial^2\phi_k(t)}{\partial t^2} + (k^2+m^2)\phi_k(t) + constant ~\delta^3(\vec k) = 0$. How would you define the creation and annihilation operators ?
Dec 10, 2013 at 16:42	comment	added	Zo the Relativist		Couldn't you also use it as a modification to your quantization conditions? $\nabla^{2}\phi + m\phi + constant = 0$ isn't all that hard to solve, and then you're free to define your new $a$ and $a^{\dagger}$ operators in terms of the space of solutions to this new pde. I get that this is almost certainly just equivalent to completing the square and then un-doing the transformation, though.
Dec 10, 2013 at 14:38	history	answered	Neuneck	CC BY-SA 3.0