Derive the Lagrangian that yields the free Schrödinger's equation from Galileian Invariance

Question

The Lagrangian Density $$L(\Psi, \Psi^*)=i \hbar \dot{\Psi} \Psi^* + \frac{\hbar^2}{2m} \Psi \Delta \Psi^*$$ will yield the schroedinger equations for $\Psi$ and $\Psi^*$. Can we derive this Lagrangian Density, if we impose only quadratic terms and Galileian invariance as conditions for the Density? Of Course the derivation can be up to total derivative terms, which do not change the physics.

Answer 1

Galilean invariance only impose the combination $i\partial_t-\frac{\triangle}{2m}$ to any power. This means that the action $$ \int_{t,x} \sum_n \psi^*\left(i\partial_t-\frac{\triangle}{2m}\right)^n\psi, $$ is invariant. You need additional constraints (like "simplicity", whatever that means, or comparison to experimental data) to obtain the OP's Lagrangian density.