# 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?

• For that Lagrangian it is in fact irrelevant. It might come into play when adding some interaction terms, where the factor $1/2$ has to even out the Leibniz rule for the derivative in the first term (but even in that case one could define the interaction terms as twice as much, sparing the $1/2$ in front). – gented Jan 15 '16 at 12:41
• @GennaroTedesco So I can commit that term? What if I want to define something that depends on the form of the lagrangian, for example the Hilbert stress tensor? – Horus Jan 15 '16 at 13:18
• Well, anything else defined from the Lagrangian will of course depend on the Lagrangian and its multiplicative constant factors (although these don't enter the equation of motion they might enter the derived quantities, which can also be a good decision criteria to consider whether or not to include them in the first place). – gented Jan 15 '16 at 13:51
• In Landau's QED Ch. 2 he constructs the complex Lagrangian and then states that because real version should be of the same form as the complex one but it has half the number of degrees of freedom, that's where the factor comes from. – bolbteppa Mar 16 '16 at 14:00

$$\langle \phi(p) \phi(q) \rangle = \frac{1}{p^2 + m^2} \delta(p+q)$$ 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.
• Just like the classical $(m\dot{x}^2-kx^2)/2$. – J.G. Sep 12 '18 at 5:14