Where does the factor of half appear from in the Klein-Gordon Lagrangian for a real scalar field? The lagangian density of a scalar field or a Klein-Gordon field has the form of
$$\begin{align}
\mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m^2 \phi^2.
\end{align}$$
Why is there a factor of half appearing in the lagrangian? Being a constant, when entered into the Euler-lagrange equation, should yield the same equation when compared to a lagrangian without the factor of half, so what is the reason for having that factor in the equation?
 A: This normalization is convenient when you work in Fourier space, since it implies that
$$ \langle \phi(p) \phi(q) \rangle = \frac{1}{p^2 + m^2} \delta(p+q)$$
which is easy to remember.  You normally derive this using path integrals/functional integration, and in that way the factor of $1/2$ comes from a Gaussian integral -- where ultimately it's more natural to have $\int dx \exp(-x^2/2)$ than $\int dx \exp(-x^2)$. Of course, it's possible to use an arbitrary normalization, and you could write
$$ \mathcal{L} = A (\partial \phi)^2 + B \phi^2 $$
for arbitrary constants $A,B$. In perturbation theory, the Feynman rules would be a bit more complicated, and all observables would depend on the ratio $B/A$. The usual normalization is nice because the Lagrangian parameter $m^2$ is actually the mass of the one-particle state in the free theory.
A: According to these Cambridge lecture notes it's "conventional". I'm guessing that choice of constant gives you the right T-V energy when you integrate the Lagrangian density over a certain volume.
