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 $$\frac{\partial L}{\partial \psi} = 0.$$

Could someone verify if the answers are correct please?

The last equation is not a typo.But something feels weird, therefore I'd appreciate a verification. '

EDIT: Solved, my mistake was that I only considered $(\frac{\partial L}{\partial (\partial_{t} \psi)}) = i \psi^*$, instead of $\partial_{t} (\frac{\partial L}{\partial (\partial_{t} \psi)}) = i \frac{\partial \psi^{*}}{\partial t}$.

  • 2
    $\begingroup$ you should show a little more effort in your exercise or at least show us your thought process behind these equations $\endgroup$ Nov 14, 2019 at 13:41
  • $\begingroup$ okay, give me a few minutes $\endgroup$ Nov 14, 2019 at 13:47
  • $\begingroup$ Okay, i added some details in my thought process. $\endgroup$ Nov 14, 2019 at 13:58
  • 2
    $\begingroup$ HINT: Your variation with respect to $\psi$ is not in fact correct. Take a careful look at where that $i \psi^*$ term came from. $\endgroup$ Nov 14, 2019 at 14:24
  • 1
    $\begingroup$ The term $\partial L/\partial (\partial_{t} \psi)$ doesn't appear in the equations of motion for $\psi$, at least not directly. $\endgroup$ Nov 14, 2019 at 14:30


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