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.
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asked Sep 13, 2017 at 23:18
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1$\begingroup$ @Qmechanic: I just found out that the equation itself is not invariant under galileian transformations, so I guess there is no way to derive the Lagrangian Density from this assumptions. $\endgroup$– QuantumwhispCommented Sep 15, 2017 at 16:21
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1$\begingroup$ What? Did you read this Phys.SE post? $\endgroup$– Qmechanic ♦Commented Sep 15, 2017 at 16:24
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1$\begingroup$ yes. The quintessence I get from it is that unless I also transform the wave function, the equation will change. ($x' = x - v t$) alone won't do the trick. $\endgroup$– QuantumwhispCommented Sep 15, 2017 at 17:28
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1$\begingroup$ $\uparrow$ Right. $\endgroup$– Qmechanic ♦Commented Sep 15, 2017 at 17:30
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$\begingroup$ Let us continue this discussion in chat. $\endgroup$– QuantumwhispCommented Sep 15, 2017 at 17:40
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1 Answer
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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.
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1$\begingroup$ Comment to the answer (v1): Are you making any assumptions about how $\psi$ behaves under Galilean transformations? $\endgroup$– Qmechanic ♦Commented Sep 14, 2017 at 11:59