# Complex Scalar Field Conjugate Momentum

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?

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 $$\pi=\frac{\partial\mathcal{L}}{\partial\dot\psi}=\dot\psi^*.$$