Why does $\phi=\phi^*$ imposed on complex scalar field Lagrangian miss out $1/2$ factors?

If we require the reality condition $$\phi=\phi^*$$ on the Lagrangian for a complex scalar field is $$\mathcal{L}=(\partial^\mu\phi^*)(\partial_\mu\phi)-m^2(\phi^*\phi),$$ two degrees of freedom $$\phi$$ and $$\phi^*$$ is reduced to one. For consistency, I would expect it should give $$\mathcal{L}=\frac{1}{2}(\partial^\mu\phi)(\partial_\mu\phi)-\frac{1}{2}m^2\phi^2.$$ But this procedure misses the $$1/2$$ factors. Why? Did I mess up some normalization?

The standard convention is to divide each term in the Lagrangian with its symmetry factor. Therefore the kinetic term for a real (complex) scalar field is with (without) a symmetry factor $$\frac{1}{2}$$, respectively. A complex scalar field $$\phi= \frac{\phi^1+i\phi^2}{\sqrt{2}}$$ can be viewed as 2 real scalar fields.
• Not sure I completely follow. I know that the Lagrangian of the sum of two real scalar fields $\phi_1$ and $\phi_2$ is equal to the Lagrangian of a complex scalar field defined as $\phi=(\phi_1+i\phi_2)/\sqrt{2}$. But my question is if you want to recover the Lagrangian of a real scalar field from that of a complex scalar field, do we need to put the factor of $1/2$ by hand after setting $\phi=\phi^*$? Jan 19, 2020 at 20:02
• What happens is that you find the Lagrangian for the real field $\phi_1$ without any adjustments, and then you drop the subscript 1. So you forget about the old complex field $\phi$ and instead let $\phi$ denote the old $\phi_1$. Jan 20, 2020 at 13:06