Consider a complex scalar field $\psi(x)$ with Lagrangian density

$$ \mathcal{L} = \partial_\mu\psi^* \partial^\mu\psi - M^2\psi^*\psi. $$

From the Lagranigan density we obtain the momentum $\pi = \dot\psi^*$. How do you obtain this? I expanded $\partial_\mu\psi = \partial _{0}\psi -\nabla\psi$ and then took the conjugate $\partial_\mu\psi^{*} = \partial _{0}\psi^{*} -\nabla\psi^{*}$ and then multiply by $\partial^\mu\psi = \partial^{0}\psi -\nabla\psi$ but at the end I get that the canonical momentum is:

$\partial _{0}\psi^{*} -\nabla\psi^{*}$

what is wrong?


1 Answer 1


Do not confuse multiplication with the idea of a dot product. The problem at hand, of constructing the kinetic term in components, is not equivalent to multiplying $(a+b)(c_d)$. Rather, we have two vectors whose components are $\langle \partial_0,\nabla\psi\rangle$ (the spatial components could be written out as well) and we are computing a dot product as defined by the Minkowski metric, which has a different sign somewhere (from what you've written it looks like you're taking the $(+,-,-,-)$ convention, so I'll use that here).

So writing out the dot product, we would actually have $$ \eta^{\mu\nu}\partial_\mu\psi^*\partial_\nu\psi=\partial_0\psi^*\partial_0\psi-\nabla\psi^*\cdot\nabla\psi. $$ From here, the conjugate momenta follows from the standard expression $$ \pi=\frac{\partial\mathcal{L}}{\partial\dot\psi}=\dot\psi^*. $$


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