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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?

Qmechanic
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Solidification
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1 Answers1

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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.

Qmechanic
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  • 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 $\phi=\phi^*$? – Solidification Jan 19 '20 at 20:02
  • We redefine in order to keep standard conventions. – Qmechanic Jan 19 '20 at 20:27
  • 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 $\phi_1$. – Elias Riedel Gårding Jan 20 '20 at 13:06