Is this lagrangian a possible one in field theory?

Question

As far as I know, there are some prerequisites for an action to be preferable in classical field theory like relativistic invariance, locality, real-valueness, derivatives up to first order and internal symmetries invariance.

There's given a lagrangian density which is called "the simplest scalar field lagrangian" in my book (and several others i've checked out): $\mathcal{L} = \frac{1}{2}(\partial_{\mu} \phi)^2 - \frac{1}{2}m^2\phi^2$. My question concerns lagrangian of the form $\mathcal{L} = \frac{1}{2}(\partial_{\mu} \phi)^2 - m^2\phi$. Isn't it a possible one? If it is not, then for what reason (it looks like it satisfies all the conditions written above)? If it is, then isn't it simpler than the first one?

It seems to be quite a stupid question, so I am sorry if ones of its kind are not welcome here. I will be thankful for any help.

Answer 1

The problem with having a linear term as the highest order term in the potential energy is that the Hamiltonian won't have a stable ground state. All realistic quantum field theories have to have a state of minimum energy about which small fluctuations can occur, otherwise the vacuum just instantly explodes in a gas of particles with amplitude of oscillations increasing in amplitude (particle number) exponentially with time. One of the Wightman axioms handles just this fact:

There is also a stability assumption which restricts the spectrum of the four-momentum to the positive light cone (and its boundary).

Incorrect old version follows: Is the only difference the factor of $\frac{1}{2}$ in the mass term? If so, follow @drN's advice and look at the equations of motion, and you'll see why the original one is regarded as simpler than the second.

Formally, both of the Lagrangian densities presented in the question are identical, up to a rescaling of the parameter $m$; do the substitution $m\rightarrow \sqrt{2}m$ in the first Lagrangian density to see. You can also redefine field strength by a scale with $\phi \rightarrow \sqrt{2} \phi$ to get a Lagrangian density with no numerical factors at all: $\partial_\mu \phi \partial^\mu \phi - m^2 \phi^2,$ and it would correspond to a change in units for how the field $\phi$ is measured, classically, and a rescaling of the units in $\hbar$ in QFT.

Answer 2

As far as I understand the difference in Lagrangians is the potential term. One is quadratic and the other is linear.

Theoretically the Lagrangian is just a fancy way of formalizing the equation of motion. Some people right out reject the veracity of the least action formalism.

Having said that the quadratic potential gives rise to the Klein-Gordon EOM. KG is largely understood as the simplest case of EOM since basically they say that E^2-p^2=m^2.

Another thing to consider is the idea that in QM eigenfunctions are defined up to a global overall phase.

The quadratic Lagrangian implements this symmetry. The linear potential doesn't.