# Verifying equations of motion of Lagrangian [closed]

Given the following Lagrangian of a scalar field $$L = +i \psi^*\partial_t\psi - \frac{1}{2m} \nabla \psi^* \cdot \nabla \psi,$$ I want to compute the equation(s) of motion. First by varying wrt $$\psi^*$$ and then wrt $$\psi$$.

First I say $$\frac{\partial L}{\partial \psi^{*}} = + i\partial_t\psi$$ and $$\partial_{\mu} (\frac{\partial L}{\partial (\partial_{\mu} \psi^*)}) = \partial_{t} (\frac{\partial L}{\partial (\partial_{t} \psi^*)}) + \nabla(\frac{\partial L}{\partial (\nabla \psi^*)}) = 0 - \frac{1}{2m}\nabla^2 \psi.$$ By Euler -Lagrange I get

$$i \frac{\partial \psi}{\partial t} = - \frac{1}{2m} \nabla^2 \psi.$$

Now if I do the same but vary to $$\psi$$ instead of $$\psi^*$$, I get $$i \psi^* = +\frac{1}{2m} \nabla^2 \psi^{*},$$ as equation of motion. The only main difference is that $$\frac{\partial L}{\partial \psi} = 0.$$

EDIT: Solved, my mistake was that I only considered $$(\frac{\partial L}{\partial (\partial_{t} \psi)}) = i \psi^*$$, instead of $$\partial_{t} (\frac{\partial L}{\partial (\partial_{t} \psi)}) = i \frac{\partial \psi^{*}}{\partial t}$$.
• HINT: Your variation with respect to $\psi$ is not in fact correct. Take a careful look at where that $i \psi^*$ term came from. Nov 14, 2019 at 14:24
• The term $\partial L/\partial (\partial_{t} \psi)$ doesn't appear in the equations of motion for $\psi$, at least not directly. Nov 14, 2019 at 14:30